1.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
     1.2 +++ b/org/bk.org	Mon Oct 17 23:17:55 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/org/bk2.org	Mon Oct 17 23:17:55 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/org/bk3.org	Mon Oct 17 23:17:55 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/org/bk4.org	Mon Oct 17 23:17:55 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/org/bk_quandary.org	Mon Oct 17 23:17:55 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/org/bkup.org	Mon Oct 17 23:17:55 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 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
     7.2 +++ b/org/quandary.org	Mon Oct 17 23:17:55 2011 -0700
     7.3 @@ -0,0 +1,631 @@
     7.4 +#+TITLE: Bugs in quantum mechanics
     7.5 +#+AUTHOR: Dylan Holmes
     7.6 +#+DESCRIPTION: I explain hermitian operators and use them to solve a confusing problem in quantum mechanics.
     7.7 +#+KEYWORDS: quantum mechanics, self-adjoint, hermitian, square well, momentum
     7.8 +#+SETUPFILE: ../../aurellem/org/setup.org
     7.9 +#+INCLUDE:   ../../aurellem/org/level-0.org
    7.10 +
    7.11 +
    7.12 +
    7.13 +#Bugs in Quantum Mechanics
    7.14 +#Bugs in the Quantum-Mechanical Momentum Operator
    7.15 +
    7.16 +
    7.17 +I studied quantum mechanics the same way I study most subjects\mdash{}
    7.18 +by collecting (and squashing) bugs in my understanding. One of these
    7.19 +bugs persisted throughout two semesters of
    7.20 +quantum mechanics coursework until I finally found
    7.21 +the paper 
    7.22 +[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
    7.23 +mechanics/]], which helped me stamp out the bug entirely. I decided to
    7.24 +write an article about the problem and its solution for a number of reasons:
    7.25 +
    7.26 +- Although the paper was not unreasonably dense, it was written for
    7.27 +  teachers. I wanted to write an article for students.
    7.28 +- I wanted to popularize the problem and its solution because other
    7.29 +  explanations are currently too hard to find. (Even Shankar's
    7.30 +  excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it, as far as I know.)
    7.31 +- Attempting an explanation is my way of making
    7.32 +  sure that the bug really /is/ gone.
    7.33 +# entirely eradicated.
    7.34 +
    7.35 +* COMMENT
    7.36 + I recommend the
    7.37 +paper not only for students who are learning
    7.38 +quantum mechanics, but especially for teachers interested in debugging
    7.39 +them. 
    7.40 +
    7.41 +* COMMENT
    7.42 +On my first exam in quantum mechanics, my professor asked us to
    7.43 +describe how certain measurements would affect a particle in a
    7.44 +box. Many of these measurement questions required routine application
    7.45 +of skills we had recently learned\mdash{}first, you recall (or
    7.46 +calculate) the eigenstates of the quantity
    7.47 +to be measured; second, you write the given state as a linear
    7.48 +sum of these eigenstates\mdash{} the coefficients on each term give
    7.49 +the probability amplitude.
    7.50 +
    7.51 +
    7.52 +* Two methods of calculation that give different results.
    7.53 +
    7.54 +In the infinitely deep well, there is a particle in the the
    7.55 +normalized state
    7.56 +
    7.57 + \(\psi(x) = \begin{cases}A\; x(x-a)&\text{for }0\lt{}x\lt{}a;\\0,&\text{for }x\lt{}0\text{ or }x>a.\end{cases}\) 
    7.58 +
    7.59 +This is apparently a perfectly respectable state: it is normalized ($A$ is a
    7.60 +normalization constant), it is zero
    7.61 +everywhere outside of the well, and it is moreover continuous.
    7.62 +
    7.63 +Even so, we will find a problem if we attempt to calculate the average
    7.64 +energy-squared of this state (that is, the quantity \(\langle \psi | H^2 | \psi \rangle\)). 
    7.65 +
    7.66 +** First method
    7.67 +
    7.68 +For short, define a new variable[fn:1] \(|\bar \psi\rangle \equiv
    7.69 +H|\psi\rangle\). We can express the state $|\bar\psi\rangle$ as a
    7.70 +function of $x$ because we know how to express $H$ and $\psi$ in terms
    7.71 +of  $x$: $H$ is the differential operator $\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}$, and
    7.72 +$\psi(x)$ is the function defined above. So, we get \(\quad\bar\psi(x) = \frac{-A\hbar^2}{m}\). For future reference, observe that $\bar\psi(x)$
    7.73 +is constant.
