1.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
     1.2 +++ b/bk/bk.org	Fri Oct 28 00:06:37 2011 -0700
     1.3 @@ -0,0 +1,88 @@
     1.4 +#+TITLE: Bugs in Quantum Mechanics
     1.5 +#+AUTHOR: Dylan Holmes
     1.6 +#+SETUPFILE: ../../aurellem/org/setup.org
     1.7 +#+INCLUDE:   ../../aurellem/org/level-0.org
     1.8 +
     1.9 +#Bugs in the Quantum-Mechanical Momentum Operator
    1.10 +
    1.11 +
    1.12 +I studied quantum mechanics the same way I study most subjects\mdash{}
    1.13 +by collecting (and squashing) bugs in my understanding. One of these
    1.14 +bugs persisted throughout two semesters of
    1.15 +quantum mechanics coursework until I finally found
    1.16 +the paper 
    1.17 +[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
    1.18 +mechanics/]], which helped me stamp out the bug entirely. I decided to
    1.19 +write an article about the problem and its solution for a number of reasons:
    1.20 +
    1.21 +- Although the paper was not unreasonably dense, it was written for
    1.22 +  teachers. I wanted to write an article for students.
    1.23 +- I wanted to popularize the problem and its solution because other
    1.24 +  explanations are currently too hard to find. (Even Shankar's
    1.25 +  excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it.)
    1.26 +- I wanted to check that the bug was indeed entirely
    1.27 +  eradicated. Attempting an explanation is my way of making
    1.28 +  sure.
    1.29 +
    1.30 +* COMMENT
    1.31 + I recommend the
    1.32 +paper not only for students who are learning
    1.33 +quantum mechanics, but especially for teachers interested in debugging
    1.34 +them. 
    1.35 +
    1.36 +* COMMENT
    1.37 +On my first exam in quantum mechanics, my professor asked us to
    1.38 +describe how certain measurements would affect a particle in a
    1.39 +box. Many of these measurement questions required routine application
    1.40 +of skills we had recently learned\mdash{}first, you recall (or
    1.41 +calculate) the eigenstates of the quantity
    1.42 +to be measured; second, you write the given state as a linear
    1.43 +sum of these eigenstates\mdash{} the coefficients on each term give
    1.44 +the probability amplitude.
    1.45 +
    1.46 +* The infinite square well potential
    1.47 +There is a particle in a one-dimensional potential well that has
    1.48 +infinitely high walls and finite width \(a\). This means that the
    1.49 +particle exists in a potential[fn:coords][fn:infinity]
    1.50 +
    1.51 +
    1.52 +\(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for
    1.53 +}\;x<0\text{ or }x>a.\end{cases}\)
    1.54 +
    1.55 +The Schr\ouml{}dinger equation describes how the particle's state 
    1.56 +\(|\psi\rangle\) will change over time in this system.
    1.57 +
    1.58 +\(\begin{eqnarray}
    1.59 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
    1.60 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
    1.61 +
    1.62 +This is a differential equation whose solutions are the physically
    1.63 +allowed states for the particle in this system. Like any differential
    1.64 +equation, 
    1.65 +
    1.66 +
    1.67 +Like any differential equation, the Schr\ouml{}dinger equation 
    1.68 +#; physically allowed states are those that change in physically
    1.69 +#allowed ways.
    1.70 +
    1.71 +
    1.72 +** Boundary conditions
    1.73 +Because the potential is infinite everywhere except within the well,
    1.74 +a realistic particle must be confined to exist only within the
    1.75 +well\mdash{}its wavefunction must be zero everywhere beyond the walls
    1.76 +of the well.
    1.77 +
    1.78 +
    1.79 +[fn:coords] I chose my coordinate system so that the well extends from
    1.80 +\(0<x<a\). Others choose a coordinate system so that the well extends from
    1.81 +\(-\frac{a}{2}<x<\frac{a}{2}\). Although both coordinate systems describe the same physical
    1.82 +situation, they give different-looking answers.
    1.83 +
    1.84 +[fn:infinity] Of course, infinite potentials are not
    1.85 +realistic. Instead, they are useful approximations to finite
    1.86 +potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
    1.87 +of the well\rdquo{} are close enough for your own practical
    1.88 +purposes. Having introduced a physical impossibility into the problem
    1.89 +already, we don't expect to get physically realistic solutions; we
    1.90 +just expect to get mathematically consistent ones. The forthcoming
    1.91 +trouble is that we don't.

     2.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
     2.2 +++ b/bk/bk2.org	Fri Oct 28 00:06:37 2011 -0700
     2.3 @@ -0,0 +1,97 @@
     2.4 +#+TITLE: Bugs in Quantum Mechanics
     2.5 +#+AUTHOR: Dylan Holmes
     2.6 +#+SETUPFILE: ../../aurellem/org/setup.org
     2.7 +#+INCLUDE:   ../../aurellem/org/level-0.org
     2.8 +
     2.9 +
    2.10 +#Bugs in the Quantum-Mechanical Momentum Operator
    2.11 +
    2.12 +
    2.13 +I studied quantum mechanics the same way I study most subjects\mdash{}
    2.14 +by collecting (and squashing) bugs in my understanding. One of these
    2.15 +bugs persisted throughout two semesters of
    2.16 +quantum mechanics coursework until I finally found
    2.17 +the paper 
    2.18 +[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
    2.19 +mechanics/]], which helped me stamp out the bug entirely. I decided to
    2.20 +write an article about the problem and its solution for a number of reasons:
    2.21 +
    2.22 +- Although the paper was not unreasonably dense, it was written for
    2.23 +  teachers. I wanted to write an article for students.
    2.24 +- I wanted to popularize the problem and its solution because other
    2.25 +  explanations are currently too hard to find. (Even Shankar's
    2.26 +  excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it.)
    2.27 +- I wanted to check that the bug was indeed entirely
    2.28 +  eradicated. Attempting an explanation is my way of making
    2.29 +  sure.
    2.30 +
    2.31 +* COMMENT
    2.32 + I recommend the
    2.33 +paper not only for students who are learning
    2.34 +quantum mechanics, but especially for teachers interested in debugging
    2.35 +them. 
    2.36 +
    2.37 +* COMMENT
    2.38 +On my first exam in quantum mechanics, my professor asked us to
    2.39 +describe how certain measurements would affect a particle in a
    2.40 +box. Many of these measurement questions required routine application
    2.41 +of skills we had recently learned\mdash{}first, you recall (or
    2.42 +calculate) the eigenstates of the quantity
    2.43 +to be measured; second, you write the given state as a linear
    2.44 +sum of these eigenstates\mdash{} the coefficients on each term give
    2.45 +the probability amplitude.
    2.46 +
    2.47 +* The infinite square well potential
    2.48 +
    2.49 +There is a particle in a one-dimensional potential well that is
    2.50 +infinite everywhere except for a well of length \(a\). This means that the
    2.51 +particle exists in a potential[fn:coords][fn:infinity]
    2.52 +
    2.53 +
    2.54 +\(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for
    2.55 +}\;x<0\text{ or }x>a.\end{cases}\)
    2.56 +
    2.57 +The Schr\ouml{}dinger equation describes how the particle's state 
    2.58 +\(|\psi\rangle\) will change over time in this system.
    2.59 +
    2.60 +\(\begin{eqnarray}
    2.61 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
    2.62 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
    2.63 +
    2.64 +This is a differential equation whose solutions are the physically
    2.65 +allowed states for the particle in this system. Physically allowed
    2.66 +states are those that change in physically allowed ways. Like any
    2.67 +differential equation, the Schr\ouml{}dinger equation can be
    2.68 +accompanied by /boundary conditions/\mdash{}conditions that
    2.69 +further restrict which states qualify as physically allowed.
    2.70 +
    2.71 +Whenever possible, physicists impose these boundary conditions:
    2.72 +- The state should be a /continuous function of/ \(x\). This means
    2.73 +  that if a particle is very likely to be /at/ a particular location,
    2.74 +  it is also very likely to be /near/ that location.
    2.75 +- 
    2.76 +
    2.77 +#; physically allowed states are those that change in physically
    2.78 +#allowed ways.
    2.79 +
    2.80 +
    2.81 +** Boundary conditions
    2.82 +Because the potential is infinite everywhere except within the well,
    2.83 +a realistic particle must be confined to exist only within the
    2.84 +well\mdash{}its wavefunction must be zero everywhere beyond the walls
    2.85 +of the well.
    2.86 +
    2.87 +
    2.88 +[fn:coords] I chose my coordinate system so that the well extends from
    2.89 +\(0<x<a\). Others choose a coordinate system so that the well extends from
    2.90 +\(-\frac{a}{2}<x<\frac{a}{2}\). Although both coordinate systems describe the same physical
    2.91 +situation, they give different-looking answers.
    2.92 +
    2.93 +[fn:infinity] Of course, infinite potentials are not
    2.94 +realistic. Instead, they are useful approximations to finite
    2.95 +potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
    2.96 +of the well\rdquo{} are close enough for your own practical
    2.97 +purposes. Having introduced a physical impossibility into the problem
    2.98 +already, we don't expect to get physically realistic solutions; we
    2.99 +just expect to get mathematically consistent ones. The forthcoming
   2.100 +trouble is that we don't.

     3.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
     3.2 +++ b/bk/bk3.org	Fri Oct 28 00:06:37 2011 -0700
     3.3 @@ -0,0 +1,257 @@
     3.4 +#+TITLE: Bugs in quantum mechanics
     3.5 +#+AUTHOR: Dylan Holmes
     3.6 +#+SETUPFILE: ../../aurellem/org/setup.org
     3.7 +#+INCLUDE:   ../../aurellem/org/level-0.org
     3.8 +
     3.9 +#Bugs in Quantum Mechanics
    3.10 +#Bugs in the Quantum-Mechanical Momentum Operator
    3.11 +
    3.12 +
    3.13 +I studied quantum mechanics the same way I study most subjects\mdash{}
    3.14 +by collecting (and squashing) bugs in my understanding. One of these
    3.15 +bugs persisted throughout two semesters of
    3.16 +quantum mechanics coursework until I finally found
    3.17 +the paper 
    3.18 +[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
    3.19 +mechanics/]], which helped me stamp out the bug entirely. I decided to
    3.20 +write an article about the problem and its solution for a number of reasons:
    3.21 +
    3.22 +- Although the paper was not unreasonably dense, it was written for
    3.23 +  teachers. I wanted to write an article for students.
    3.24 +- I wanted to popularize the problem and its solution because other
    3.25 +  explanations are currently too hard to find. (Even Shankar's
    3.26 +  excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it.)
    3.27 +- I wanted to check that the bug was indeed entirely
    3.28 +  eradicated. Attempting an explanation is my way of making
    3.29 +  sure.
    3.30 +
    3.31 +* COMMENT
    3.32 + I recommend the
    3.33 +paper not only for students who are learning
    3.34 +quantum mechanics, but especially for teachers interested in debugging
    3.35 +them. 
    3.36 +
    3.37 +* COMMENT
    3.38 +On my first exam in quantum mechanics, my professor asked us to
    3.39 +describe how certain measurements would affect a particle in a
    3.40 +box. Many of these measurement questions required routine application
    3.41 +of skills we had recently learned\mdash{}first, you recall (or
    3.42 +calculate) the eigenstates of the quantity
    3.43 +to be measured; second, you write the given state as a linear
    3.44 +sum of these eigenstates\mdash{} the coefficients on each term give
    3.45 +the probability amplitude.
    3.46 +
    3.47 +
    3.48 +* What I thought I knew
    3.49 +
    3.50 +The following is a list of things I thought were true of quantum
    3.51 +mechanics; the catch is that the list contradicts itself.
    3.52 +
    3.53 +- For any hermitian operator: Eigenstates with different eigenvalues are orthogonal.
    3.54 +- For any hermitian operator: Any physically allowed state can be
    3.55 +  written as a linear sum of eigenstates of the operator.
    3.56 +- The momentum operator and energy operator are hermitian, because
    3.57 +  momentum and energy are measureable quantities.
    3.58 +- In vacuum,
    3.59 +  - the momentum operator has an eigenstate
    3.60 +    \(p(x)=\exp{(ipx/\hbar)}\) for each value of $p$.
    3.61 +  - the energy operator has an eigenstate \(|E\rangle =
    3.62 +    \alpha|p\rangle + \beta|-p\rangle\) for any \(\alpha,\beta\) and
    3.63 +    the particular choice of momentum $p=\sqrt{2mE}$.
    3.64 +- In the infinitely deep potential well,
    3.65 +  - the momentum operator has eigenstates with the same form $p(x) =
    3.66 +    \exp{(ipx/\hbar)}$, but because of the boundary conditions on the
    3.67 +    well, the following modifications are required.
    3.68 +    - The wavefunction must be zero everywhere outside the well. That
    3.69 +      is, \(p(x) = \begin{cases}\exp{(ipx/\hbar)},& 0\lt{}x\lt{}a;
    3.70 +      \\0, & \text{for }x<0\text{ or }x>a \\ \end{cases}\)
    3.71 +#0,&\text{for }x\lt{}0\text{ or }x\gt{}a\end{cases}\)
    3.72 +    - no longer has an eigenstate for each value
    3.73 +      of $p$. Instead, only values of $p$ that are integer multiples of
    3.74 +      $\pi a/\hbar$ are physically realistic. 
    3.75 +
    3.76 +
    3.77 +
    3.78 +* COMMENT: 
    3.79 +
    3.80 +** Eigenstates with different eigenvalues are orthogonal
    3.81 +
    3.82 +#+begin_quote
    3.83 +*Theorem:* Eigenstates with different eigenvalues are orthogonal.
    3.84 +#+end_quote
    3.85 +
    3.86 +** COMMENT :
    3.87 +I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$
    3.88 +and $|b\rangle$ are eigenstates of $\Lambda$. This means that
    3.89 +
    3.90 +
    3.91 +\(
    3.92 +\begin{eqnarray}
    3.93 +\Lambda |a\rangle&=& a|a\rangle,\\
    3.94 +\Lambda|b\rangle&=& b|b\rangle.\\
    3.95 +\end{eqnarray}
    3.96 +\)
    3.97 +
    3.98 +If we take the difference of these eigenstates, we find that
    3.99 +
   3.100 +\(
   3.101 +\begin{eqnarray}
   3.102 +\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle
   3.103 +\qquad \text{(because $\Lambda$ is linear.)}\\
   3.104 +&=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and
   3.105 +$|b\rangle$ are eigenstates of $\Lambda$)}
   3.106 +\end{eqnarray}\)
   3.107 +
   3.108 +
   3.109 +which means that $a\neq b$.
   3.110 +
   3.111 +** Eigenvectors of hermitian operators span the space of solutions
   3.112 +
   3.113 +#+begin_quote
   3.114 +*Theorem:* If $\Omega$ is a hermitian operator, then every physically
   3.115 + allowed state can be written as a linear sum of eigenstates of
   3.116 + $\Omega$.
   3.117 +#+end_quote
   3.118 +
   3.119 +
   3.120 +
   3.121 +** Momentum and energy are hermitian operators
   3.122 +This ought to be true because hermitian operators correspond to
   3.123 +observable quantities. Since we expect momentum and energy to be
   3.124 +measureable quantities, we expect that there are hermitian operators
   3.125 +to represent them.
   3.126 +
   3.127 +
   3.128 +** Momentum and energy eigenstates in vacuum
   3.129 +An eigenstate of the momentum operator $P$ would be a state
   3.130 +\(|p\rangle\) such that \(P|p\rangle=p|p\rangle\).
   3.131 +
   3.132 +** Momentum and energy eigenstates in the infinitely deep well
   3.133 +
   3.134 +
   3.135 +
   3.136 +* Can you measure momentum in the infinite square well?
   3.137 +
   3.138 +
   3.139 +
   3.140 +** COMMENT  Momentum eigenstates
   3.141 +
   3.142 +In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the
   3.143 +momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\).
   3.144 +
   3.145 +In the infinitely deep potential well, the Hamiltonian is the same but
   3.146 +there is a new condition in order for states to qualify as physically
   3.147 +allowed: the states must not exist anywhere outside of well, as it
   3.148 +takes an infinite amount of energy to do so. 
   3.149 +
   3.150 +Notice that the momentum eigenstates defined above do /not/ satisfy
   3.151 +this condition.
   3.152 +
   3.153 +
   3.154 +
   3.155 +* COMMENT
   3.156 +For each physical system, there is a Schr\ouml{}dinger equation that
   3.157 +describes how a particle's state $|\psi\rangle$  will change over
   3.158 +time.
   3.159 +
   3.160 +\(\begin{eqnarray}
   3.161 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
   3.162 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
   3.163 +
   3.164 +This is a differential equation; each solution to the
   3.165 +Schr\ouml{}dinger equation is a state that is physically allowed for
   3.166 +our particle. Here, physically allowed states are
   3.167 +those that change in physically allowed ways. However, like any differential
   3.168 +equation, the Schr\ouml{}dinger equation can be accompanied by
   3.169 +/boundary conditions/\mdash{}conditions that further restrict which
   3.170 +states qualify as physically allowed.
   3.171 +
   3.172 +
   3.173 +
   3.174 +
   3.175 +** Eigenstates of momentum
   3.176 +
   3.177 +
   3.178 +
   3.179 +
   3.180 +#In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger
   3.181 +
   3.182 +#\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\)
   3.183 +
   3.184 +
   3.185 +
   3.186 +
   3.187 +
   3.188 +
   3.189 +
   3.190 +* COMMENT
   3.191 +
   3.192 +#* The infinite square well potential
   3.193 +
   3.194 +A particle exists in a potential that is
   3.195 +infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the
   3.196 +particle exists in a potential[fn:coords][fn:infinity]
   3.197 +
   3.198 +
   3.199 +\(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for
   3.200 +}\;x<0\text{ or }x>a.\end{cases}\)
   3.201 +
   3.202 +The Schr\ouml{}dinger equation describes how the particle's state 
   3.203 +\(|\psi\rangle\) will change over time in this system.
   3.204 +
   3.205 +\(\begin{eqnarray}
   3.206 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
   3.207 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
   3.208 +
   3.209 +This is a differential equation; each solution to the
   3.210 +Schr\ouml{}dinger equation is a state that is physically allowed for
   3.211 +our particle. Here, physically allowed states are
   3.212 +those that change in physically allowed ways. However, like any differential
   3.213 +equation, the Schr\ouml{}dinger equation can be accompanied by
   3.214 +/boundary conditions/\mdash{}conditions that further restrict which
   3.215 +states qualify as physically allowed.
   3.216 +
   3.217 +
   3.218 +Whenever possible, physicists impose these boundary conditions:
   3.219 +- A physically allowed state ought to be a /smoothly-varying function of position./ This means
   3.220 +  that if a particle in the state  is likely to be /at/ a particular location,
   3.221 +  it is also likely to be /near/ that location.
   3.222 +
   3.223 +These boundary conditions imply that for the square well potential in
   3.224 +this problem,
   3.225 +
   3.226 +- Physically allowed states must be totally confined to the well,
   3.227 +  because it takes an infinite amount of energy to exist anywhere
   3.228 +  outside of the well (and physically allowed states ought to have
   3.229 +  only finite energy).
   3.230 +- Physically allowed states must be increasingly unlikely to find very
   3.231 +  close to the walls of the well. This is because of two conditions: the above
   3.232 +  condition says that the particle is /impossible/ to find
   3.233 +  outside of the well, and the smoothly-varying condition says
   3.234 +  that if a particle is impossible to find at a particular location,
   3.235 +  it must be unlikely to be found nearby that location.
   3.236 +
   3.237 +#; physically allowed states are those that change in physically
   3.238 +#allowed ways.
   3.239 +
   3.240 +
   3.241 +#** Boundary conditions
   3.242 +Because the potential is infinite everywhere except within the well,
   3.243 +a realistic particle must be confined to exist only within the
   3.244 +well\mdash{}its wavefunction must be zero everywhere beyond the walls
   3.245 +of the well.
   3.246 +
   3.247 +
   3.248 +[fn:coords] I chose my coordinate system so that the well extends from
   3.249 +\(0<x<a\). Others choose a coordinate system so that the well extends from
   3.250 +\(-\frac{a}{2}<x<\frac{a}{2}\). Although both coordinate systems describe the same physical
   3.251 +situation, they give different-looking answers.
   3.252 +
   3.253 +[fn:infinity] Of course, infinite potentials are not
   3.254 +realistic. Instead, they are useful approximations to finite
   3.255 +potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
   3.256 +of the well\rdquo{} are close enough for your own practical
   3.257 +purposes. Having introduced a physical impossibility into the problem
   3.258 +already, we don't expect to get physically realistic solutions; we
   3.259 +just expect to get mathematically consistent ones. The forthcoming
   3.260 +trouble is that we don't.

     4.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
     4.2 +++ b/bk/bk4.org	Fri Oct 28 00:06:37 2011 -0700
     4.3 @@ -0,0 +1,309 @@
     4.4 +#+TITLE: Bugs in quantum mechanics
     4.5 +#+AUTHOR: Dylan Holmes
     4.6 +#+SETUPFILE: ../../aurellem/org/setup.org
     4.7 +#+INCLUDE:   ../../aurellem/org/level-0.org
     4.8 +
     4.9 +#Bugs in Quantum Mechanics
    4.10 +#Bugs in the Quantum-Mechanical Momentum Operator
    4.11 +
    4.12 +
    4.13 +I studied quantum mechanics the same way I study most subjects\mdash{}
    4.14 +by collecting (and squashing) bugs in my understanding. One of these
    4.15 +bugs persisted throughout two semesters of
    4.16 +quantum mechanics coursework until I finally found
    4.17 +the paper 
    4.18 +[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
    4.19 +mechanics/]], which helped me stamp out the bug entirely. I decided to
    4.20 +write an article about the problem and its solution for a number of reasons:
    4.21 +
    4.22 +- Although the paper was not unreasonably dense, it was written for
    4.23 +  teachers. I wanted to write an article for students.
    4.24 +- I wanted to popularize the problem and its solution because other
    4.25 +  explanations are currently too hard to find. (Even Shankar's
    4.26 +  excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it.)
    4.27 +- I wanted to check that the bug was indeed entirely
    4.28 +  eradicated. Attempting an explanation is my way of making
    4.29 +  sure.
    4.30 +
    4.31 +* COMMENT
    4.32 + I recommend the
    4.33 +paper not only for students who are learning
    4.34 +quantum mechanics, but especially for teachers interested in debugging
    4.35 +them. 
    4.36 +
    4.37 +* COMMENT
    4.38 +On my first exam in quantum mechanics, my professor asked us to
    4.39 +describe how certain measurements would affect a particle in a
    4.40 +box. Many of these measurement questions required routine application
    4.41 +of skills we had recently learned\mdash{}first, you recall (or
    4.42 +calculate) the eigenstates of the quantity
    4.43 +to be measured; second, you write the given state as a linear
    4.44 +sum of these eigenstates\mdash{} the coefficients on each term give
    4.45 +the probability amplitude.
    4.46 +
    4.47 +
    4.48 +* What I thought I knew
    4.49 +
    4.50 +The following is a list of things I thought were true of quantum
    4.51 +mechanics; the catch is that the list contradicts itself.
    4.52 +
    4.53 +1. For any hermitian operator: Eigenstates with different eigenvalues are orthogonal.
    4.54 +2. For any hermitian operator: Any physically allowed state can be
    4.55 +   written as a linear sum of eigenstates of the operator.
    4.56 +3. The momentum operator and energy operator are hermitian, because
    4.57 +   momentum and energy are measureable quantities.
    4.58 +4. In the vacuum potential, the momentum and energy operators have these eigenstates:
    4.59 +   - the momentum operator has an eigenstate
    4.60 +     \(p(x)=\exp{(ipx/\hbar)}\) for each value of $p$.
    4.61 +   - the energy operator has an eigenstate \(|E\rangle =
    4.62 +     \alpha|p\rangle + \beta|-p\rangle\) for any \(\alpha,\beta\) and
    4.63 +     the particular choice of momentum $p=\sqrt{2mE}$.
    4.64 +5. In the infinitely deep potential well, the momentum and energy
    4.65 +   operators have these eigenstates:
    4.66 +   - The momentum eigenstates and energy eigenstates have the same form
    4.67 +     as in the vacuum potential: $p(x) =
    4.68 +     \exp{(ipx/\hbar)}$ and $|E\rangle = \alpha|p\rangle + \beta|-p\rangle$.
    4.69 +   - Even so, because of the boundary conditions on the
    4.70 +     well, we must make the following modifications:
    4.71 +     + Physically realistic states must be impossible to find outside the well. (Only a state of infinite
    4.72 +       energy could exist outside the well, and infinite energy is not
    4.73 +       realistic.) This requirement means, for example, that momentum
    4.74 +       eigenstates in the infinitely deep well must be
    4.75 +       \(p(x)
    4.76 +       = \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;
    4.77 +       \\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
    4.78 +     + Physically realistic states must vary smoothly throughout
    4.79 +       space. This means that if a particle in some state is very unlikely to be
    4.80 +       /at/ a particular location, it is also very unlikely be /near/
    4.81 +       that location. Combining this requirement with the above
    4.82 +       requirement, we find that the momentum operator no longer has
    4.83 +       an eigenstate for each value of $p$; instead, only values of
    4.84 +       $p$ that are integer multiples of $\pi a/\hbar$ are physically
    4.85 +       realistic. Similarly, the energy operator no longer has an
    4.86 +       eigenstate for each value of $E$; instead, the only energy
    4.87 +       eigenstates in the infinitely deep well
    4.88 +       are $E_n(x)=\sin(n\pi x/ a)$ for positive integers $n$.
    4.89 +
    4.90 +* COMMENT: 
    4.91 +
    4.92 +** Eigenstates with different eigenvalues are orthogonal
    4.93 +
    4.94 +#+begin_quote
    4.95 +*Theorem:* Eigenstates with different eigenvalues are orthogonal.
    4.96 +#+end_quote
    4.97 +
    4.98 +** COMMENT :
    4.99 +I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$
   4.100 +and $|b\rangle$ are eigenstates of $\Lambda$. This means that
   4.101 +
   4.102 +
   4.103 +\(
   4.104 +\begin{eqnarray}
   4.105 +\Lambda |a\rangle&=& a|a\rangle,\\
   4.106 +\Lambda|b\rangle&=& b|b\rangle.\\
   4.107 +\end{eqnarray}
   4.108 +\)
   4.109 +
   4.110 +If we take the difference of these eigenstates, we find that
   4.111 +
   4.112 +\(
   4.113 +\begin{eqnarray}
   4.114 +\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle
   4.115 +\qquad \text{(because $\Lambda$ is linear.)}\\
   4.116 +&=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and
   4.117 +$|b\rangle$ are eigenstates of $\Lambda$)}
   4.118 +\end{eqnarray}\)
   4.119 +
   4.120 +
   4.121 +which means that $a\neq b$.
   4.122 +
   4.123 +** Eigenvectors of hermitian operators span the space of solutions
   4.124 +
   4.125 +#+begin_quote
   4.126 +*Theorem:* If $\Omega$ is a hermitian operator, then every physically
   4.127 + allowed state can be written as a linear sum of eigenstates of
   4.128 + $\Omega$.
   4.129 +#+end_quote
   4.130 +
   4.131 +
   4.132 +
   4.133 +** Momentum and energy are hermitian operators
   4.134 +This ought to be true because hermitian operators correspond to
   4.135 +observable quantities. Since we expect momentum and energy to be
   4.136 +measureable quantities, we expect that there are hermitian operators
   4.137 +to represent them.
   4.138 +
   4.139 +
   4.140 +** Momentum and energy eigenstates in vacuum
   4.141 +An eigenstate of the momentum operator $P$ would be a state
   4.142 +\(|p\rangle\) such that \(P|p\rangle=p|p\rangle\).
   4.143 +
   4.144 +** Momentum and energy eigenstates in the infinitely deep well
   4.145 +
   4.146 +
   4.147 +
   4.148 +* Can you measure momentum in the infinitely deep well?
   4.149 +In summary, I thought I knew:
   4.150 +1. For any hermitian operator: eigenstates with different eigenvalues
   4.151 +   are orthogonal.
   4.152 +2. For any hermitian operator: any physically realistic state can be
   4.153 +   written as a linear sum of eigenstates of the operator.
   4.154 +3. The momentum operator and energy operator are hermitian, because
   4.155 +   momentum and energy are observable quantities. 
   4.156 +4. (The form of the momentum and energy eigenstates in the vacuum potential)
   4.157 +5. (The form of the momentum and energy eigenstates in the infinitely deep well potential)
   4.158 +
   4.159 +Additionally, I understood that because the infinitely deep potential
   4.160 +well is not realistic, states of such a system  are not necessarily
   4.161 +physically realistic. Instead, I understood
   4.162 +\ldquo{}realistic states\rdquo{} to be those that satisfy the physically
   4.163 +unrealistic Schr\ouml{}dinger equation and its boundary conditions.
   4.164 +
   4.165 +With that final caveat, here is the problem:
   4.166 +
   4.167 +According to (5), the momentum eigenstates in the well are 
   4.168 +
   4.169 +\(p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
   4.170 +
   4.171 +However, /these/ states are not orthogonal, which contradicts the
   4.172 +assumption that (3) the momentum operator is hermitian and (2)
   4.173 +eigenstates of a hermitian are orthogonal if they have different eigenvalues.
   4.174 +
   4.175 +#+begin_quote 
   4.176 +*Problem 1. The momentum eigenstates of the well are not orthogonal*
   4.177 +
   4.178 +/Proof./ If $p_1\neq p_2$, then 
   4.179 +
   4.180 +\(\begin{eqnarray}
   4.181 +\langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)dx\\
   4.182 +&=& \int_0^a p_1^*(x)p_2(x)dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{
   4.183 +outside the well.}\\
   4.184 +&=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)dx}}
   4.185 +\end{eqnarray}\)
   4.186 +$\square$
   4.187 +
   4.188 +#+end_quote
   4.189 +
   4.190 +
   4.191 +
   4.192 +** COMMENT  Momentum eigenstates
   4.193 +
   4.194 +In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the
   4.195 +momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\).
   4.196 +
   4.197 +In the infinitely deep potential well, the Hamiltonian is the same but
   4.198 +there is a new condition in order for states to qualify as physically
   4.199 +allowed: the states must not exist anywhere outside of well, as it
   4.200 +takes an infinite amount of energy to do so. 
   4.201 +
   4.202 +Notice that the momentum eigenstates defined above do /not/ satisfy
   4.203 +this condition.
   4.204 +
   4.205 +
   4.206 +
   4.207 +* COMMENT
   4.208 +For each physical system, there is a Schr\ouml{}dinger equation that
   4.209 +describes how a particle's state $|\psi\rangle$  will change over
   4.210 +time.
   4.211 +
   4.212 +\(\begin{eqnarray}
   4.213 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
   4.214 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
   4.215 +
   4.216 +This is a differential equation; each solution to the
   4.217 +Schr\ouml{}dinger equation is a state that is physically allowed for
   4.218 +our particle. Here, physically allowed states are
   4.219 +those that change in physically allowed ways. However, like any differential
   4.220 +equation, the Schr\ouml{}dinger equation can be accompanied by
   4.221 +/boundary conditions/\mdash{}conditions that further restrict which
   4.222 +states qualify as physically allowed.
   4.223 +
   4.224 +
   4.225 +
   4.226 +
   4.227 +** Eigenstates of momentum
   4.228 +
   4.229 +
   4.230 +
   4.231 +
   4.232 +#In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger
   4.233 +
   4.234 +#\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\)
   4.235 +
   4.236 +
   4.237 +
   4.238 +
   4.239 +
   4.240 +
   4.241 +
   4.242 +* COMMENT
   4.243 +
   4.244 +#* The infinite square well potential
   4.245 +
   4.246 +A particle exists in a potential that is
   4.247 +infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the
   4.248 +particle exists in a potential[fn:coords][fn:infinity]
   4.249 +
   4.250 +
   4.251 +\(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for
   4.252 +}\;x<0\text{ or }x>a.\end{cases}\)
   4.253 +
   4.254 +The Schr\ouml{}dinger equation describes how the particle's state 
   4.255 +\(|\psi\rangle\) will change over time in this system.
   4.256 +
   4.257 +\(\begin{eqnarray}
   4.258 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
   4.259 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
   4.260 +
   4.261 +This is a differential equation; each solution to the
   4.262 +Schr\ouml{}dinger equation is a state that is physically allowed for
   4.263 +our particle. Here, physically allowed states are
   4.264 +those that change in physically allowed ways. However, like any differential
   4.265 +equation, the Schr\ouml{}dinger equation can be accompanied by
   4.266 +/boundary conditions/\mdash{}conditions that further restrict which
   4.267 +states qualify as physically allowed.
   4.268 +
   4.269 +
   4.270 +Whenever possible, physicists impose these boundary conditions:
   4.271 +- A physically allowed state ought to be a /smoothly-varying function of position./ This means
   4.272 +  that if a particle in the state  is likely to be /at/ a particular location,
   4.273 +  it is also likely to be /near/ that location.