    7.74 +
    7.75 +Having introduced $|\bar\psi\rangle$, we can calculate the average energy-squared of $|\psi\rangle$ in the
    7.76 +following way.
    7.77 +
    7.78 +\(\begin{eqnarray}
    7.79 +\langle \psi | H^2 |\psi\rangle &=& \langle \psi H^\dagger | H
    7.80 +\psi\rangle\\
    7.81 +&=& \langle \psi H | H\psi \rangle\\
    7.82 +&=& \langle \bar\psi | \bar\psi \rangle\\
    7.83 +&=& \int_0^a \left(\frac{A\hbar^2}{m}\right)^2\;dx\\
    7.84 +&=& \frac{A^2\hbar^4 a}{m^2}\\
    7.85 +\end{eqnarray}\)
    7.86 + 
    7.87 +For future reference, observe that this value is  nonzero
    7.88 +(which makes sense).
    7.89 +
    7.90 +** Second method
    7.91 +We can also calculate the average energy-squared of $|\psi\rangle$ in the
    7.92 +following way.
    7.93 +
    7.94 +\begin{eqnarray}
    7.95 +\langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\
    7.96 +&=& \langle \psi |H \bar\psi \rangle\\
    7.97 +&=&\int_0^a Ax(x-a)
    7.98 +\left(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\right)\frac{-A\hbar^2}{m}\;dx\\
    7.99 +&=& 0\quad (!)\\
   7.100 +\end{eqnarray}
   7.101 +
   7.102 +The second-to-last term must be zero because the second derivative
   7.103 +of a constant \(\left(\frac{-A\hbar^2}{m}\right)\) is zero.
   7.104 +
   7.105 +* What is the problem?
   7.106 +
   7.107 +To recap: We used two different methods to calculate the average
   7.108 +energy-squared of a state $|\psi\rangle$. For the first method, we
   7.109 +used the fact that $H$ is a hermitian operator, replacing \(\langle
   7.110 +\psi H^\dagger | H \psi\rangle\) with \(\langle \psi H | H
   7.111 +\psi\rangle\). Using this substitution rule, we calculated the answer.
   7.112 +
   7.113 +For the second method, 
   7.114 +#we didn't use the fact that $H$ was hermitian;
   7.115 +we instead used the fact that we know how to represent $H$ and $\psi$
   7.116 +as functions of $x$: $H$ is a differential operator
   7.117 +\(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\) and $\psi$ is a quadratic
   7.118 +function of $x$. By applying $H$ to $\psi$, we took several
   7.119 +derivatives and arrived at our answer.
   7.120 +
   7.121 +These two methods gave different results. In the following sections,
   7.122 +I'll describe and analyze the source of this difference.
   7.123 +
   7.124 +** Physical operators only act on physical wavefunctions
   7.125 +   :PROPERTIES:
   7.126 +   :ORDERED:  t
   7.127 +   :END:
   7.128 +#In quantum mechanics, an operator is a function that takes in a
   7.129 +#physical state and produces another physical state as ouput. Some
   7.130 +#operators correspond to physical quantities such as energy,
   7.131 +#momentum, or position; the mathematical properties of these operators correspond to
   7.132 +#physical properties of the system.
   7.133 +
   7.134 +#Eigenstates are an example of this correspondence: an 
   7.135 +
   7.136 +Physical states are represented as wavefunctions in quantum
   7.137 +mechanics. Just as we disallow certain physically nonsensical states
   7.138 +in classical mechanics (for example, we consider it to be nonphysical
   7.139 +for an object to spontaneously disappear from one place and reappear
   7.140 +in another), we also disallow certain wavefunctions in quantum
   7.141 +mechanics.
   7.142 +
   7.143 +For example, since wavefunctions are supposed to correspond to
   7.144 +probability amplitudes, we require wavefunctions to be normalized
   7.145 +\((\langle \psi |\psi \rangle = 1)\). Generally, we disallow
   7.146 +wavefunctions that do not satisfy this property (although there are
   7.147 +some exceptions[fn:2]).