   4.274 +
   4.275 +These boundary conditions imply that for the square well potential in
   4.276 +this problem,
   4.277 +
   4.278 +- Physically allowed states must be totally confined to the well,
   4.279 +  because it takes an infinite amount of energy to exist anywhere
   4.280 +  outside of the well (and physically allowed states ought to have
   4.281 +  only finite energy).
   4.282 +- Physically allowed states must be increasingly unlikely to find very
   4.283 +  close to the walls of the well. This is because of two conditions: the above
   4.284 +  condition says that the particle is /impossible/ to find
   4.285 +  outside of the well, and the smoothly-varying condition says
   4.286 +  that if a particle is impossible to find at a particular location,
   4.287 +  it must be unlikely to be found nearby that location.
   4.288 +
   4.289 +#; physically allowed states are those that change in physically
   4.290 +#allowed ways.
   4.291 +
   4.292 +
   4.293 +#** Boundary conditions
   4.294 +Because the potential is infinite everywhere except within the well,
   4.295 +a realistic particle must be confined to exist only within the
   4.296 +well\mdash{}its wavefunction must be zero everywhere beyond the walls
   4.297 +of the well.
   4.298 +
   4.299 +
   4.300 +[fn:coords] I chose my coordinate system so that the well extends from
   4.301 +\(0<x<a\). Others choose a coordinate system so that the well extends from
   4.302 +\(-\frac{a}{2}<x<\frac{a}{2}\). Although both coordinate systems describe the same physical
   4.303 +situation, they give different-looking answers.
   4.304 +
   4.305 +[fn:infinity] Of course, infinite potentials are not
   4.306 +realistic. Instead, they are useful approximations to finite
   4.307 +potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
   4.308 +of the well\rdquo{} are close enough for your own practical
   4.309 +purposes. Having introduced a physical impossibility into the problem
   4.310 +already, we don't expect to get physically realistic solutions; we
   4.311 +just expect to get mathematically consistent ones. The forthcoming
   4.312 +trouble is that we don't.

     5.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
     5.2 +++ b/bk/bk_quandary.org	Fri Oct 28 00:06:37 2011 -0700
     5.3 @@ -0,0 +1,566 @@
     5.4 +#+TITLE: Bugs in quantum mechanics
     5.5 +#+AUTHOR: Dylan Holmes
     5.6 +#+DESCRIPTION: I explain hermitian operators and use them to solve a confusing problem in quantum mechanics.
     5.7 +#+KEYWORDS: quantum mechanics, self-adjoint, hermitian, square well, momentum
     5.8 +#+SETUPFILE: ../../aurellem/org/setup.org
     5.9 +#+INCLUDE:   ../../aurellem/org/level-0.org
    5.10 +
    5.11 +
    5.12 +
    5.13 +#Bugs in Quantum Mechanics
    5.14 +#Bugs in the Quantum-Mechanical Momentum Operator
    5.15 +
    5.16 +
    5.17 +I studied quantum mechanics the same way I study most subjects\mdash{}
    5.18 +by collecting (and squashing) bugs in my understanding. One of these
    5.19 +bugs persisted throughout two semesters of
    5.20 +quantum mechanics coursework until I finally found
    5.21 +the paper 
    5.22 +[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
    5.23 +mechanics/]], which helped me stamp out the bug entirely. I decided to
    5.24 +write an article about the problem and its solution for a number of reasons:
    5.25 +
    5.26 +- Although the paper was not unreasonably dense, it was written for
    5.27 +  teachers. I wanted to write an article for students.
    5.28 +- I wanted to popularize the problem and its solution because other
    5.29 +  explanations are currently too hard to find. (Even Shankar's
    5.30 +  excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it, as far as I know.)
    5.31 +- Attempting an explanation is my way of making
    5.32 +  sure that the bug is /really/ gone.
    5.33 +# entirely eradicated.
    5.34 +
    5.35 +* COMMENT
    5.36 + I recommend the
    5.37 +paper not only for students who are learning
    5.38 +quantum mechanics, but especially for teachers interested in debugging
    5.39 +them. 
    5.40 +
    5.41 +* COMMENT
    5.42 +On my first exam in quantum mechanics, my professor asked us to
    5.43 +describe how certain measurements would affect a particle in a
    5.44 +box. Many of these measurement questions required routine application
    5.45 +of skills we had recently learned\mdash{}first, you recall (or
    5.46 +calculate) the eigenstates of the quantity
    5.47 +to be measured; second, you write the given state as a linear
    5.48 +sum of these eigenstates\mdash{} the coefficients on each term give
    5.49 +the probability amplitude.
    5.50 +
    5.51 +
    5.52 +* Two methods of calculation that give different results.
    5.53 +
    5.54 +In the infinitely deep well, there is a particle in the the
    5.55 +normalized state
    5.56 +
    5.57 + \(\psi(x) = \begin{cases}A\; x(x-a)&\text{for }0\lt{}x\lt{}a;\\0,&\text{for }x\lt{}0\text{ or }x>a.\end{cases}\) 
    5.58 +
    5.59 +This is apparently a perfectly respectable state: it is normalized ($A$ is a
    5.60 +normalization constant), it is zero
    5.61 +everywhere outside of the well, and it is moreover continuous.
    5.62 +
    5.63 +Even so, we will find a problem if we attempt to calculate the average
    5.64 +energy-squared of this state (that is, the quantity \(\langle \psi | H^2 | \psi \rangle\)). 
    5.65 +
    5.66 +** First method
    5.67 +
    5.68 +For short, define a new variable[fn:1] \(|\bar \psi\rangle \equiv
    5.69 +H|\psi\rangle\). We can express the state $|\bar\psi\rangle$ as a
    5.70 +function of $x$ because we know how to express $H$ and $\psi$ in terms
    5.71 +of  $x$: $H$ is the differential operator $\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}$, and
    5.72 +$\psi(x)$ is the function defined above. So, we get \(\quad\bar\psi(x) = \frac{-A\hbar^2}{m}\). For future reference, observe that $\bar\psi(x)$
    5.73 +is constant.
    5.74 +
    5.75 +Having introduced $|\bar\psi\rangle$, we can calculate the average energy-squared of $|\psi\rangle$ in the
    5.76 +following way.
    5.77 +
    5.78 +\(\begin{eqnarray}
    5.79 +\langle \psi | H^2 |\psi\rangle &=& \langle \psi H^\dagger | H
    5.80 +\psi\rangle\\
    5.81 +&=& \langle \psi H | H\psi \rangle\\
    5.82 +&=& \langle \bar\psi | \bar\psi \rangle\\
    5.83 +&=& \int_0^a \left(\frac{A\hbar^2}{m}\right)^2\;dx\\
    5.84 +&=& \frac{A^2\hbar^4 a}{m^2}\\
    5.85 +\end{eqnarray}\)
    5.86 + 
    5.87 +For future reference, observe that this value is  nonzero
    5.88 +(which makes sense).
    5.89 +
    5.90 +** Second method
    5.91 +We can also calculate the average energy-squared of $|\psi\rangle$ in the
    5.92 +following way.
    5.93 +
    5.94 +\begin{eqnarray}
    5.95 +\langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\
    5.96 +&=& \langle \psi |H \bar\psi \rangle\\
    5.97 +&=&\int_0^a Ax(x-a)
    5.98 +\left(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\right)\frac{-A\hbar^2}{m}\;dx\\
    5.99 +&=& 0\quad (!)\\
   5.100 +\end{eqnarray}
   5.101 +
   5.102 +The second-to-last term must be zero because the second derivative
   5.103 +of a constant \(\left(\frac{-A\hbar^2}{m}\right)\) is zero.
   5.104 +
   5.105 +* What is the problem?
   5.106 +
   5.107 +To recap: We used two different methods to calculate the average
   5.108 +energy-squared of a state $|\psi\rangle$. For the first method, we
   5.109 +used the fact that $H$ is a hermitian operator, replacing \(\langle
   5.110 +\psi H^\dagger | H \psi\rangle\) with \(\langle \psi H | H
   5.111 +\psi\rangle\). Using this substitution rule, we calculated the answer.
   5.112 +
   5.113 +For the second method, we didn't use the fact that $H$ was hermitian;
   5.114 +instead, we used the fact that we know how to represent $H$ and $\psi$
   5.115 +as functions of $x$: $H$ is a differential operator
   5.116 +\(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\) and $\psi$ is a quadratic
   5.117 +function of $x$. By applying $H$ to $\psi$, we took several
   5.118 +derivatives and arrived at our answer.
   5.119 +
   5.120 +These two methods gave different results. In the following sections,
   5.121 +I'll describe and analyze the source of this difference.
   5.122 +
   5.123 +** Physical operators only act on physical wavefunctions
   5.124 +   :PROPERTIES:
   5.125 +   :ORDERED:  t
   5.126 +   :END:
   5.127 +#In quantum mechanics, an operator is a function that takes in a
   5.128 +#physical state and produces another physical state as ouput. Some
   5.129 +#operators correspond to physical quantities such as energy,
   5.130 +#momentum, or position; the mathematical properties of these operators correspond to
   5.131 +#physical properties of the system.
   5.132 +
   5.133 +#Eigenstates are an example of this correspondence: an 
   5.134 +
   5.135 +Physical states are represented as wavefunctions in quantum
   5.136 +mechanics. Just as we disallow certain physically nonsensical states
   5.137 +in classical mechanics (for example, we consider it to be nonphysical
   5.138 +for an object to spontaneously disappear from one place and reappear
   5.139 +in another), we also disallow certain wavefunctions in quantum
   5.140 +mechanics.
   5.141 +
   5.142 +For example, since wavefunctions are supposed to correspond to
   5.143 +probability amplitudes, we require wavefunctions to be normalized
   5.144 +\((\langle \psi |\psi \rangle = 1)\). Generally, we disallow
   5.145 +wavefunctions that do not satisfy this property (although there are
   5.146 +some exceptions[fn:2]).
   5.147 +
   5.148 +As another example, we generally expect probability to vary smoothly\mdash{}if
   5.149 +a particle is very likely or very unlikely to be found at a particular
   5.150 +location, it should also be somewhat likely or somewhat unlikely to be
   5.151 +found /near/ that location. In more precise terms, we expect that for
   5.152 +physically meaningful wavefunctions, the probability 
   5.153 +\(\text{Pr}(x)=\int^x \psi^*\psi\) should be a continuous function of
   5.154 +$x$ and, again, we disallow wavefunctions that do not satisfy this
   5.155 +property because we consider them to be physically nonsensical.
   5.156 +
   5.157 +So, physical wavefunctions must satisfy certain properties
   5.158 +like the two just described. Wavefunctions that do not satisfy these properties are
   5.159 +rejected for being physically nonsensical: even though we can perform
   5.160 +calculations with them, the mathematical results we obtain do not mean
   5.161 +anything physically.
   5.162 +
   5.163 +Now, in quantum mechanics, an *operator* is a function that converts
   5.164 +states into other states. Some operators correspond to
   5.165 +physical quantities such as energy, momentum, or position, and as a
   5.166 +result, the mathematical properties of these operators correspond to
   5.167 +physical properties of the system. Physical operators are furthermore
   5.168 +subject to the following rule: they are only allowed to operate on 
   5.169 +#physical wavefunctions, and they are only allowed to produce
   5.170 +#physical wavefunctions[fn:why].
   5.171 +the wavefunctions we have deemed physical, and they only produce physical wavefunctions[fn::
   5.172 +
   5.173 + If you require a hermitian operator to have physical
   5.174 +  eigenstates, you get a very strong result: you guarantee that the
   5.175 +  operator will convert /every/ physical wavefunction into another
   5.176 +  physical wavefunction:
   5.177 +
   5.178 +  For any linear operator $\Omega$, the eigenvalue equation is
   5.179 +\(\Omega|\omega\rangle = \omega |\omega\rangle\). Notice that if an
   5.180 +eigenstate $|\omega\rangle$ is a physical wavefunction, the
   5.181 +eigenvalue equation equation forces $\Omega|\omega\rangle$ to be a
   5.182 +physical wavefunction as well. To elaborate, if the eigenstates of
   5.183 +$\Omega$ are physical functions, then $\Omega$ is guaranteed to
   5.184 +convert them into other physical functions.  Even more is true if the
   5.185 +operator $\Omega$ is also hermitian: there is a theorem which states
   5.186 +that \ldquo{}If \Omega is hermitian, then every physical wavefunction
   5.187 +can be written as a weighted sum of eigenstates of \Omega.\rdquo{} This
   5.188 +theorem implies that /if $\Omega$ is hermitian/, and /if the eigenstates
   5.189 +of \Omega are physically allowed/, then \Omega is guaranteed to
   5.190 +convert every physically allowed wavefunction into another physically
   5.191 +allowed wavefunction.].
   5.192 +
   5.193 +In fact, this rule for physical operators is the source of our
   5.194 +problem, as we unknowingly violated it when applying our second
   5.195 +method!
   5.196 +
   5.197 +** The violation
   5.198 +
   5.199 +I'll start explaining this violation by being more specific about the
   5.200 +infinitely deep well potential. We have said already that physicists
   5.201 +require wavefunctions to satisfy certain properties in order to be
   5.202 +deemed \ldquo{}physical\rdquo{}. To be specific, wavefunctions in the
   5.203 +infinitely deep well
   5.204 +- Must be *normalizable*, because they correspond to
   5.205 +  probability amplitudes.
   5.206 +- Must have *smoothly-varying probability*, because if a particle is very
   5.207 +  likely to be at a location, it ought to be likely to be /near/
   5.208 +  it as well.
   5.209 +- Must *not exist outside the well*, because it
   5.210 +  would take an infinite amount of energy to do so.
   5.211 +
   5.212 +Additionally, by combining the second and third conditions, some
   5.213 +physicists reason that wavefunctions in the infinitely deep well
   5.214 +
   5.215 +- Must *become zero* towards the edges of the well.
   5.216 +
   5.217 +
   5.218 +
   5.219 +
   5.220 +You'll remember we had
   5.221 +
   5.222 +\(
   5.223 +\begin{eqnarray}
   5.224 +\psi(x) &=& A\;x(x-a)\\
   5.225 +\bar\psi(x)&=& \frac{-A\hbar^2}{m}\\
   5.226 +&&\text{for }0\lt{}x\lt{}a\\
   5.227 +\end{eqnarray}
   5.228 +\)
   5.229 +
   5.230 +In our second method, we wrote 
   5.231 +
   5.232 +
   5.233 +\(\begin{eqnarray}
   5.234 +\langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\
   5.235 +&=& \langle \psi |H \bar\psi \rangle\\
   5.236 +& \vdots&\\
   5.237 +&=& 0\\
   5.238 +\end{eqnarray}\)
   5.239 +
   5.240 +However, there are two issues here. First, we were not allowed to operate with $H$ on the wavefunction
   5.241 +$|\bar \psi\rangle$, because $H$ is a physical operator and $|\bar
   5.242 +\psi\rangle$ is a nonphysical state: in the infinite square well,
   5.243 +physical wavefunctions must approach zero at the edges of the well,
   5.244 +which the constant function $|\bar\psi\rangle$ does not do. By
   5.245 +feeding $H$ a nonphysical wavefunction, we obtained nonsensical
   5.246 +results.
   5.247 +
   5.248 +Second, we claimed that $H$ was a physical operator\mdash{}that $H$
   5.249 +was hermitian. According to the rule, this means $H$ must convert physical states into other
   5.250 +physical states. But $H$ converts the physical state $|\psi\rangle$
   5.251 +into the nonphysical state $|\bar\psi\rangle = H|\psi\rangle$ is not. Because $H$ converts some
   5.252 +physical states into nonphysical states, it cannot be a hermitian operator.
   5.253 +
   5.254 +** Boundary conditions affect hermiticity
   5.255 +We have now discovered a flaw: when applied to the state
   5.256 +$|\psi\rangle$, the second method violates the rule that physical
   5.257 +operators must only take in physical states and must only produce
   5.258 +physical states. This suggests that the problem was with the state
   5.259 +$|\bar\psi\rangle$ that we wrongly fed into $H$. However, the problem
   5.260 +is more serious still: the state $|\psi\rangle 
   5.261 +
   5.262 +** COMMENT Re-examining physical constraints
   5.263 +
   5.264 +We have now discovered a flaw: when applied to the state
   5.265 +$|\psi\rangle$, the second method violates the rule that physical
   5.266 +operators must only take in physical states and must only produce
   5.267 +physical states. Let's examine the problem more closely.
   5.268 +
   5.269 +We have said already that physicists require wavefunctions to satisfy
   5.270 +certain properties in order to be deemed \ldquo{}physical\rdquo{}. To
   5.271 +be specific, wavefunctions in the infinitely deep well
   5.272 +- Must be *normalizable*, because they correspond to
   5.273 +  probability amplitudes.
   5.274 +- Must have *smoothly-varying probability*, because if a particle is very
   5.275 +  likely to be at a location, it ought to be likely to be /near/
   5.276 +  it as well.
   5.277 +- Must *not exist outside the well*, because it
   5.278 +  would take an infinite amount of energy to do so.
   5.279 +
   5.280 +We now have discovered an important flaw in the second method: when
   5.281 +applied to the state $|\bar\psi\rangle$, the second method violates
   5.282 +the rule that physical operators must only take in
   5.283 +physical states and must only produce physical states. The problem is
   5.284 +even more serious, however
   5.285 +
   5.286 +
   5.287 +
   5.288 +[fn:1] I'm defining a new variable just to make certain expressions
   5.289 +  look shorter; this cannot affect the content of the answer we'll
   5.290 +  get. 
   5.291 +
   5.292 +[fn:2] For example, in vaccuum (i.e., when the potential of the
   5.293 +  physical system is $V(x)=0$ throughout all space), the momentum
   5.294 +  eigenstates are not normalizable\mdash{}the relevant integral blows
   5.295 +  up to infinity instead of converging to a number. Physicists modify
   5.296 +  the definition of normalization slightly so that
   5.297 +  \ldquo{}delta-normalizable \rdquo{} functions like these are included
   5.298 +  among the physical wavefunctions.
   5.299 +
   5.300 +
   5.301 +
   5.302 +* COMMENT: What I thought I knew
   5.303 +
   5.304 +The following is a list of things I thought were true of quantum
   5.305 +mechanics; the catch is that the list contradicts itself.
   5.306 +
   5.307 +1. For any hermitian operator: Eigenstates with different eigenvalues are orthogonal.
   5.308 +2. For any hermitian operator: Any physically allowed state can be
   5.309 +   written as a linear sum of eigenstates of the operator.
   5.310 +3. The momentum operator and energy operator are hermitian, because
   5.311 +   momentum and energy are measureable quantities.
   5.312 +4. In the vacuum potential, the momentum and energy operators have these eigenstates:
   5.313 +   - the momentum operator has an eigenstate
   5.314 +     \(p(x)=\exp{(ipx/\hbar)}\) for each value of $p$.
   5.315 +   - the energy operator has an eigenstate \(|E\rangle =
   5.316 +     \alpha|p\rangle + \beta|-p\rangle\) for any \(\alpha,\beta\) and
   5.317 +     the particular choice of momentum $p=\sqrt{2mE}$.
   5.318 +5. In the infinitely deep potential well, the momentum and energy
   5.319 +   operators have these eigenstates:
   5.320 +   - The momentum eigenstates and energy eigenstates have the same form
   5.321 +     as in the vacuum potential: $p(x) =
   5.322 +     \exp{(ipx/\hbar)}$ and $|E\rangle = \alpha|p\rangle + \beta|-p\rangle$.
   5.323 +   - Even so, because of the boundary conditions on the
   5.324 +     well, we must make the following modifications:
   5.325 +     + Physically realistic states must be impossible to find outside the well. (Only a state of infinite
   5.326 +       energy could exist outside the well, and infinite energy is not
   5.327 +       realistic.) This requirement means, for example, that momentum
   5.328 +       eigenstates in the infinitely deep well must be
   5.329 +       \(p(x)
   5.330 +       = \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;
   5.331 +       \\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
   5.332 +     + Physically realistic states must vary smoothly throughout
   5.333 +       space. This means that if a particle in some state is very unlikely to be
   5.334 +       /at/ a particular location, it is also very unlikely be /near/
   5.335 +       that location. Combining this requirement with the above
   5.336 +       requirement, we find that the momentum operator no longer has
   5.337 +       an eigenstate for each value of $p$; instead, only values of
   5.338 +       $p$ that are integer multiples of $\pi \hbar/a$ are physically
   5.339 +       realistic. Similarly, the energy operator no longer has an
   5.340 +       eigenstate for each value of $E$; instead, the only energy
   5.341 +       eigenstates in the infinitely deep well
   5.342 +       are $E_n(x)=\sin(n\pi x/ a)$ for positive integers $n$.
   5.343 +
   5.344 +* COMMENT: 
   5.345 +
   5.346 +** Eigenstates with different eigenvalues are orthogonal
   5.347 +
   5.348 +#+begin_quote
   5.349 +*Theorem:* Eigenstates with different eigenvalues are orthogonal.
   5.350 +#+end_quote
   5.351 +
   5.352 +** COMMENT :
   5.353 +I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$
   5.354 +and $|b\rangle$ are eigenstates of $\Lambda$. This means that
   5.355 +
   5.356 +
   5.357 +\(
   5.358 +\begin{eqnarray}
   5.359 +\Lambda |a\rangle&=& a|a\rangle,\\
   5.360 +\Lambda|b\rangle&=& b|b\rangle.\\
   5.361 +\end{eqnarray}
   5.362 +\)
   5.363 +
   5.364 +If we take the difference of these eigenstates, we find that
   5.365 +
   5.366 +\(
   5.367 +\begin{eqnarray}
   5.368 +\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle
   5.369 +\qquad \text{(because $\Lambda$ is linear.)}\\
   5.370 +&=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and
   5.371 +$|b\rangle$ are eigenstates of $\Lambda$)}
   5.372 +\end{eqnarray}\)
   5.373 +
   5.374 +
   5.375 +which means that $a\neq b$.
   5.376 +
   5.377 +** Eigenvectors of hermitian operators span the space of solutions
   5.378 +
   5.379 +#+begin_quote
   5.380 +*Theorem:* If $\Omega$ is a hermitian operator, then every physically
   5.381 + allowed state can be written as a linear sum of eigenstates of
   5.382 + $\Omega$.
   5.383 +#+end_quote
   5.384 +
   5.385 +
   5.386 +
   5.387 +** Momentum and energy are hermitian operators
   5.388 +This ought to be true because hermitian operators correspond to
   5.389 +observable quantities. Since we expect momentum and energy to be
   5.390 +measureable quantities, we expect that there are hermitian operators
   5.391 +to represent them.
   5.392 +
   5.393 +
   5.394 +** Momentum and energy eigenstates in vacuum
   5.395 +An eigenstate of the momentum operator $P$ would be a state
   5.396 +\(|p\rangle\) such that \(P|p\rangle=p|p\rangle\).
   5.397 +
   5.398 +** Momentum and energy eigenstates in the infinitely deep well
   5.399 +
   5.400 +
   5.401 +
   5.402 +* COMMENT Can you measure momentum in the infinitely deep well?
   5.403 +In summary, I thought I knew:
   5.404 +1. For any hermitian operator: eigenstates with different eigenvalues
   5.405 +   are orthogonal.
   5.406 +2. For any hermitian operator: any physically realistic state can be
   5.407 +   written as a linear sum of eigenstates of the operator.
   5.408 +3. The momentum operator and energy operator are hermitian, because
   5.409 +   momentum and energy are observable quantities. 
   5.410 +4. (The form of the momentum and energy eigenstates in the vacuum potential)
   5.411 +5. (The form of the momentum and energy eigenstates in the infinitely deep well potential)
   5.412 +
   5.413 +Additionally, I understood that because the infinitely deep potential
   5.414 +well is not realistic, states of such a system  are not necessarily
   5.415 +physically realistic. Instead, I understood
   5.416 +\ldquo{}realistic states\rdquo{} to be those that satisfy the physically
   5.417 +unrealistic Schr\ouml{}dinger equation and its boundary conditions.
   5.418 +
   5.419 +With that final caveat, here is the problem:
   5.420 +
   5.421 +According to (5), the momentum eigenstates in the well are 
   5.422 +
   5.423 +\(p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
   5.424 +
   5.425 +(Additionally, we require that $p$ be an integer multiple of $\pi\hbar/a$.) 
   5.426 +
   5.427 +However, /these/ states are not orthogonal, which contradicts the
   5.428 +assumption that (3) the momentum operator is hermitian and (2)
   5.429 +eigenstates of a hermitian are orthogonal if they have different eigenvalues.
   5.430 +
   5.431 +#+begin_quote 
   5.432 +*Problem 1. The momentum eigenstates of the well are not orthogonal*
   5.433 +
   5.434 +/Proof./ If $p_1\neq p_2$, then 
   5.435 +
   5.436 +\(\begin{eqnarray}
   5.437 +\langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)\,dx\\
   5.438 +&=& \int_0^a p_1^*(x)p_2(x)\,dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{
   5.439 +outside the well.}\\
   5.440 +&=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)\,dx}}\\
   5.441 +&=& \int_0^a \exp{\left[i(p_1-p_2)x/\hbar\right]}\,dx\\
   5.442 +\end{eqnarray}\)
   5.443 +$\square$
   5.444 +
   5.445 +#+end_quote
   5.446 +
   5.447 +
   5.448 +
   5.449 +** COMMENT  Momentum eigenstates
   5.450 +
   5.451 +In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the
   5.452 +momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\).
   5.453 +
   5.454 +In the infinitely deep potential well, the Hamiltonian is the same but
   5.455 +there is a new condition in order for states to qualify as physically
   5.456 +allowed: the states must not exist anywhere outside of well, as it
   5.457 +takes an infinite amount of energy to do so. 
   5.458 +
   5.459 +Notice that the momentum eigenstates defined above do /not/ satisfy
   5.460 +this condition.
   5.461 +
   5.462 +
   5.463 +
   5.464 +* COMMENT
   5.465 +For each physical system, there is a Schr\ouml{}dinger equation that
   5.466 +describes how a particle's state $|\psi\rangle$  will change over
   5.467 +time.
   5.468 +
   5.469 +\(\begin{eqnarray}
   5.470 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
   5.471 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
   5.472 +
   5.473 +This is a differential equation; each solution to the
   5.474 +Schr\ouml{}dinger equation is a state that is physically allowed for
   5.475 +our particle. Here, physically allowed states are
   5.476 +those that change in physically allowed ways. However, like any differential
   5.477 +equation, the Schr\ouml{}dinger equation can be accompanied by
   5.478 +/boundary conditions/\mdash{}conditions that further restrict which
   5.479 +states qualify as physically allowed.
   5.480 +
   5.481 +
   5.482 +
   5.483 +
   5.484 +** Eigenstates of momentum
   5.485 +
   5.486 +
   5.487 +
   5.488 +
   5.489 +#In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger
   5.490 +
   5.491 +#\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\)
   5.492 +
   5.493 +
   5.494 +
   5.495 +
   5.496 +
   5.497 +
   5.498 +
   5.499 +* COMMENT
   5.500 +
   5.501 +#* The infinite square well potential
   5.502 +
   5.503 +A particle exists in a potential that is
   5.504 +infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the
   5.505 +particle exists in a potential[fn:coords][fn:infinity]
   5.506 +
   5.507 +
   5.508 +\(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for
   5.509 +}\;x<0\text{ or }x>a.\end{cases}\)
   5.510 +
   5.511 +The Schr\ouml{}dinger equation describes how the particle's state 
   5.512 +\(|\psi\rangle\) will change over time in this system.
   5.513 +
   5.514 +\(\begin{eqnarray}
   5.515 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
   5.516 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
   5.517 +
   5.518 +This is a differential equation; each solution to the
   5.519 +Schr\ouml{}dinger equation is a state that is physically allowed for
   5.520 +our particle. Here, physically allowed states are
   5.521 +those that change in physically allowed ways. However, like any differential
   5.522 +equation, the Schr\ouml{}dinger equation can be accompanied by
   5.523 +/boundary conditions/\mdash{}conditions that further restrict which
   5.524 +states qualify as physically allowed.
   5.525 +
   5.526 +
   5.527 +Whenever possible, physicists impose these boundary conditions:
   5.528 +- A physically allowed state ought to be a /smoothly-varying function of position./ This means
   5.529 +  that if a particle in the state  is likely to be /at/ a particular location,
   5.530 +  it is also likely to be /near/ that location.
   5.531 +
   5.532 +These boundary conditions imply that for the square well potential in
   5.533 +this problem,
   5.534 +
   5.535 +- Physically allowed states must be totally confined to the well,
   5.536 +  because it takes an infinite amount of energy to exist anywhere
   5.537 +  outside of the well (and physically allowed states ought to have
   5.538 +  only finite energy).
   5.539 +- Physically allowed states must be increasingly unlikely to find very
   5.540 +  close to the walls of the well. This is because of two conditions: the above
   5.541 +  condition says that the particle is /impossible/ to find
   5.542 +  outside of the well, and the smoothly-varying condition says
   5.543 +  that if a particle is impossible to find at a particular location,
   5.544 +  it must be unlikely to be found nearby that location.
   5.545 +
   5.546 +#; physically allowed states are those that change in physically
   5.547 +#allowed ways.
   5.548 +
   5.549 +
   5.550 +#** Boundary conditions
   5.551 +Because the potential is infinite everywhere except within the well,
   5.552 +a realistic particle must be confined to exist only within the
   5.553 +well\mdash{}its wavefunction must be zero everywhere beyond the walls
   5.554 +of the well.
   5.555 +
   5.556 +
   5.557 +[fn:coords] I chose my coordinate system so that the well extends from
   5.558 +\(0<x<a\). Others choose a coordinate system so that the well extends from
   5.559 +\(-\frac{a}{2}<x<\frac{a}{2}\). Although both coordinate systems describe the same physical
   5.560 +situation, they give different-looking answers.
   5.561 +
   5.562 +[fn:infinity] Of course, infinite potentials are not
   5.563 +realistic. Instead, they are useful approximations to finite
   5.564 +potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
   5.565 +of the well\rdquo{} are close enough for your own practical
   5.566 +purposes. Having introduced a physical impossibility into the problem
   5.567 +already, we don't expect to get physically realistic solutions; we
   5.568 +just expect to get mathematically consistent ones. The forthcoming
   5.569 +trouble is that we don't.

     6.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
     6.2 +++ b/bk/bkup.org	Fri Oct 28 00:06:37 2011 -0700
     6.3 @@ -0,0 +1,49 @@
     6.4 +#+TITLE: Bugs in Quantum Mechanics
     6.5 +#+AUTHOR: Dylan Holmes
     6.6 +#+SETUPFILE: ../../aurellem/org/setup.org
     6.7 +#+INCLUDE:   ../../aurellem/org/level-0.org
     6.8 +
     6.9 +
    6.10 +#Bugs in the Quantum-Mechanical Momentum Operator
    6.11 +
    6.12 +
    6.13 +I studied quantum mechanics the same way I study most subjects\mdash{}
    6.14 +by collecting (and squashing) bugs in my understanding. One of these
    6.15 +bugs persisted throughout two semesters of
    6.16 +quantum mechanics coursework until I finally found
    6.17 +the paper 
    6.18 +[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
    6.19 +mechanics/]], which helped me stamp out the bug entirely. I decided to
    6.20 +write an article about the problem and its solution for a number of reasons:
    6.21 +
    6.22 +- Although the paper was not unreasonably dense, it was written for
    6.23 +  teachers. I wanted to write an article for students.
    6.24 +- I wanted to popularize the problem and its solution because
    6.25 +  other explanations are currently too hard to find.
    6.26 +- I wanted to check that the bug was indeed entirely
    6.27 +  eradicated. Attempting an explanation is my way of making
    6.28 +  sure.
    6.29 +
    6.30 +* COMMENT
    6.31 + I recommend the
    6.32 +paper not only for students who are learning
    6.33 +quantum mechanics, but especially for teachers interested in debugging
    6.34 +them. 
    6.35 +
    6.36 +* COMMENT
    6.37 +On my first exam in quantum mechanics, my professor asked us to
    6.38 +describe how certain measurements would affect a particle in a
    6.39 +box. Many of these measurement questions required routine application
    6.40 +of skills we had recently learned\mdash{}first, you recall (or
    6.41 +calculate) the eigenstates of the quantity
    6.42 +to be measured; second, you write the given state as a linear
    6.43 +sum of these eigenstates\mdash{} the coefficients on each term give
    6.44 +the probability amplitude.