   7.148 +
   7.149 +As another example, we generally expect probability to vary smoothly\mdash{}if
   7.150 +a particle is very likely or very unlikely to be found at a particular
   7.151 +location, it should also be somewhat likely or somewhat unlikely to be
   7.152 +found /near/ that location. In more precise terms, we expect that for
   7.153 +physically meaningful wavefunctions, the probability 
   7.154 +\(\text{Pr}(x)=\int^x \psi^*\psi\) should be a continuous function of
   7.155 +$x$ and, again, we disallow wavefunctions that do not satisfy this
   7.156 +property because we consider them to be physically nonsensical.
   7.157 +
   7.158 +So, physical wavefunctions must satisfy certain properties
   7.159 +like the two just described. Wavefunctions that do not satisfy these properties are
   7.160 +rejected for being physically nonsensical: even though we can perform
   7.161 +calculations with them, the mathematical results we obtain do not mean
   7.162 +anything physically.
   7.163 +
   7.164 +Now, in quantum mechanics, an *operator* is a function that converts
   7.165 +states into other states. Some operators correspond to
   7.166 +physical quantities such as energy, momentum, or position, and as a
   7.167 +result, the mathematical properties of these operators correspond to
   7.168 +physical properties of the system. Such operators are called
   7.169 +/hermitian operators/; one important property of hermitian operators
   7.170 +is this rule: 
   7.171 +
   7.172 +#+begin_quote 
   7.173 +*Hermitian operator rule:* A hermitian operator must only operate on
   7.174 +the wavefunctions we have deemed physical, and must only produce
   7.175 +physical wavefunctions[fn:: If you require a hermitian operator to
   7.176 +have physical eigenstates (which is reasonable), you get a very strong result: you guarantee
   7.177 +that the operator will convert every physical wavefunction into
   7.178 +another physical wavefunction: 
   7.179 +
   7.180 +  For any linear operator $\Omega$, the eigenvalue equation is
   7.181 +\(\Omega|\omega\rangle = \omega |\omega\rangle\). Notice that if an
   7.182 +eigenstate $|\omega\rangle$ is a physical wavefunction, the
   7.183 +eigenvalue equation forces $\Omega|\omega\rangle$ to be a
   7.184 +physical wavefunction as well. To elaborate, if the eigenstates of
   7.185 +$\Omega$ are physical functions, then $\Omega$ is guaranteed to
   7.186 +convert them into other physical functions.  Even more is true if the
   7.187 +operator $\Omega$ is also hermitian: there is a theorem which states
   7.188 +that \ldquo{}If \Omega is hermitian, then every physical wavefunction
   7.189 +can be written as a weighted sum of eigenstates of \Omega.\rdquo{} This
   7.190 +theorem implies that /if $\Omega$ is hermitian/, and /if the eigenstates
   7.191 +of \Omega are physically allowed/, then \Omega is guaranteed to
   7.192 +convert every physically allowed wavefunction into another physically
   7.193 +allowed wavefunction.].
   7.194 +#+end_quote
   7.195 +
   7.196 +As you can see, this rule comes in two pieces. The first part is a
   7.197 +constraint on *you*, the physicist: you must never feed a nonphysical
   7.198 +state into a Hermitian operator, as it may produce nonsense. The
   7.199 +second part is a constraint on the *operator*: the operator is
   7.200 +guaranteed only to produce physical wavefunctions.
   7.201 +
   7.202 +In fact, this rule for hermitian operators is the source of our
   7.203 +problem, as we unknowingly violated it when applying our second
   7.204 +method!