    6.45 +
    6.46 +* Statement of the Problem
    6.47 +A particle is 
    6.48 +
    6.49 +
    6.50 +
    6.51 +
    6.52 +* COMMENT [TABLE-OF-CONTENTS]

     7.1 --- a/org/bk.org	Fri Oct 28 00:03:05 2011 -0700
     7.2 +++ /dev/null	Thu Jan 01 00:00:00 1970 +0000
     7.3 @@ -1,88 +0,0 @@
     7.4 -#+TITLE: Bugs in Quantum Mechanics
     7.5 -#+AUTHOR: Dylan Holmes
     7.6 -#+SETUPFILE: ../../aurellem/org/setup.org
     7.7 -#+INCLUDE:   ../../aurellem/org/level-0.org
     7.8 -
     7.9 -#Bugs in the Quantum-Mechanical Momentum Operator
    7.10 -
    7.11 -
    7.12 -I studied quantum mechanics the same way I study most subjects\mdash{}
    7.13 -by collecting (and squashing) bugs in my understanding. One of these
    7.14 -bugs persisted throughout two semesters of
    7.15 -quantum mechanics coursework until I finally found
    7.16 -the paper 
    7.17 -[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
    7.18 -mechanics/]], which helped me stamp out the bug entirely. I decided to
    7.19 -write an article about the problem and its solution for a number of reasons:
    7.20 -
    7.21 -- Although the paper was not unreasonably dense, it was written for
    7.22 -  teachers. I wanted to write an article for students.
    7.23 -- I wanted to popularize the problem and its solution because other
    7.24 -  explanations are currently too hard to find. (Even Shankar's
    7.25 -  excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it.)
    7.26 -- I wanted to check that the bug was indeed entirely
    7.27 -  eradicated. Attempting an explanation is my way of making
    7.28 -  sure.
    7.29 -
    7.30 -* COMMENT
    7.31 - I recommend the
    7.32 -paper not only for students who are learning
    7.33 -quantum mechanics, but especially for teachers interested in debugging
    7.34 -them. 
    7.35 -
    7.36 -* COMMENT
    7.37 -On my first exam in quantum mechanics, my professor asked us to
    7.38 -describe how certain measurements would affect a particle in a
    7.39 -box. Many of these measurement questions required routine application
    7.40 -of skills we had recently learned\mdash{}first, you recall (or
    7.41 -calculate) the eigenstates of the quantity
    7.42 -to be measured; second, you write the given state as a linear
    7.43 -sum of these eigenstates\mdash{} the coefficients on each term give
    7.44 -the probability amplitude.
    7.45 -
    7.46 -* The infinite square well potential
    7.47 -There is a particle in a one-dimensional potential well that has
    7.48 -infinitely high walls and finite width \(a\). This means that the
    7.49 -particle exists in a potential[fn:coords][fn:infinity]
    7.50 -
    7.51 -
    7.52 -\(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for
    7.53 -}\;x<0\text{ or }x>a.\end{cases}\)
    7.54 -
    7.55 -The Schr\ouml{}dinger equation describes how the particle's state 
    7.56 -\(|\psi\rangle\) will change over time in this system.
    7.57 -
    7.58 -\(\begin{eqnarray}
    7.59 -i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
    7.60 -H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
    7.61 -
    7.62 -This is a differential equation whose solutions are the physically
    7.63 -allowed states for the particle in this system. Like any differential
    7.64 -equation, 
    7.65 -
    7.66 -
    7.67 -Like any differential equation, the Schr\ouml{}dinger equation 
    7.68 -#; physically allowed states are those that change in physically
    7.69 -#allowed ways.
    7.70 -
    7.71 -
    7.72 -** Boundary conditions
    7.73 -Because the potential is infinite everywhere except within the well,
    7.74 -a realistic particle must be confined to exist only within the
    7.75 -well\mdash{}its wavefunction must be zero everywhere beyond the walls
    7.76 -of the well.
    7.77 -
    7.78 -
    7.79 -[fn:coords] I chose my coordinate system so that the well extends from
    7.80 -\(0<x<a\). Others choose a coordinate system so that the well extends from
    7.81 -\(-\frac{a}{2}<x<\frac{a}{2}\). Although both coordinate systems describe the same physical
    7.82 -situation, they give different-looking answers.
    7.83 -
    7.84 -[fn:infinity] Of course, infinite potentials are not
    7.85 -realistic. Instead, they are useful approximations to finite
    7.86 -potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
    7.87 -of the well\rdquo{} are close enough for your own practical
    7.88 -purposes. Having introduced a physical impossibility into the problem
    7.89 -already, we don't expect to get physically realistic solutions; we
    7.90 -just expect to get mathematically consistent ones. The forthcoming
    7.91 -trouble is that we don't.

     8.1 --- a/org/bk2.org	Fri Oct 28 00:03:05 2011 -0700
     8.2 +++ /dev/null	Thu Jan 01 00:00:00 1970 +0000
     8.3 @@ -1,97 +0,0 @@
     8.4 -#+TITLE: Bugs in Quantum Mechanics
     8.5 -#+AUTHOR: Dylan Holmes
     8.6 -#+SETUPFILE: ../../aurellem/org/setup.org
     8.7 -#+INCLUDE:   ../../aurellem/org/level-0.org
     8.8 -
     8.9 -
    8.10 -#Bugs in the Quantum-Mechanical Momentum Operator
    8.11 -
    8.12 -
    8.13 -I studied quantum mechanics the same way I study most subjects\mdash{}
    8.14 -by collecting (and squashing) bugs in my understanding. One of these
    8.15 -bugs persisted throughout two semesters of
    8.16 -quantum mechanics coursework until I finally found
    8.17 -the paper 
    8.18 -[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
    8.19 -mechanics/]], which helped me stamp out the bug entirely. I decided to
    8.20 -write an article about the problem and its solution for a number of reasons:
    8.21 -
    8.22 -- Although the paper was not unreasonably dense, it was written for
    8.23 -  teachers. I wanted to write an article for students.
    8.24 -- I wanted to popularize the problem and its solution because other
    8.25 -  explanations are currently too hard to find. (Even Shankar's
    8.26 -  excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it.)
    8.27 -- I wanted to check that the bug was indeed entirely
    8.28 -  eradicated. Attempting an explanation is my way of making
    8.29 -  sure.
    8.30 -
    8.31 -* COMMENT
    8.32 - I recommend the
    8.33 -paper not only for students who are learning
    8.34 -quantum mechanics, but especially for teachers interested in debugging
    8.35 -them. 
    8.36 -
    8.37 -* COMMENT
    8.38 -On my first exam in quantum mechanics, my professor asked us to
    8.39 -describe how certain measurements would affect a particle in a
    8.40 -box. Many of these measurement questions required routine application
    8.41 -of skills we had recently learned\mdash{}first, you recall (or
    8.42 -calculate) the eigenstates of the quantity
    8.43 -to be measured; second, you write the given state as a linear
    8.44 -sum of these eigenstates\mdash{} the coefficients on each term give
    8.45 -the probability amplitude.
    8.46 -
    8.47 -* The infinite square well potential
    8.48 -
    8.49 -There is a particle in a one-dimensional potential well that is
    8.50 -infinite everywhere except for a well of length \(a\). This means that the
    8.51 -particle exists in a potential[fn:coords][fn:infinity]
    8.52 -
    8.53 -
    8.54 -\(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for
    8.55 -}\;x<0\text{ or }x>a.\end{cases}\)
    8.56 -
    8.57 -The Schr\ouml{}dinger equation describes how the particle's state 
    8.58 -\(|\psi\rangle\) will change over time in this system.
    8.59 -
    8.60 -\(\begin{eqnarray}
    8.61 -i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
    8.62 -H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
    8.63 -
    8.64 -This is a differential equation whose solutions are the physically
    8.65 -allowed states for the particle in this system. Physically allowed
    8.66 -states are those that change in physically allowed ways. Like any
    8.67 -differential equation, the Schr\ouml{}dinger equation can be
    8.68 -accompanied by /boundary conditions/\mdash{}conditions that
    8.69 -further restrict which states qualify as physically allowed.
    8.70 -
    8.71 -Whenever possible, physicists impose these boundary conditions:
    8.72 -- The state should be a /continuous function of/ \(x\). This means
    8.73 -  that if a particle is very likely to be /at/ a particular location,
    8.74 -  it is also very likely to be /near/ that location.
    8.75 -- 
    8.76 -
    8.77 -#; physically allowed states are those that change in physically
    8.78 -#allowed ways.
    8.79 -
    8.80 -
    8.81 -** Boundary conditions
    8.82 -Because the potential is infinite everywhere except within the well,
    8.83 -a realistic particle must be confined to exist only within the
    8.84 -well\mdash{}its wavefunction must be zero everywhere beyond the walls
    8.85 -of the well.
    8.86 -
    8.87 -
    8.88 -[fn:coords] I chose my coordinate system so that the well extends from
    8.89 -\(0<x<a\). Others choose a coordinate system so that the well extends from
    8.90 -\(-\frac{a}{2}<x<\frac{a}{2}\). Although both coordinate systems describe the same physical
    8.91 -situation, they give different-looking answers.
    8.92 -
    8.93 -[fn:infinity] Of course, infinite potentials are not
    8.94 -realistic. Instead, they are useful approximations to finite
    8.95 -potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
    8.96 -of the well\rdquo{} are close enough for your own practical
    8.97 -purposes. Having introduced a physical impossibility into the problem
    8.98 -already, we don't expect to get physically realistic solutions; we
    8.99 -just expect to get mathematically consistent ones. The forthcoming
   8.100 -trouble is that we don't.

     9.1 --- a/org/bk3.org	Fri Oct 28 00:03:05 2011 -0700
     9.2 +++ /dev/null	Thu Jan 01 00:00:00 1970 +0000
     9.3 @@ -1,257 +0,0 @@
     9.4 -#+TITLE: Bugs in quantum mechanics
     9.5 -#+AUTHOR: Dylan Holmes
     9.6 -#+SETUPFILE: ../../aurellem/org/setup.org
     9.7 -#+INCLUDE:   ../../aurellem/org/level-0.org
     9.8 -
     9.9 -#Bugs in Quantum Mechanics
    9.10 -#Bugs in the Quantum-Mechanical Momentum Operator
    9.11 -
    9.12 -
    9.13 -I studied quantum mechanics the same way I study most subjects\mdash{}
    9.14 -by collecting (and squashing) bugs in my understanding. One of these
    9.15 -bugs persisted throughout two semesters of
    9.16 -quantum mechanics coursework until I finally found
    9.17 -the paper 
    9.18 -[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
    9.19 -mechanics/]], which helped me stamp out the bug entirely. I decided to
    9.20 -write an article about the problem and its solution for a number of reasons:
    9.21 -
    9.22 -- Although the paper was not unreasonably dense, it was written for
    9.23 -  teachers. I wanted to write an article for students.
    9.24 -- I wanted to popularize the problem and its solution because other
    9.25 -  explanations are currently too hard to find. (Even Shankar's
    9.26 -  excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it.)
    9.27 -- I wanted to check that the bug was indeed entirely
    9.28 -  eradicated. Attempting an explanation is my way of making
    9.29 -  sure.
    9.30 -
    9.31 -* COMMENT
    9.32 - I recommend the
    9.33 -paper not only for students who are learning
    9.34 -quantum mechanics, but especially for teachers interested in debugging
    9.35 -them. 
    9.36 -
    9.37 -* COMMENT
    9.38 -On my first exam in quantum mechanics, my professor asked us to
    9.39 -describe how certain measurements would affect a particle in a
    9.40 -box. Many of these measurement questions required routine application
    9.41 -of skills we had recently learned\mdash{}first, you recall (or
    9.42 -calculate) the eigenstates of the quantity
    9.43 -to be measured; second, you write the given state as a linear
    9.44 -sum of these eigenstates\mdash{} the coefficients on each term give
    9.45 -the probability amplitude.
    9.46 -
    9.47 -
    9.48 -* What I thought I knew
    9.49 -
    9.50 -The following is a list of things I thought were true of quantum
    9.51 -mechanics; the catch is that the list contradicts itself.
    9.52 -
    9.53 -- For any hermitian operator: Eigenstates with different eigenvalues are orthogonal.
    9.54 -- For any hermitian operator: Any physically allowed state can be
    9.55 -  written as a linear sum of eigenstates of the operator.
    9.56 -- The momentum operator and energy operator are hermitian, because
    9.57 -  momentum and energy are measureable quantities.
    9.58 -- In vacuum,
    9.59 -  - the momentum operator has an eigenstate
    9.60 -    \(p(x)=\exp{(ipx/\hbar)}\) for each value of $p$.
    9.61 -  - the energy operator has an eigenstate \(|E\rangle =
    9.62 -    \alpha|p\rangle + \beta|-p\rangle\) for any \(\alpha,\beta\) and
    9.63 -    the particular choice of momentum $p=\sqrt{2mE}$.
    9.64 -- In the infinitely deep potential well,
    9.65 -  - the momentum operator has eigenstates with the same form $p(x) =
    9.66 -    \exp{(ipx/\hbar)}$, but because of the boundary conditions on the
    9.67 -    well, the following modifications are required.
    9.68 -    - The wavefunction must be zero everywhere outside the well. That
    9.69 -      is, \(p(x) = \begin{cases}\exp{(ipx/\hbar)},& 0\lt{}x\lt{}a;
    9.70 -      \\0, & \text{for }x<0\text{ or }x>a \\ \end{cases}\)
    9.71 -#0,&\text{for }x\lt{}0\text{ or }x\gt{}a\end{cases}\)
    9.72 -    - no longer has an eigenstate for each value
    9.73 -      of $p$. Instead, only values of $p$ that are integer multiples of
    9.74 -      $\pi a/\hbar$ are physically realistic. 
    9.75 -
    9.76 -
    9.77 -
    9.78 -* COMMENT: 
    9.79 -
    9.80 -** Eigenstates with different eigenvalues are orthogonal
    9.81 -
    9.82 -#+begin_quote
    9.83 -*Theorem:* Eigenstates with different eigenvalues are orthogonal.
    9.84 -#+end_quote
    9.85 -
    9.86 -** COMMENT :
    9.87 -I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$
    9.88 -and $|b\rangle$ are eigenstates of $\Lambda$. This means that
    9.89 -
    9.90 -
    9.91 -\(
    9.92 -\begin{eqnarray}
    9.93 -\Lambda |a\rangle&=& a|a\rangle,\\
    9.94 -\Lambda|b\rangle&=& b|b\rangle.\\
    9.95 -\end{eqnarray}
    9.96 -\)
    9.97 -
    9.98 -If we take the difference of these eigenstates, we find that
    9.99 -
   9.100 -\(
   9.101 -\begin{eqnarray}
   9.102 -\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle
   9.103 -\qquad \text{(because $\Lambda$ is linear.)}\\
   9.104 -&=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and
   9.105 -$|b\rangle$ are eigenstates of $\Lambda$)}
   9.106 -\end{eqnarray}\)
   9.107 -
   9.108 -
   9.109 -which means that $a\neq b$.
   9.110 -
   9.111 -** Eigenvectors of hermitian operators span the space of solutions
   9.112 -
   9.113 -#+begin_quote
   9.114 -*Theorem:* If $\Omega$ is a hermitian operator, then every physically
   9.115 - allowed state can be written as a linear sum of eigenstates of
   9.116 - $\Omega$.
   9.117 -#+end_quote
   9.118 -
   9.119 -
   9.120 -
   9.121 -** Momentum and energy are hermitian operators
   9.122 -This ought to be true because hermitian operators correspond to
   9.123 -observable quantities. Since we expect momentum and energy to be
   9.124 -measureable quantities, we expect that there are hermitian operators
   9.125 -to represent them.
   9.126 -
   9.127 -
   9.128 -** Momentum and energy eigenstates in vacuum
   9.129 -An eigenstate of the momentum operator $P$ would be a state
   9.130 -\(|p\rangle\) such that \(P|p\rangle=p|p\rangle\).
   9.131 -
   9.132 -** Momentum and energy eigenstates in the infinitely deep well
   9.133 -
   9.134 -
   9.135 -
   9.136 -* Can you measure momentum in the infinite square well?
   9.137 -
   9.138 -
   9.139 -
   9.140 -** COMMENT  Momentum eigenstates
   9.141 -
   9.142 -In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the
   9.143 -momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\).
   9.144 -
   9.145 -In the infinitely deep potential well, the Hamiltonian is the same but
   9.146 -there is a new condition in order for states to qualify as physically
   9.147 -allowed: the states must not exist anywhere outside of well, as it
   9.148 -takes an infinite amount of energy to do so. 
   9.149 -
   9.150 -Notice that the momentum eigenstates defined above do /not/ satisfy
   9.151 -this condition.
   9.152 -
   9.153 -
   9.154 -
   9.155 -* COMMENT
   9.156 -For each physical system, there is a Schr\ouml{}dinger equation that
   9.157 -describes how a particle's state $|\psi\rangle$  will change over
   9.158 -time.