   7.205 +
   7.206 +** The Hamiltonian is nonphysical 
   7.207 +You'll remember that in the second method we had wavefunctions within
   7.208 +the well
   7.209 +
   7.210 +\(
   7.211 +\begin{eqnarray}
   7.212 +\psi(x) &=& A\;x(x-a)\\
   7.213 +\bar\psi(x)&\equiv& H|\psi\rangle = \frac{-A\hbar^2}{m}\\
   7.214 +\end{eqnarray}
   7.215 +\)
   7.216 +
   7.217 + Using this, we wrote
   7.218 +
   7.219 +
   7.220 +\(\begin{eqnarray}
   7.221 +\langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\
   7.222 +&=& \langle \psi |H \bar\psi \rangle\\
   7.223 +& \vdots&\\
   7.224 +&=& 0\\
   7.225 +\end{eqnarray}\)
   7.226 +
   7.227 +However, there are two issues here. First, we were not allowed to operate with $H$ on the wavefunction
   7.228 +$|\bar \psi\rangle$, because $H$ is supposed to be a physical operator and $|\bar
   7.229 +\psi\rangle$ is a nonphysical state. Indeed, the constant function $|\bar\psi\rangle$ does
   7.230 +not approach zero at the edges of the well. By
   7.231 +feeding $H$ a nonphysical wavefunction, we obtained nonsensical
   7.232 +results.
   7.233 +
   7.234 +Second, and more importantly, we were wrong to claim that $H$ was a
   7.235 +physical operator\mdash{}that $H$ was hermitian. According to the
   7.236 +rule, a hermitian operator must convert physical states into other
   7.237 +physical states. But $|\psi\rangle$ is a physical state, as we said
   7.238 +when we first introduced it \mdash{}it is a normalized, continuous
   7.239 +function which approaches zero at the edges of the well and doesn't
   7.240 +exist outside it. On the other hand, $|\bar\psi\rangle$ is nonphysical
   7.241 +because it does not go to zero at the edges of the well. It is
   7.242 +therefore impermissible for $H$ to transform the physical state
   7.243 +$|\psi\rangle$ into the nonphysical state $|\bar\psi\rangle$.  Because
   7.244 +$H$ converts some physical states into nonphysical states, it cannot
   7.245 +be a hermitian operator as we assumed.
   7.246 +
   7.247 +# Boundary conditions affect hermiticity
   7.248 +** Boundary conditions alter hermiticity
   7.249 +It may surprise you (and it certainly surprised me) to find that the
   7.250 +Hamiltonian is not hermitian. One of the fundamental principles of
   7.251 +quantum mechanics is that hermitian operators correspond to physically
   7.252 +observable quantities; for this reason, surely the
   7.253 +Hamiltonian\mdash{}the operator corresponding to energy\mdash{}ought to be hermitian?
   7.254 +
   7.255 +But we must understand the correspondence between physically
   7.256 +observable quantities and hermitian operators: every hermitian
   7.257 +operator corresponds to a physically observable quantity, but not
   7.258 +every quantity that intuitively \ldquo{}ought\rdquo{} to be observable
   7.259 +will correspond to a hermitian operator[fn::For a simple example,
   7.260 +consider the differential operator \(D=\frac{d}{dx}\); although our
   7.261 +intuitions might suggest that $D$ is observable which leads us to
   7.262 +guess that $D$ is hermitian, it isn't. Still, the very closely related
   7.263 +operator $P=i\hbar\frac{d}{dx}$ /is/ hermitian. The point is that we
   7.264 +ought to validate our intuitions by checking the definitions.]. The
   7.265 +true definition of a hermitian operator imply that the Hamiltonian
   7.266 +stops being hermitian in the infinitely deep well. Here we arrive at a
   7.267 +crucial point:
   7.268 +
   7.269 +Operators do not change /form/ between problems: the one-dimensional
   7.270 +Hamiltonian is always $H=\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}+V$, the
   7.271 +one-dimensional canonical momentum is always $P=i\hbar\frac{d}{dx}$,
   7.272 +and so on.
   7.273 +
   7.274 +However, operators do change in this respect: hermitian operators must
   7.275 +only take in physical states, and must only produce physical states; because
   7.276 +in different problems we /do/ change the requirements for being a
   7.277 +physical state, we also change what it takes for
   7.278 +an operator to be called hermitian.  As a result, an operator that
   7.279 +is hermitian in one setting may fail to be hermitian in another.
   7.280 +
   7.281 +Having seen how boundary conditions can affect hermiticity, we
   7.282 +ought to be extra careful about which conditions we impose on our
   7.283 +wavefunctions.
   7.284 +
   7.285 +** Choosing the right constraints 
   7.286 +
   7.287 + We have said already that physicists
   7.288 +require wavefunctions to satisfy certain properties in order to be
   7.289 +deemed \ldquo{}physical\rdquo{}. To be specific, wavefunctions in the
   7.290 +infinitely deep well
   7.291 +- Must be *normalizable*, because they correspond to
   7.292 +  probability amplitudes.