   9.159 -
   9.160 -\(\begin{eqnarray}
   9.161 -i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
   9.162 -H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
   9.163 -
   9.164 -This is a differential equation; each solution to the
   9.165 -Schr\ouml{}dinger equation is a state that is physically allowed for
   9.166 -our particle. Here, physically allowed states are
   9.167 -those that change in physically allowed ways. However, like any differential
   9.168 -equation, the Schr\ouml{}dinger equation can be accompanied by
   9.169 -/boundary conditions/\mdash{}conditions that further restrict which
   9.170 -states qualify as physically allowed.
   9.171 -
   9.172 -
   9.173 -
   9.174 -
   9.175 -** Eigenstates of momentum
   9.176 -
   9.177 -
   9.178 -
   9.179 -
   9.180 -#In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger
   9.181 -
   9.182 -#\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\)
   9.183 -
   9.184 -
   9.185 -
   9.186 -
   9.187 -
   9.188 -
   9.189 -
   9.190 -* COMMENT
   9.191 -
   9.192 -#* The infinite square well potential
   9.193 -
   9.194 -A particle exists in a potential that is
   9.195 -infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the
   9.196 -particle exists in a potential[fn:coords][fn:infinity]
   9.197 -
   9.198 -
   9.199 -\(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for
   9.200 -}\;x<0\text{ or }x>a.\end{cases}\)
   9.201 -
   9.202 -The Schr\ouml{}dinger equation describes how the particle's state 
   9.203 -\(|\psi\rangle\) will change over time in this system.
   9.204 -
   9.205 -\(\begin{eqnarray}
   9.206 -i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
   9.207 -H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
   9.208 -
   9.209 -This is a differential equation; each solution to the
   9.210 -Schr\ouml{}dinger equation is a state that is physically allowed for
   9.211 -our particle. Here, physically allowed states are
   9.212 -those that change in physically allowed ways. However, like any differential
   9.213 -equation, the Schr\ouml{}dinger equation can be accompanied by
   9.214 -/boundary conditions/\mdash{}conditions that further restrict which
   9.215 -states qualify as physically allowed.
   9.216 -
   9.217 -
   9.218 -Whenever possible, physicists impose these boundary conditions:
   9.219 -- A physically allowed state ought to be a /smoothly-varying function of position./ This means
   9.220 -  that if a particle in the state  is likely to be /at/ a particular location,
   9.221 -  it is also likely to be /near/ that location.
   9.222 -
   9.223 -These boundary conditions imply that for the square well potential in
   9.224 -this problem,
   9.225 -
   9.226 -- Physically allowed states must be totally confined to the well,
   9.227 -  because it takes an infinite amount of energy to exist anywhere
   9.228 -  outside of the well (and physically allowed states ought to have
   9.229 -  only finite energy).
   9.230 -- Physically allowed states must be increasingly unlikely to find very
   9.231 -  close to the walls of the well. This is because of two conditions: the above
   9.232 -  condition says that the particle is /impossible/ to find
   9.233 -  outside of the well, and the smoothly-varying condition says
   9.234 -  that if a particle is impossible to find at a particular location,
   9.235 -  it must be unlikely to be found nearby that location.
   9.236 -
   9.237 -#; physically allowed states are those that change in physically
   9.238 -#allowed ways.
   9.239 -
   9.240 -
   9.241 -#** Boundary conditions
   9.242 -Because the potential is infinite everywhere except within the well,
   9.243 -a realistic particle must be confined to exist only within the
   9.244 -well\mdash{}its wavefunction must be zero everywhere beyond the walls
   9.245 -of the well.
   9.246 -
   9.247 -
   9.248 -[fn:coords] I chose my coordinate system so that the well extends from
   9.249 -\(0<x<a\). Others choose a coordinate system so that the well extends from
   9.250 -\(-\frac{a}{2}<x<\frac{a}{2}\). Although both coordinate systems describe the same physical
   9.251 -situation, they give different-looking answers.
   9.252 -
   9.253 -[fn:infinity] Of course, infinite potentials are not
   9.254 -realistic. Instead, they are useful approximations to finite
   9.255 -potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
   9.256 -of the well\rdquo{} are close enough for your own practical
   9.257 -purposes. Having introduced a physical impossibility into the problem
   9.258 -already, we don't expect to get physically realistic solutions; we
   9.259 -just expect to get mathematically consistent ones. The forthcoming
   9.260 -trouble is that we don't.

    10.1 --- a/org/bk4.org	Fri Oct 28 00:03:05 2011 -0700
    10.2 +++ /dev/null	Thu Jan 01 00:00:00 1970 +0000
    10.3 @@ -1,309 +0,0 @@
    10.4 -#+TITLE: Bugs in quantum mechanics
    10.5 -#+AUTHOR: Dylan Holmes
    10.6 -#+SETUPFILE: ../../aurellem/org/setup.org
    10.7 -#+INCLUDE:   ../../aurellem/org/level-0.org
    10.8 -
    10.9 -#Bugs in Quantum Mechanics
   10.10 -#Bugs in the Quantum-Mechanical Momentum Operator
   10.11 -
   10.12 -
   10.13 -I studied quantum mechanics the same way I study most subjects\mdash{}
   10.14 -by collecting (and squashing) bugs in my understanding. One of these
   10.15 -bugs persisted throughout two semesters of
   10.16 -quantum mechanics coursework until I finally found
   10.17 -the paper 
   10.18 -[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
   10.19 -mechanics/]], which helped me stamp out the bug entirely. I decided to
   10.20 -write an article about the problem and its solution for a number of reasons:
   10.21 -
   10.22 -- Although the paper was not unreasonably dense, it was written for
   10.23 -  teachers. I wanted to write an article for students.
   10.24 -- I wanted to popularize the problem and its solution because other
   10.25 -  explanations are currently too hard to find. (Even Shankar's
   10.26 -  excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it.)
   10.27 -- I wanted to check that the bug was indeed entirely
   10.28 -  eradicated. Attempting an explanation is my way of making
   10.29 -  sure.
   10.30 -
   10.31 -* COMMENT
   10.32 - I recommend the
   10.33 -paper not only for students who are learning
   10.34 -quantum mechanics, but especially for teachers interested in debugging
   10.35 -them. 
   10.36 -
   10.37 -* COMMENT
   10.38 -On my first exam in quantum mechanics, my professor asked us to
   10.39 -describe how certain measurements would affect a particle in a
   10.40 -box. Many of these measurement questions required routine application
   10.41 -of skills we had recently learned\mdash{}first, you recall (or
   10.42 -calculate) the eigenstates of the quantity
   10.43 -to be measured; second, you write the given state as a linear
   10.44 -sum of these eigenstates\mdash{} the coefficients on each term give
   10.45 -the probability amplitude.
   10.46 -
   10.47 -
   10.48 -* What I thought I knew
   10.49 -
   10.50 -The following is a list of things I thought were true of quantum
   10.51 -mechanics; the catch is that the list contradicts itself.
   10.52 -
   10.53 -1. For any hermitian operator: Eigenstates with different eigenvalues are orthogonal.
   10.54 -2. For any hermitian operator: Any physically allowed state can be
   10.55 -   written as a linear sum of eigenstates of the operator.
   10.56 -3. The momentum operator and energy operator are hermitian, because
   10.57 -   momentum and energy are measureable quantities.
   10.58 -4. In the vacuum potential, the momentum and energy operators have these eigenstates:
   10.59 -   - the momentum operator has an eigenstate
   10.60 -     \(p(x)=\exp{(ipx/\hbar)}\) for each value of $p$.
   10.61 -   - the energy operator has an eigenstate \(|E\rangle =
   10.62 -     \alpha|p\rangle + \beta|-p\rangle\) for any \(\alpha,\beta\) and
   10.63 -     the particular choice of momentum $p=\sqrt{2mE}$.
   10.64 -5. In the infinitely deep potential well, the momentum and energy
   10.65 -   operators have these eigenstates:
   10.66 -   - The momentum eigenstates and energy eigenstates have the same form
   10.67 -     as in the vacuum potential: $p(x) =
   10.68 -     \exp{(ipx/\hbar)}$ and $|E\rangle = \alpha|p\rangle + \beta|-p\rangle$.
   10.69 -   - Even so, because of the boundary conditions on the
   10.70 -     well, we must make the following modifications:
   10.71 -     + Physically realistic states must be impossible to find outside the well. (Only a state of infinite
   10.72 -       energy could exist outside the well, and infinite energy is not
   10.73 -       realistic.) This requirement means, for example, that momentum
   10.74 -       eigenstates in the infinitely deep well must be
   10.75 -       \(p(x)
   10.76 -       = \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;
   10.77 -       \\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
   10.78 -     + Physically realistic states must vary smoothly throughout
   10.79 -       space. This means that if a particle in some state is very unlikely to be
   10.80 -       /at/ a particular location, it is also very unlikely be /near/
   10.81 -       that location. Combining this requirement with the above
   10.82 -       requirement, we find that the momentum operator no longer has
   10.83 -       an eigenstate for each value of $p$; instead, only values of
   10.84 -       $p$ that are integer multiples of $\pi a/\hbar$ are physically
   10.85 -       realistic. Similarly, the energy operator no longer has an
   10.86 -       eigenstate for each value of $E$; instead, the only energy
   10.87 -       eigenstates in the infinitely deep well
   10.88 -       are $E_n(x)=\sin(n\pi x/ a)$ for positive integers $n$.
   10.89 -
   10.90 -* COMMENT: 
   10.91 -
   10.92 -** Eigenstates with different eigenvalues are orthogonal
   10.93 -
   10.94 -#+begin_quote
   10.95 -*Theorem:* Eigenstates with different eigenvalues are orthogonal.
   10.96 -#+end_quote
   10.97 -
   10.98 -** COMMENT :
   10.99 -I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$
  10.100 -and $|b\rangle$ are eigenstates of $\Lambda$. This means that
  10.101 -
  10.102 -
  10.103 -\(
  10.104 -\begin{eqnarray}
  10.105 -\Lambda |a\rangle&=& a|a\rangle,\\
  10.106 -\Lambda|b\rangle&=& b|b\rangle.\\
  10.107 -\end{eqnarray}
  10.108 -\)
  10.109 -
  10.110 -If we take the difference of these eigenstates, we find that
  10.111 -
  10.112 -\(
  10.113 -\begin{eqnarray}
  10.114 -\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle
  10.115 -\qquad \text{(because $\Lambda$ is linear.)}\\
  10.116 -&=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and
  10.117 -$|b\rangle$ are eigenstates of $\Lambda$)}
  10.118 -\end{eqnarray}\)
  10.119 -
  10.120 -
  10.121 -which means that $a\neq b$.
  10.122 -
  10.123 -** Eigenvectors of hermitian operators span the space of solutions
  10.124 -
  10.125 -#+begin_quote
  10.126 -*Theorem:* If $\Omega$ is a hermitian operator, then every physically
  10.127 - allowed state can be written as a linear sum of eigenstates of
  10.128 - $\Omega$.
  10.129 -#+end_quote
  10.130 -
  10.131 -
  10.132 -
  10.133 -** Momentum and energy are hermitian operators
  10.134 -This ought to be true because hermitian operators correspond to
  10.135 -observable quantities. Since we expect momentum and energy to be
  10.136 -measureable quantities, we expect that there are hermitian operators
  10.137 -to represent them.
  10.138 -
  10.139 -
  10.140 -** Momentum and energy eigenstates in vacuum
  10.141 -An eigenstate of the momentum operator $P$ would be a state
  10.142 -\(|p\rangle\) such that \(P|p\rangle=p|p\rangle\).
  10.143 -
  10.144 -** Momentum and energy eigenstates in the infinitely deep well
  10.145 -
  10.146 -
  10.147 -
  10.148 -* Can you measure momentum in the infinitely deep well?
  10.149 -In summary, I thought I knew:
  10.150 -1. For any hermitian operator: eigenstates with different eigenvalues
  10.151 -   are orthogonal.
  10.152 -2. For any hermitian operator: any physically realistic state can be
  10.153 -   written as a linear sum of eigenstates of the operator.
  10.154 -3. The momentum operator and energy operator are hermitian, because
  10.155 -   momentum and energy are observable quantities. 
  10.156 -4. (The form of the momentum and energy eigenstates in the vacuum potential)
  10.157 -5. (The form of the momentum and energy eigenstates in the infinitely deep well potential)
  10.158 -
  10.159 -Additionally, I understood that because the infinitely deep potential
  10.160 -well is not realistic, states of such a system  are not necessarily
  10.161 -physically realistic. Instead, I understood
  10.162 -\ldquo{}realistic states\rdquo{} to be those that satisfy the physically
  10.163 -unrealistic Schr\ouml{}dinger equation and its boundary conditions.
  10.164 -
  10.165 -With that final caveat, here is the problem:
  10.166 -
  10.167 -According to (5), the momentum eigenstates in the well are 
  10.168 -
  10.169 -\(p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
  10.170 -
  10.171 -However, /these/ states are not orthogonal, which contradicts the
  10.172 -assumption that (3) the momentum operator is hermitian and (2)
  10.173 -eigenstates of a hermitian are orthogonal if they have different eigenvalues.
  10.174 -
  10.175 -#+begin_quote 
  10.176 -*Problem 1. The momentum eigenstates of the well are not orthogonal*
  10.177 -
  10.178 -/Proof./ If $p_1\neq p_2$, then 
  10.179 -
  10.180 -\(\begin{eqnarray}
  10.181 -\langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)dx\\
  10.182 -&=& \int_0^a p_1^*(x)p_2(x)dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{
  10.183 -outside the well.}\\
  10.184 -&=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)dx}}
  10.185 -\end{eqnarray}\)
  10.186 -$\square$
  10.187 -
  10.188 -#+end_quote
  10.189 -
  10.190 -
  10.191 -
  10.192 -** COMMENT  Momentum eigenstates
  10.193 -
  10.194 -In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the
  10.195 -momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\).
  10.196 -
  10.197 -In the infinitely deep potential well, the Hamiltonian is the same but
  10.198 -there is a new condition in order for states to qualify as physically
  10.199 -allowed: the states must not exist anywhere outside of well, as it
  10.200 -takes an infinite amount of energy to do so. 
  10.201 -
  10.202 -Notice that the momentum eigenstates defined above do /not/ satisfy
  10.203 -this condition.
  10.204 -
  10.205 -
  10.206 -
  10.207 -* COMMENT
  10.208 -For each physical system, there is a Schr\ouml{}dinger equation that
  10.209 -describes how a particle's state $|\psi\rangle$  will change over
  10.210 -time.
  10.211 -
  10.212 -\(\begin{eqnarray}
  10.213 -i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
  10.214 -H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
  10.215 -
  10.216 -This is a differential equation; each solution to the
  10.217 -Schr\ouml{}dinger equation is a state that is physically allowed for
  10.218 -our particle. Here, physically allowed states are
  10.219 -those that change in physically allowed ways. However, like any differential
  10.220 -equation, the Schr\ouml{}dinger equation can be accompanied by
  10.221 -/boundary conditions/\mdash{}conditions that further restrict which
  10.222 -states qualify as physically allowed.
  10.223 -
  10.224 -
  10.225 -
  10.226 -
  10.227 -** Eigenstates of momentum
  10.228 -
  10.229 -
  10.230 -
  10.231 -
  10.232 -#In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger
  10.233 -
  10.234 -#\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\)
  10.235 -
  10.236 -
  10.237 -
  10.238 -
  10.239 -
  10.240 -
  10.241 -
  10.242 -* COMMENT
  10.243 -
  10.244 -#* The infinite square well potential
  10.245 -
  10.246 -A particle exists in a potential that is
  10.247 -infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the
  10.248 -particle exists in a potential[fn:coords][fn:infinity]
  10.249 -
  10.250 -
  10.251 -\(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for
  10.252 -}\;x<0\text{ or }x>a.\end{cases}\)
  10.253 -
  10.254 -The Schr\ouml{}dinger equation describes how the particle's state 
  10.255 -\(|\psi\rangle\) will change over time in this system.
  10.256 -
  10.257 -\(\begin{eqnarray}
  10.258 -i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
  10.259 -H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
  10.260 -
  10.261 -This is a differential equation; each solution to the
  10.262 -Schr\ouml{}dinger equation is a state that is physically allowed for
  10.263 -our particle. Here, physically allowed states are
  10.264 -those that change in physically allowed ways. However, like any differential
  10.265 -equation, the Schr\ouml{}dinger equation can be accompanied by
  10.266 -/boundary conditions/\mdash{}conditions that further restrict which
  10.267 -states qualify as physically allowed.
  10.268 -
  10.269 -
  10.270 -Whenever possible, physicists impose these boundary conditions:
  10.271 -- A physically allowed state ought to be a /smoothly-varying function of position./ This means
  10.272 -  that if a particle in the state  is likely to be /at/ a particular location,
  10.273 -  it is also likely to be /near/ that location.