   7.293 +- Must have *smoothly-varying probability*, because if a particle is very
   7.294 +  likely to be at a location, it ought to be likely to be /near/
   7.295 +  it as well.
   7.296 +- Must *not exist outside the well*, because it
   7.297 +  would take an infinite amount of energy to do so.
   7.298 +
   7.299 +These conditions are surely reasonable. However, physicists sometimes
   7.300 +assert that in order to satisfy the second and third conditions,
   7.301 +physical wavefunctions
   7.302 +
   7.303 +- (?) Must *smoothly approach zero* towards the edges of the well.
   7.304 +
   7.305 +This final constraint is our reason for rejecting $|\bar\psi\rangle$
   7.306 +as nonphysical and is consequently the reason why $H$ is not hermitian. If
   7.307 +we can convince ourselves that the final constraint is unnecessary,
   7.308 +$H$ may again be hermitian. This will satisfy our intuitions that the
   7.309 +energy operator /ought/ to be hermitian.
   7.310 +
   7.311 +But in fact, we have the following mathematical observation to save
   7.312 +us: a function $f$ does not need to be continuous in order for the
   7.313 +integral \(\int^x f\) to be continuous. As a particularly relevant
   7.314 +example, you may now notice that the function $\bar\psi(x)$ is not
   7.315 +itself continuous, although the integral $\int_0^x \bar\psi$ /is/
   7.316 +continuous. Evidently, it doesn't matter that the wavefunction
   7.317 +$\bar\psi$ itself is not continuous; the probability corresponding to
   7.318 +$\bar\psi$ /does/ manage to vary continuously anyways. Because the
   7.319 +probability corresponding to $\bar\psi$ is the only aspect of
   7.320 +$\bar\psi$ which we can detect physically, we /can/ safely omit the
   7.321 +final constraint while keeping the other three.
   7.322 +
   7.323 +** Symmetric operators look like hermitian operators, but sometimes aren't.
   7.324 +
   7.325 +
   7.326 +#+end_quote
   7.327 +** COMMENT Re-examining physical constraints
   7.328 +
   7.329 +We have now discovered a flaw: when applied to the state
   7.330 +$|\psi\rangle$, the second method violates the rule that physical
   7.331 +operators must only take in physical states and must only produce
   7.332 +physical states. Let's examine the problem more closely.
   7.333 +
   7.334 +We have said already that physicists require wavefunctions to satisfy
   7.335 +certain properties in order to be deemed \ldquo{}physical\rdquo{}. To
   7.336 +be specific, wavefunctions in the infinitely deep well
   7.337 +- Must be *normalizable*, because they correspond to
   7.338 +  probability amplitudes.
   7.339 +- Must have *smoothly-varying probability*, because if a particle is very
   7.340 +  likely to be at a location, it ought to be likely to be /near/
   7.341 +  it as well.
   7.342 +- Must *not exist outside the well*, because it
   7.343 +  would take an infinite amount of energy to do so.
   7.344 +
   7.345 +We now have discovered an important flaw in the second method: when
   7.346 +applied to the state $|\bar\psi\rangle$, the second method violates
   7.347 +the rule that physical operators must only take in
   7.348 +physical states and must only produce physical states. The problem is
   7.349 +even more serious, however
   7.350 +
   7.351 +
   7.352 +
   7.353 +[fn:1] I'm defining a new variable just to make certain expressions
   7.354 +  look shorter; this cannot affect the content of the answer we'll
   7.355 +  get. 
   7.356 +
   7.357 +[fn:2] For example, in vaccuum (i.e., when the potential of the
   7.358 +  physical system is $V(x)=0$ throughout all space), the momentum
   7.359 +  eigenstates are not normalizable\mdash{}the relevant integral blows
   7.360 +  up to infinity instead of converging to a number. Physicists modify
   7.361 +  the definition of normalization slightly so that
   7.362 +  \ldquo{}delta-normalizable \rdquo{} functions like these are included
   7.363 +  among the physical wavefunctions.
   7.364 +
   7.365 +
   7.366 +
   7.367 +* COMMENT: What I thought I knew
   7.368 +
   7.369 +The following is a list of things I thought were true of quantum
   7.370 +mechanics; the catch is that the list contradicts itself.
   7.371 +
   7.372 +1. For any hermitian operator: Eigenstates with different eigenvalues are orthogonal.
   7.373 +2. For any hermitian operator: Any physically allowed state can be
   7.374 +   written as a linear sum of eigenstates of the operator.
   7.375 +3. The momentum operator and energy operator are hermitian, because
   7.376 +   momentum and energy are measureable quantities.
   7.377 +4. In the vacuum potential, the momentum and energy operators have these eigenstates:
   7.378 +   - the momentum operator has an eigenstate
   7.379 +     \(p(x)=\exp{(ipx/\hbar)}\) for each value of $p$.
   7.380 +   - the energy operator has an eigenstate \(|E\rangle =
   7.381 +     \alpha|p\rangle + \beta|-p\rangle\) for any \(\alpha,\beta\) and
   7.382 +     the particular choice of momentum $p=\sqrt{2mE}$.
   7.383 +5. In the infinitely deep potential well, the momentum and energy
   7.384 +   operators have these eigenstates:
   7.385 +   - The momentum eigenstates and energy eigenstates have the same form
   7.386 +     as in the vacuum potential: $p(x) =
   7.387 +     \exp{(ipx/\hbar)}$ and $|E\rangle = \alpha|p\rangle + \beta|-p\rangle$.
   7.388 +   - Even so, because of the boundary conditions on the
   7.389 +     well, we must make the following modifications:
   7.390 +     + Physically realistic states must be impossible to find outside the well. (Only a state of infinite
   7.391 +       energy could exist outside the well, and infinite energy is not
   7.392 +       realistic.) This requirement means, for example, that momentum
   7.393 +       eigenstates in the infinitely deep well must be
   7.394 +       \(p(x)
   7.395 +       = \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;
   7.396 +       \\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
   7.397 +     + Physically realistic states must vary smoothly throughout
   7.398 +       space. This means that if a particle in some state is very unlikely to be
   7.399 +       /at/ a particular location, it is also very unlikely be /near/
   7.400 +       that location. Combining this requirement with the above
   7.401 +       requirement, we find that the momentum operator no longer has
   7.402 +       an eigenstate for each value of $p$; instead, only values of
   7.403 +       $p$ that are integer multiples of $\pi \hbar/a$ are physically
   7.404 +       realistic. Similarly, the energy operator no longer has an
   7.405 +       eigenstate for each value of $E$; instead, the only energy
   7.406 +       eigenstates in the infinitely deep well
   7.407 +       are $E_n(x)=\sin(n\pi x/ a)$ for positive integers $n$.
   7.408 +
   7.409 +* COMMENT: 
   7.410 +
   7.411 +** Eigenstates with different eigenvalues are orthogonal
   7.412 +
   7.413 +#+begin_quote
   7.414 +*Theorem:* Eigenstates with different eigenvalues are orthogonal.
   7.415 +#+end_quote
   7.416 +
   7.417 +** COMMENT :
   7.418 +I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$
   7.419 +and $|b\rangle$ are eigenstates of $\Lambda$. This means that
   7.420 +
   7.421 +
   7.422 +\(
   7.423 +\begin{eqnarray}
   7.424 +\Lambda |a\rangle&=& a|a\rangle,\\
   7.425 +\Lambda|b\rangle&=& b|b\rangle.\\
   7.426 +\end{eqnarray}
   7.427 +\)
   7.428 +
   7.429 +If we take the difference of these eigenstates, we find that
   7.430 +
   7.431 +\(
   7.432 +\begin{eqnarray}
   7.433 +\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle
   7.434 +\qquad \text{(because $\Lambda$ is linear.)}\\
   7.435 +&=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and
   7.436 +$|b\rangle$ are eigenstates of $\Lambda$)}
   7.437 +\end{eqnarray}\)
   7.438 +
   7.439 +
   7.440 +which means that $a\neq b$.
   7.441 +
   7.442 +** Eigenvectors of hermitian operators span the space of solutions
   7.443 +
   7.444 +#+begin_quote
   7.445 +*Theorem:* If $\Omega$ is a hermitian operator, then every physically
   7.446 + allowed state can be written as a linear sum of eigenstates of
   7.447 + $\Omega$.