  10.274 -
  10.275 -These boundary conditions imply that for the square well potential in
  10.276 -this problem,
  10.277 -
  10.278 -- Physically allowed states must be totally confined to the well,
  10.279 -  because it takes an infinite amount of energy to exist anywhere
  10.280 -  outside of the well (and physically allowed states ought to have
  10.281 -  only finite energy).
  10.282 -- Physically allowed states must be increasingly unlikely to find very
  10.283 -  close to the walls of the well. This is because of two conditions: the above
  10.284 -  condition says that the particle is /impossible/ to find
  10.285 -  outside of the well, and the smoothly-varying condition says
  10.286 -  that if a particle is impossible to find at a particular location,
  10.287 -  it must be unlikely to be found nearby that location.
  10.288 -
  10.289 -#; physically allowed states are those that change in physically
  10.290 -#allowed ways.
  10.291 -
  10.292 -
  10.293 -#** Boundary conditions
  10.294 -Because the potential is infinite everywhere except within the well,
  10.295 -a realistic particle must be confined to exist only within the
  10.296 -well\mdash{}its wavefunction must be zero everywhere beyond the walls
  10.297 -of the well.
  10.298 -
  10.299 -
  10.300 -[fn:coords] I chose my coordinate system so that the well extends from
  10.301 -\(0<x<a\). Others choose a coordinate system so that the well extends from
  10.302 -\(-\frac{a}{2}<x<\frac{a}{2}\). Although both coordinate systems describe the same physical
  10.303 -situation, they give different-looking answers.
  10.304 -
  10.305 -[fn:infinity] Of course, infinite potentials are not
  10.306 -realistic. Instead, they are useful approximations to finite
  10.307 -potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
  10.308 -of the well\rdquo{} are close enough for your own practical
  10.309 -purposes. Having introduced a physical impossibility into the problem
  10.310 -already, we don't expect to get physically realistic solutions; we
  10.311 -just expect to get mathematically consistent ones. The forthcoming
  10.312 -trouble is that we don't.

    11.1 --- a/org/bk_quandary.org	Fri Oct 28 00:03:05 2011 -0700
    11.2 +++ /dev/null	Thu Jan 01 00:00:00 1970 +0000
    11.3 @@ -1,566 +0,0 @@
    11.4 -#+TITLE: Bugs in quantum mechanics
    11.5 -#+AUTHOR: Dylan Holmes
    11.6 -#+DESCRIPTION: I explain hermitian operators and use them to solve a confusing problem in quantum mechanics.
    11.7 -#+KEYWORDS: quantum mechanics, self-adjoint, hermitian, square well, momentum
    11.8 -#+SETUPFILE: ../../aurellem/org/setup.org
    11.9 -#+INCLUDE:   ../../aurellem/org/level-0.org
   11.10 -
   11.11 -
   11.12 -
   11.13 -#Bugs in Quantum Mechanics
   11.14 -#Bugs in the Quantum-Mechanical Momentum Operator
   11.15 -
   11.16 -
   11.17 -I studied quantum mechanics the same way I study most subjects\mdash{}
   11.18 -by collecting (and squashing) bugs in my understanding. One of these
   11.19 -bugs persisted throughout two semesters of
   11.20 -quantum mechanics coursework until I finally found
   11.21 -the paper 
   11.22 -[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
   11.23 -mechanics/]], which helped me stamp out the bug entirely. I decided to
   11.24 -write an article about the problem and its solution for a number of reasons:
   11.25 -
   11.26 -- Although the paper was not unreasonably dense, it was written for
   11.27 -  teachers. I wanted to write an article for students.
   11.28 -- I wanted to popularize the problem and its solution because other
   11.29 -  explanations are currently too hard to find. (Even Shankar's
   11.30 -  excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it, as far as I know.)
   11.31 -- Attempting an explanation is my way of making
   11.32 -  sure that the bug is /really/ gone.
   11.33 -# entirely eradicated.
   11.34 -
   11.35 -* COMMENT
   11.36 - I recommend the
   11.37 -paper not only for students who are learning
   11.38 -quantum mechanics, but especially for teachers interested in debugging
   11.39 -them. 
   11.40 -
   11.41 -* COMMENT
   11.42 -On my first exam in quantum mechanics, my professor asked us to
   11.43 -describe how certain measurements would affect a particle in a
   11.44 -box. Many of these measurement questions required routine application
   11.45 -of skills we had recently learned\mdash{}first, you recall (or
   11.46 -calculate) the eigenstates of the quantity
   11.47 -to be measured; second, you write the given state as a linear
   11.48 -sum of these eigenstates\mdash{} the coefficients on each term give
   11.49 -the probability amplitude.
   11.50 -
   11.51 -
   11.52 -* Two methods of calculation that give different results.
   11.53 -
   11.54 -In the infinitely deep well, there is a particle in the the
   11.55 -normalized state
   11.56 -
   11.57 - \(\psi(x) = \begin{cases}A\; x(x-a)&\text{for }0\lt{}x\lt{}a;\\0,&\text{for }x\lt{}0\text{ or }x>a.\end{cases}\) 
   11.58 -
   11.59 -This is apparently a perfectly respectable state: it is normalized ($A$ is a
   11.60 -normalization constant), it is zero
   11.61 -everywhere outside of the well, and it is moreover continuous.
   11.62 -
   11.63 -Even so, we will find a problem if we attempt to calculate the average
   11.64 -energy-squared of this state (that is, the quantity \(\langle \psi | H^2 | \psi \rangle\)). 
   11.65 -
   11.66 -** First method
   11.67 -
   11.68 -For short, define a new variable[fn:1] \(|\bar \psi\rangle \equiv
   11.69 -H|\psi\rangle\). We can express the state $|\bar\psi\rangle$ as a
   11.70 -function of $x$ because we know how to express $H$ and $\psi$ in terms
   11.71 -of  $x$: $H$ is the differential operator $\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}$, and
   11.72 -$\psi(x)$ is the function defined above. So, we get \(\quad\bar\psi(x) = \frac{-A\hbar^2}{m}\). For future reference, observe that $\bar\psi(x)$
   11.73 -is constant.
   11.74 -
   11.75 -Having introduced $|\bar\psi\rangle$, we can calculate the average energy-squared of $|\psi\rangle$ in the
   11.76 -following way.
   11.77 -
   11.78 -\(\begin{eqnarray}
   11.79 -\langle \psi | H^2 |\psi\rangle &=& \langle \psi H^\dagger | H
   11.80 -\psi\rangle\\
   11.81 -&=& \langle \psi H | H\psi \rangle\\
   11.82 -&=& \langle \bar\psi | \bar\psi \rangle\\
   11.83 -&=& \int_0^a \left(\frac{A\hbar^2}{m}\right)^2\;dx\\
   11.84 -&=& \frac{A^2\hbar^4 a}{m^2}\\
   11.85 -\end{eqnarray}\)
   11.86 - 
   11.87 -For future reference, observe that this value is  nonzero
   11.88 -(which makes sense).
   11.89 -
   11.90 -** Second method
   11.91 -We can also calculate the average energy-squared of $|\psi\rangle$ in the
   11.92 -following way.
   11.93 -
   11.94 -\begin{eqnarray}
   11.95 -\langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\
   11.96 -&=& \langle \psi |H \bar\psi \rangle\\
   11.97 -&=&\int_0^a Ax(x-a)
   11.98 -\left(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\right)\frac{-A\hbar^2}{m}\;dx\\
   11.99 -&=& 0\quad (!)\\
  11.100 -\end{eqnarray}
  11.101 -
  11.102 -The second-to-last term must be zero because the second derivative
  11.103 -of a constant \(\left(\frac{-A\hbar^2}{m}\right)\) is zero.
  11.104 -
  11.105 -* What is the problem?
  11.106 -
  11.107 -To recap: We used two different methods to calculate the average
  11.108 -energy-squared of a state $|\psi\rangle$. For the first method, we
  11.109 -used the fact that $H$ is a hermitian operator, replacing \(\langle
  11.110 -\psi H^\dagger | H \psi\rangle\) with \(\langle \psi H | H
  11.111 -\psi\rangle\). Using this substitution rule, we calculated the answer.
  11.112 -
  11.113 -For the second method, we didn't use the fact that $H$ was hermitian;
  11.114 -instead, we used the fact that we know how to represent $H$ and $\psi$
  11.115 -as functions of $x$: $H$ is a differential operator
  11.116 -\(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\) and $\psi$ is a quadratic
  11.117 -function of $x$. By applying $H$ to $\psi$, we took several
  11.118 -derivatives and arrived at our answer.
  11.119 -
  11.120 -These two methods gave different results. In the following sections,
  11.121 -I'll describe and analyze the source of this difference.
  11.122 -
  11.123 -** Physical operators only act on physical wavefunctions
  11.124 -   :PROPERTIES:
  11.125 -   :ORDERED:  t
  11.126 -   :END:
  11.127 -#In quantum mechanics, an operator is a function that takes in a
  11.128 -#physical state and produces another physical state as ouput. Some
  11.129 -#operators correspond to physical quantities such as energy,
  11.130 -#momentum, or position; the mathematical properties of these operators correspond to
  11.131 -#physical properties of the system.
  11.132 -
  11.133 -#Eigenstates are an example of this correspondence: an 
  11.134 -
  11.135 -Physical states are represented as wavefunctions in quantum
  11.136 -mechanics. Just as we disallow certain physically nonsensical states
  11.137 -in classical mechanics (for example, we consider it to be nonphysical
  11.138 -for an object to spontaneously disappear from one place and reappear
  11.139 -in another), we also disallow certain wavefunctions in quantum
  11.140 -mechanics.
  11.141 -
  11.142 -For example, since wavefunctions are supposed to correspond to
  11.143 -probability amplitudes, we require wavefunctions to be normalized
  11.144 -\((\langle \psi |\psi \rangle = 1)\). Generally, we disallow
  11.145 -wavefunctions that do not satisfy this property (although there are
  11.146 -some exceptions[fn:2]).
  11.147 -
  11.148 -As another example, we generally expect probability to vary smoothly\mdash{}if
  11.149 -a particle is very likely or very unlikely to be found at a particular
  11.150 -location, it should also be somewhat likely or somewhat unlikely to be
  11.151 -found /near/ that location. In more precise terms, we expect that for
  11.152 -physically meaningful wavefunctions, the probability 
  11.153 -\(\text{Pr}(x)=\int^x \psi^*\psi\) should be a continuous function of
  11.154 -$x$ and, again, we disallow wavefunctions that do not satisfy this
  11.155 -property because we consider them to be physically nonsensical.
  11.156 -
  11.157 -So, physical wavefunctions must satisfy certain properties
  11.158 -like the two just described. Wavefunctions that do not satisfy these properties are
  11.159 -rejected for being physically nonsensical: even though we can perform
  11.160 -calculations with them, the mathematical results we obtain do not mean
  11.161 -anything physically.
  11.162 -
  11.163 -Now, in quantum mechanics, an *operator* is a function that converts
  11.164 -states into other states. Some operators correspond to
  11.165 -physical quantities such as energy, momentum, or position, and as a
  11.166 -result, the mathematical properties of these operators correspond to
  11.167 -physical properties of the system. Physical operators are furthermore
  11.168 -subject to the following rule: they are only allowed to operate on 
  11.169 -#physical wavefunctions, and they are only allowed to produce
  11.170 -#physical wavefunctions[fn:why].
  11.171 -the wavefunctions we have deemed physical, and they only produce physical wavefunctions[fn::
  11.172 -
  11.173 - If you require a hermitian operator to have physical
  11.174 -  eigenstates, you get a very strong result: you guarantee that the
  11.175 -  operator will convert /every/ physical wavefunction into another
  11.176 -  physical wavefunction:
  11.177 -
  11.178 -  For any linear operator $\Omega$, the eigenvalue equation is
  11.179 -\(\Omega|\omega\rangle = \omega |\omega\rangle\). Notice that if an
  11.180 -eigenstate $|\omega\rangle$ is a physical wavefunction, the
  11.181 -eigenvalue equation equation forces $\Omega|\omega\rangle$ to be a
  11.182 -physical wavefunction as well. To elaborate, if the eigenstates of
  11.183 -$\Omega$ are physical functions, then $\Omega$ is guaranteed to
  11.184 -convert them into other physical functions.  Even more is true if the
  11.185 -operator $\Omega$ is also hermitian: there is a theorem which states
  11.186 -that \ldquo{}If \Omega is hermitian, then every physical wavefunction
  11.187 -can be written as a weighted sum of eigenstates of \Omega.\rdquo{} This
  11.188 -theorem implies that /if $\Omega$ is hermitian/, and /if the eigenstates
  11.189 -of \Omega are physically allowed/, then \Omega is guaranteed to
  11.190 -convert every physically allowed wavefunction into another physically
  11.191 -allowed wavefunction.].
  11.192 -
  11.193 -In fact, this rule for physical operators is the source of our
  11.194 -problem, as we unknowingly violated it when applying our second
  11.195 -method!
  11.196 -
  11.197 -** The violation
  11.198 -
  11.199 -I'll start explaining this violation by being more specific about the
  11.200 -infinitely deep well potential. We have said already that physicists
  11.201 -require wavefunctions to satisfy certain properties in order to be
  11.202 -deemed \ldquo{}physical\rdquo{}. To be specific, wavefunctions in the
  11.203 -infinitely deep well
  11.204 -- Must be *normalizable*, because they correspond to
  11.205 -  probability amplitudes.
  11.206 -- Must have *smoothly-varying probability*, because if a particle is very
  11.207 -  likely to be at a location, it ought to be likely to be /near/
  11.208 -  it as well.
  11.209 -- Must *not exist outside the well*, because it
  11.210 -  would take an infinite amount of energy to do so.
  11.211 -
  11.212 -Additionally, by combining the second and third conditions, some
  11.213 -physicists reason that wavefunctions in the infinitely deep well
  11.214 -
  11.215 -- Must *become zero* towards the edges of the well.
  11.216 -
  11.217 -
  11.218 -
  11.219 -
  11.220 -You'll remember we had
  11.221 -
  11.222 -\(
  11.223 -\begin{eqnarray}
  11.224 -\psi(x) &=& A\;x(x-a)\\
  11.225 -\bar\psi(x)&=& \frac{-A\hbar^2}{m}\\
  11.226 -&&\text{for }0\lt{}x\lt{}a\\
  11.227 -\end{eqnarray}
  11.228 -\)
  11.229 -
  11.230 -In our second method, we wrote 
  11.231 -
  11.232 -
  11.233 -\(\begin{eqnarray}
  11.234 -\langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\
  11.235 -&=& \langle \psi |H \bar\psi \rangle\\
  11.236 -& \vdots&\\
  11.237 -&=& 0\\
  11.238 -\end{eqnarray}\)
  11.239 -
  11.240 -However, there are two issues here. First, we were not allowed to operate with $H$ on the wavefunction
  11.241 -$|\bar \psi\rangle$, because $H$ is a physical operator and $|\bar
  11.242 -\psi\rangle$ is a nonphysical state: in the infinite square well,
  11.243 -physical wavefunctions must approach zero at the edges of the well,
  11.244 -which the constant function $|\bar\psi\rangle$ does not do. By
  11.245 -feeding $H$ a nonphysical wavefunction, we obtained nonsensical
  11.246 -results.
  11.247 -
  11.248 -Second, we claimed that $H$ was a physical operator\mdash{}that $H$
  11.249 -was hermitian. According to the rule, this means $H$ must convert physical states into other
  11.250 -physical states. But $H$ converts the physical state $|\psi\rangle$
  11.251 -into the nonphysical state $|\bar\psi\rangle = H|\psi\rangle$ is not. Because $H$ converts some
  11.252 -physical states into nonphysical states, it cannot be a hermitian operator.
  11.253 -
  11.254 -** Boundary conditions affect hermiticity
  11.255 -We have now discovered a flaw: when applied to the state
  11.256 -$|\psi\rangle$, the second method violates the rule that physical
  11.257 -operators must only take in physical states and must only produce
  11.258 -physical states. This suggests that the problem was with the state
  11.259 -$|\bar\psi\rangle$ that we wrongly fed into $H$. However, the problem
  11.260 -is more serious still: the state $|\psi\rangle 
  11.261 -
  11.262 -** COMMENT Re-examining physical constraints
  11.263 -
  11.264 -We have now discovered a flaw: when applied to the state
  11.265 -$|\psi\rangle$, the second method violates the rule that physical
  11.266 -operators must only take in physical states and must only produce
  11.267 -physical states. Let's examine the problem more closely.