   7.448 +#+end_quote
   7.449 +
   7.450 +
   7.451 +
   7.452 +** Momentum and energy are hermitian operators
   7.453 +This ought to be true because hermitian operators correspond to
   7.454 +observable quantities. Since we expect momentum and energy to be
   7.455 +measureable quantities, we expect that there are hermitian operators
   7.456 +to represent them.
   7.457 +
   7.458 +
   7.459 +** Momentum and energy eigenstates in vacuum
   7.460 +An eigenstate of the momentum operator $P$ would be a state
   7.461 +\(|p\rangle\) such that \(P|p\rangle=p|p\rangle\).
   7.462 +
   7.463 +** Momentum and energy eigenstates in the infinitely deep well
   7.464 +
   7.465 +
   7.466 +
   7.467 +* COMMENT Can you measure momentum in the infinitely deep well?
   7.468 +In summary, I thought I knew:
   7.469 +1. For any hermitian operator: eigenstates with different eigenvalues
   7.470 +   are orthogonal.
   7.471 +2. For any hermitian operator: any physically realistic state can be
   7.472 +   written as a linear sum of eigenstates of the operator.
   7.473 +3. The momentum operator and energy operator are hermitian, because
   7.474 +   momentum and energy are observable quantities. 
   7.475 +4. (The form of the momentum and energy eigenstates in the vacuum potential)
   7.476 +5. (The form of the momentum and energy eigenstates in the infinitely deep well potential)
   7.477 +
   7.478 +Additionally, I understood that because the infinitely deep potential
   7.479 +well is not realistic, states of such a system  are not necessarily
   7.480 +physically realistic. Instead, I understood
   7.481 +\ldquo{}realistic states\rdquo{} to be those that satisfy the physically
   7.482 +unrealistic Schr\ouml{}dinger equation and its boundary conditions.
   7.483 +
   7.484 +With that final caveat, here is the problem:
   7.485 +
   7.486 +According to (5), the momentum eigenstates in the well are 
   7.487 +
   7.488 +\(p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
   7.489 +
   7.490 +(Additionally, we require that $p$ be an integer multiple of $\pi\hbar/a$.) 
   7.491 +
   7.492 +However, /these/ states are not orthogonal, which contradicts the
   7.493 +assumption that (3) the momentum operator is hermitian and (2)
   7.494 +eigenstates of a hermitian are orthogonal if they have different eigenvalues.
   7.495 +
   7.496 +#+begin_quote 
   7.497 +*Problem 1. The momentum eigenstates of the well are not orthogonal*
   7.498 +
   7.499 +/Proof./ If $p_1\neq p_2$, then 
   7.500 +
   7.501 +\(\begin{eqnarray}
   7.502 +\langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)\,dx\\
   7.503 +&=& \int_0^a p_1^*(x)p_2(x)\,dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{
   7.504 +outside the well.}\\
   7.505 +&=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)\,dx}}\\
   7.506 +&=& \int_0^a \exp{\left[i(p_1-p_2)x/\hbar\right]}\,dx\\
   7.507 +\end{eqnarray}\)
   7.508 +$\square$
   7.509 +
   7.510 +#+end_quote
   7.511 +
   7.512 +
   7.513 +
   7.514 +** COMMENT  Momentum eigenstates
   7.515 +
   7.516 +In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the
   7.517 +momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\).
   7.518 +
   7.519 +In the infinitely deep potential well, the Hamiltonian is the same but
   7.520 +there is a new condition in order for states to qualify as physically
   7.521 +allowed: the states must not exist anywhere outside of well, as it
   7.522 +takes an infinite amount of energy to do so. 
   7.523 +
   7.524 +Notice that the momentum eigenstates defined above do /not/ satisfy
   7.525 +this condition.
   7.526 +
   7.527 +
   7.528 +
   7.529 +* COMMENT
   7.530 +For each physical system, there is a Schr\ouml{}dinger equation that
   7.531 +describes how a particle's state $|\psi\rangle$  will change over
   7.532 +time.
   7.533 +
   7.534 +\(\begin{eqnarray}
   7.535 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
   7.536 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
   7.537 +
   7.538 +This is a differential equation; each solution to the
   7.539 +Schr\ouml{}dinger equation is a state that is physically allowed for
   7.540 +our particle. Here, physically allowed states are
   7.541 +those that change in physically allowed ways. However, like any differential
   7.542 +equation, the Schr\ouml{}dinger equation can be accompanied by
   7.543 +/boundary conditions/\mdash{}conditions that further restrict which
   7.544 +states qualify as physically allowed.
   7.545 +
   7.546 +
   7.547 +
   7.548 +
   7.549 +** Eigenstates of momentum
   7.550 +
   7.551 +
   7.552 +
   7.553 +
   7.554 +#In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger
   7.555 +
   7.556 +#\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\)
   7.557 +
   7.558 +
   7.559 +
   7.560 +
   7.561 +
   7.562 +
   7.563 +
   7.564 +* COMMENT
   7.565 +
   7.566 +#* The infinite square well potential
   7.567 +
   7.568 +A particle exists in a potential that is
   7.569 +infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the
   7.570 +particle exists in a potential[fn:coords][fn:infinity]
   7.571 +
   7.572 +
   7.573 +\(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for
   7.574 +}\;x<0\text{ or }x>a.\end{cases}\)
   7.575 +
   7.576 +The Schr\ouml{}dinger equation describes how the particle's state 
   7.577 +\(|\psi\rangle\) will change over time in this system.
   7.578 +
   7.579 +\(\begin{eqnarray}
   7.580 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
   7.581 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
   7.582 +
   7.583 +This is a differential equation; each solution to the
   7.584 +Schr\ouml{}dinger equation is a state that is physically allowed for
   7.585 +our particle. Here, physically allowed states are
   7.586 +those that change in physically allowed ways. However, like any differential
   7.587 +equation, the Schr\ouml{}dinger equation can be accompanied by
   7.588 +/boundary conditions/\mdash{}conditions that further restrict which
   7.589 +states qualify as physically allowed.
   7.590 +
   7.591 +
   7.592 +Whenever possible, physicists impose these boundary conditions:
   7.593 +- A physically allowed state ought to be a /smoothly-varying function of position./ This means
   7.594 +  that if a particle in the state  is likely to be /at/ a particular location,
   7.595 +  it is also likely to be /near/ that location.
   7.596 +
   7.597 +These boundary conditions imply that for the square well potential in
   7.598 +this problem,
   7.599 +
   7.600 +- Physically allowed states must be totally confined to the well,
   7.601 +  because it takes an infinite amount of energy to exist anywhere
   7.602 +  outside of the well (and physically allowed states ought to have
   7.603 +  only finite energy).
   7.604 +- Physically allowed states must be increasingly unlikely to find very
   7.605 +  close to the walls of the well. This is because of two conditions: the above
   7.606 +  condition says that the particle is /impossible/ to find
   7.607 +  outside of the well, and the smoothly-varying condition says
   7.608 +  that if a particle is impossible to find at a particular location,
   7.609 +  it must be unlikely to be found nearby that location.
   7.610 +
   7.611 +#; physically allowed states are those that change in physically
   7.612 +#allowed ways.
   7.613 +
   7.614 +
   7.615 +#** Boundary conditions
   7.616 +Because the potential is infinite everywhere except within the well,
   7.617 +a realistic particle must be confined to exist only within the
   7.618 +well\mdash{}its wavefunction must be zero everywhere beyond the walls
   7.619 +of the well.
   7.620 +
   7.621 +
   7.622 +[fn:coords] I chose my coordinate system so that the well extends from
   7.623 +\(0<x<a\). Others choose a coordinate system so that the well extends from
   7.624 +\(-\frac{a}{2}<x<\frac{a}{2}\). Although both coordinate systems describe the same physical
   7.625 +situation, they give different-looking answers.
   7.626 +
   7.627 +[fn:infinity] Of course, infinite potentials are not
   7.628 +realistic. Instead, they are useful approximations to finite
   7.629 +potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
   7.630 +of the well\rdquo{} are close enough for your own practical
   7.631 +purposes. Having introduced a physical impossibility into the problem
   7.632 +already, we don't expect to get physically realistic solutions; we
   7.633 +just expect to get mathematically consistent ones. The forthcoming
   7.634 +trouble is that we don't.