  11.268 -
  11.269 -We have said already that physicists require wavefunctions to satisfy
  11.270 -certain properties in order to be deemed \ldquo{}physical\rdquo{}. To
  11.271 -be specific, wavefunctions in the infinitely deep well
  11.272 -- Must be *normalizable*, because they correspond to
  11.273 -  probability amplitudes.
  11.274 -- Must have *smoothly-varying probability*, because if a particle is very
  11.275 -  likely to be at a location, it ought to be likely to be /near/
  11.276 -  it as well.
  11.277 -- Must *not exist outside the well*, because it
  11.278 -  would take an infinite amount of energy to do so.
  11.279 -
  11.280 -We now have discovered an important flaw in the second method: when
  11.281 -applied to the state $|\bar\psi\rangle$, the second method violates
  11.282 -the rule that physical operators must only take in
  11.283 -physical states and must only produce physical states. The problem is
  11.284 -even more serious, however
  11.285 -
  11.286 -
  11.287 -
  11.288 -[fn:1] I'm defining a new variable just to make certain expressions
  11.289 -  look shorter; this cannot affect the content of the answer we'll
  11.290 -  get. 
  11.291 -
  11.292 -[fn:2] For example, in vaccuum (i.e., when the potential of the
  11.293 -  physical system is $V(x)=0$ throughout all space), the momentum
  11.294 -  eigenstates are not normalizable\mdash{}the relevant integral blows
  11.295 -  up to infinity instead of converging to a number. Physicists modify
  11.296 -  the definition of normalization slightly so that
  11.297 -  \ldquo{}delta-normalizable \rdquo{} functions like these are included
  11.298 -  among the physical wavefunctions.
  11.299 -
  11.300 -
  11.301 -
  11.302 -* COMMENT: What I thought I knew
  11.303 -
  11.304 -The following is a list of things I thought were true of quantum
  11.305 -mechanics; the catch is that the list contradicts itself.
  11.306 -
  11.307 -1. For any hermitian operator: Eigenstates with different eigenvalues are orthogonal.
  11.308 -2. For any hermitian operator: Any physically allowed state can be
  11.309 -   written as a linear sum of eigenstates of the operator.
  11.310 -3. The momentum operator and energy operator are hermitian, because
  11.311 -   momentum and energy are measureable quantities.
  11.312 -4. In the vacuum potential, the momentum and energy operators have these eigenstates:
  11.313 -   - the momentum operator has an eigenstate
  11.314 -     \(p(x)=\exp{(ipx/\hbar)}\) for each value of $p$.
  11.315 -   - the energy operator has an eigenstate \(|E\rangle =
  11.316 -     \alpha|p\rangle + \beta|-p\rangle\) for any \(\alpha,\beta\) and
  11.317 -     the particular choice of momentum $p=\sqrt{2mE}$.
  11.318 -5. In the infinitely deep potential well, the momentum and energy
  11.319 -   operators have these eigenstates:
  11.320 -   - The momentum eigenstates and energy eigenstates have the same form
  11.321 -     as in the vacuum potential: $p(x) =
  11.322 -     \exp{(ipx/\hbar)}$ and $|E\rangle = \alpha|p\rangle + \beta|-p\rangle$.
  11.323 -   - Even so, because of the boundary conditions on the
  11.324 -     well, we must make the following modifications:
  11.325 -     + Physically realistic states must be impossible to find outside the well. (Only a state of infinite
  11.326 -       energy could exist outside the well, and infinite energy is not
  11.327 -       realistic.) This requirement means, for example, that momentum
  11.328 -       eigenstates in the infinitely deep well must be
  11.329 -       \(p(x)
  11.330 -       = \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;
  11.331 -       \\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
  11.332 -     + Physically realistic states must vary smoothly throughout
  11.333 -       space. This means that if a particle in some state is very unlikely to be
  11.334 -       /at/ a particular location, it is also very unlikely be /near/
  11.335 -       that location. Combining this requirement with the above
  11.336 -       requirement, we find that the momentum operator no longer has
  11.337 -       an eigenstate for each value of $p$; instead, only values of
  11.338 -       $p$ that are integer multiples of $\pi \hbar/a$ are physically
  11.339 -       realistic. Similarly, the energy operator no longer has an
  11.340 -       eigenstate for each value of $E$; instead, the only energy
  11.341 -       eigenstates in the infinitely deep well
  11.342 -       are $E_n(x)=\sin(n\pi x/ a)$ for positive integers $n$.
  11.343 -
  11.344 -* COMMENT: 
  11.345 -
  11.346 -** Eigenstates with different eigenvalues are orthogonal
  11.347 -
  11.348 -#+begin_quote
  11.349 -*Theorem:* Eigenstates with different eigenvalues are orthogonal.
  11.350 -#+end_quote
  11.351 -
  11.352 -** COMMENT :
  11.353 -I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$
  11.354 -and $|b\rangle$ are eigenstates of $\Lambda$. This means that
  11.355 -
  11.356 -
  11.357 -\(
  11.358 -\begin{eqnarray}
  11.359 -\Lambda |a\rangle&=& a|a\rangle,\\
  11.360 -\Lambda|b\rangle&=& b|b\rangle.\\
  11.361 -\end{eqnarray}
  11.362 -\)
  11.363 -
  11.364 -If we take the difference of these eigenstates, we find that
  11.365 -
  11.366 -\(
  11.367 -\begin{eqnarray}
  11.368 -\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle
  11.369 -\qquad \text{(because $\Lambda$ is linear.)}\\
  11.370 -&=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and
  11.371 -$|b\rangle$ are eigenstates of $\Lambda$)}
  11.372 -\end{eqnarray}\)
  11.373 -
  11.374 -
  11.375 -which means that $a\neq b$.
  11.376 -
  11.377 -** Eigenvectors of hermitian operators span the space of solutions
  11.378 -
  11.379 -#+begin_quote
  11.380 -*Theorem:* If $\Omega$ is a hermitian operator, then every physically
  11.381 - allowed state can be written as a linear sum of eigenstates of
  11.382 - $\Omega$.
  11.383 -#+end_quote
  11.384 -
  11.385 -
  11.386 -
  11.387 -** Momentum and energy are hermitian operators
  11.388 -This ought to be true because hermitian operators correspond to
  11.389 -observable quantities. Since we expect momentum and energy to be
  11.390 -measureable quantities, we expect that there are hermitian operators
  11.391 -to represent them.
  11.392 -
  11.393 -
  11.394 -** Momentum and energy eigenstates in vacuum
  11.395 -An eigenstate of the momentum operator $P$ would be a state
  11.396 -\(|p\rangle\) such that \(P|p\rangle=p|p\rangle\).
  11.397 -
  11.398 -** Momentum and energy eigenstates in the infinitely deep well
  11.399 -
  11.400 -
  11.401 -
  11.402 -* COMMENT Can you measure momentum in the infinitely deep well?
  11.403 -In summary, I thought I knew:
  11.404 -1. For any hermitian operator: eigenstates with different eigenvalues
  11.405 -   are orthogonal.
  11.406 -2. For any hermitian operator: any physically realistic state can be
  11.407 -   written as a linear sum of eigenstates of the operator.
  11.408 -3. The momentum operator and energy operator are hermitian, because
  11.409 -   momentum and energy are observable quantities. 
  11.410 -4. (The form of the momentum and energy eigenstates in the vacuum potential)
  11.411 -5. (The form of the momentum and energy eigenstates in the infinitely deep well potential)
  11.412 -
  11.413 -Additionally, I understood that because the infinitely deep potential
  11.414 -well is not realistic, states of such a system  are not necessarily
  11.415 -physically realistic. Instead, I understood
  11.416 -\ldquo{}realistic states\rdquo{} to be those that satisfy the physically
  11.417 -unrealistic Schr\ouml{}dinger equation and its boundary conditions.
  11.418 -
  11.419 -With that final caveat, here is the problem:
  11.420 -
  11.421 -According to (5), the momentum eigenstates in the well are 
  11.422 -
  11.423 -\(p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
  11.424 -
  11.425 -(Additionally, we require that $p$ be an integer multiple of $\pi\hbar/a$.) 
  11.426 -
  11.427 -However, /these/ states are not orthogonal, which contradicts the
  11.428 -assumption that (3) the momentum operator is hermitian and (2)
  11.429 -eigenstates of a hermitian are orthogonal if they have different eigenvalues.
  11.430 -
  11.431 -#+begin_quote 
  11.432 -*Problem 1. The momentum eigenstates of the well are not orthogonal*
  11.433 -
  11.434 -/Proof./ If $p_1\neq p_2$, then 
  11.435 -
  11.436 -\(\begin{eqnarray}
  11.437 -\langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)\,dx\\
  11.438 -&=& \int_0^a p_1^*(x)p_2(x)\,dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{
  11.439 -outside the well.}\\
  11.440 -&=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)\,dx}}\\
  11.441 -&=& \int_0^a \exp{\left[i(p_1-p_2)x/\hbar\right]}\,dx\\
  11.442 -\end{eqnarray}\)
  11.443 -$\square$
  11.444 -
  11.445 -#+end_quote
  11.446 -
  11.447 -
  11.448 -
  11.449 -** COMMENT  Momentum eigenstates
  11.450 -
  11.451 -In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the
  11.452 -momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\).
  11.453 -
  11.454 -In the infinitely deep potential well, the Hamiltonian is the same but
  11.455 -there is a new condition in order for states to qualify as physically
  11.456 -allowed: the states must not exist anywhere outside of well, as it
  11.457 -takes an infinite amount of energy to do so. 
  11.458 -
  11.459 -Notice that the momentum eigenstates defined above do /not/ satisfy
  11.460 -this condition.
  11.461 -
  11.462 -
  11.463 -
  11.464 -* COMMENT
  11.465 -For each physical system, there is a Schr\ouml{}dinger equation that
  11.466 -describes how a particle's state $|\psi\rangle$  will change over
  11.467 -time.
  11.468 -
  11.469 -\(\begin{eqnarray}
  11.470 -i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
  11.471 -H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
  11.472 -
  11.473 -This is a differential equation; each solution to the
  11.474 -Schr\ouml{}dinger equation is a state that is physically allowed for
  11.475 -our particle. Here, physically allowed states are
  11.476 -those that change in physically allowed ways. However, like any differential
  11.477 -equation, the Schr\ouml{}dinger equation can be accompanied by
  11.478 -/boundary conditions/\mdash{}conditions that further restrict which
  11.479 -states qualify as physically allowed.
  11.480 -
  11.481 -
  11.482 -
  11.483 -
  11.484 -** Eigenstates of momentum
  11.485 -
  11.486 -
  11.487 -
  11.488 -
  11.489 -#In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger
  11.490 -
  11.491 -#\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\)
  11.492 -
  11.493 -
  11.494 -
  11.495 -
  11.496 -
  11.497 -
  11.498 -
  11.499 -* COMMENT
  11.500 -
  11.501 -#* The infinite square well potential
  11.502 -
  11.503 -A particle exists in a potential that is
  11.504 -infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the
  11.505 -particle exists in a potential[fn:coords][fn:infinity]
  11.506 -
  11.507 -
  11.508 -\(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for
  11.509 -}\;x<0\text{ or }x>a.\end{cases}\)
  11.510 -
  11.511 -The Schr\ouml{}dinger equation describes how the particle's state 
  11.512 -\(|\psi\rangle\) will change over time in this system.
  11.513 -
  11.514 -\(\begin{eqnarray}
  11.515 -i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
  11.516 -H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
  11.517 -
  11.518 -This is a differential equation; each solution to the
  11.519 -Schr\ouml{}dinger equation is a state that is physically allowed for
  11.520 -our particle. Here, physically allowed states are
  11.521 -those that change in physically allowed ways. However, like any differential
  11.522 -equation, the Schr\ouml{}dinger equation can be accompanied by
  11.523 -/boundary conditions/\mdash{}conditions that further restrict which
  11.524 -states qualify as physically allowed.
  11.525 -
  11.526 -
  11.527 -Whenever possible, physicists impose these boundary conditions:
  11.528 -- A physically allowed state ought to be a /smoothly-varying function of position./ This means
  11.529 -  that if a particle in the state  is likely to be /at/ a particular location,
  11.530 -  it is also likely to be /near/ that location.
  11.531 -
  11.532 -These boundary conditions imply that for the square well potential in
  11.533 -this problem,
  11.534 -
  11.535 -- Physically allowed states must be totally confined to the well,
  11.536 -  because it takes an infinite amount of energy to exist anywhere
  11.537 -  outside of the well (and physically allowed states ought to have
  11.538 -  only finite energy).
  11.539 -- Physically allowed states must be increasingly unlikely to find very
  11.540 -  close to the walls of the well. This is because of two conditions: the above
  11.541 -  condition says that the particle is /impossible/ to find
  11.542 -  outside of the well, and the smoothly-varying condition says
  11.543 -  that if a particle is impossible to find at a particular location,
  11.544 -  it must be unlikely to be found nearby that location.
  11.545 -
  11.546 -#; physically allowed states are those that change in physically
  11.547 -#allowed ways.
  11.548 -
  11.549 -
  11.550 -#** Boundary conditions
  11.551 -Because the potential is infinite everywhere except within the well,
  11.552 -a realistic particle must be confined to exist only within the
  11.553 -well\mdash{}its wavefunction must be zero everywhere beyond the walls
  11.554 -of the well.
  11.555 -
  11.556 -
  11.557 -[fn:coords] I chose my coordinate system so that the well extends from
  11.558 -\(0<x<a\). Others choose a coordinate system so that the well extends from
  11.559 -\(-\frac{a}{2}<x<\frac{a}{2}\). Although both coordinate systems describe the same physical
  11.560 -situation, they give different-looking answers.
  11.561 -
  11.562 -[fn:infinity] Of course, infinite potentials are not
  11.563 -realistic. Instead, they are useful approximations to finite
  11.564 -potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
  11.565 -of the well\rdquo{} are close enough for your own practical
  11.566 -purposes. Having introduced a physical impossibility into the problem
  11.567 -already, we don't expect to get physically realistic solutions; we
  11.568 -just expect to get mathematically consistent ones. The forthcoming
  11.569 -trouble is that we don't.

    12.1 --- a/org/bkup.org	Fri Oct 28 00:03:05 2011 -0700
    12.2 +++ /dev/null	Thu Jan 01 00:00:00 1970 +0000
    12.3 @@ -1,49 +0,0 @@
    12.4 -#+TITLE: Bugs in Quantum Mechanics
    12.5 -#+AUTHOR: Dylan Holmes
    12.6 -#+SETUPFILE: ../../aurellem/org/setup.org
    12.7 -#+INCLUDE:   ../../aurellem/org/level-0.org
    12.8 -
    12.9 -
   12.10 -#Bugs in the Quantum-Mechanical Momentum Operator
   12.11 -
   12.12 -
   12.13 -I studied quantum mechanics the same way I study most subjects\mdash{}
   12.14 -by collecting (and squashing) bugs in my understanding. One of these
   12.15 -bugs persisted throughout two semesters of
   12.16 -quantum mechanics coursework until I finally found
   12.17 -the paper 
   12.18 -[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
   12.19 -mechanics/]], which helped me stamp out the bug entirely. I decided to
   12.20 -write an article about the problem and its solution for a number of reasons:
   12.21 -
   12.22 -- Although the paper was not unreasonably dense, it was written for
   12.23 -  teachers. I wanted to write an article for students.
   12.24 -- I wanted to popularize the problem and its solution because
   12.25 -  other explanations are currently too hard to find.
   12.26 -- I wanted to check that the bug was indeed entirely
   12.27 -  eradicated. Attempting an explanation is my way of making
   12.28 -  sure.
   12.29 -
   12.30 -* COMMENT
   12.31 - I recommend the
   12.32 -paper not only for students who are learning
   12.33 -quantum mechanics, but especially for teachers interested in debugging
   12.34 -them. 
   12.35 -
   12.36 -* COMMENT
   12.37 -On my first exam in quantum mechanics, my professor asked us to
   12.38 -describe how certain measurements would affect a particle in a
   12.39 -box. Many of these measurement questions required routine application
   12.40 -of skills we had recently learned\mdash{}first, you recall (or
   12.41 -calculate) the eigenstates of the quantity
   12.42 -to be measured; second, you write the given state as a linear
   12.43 -sum of these eigenstates\mdash{} the coefficients on each term give
   12.44 -the probability amplitude.
   12.45 -
   12.46 -* Statement of the Problem
   12.47 -A particle is 
   12.48 -
   12.49 -
   12.50 -
   12.51 -
   12.52 -* COMMENT [TABLE-OF-CONTENTS]