### changeset 3:44d3dc936f6a

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author Robert McIntyre Fri, 28 Oct 2011 00:06:37 -0700 b4de894a1e2e 10c30f787f4b bk/bk.org bk/bk2.org bk/bk3.org bk/bk4.org bk/bk_quandary.org bk/bkup.org org/bk.org org/bk2.org org/bk3.org org/bk4.org org/bk_quandary.org org/bkup.org 12 files changed, 1366 insertions(+), 1366 deletions(-) [+]
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     1.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
1.2 +++ b/bk/bk.org	Fri Oct 28 00:06:37 2011 -0700
1.3 @@ -0,0 +1,88 @@
1.4 +#+TITLE: Bugs in Quantum Mechanics
1.5 +#+AUTHOR: Dylan Holmes
1.6 +#+SETUPFILE: ../../aurellem/org/setup.org
1.7 +#+INCLUDE:   ../../aurellem/org/level-0.org
1.8 +
1.9 +#Bugs in the Quantum-Mechanical Momentum Operator
1.10 +
1.11 +
1.12 +I studied quantum mechanics the same way I study most subjects\mdash{}
1.13 +by collecting (and squashing) bugs in my understanding. One of these
1.14 +bugs persisted throughout two semesters of
1.15 +quantum mechanics coursework until I finally found
1.16 +the paper
1.17 +[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
1.18 +mechanics/]], which helped me stamp out the bug entirely. I decided to
1.19 +write an article about the problem and its solution for a number of reasons:
1.20 +
1.21 +- Although the paper was not unreasonably dense, it was written for
1.22 +  teachers. I wanted to write an article for students.
1.23 +- I wanted to popularize the problem and its solution because other
1.24 +  explanations are currently too hard to find. (Even Shankar's
1.25 +  excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it.)
1.26 +- I wanted to check that the bug was indeed entirely
1.27 +  eradicated. Attempting an explanation is my way of making
1.28 +  sure.
1.29 +
1.30 +* COMMENT
1.31 + I recommend the
1.32 +paper not only for students who are learning
1.33 +quantum mechanics, but especially for teachers interested in debugging
1.34 +them.
1.35 +
1.36 +* COMMENT
1.37 +On my first exam in quantum mechanics, my professor asked us to
1.38 +describe how certain measurements would affect a particle in a
1.39 +box. Many of these measurement questions required routine application
1.40 +of skills we had recently learned\mdash{}first, you recall (or
1.41 +calculate) the eigenstates of the quantity
1.42 +to be measured; second, you write the given state as a linear
1.43 +sum of these eigenstates\mdash{} the coefficients on each term give
1.44 +the probability amplitude.
1.45 +
1.46 +* The infinite square well potential
1.47 +There is a particle in a one-dimensional potential well that has
1.48 +infinitely high walls and finite width $$a$$. This means that the
1.49 +particle exists in a potential[fn:coords][fn:infinity]
1.50 +
1.51 +
1.52 +$$V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for 1.53 +}\;x<0\text{ or }x>a.\end{cases}$$
1.54 +
1.55 +The Schr\ouml{}dinger equation describes how the particle's state
1.56 +$$|\psi\rangle$$ will change over time in this system.
1.57 +
1.58 +$$\begin{eqnarray} 1.59 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& 1.60 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}$$
1.61 +
1.62 +This is a differential equation whose solutions are the physically
1.63 +allowed states for the particle in this system. Like any differential
1.64 +equation,
1.65 +
1.66 +
1.67 +Like any differential equation, the Schr\ouml{}dinger equation
1.68 +#; physically allowed states are those that change in physically
1.69 +#allowed ways.
1.70 +
1.71 +
1.72 +** Boundary conditions
1.73 +Because the potential is infinite everywhere except within the well,
1.74 +a realistic particle must be confined to exist only within the
1.75 +well\mdash{}its wavefunction must be zero everywhere beyond the walls
1.76 +of the well.
1.77 +
1.78 +
1.79 +[fn:coords] I chose my coordinate system so that the well extends from
1.80 +$$0<x<a$$. Others choose a coordinate system so that the well extends from
1.81 +$$-\frac{a}{2}<x<\frac{a}{2}$$. Although both coordinate systems describe the same physical
1.82 +situation, they give different-looking answers.
1.83 +
1.84 +[fn:infinity] Of course, infinite potentials are not
1.85 +realistic. Instead, they are useful approximations to finite
1.86 +potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
1.87 +of the well\rdquo{} are close enough for your own practical
1.88 +purposes. Having introduced a physical impossibility into the problem
1.89 +already, we don't expect to get physically realistic solutions; we
1.90 +just expect to get mathematically consistent ones. The forthcoming
1.91 +trouble is that we don't.

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

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

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

     5.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
5.2 +++ b/bk/bk_quandary.org	Fri Oct 28 00:06:37 2011 -0700
5.3 @@ -0,0 +1,566 @@
5.4 +#+TITLE: Bugs in quantum mechanics
5.5 +#+AUTHOR: Dylan Holmes
5.6 +#+DESCRIPTION: I explain hermitian operators and use them to solve a confusing problem in quantum mechanics.
5.7 +#+KEYWORDS: quantum mechanics, self-adjoint, hermitian, square well, momentum
5.8 +#+SETUPFILE: ../../aurellem/org/setup.org
5.9 +#+INCLUDE:   ../../aurellem/org/level-0.org
5.10 +
5.11 +
5.12 +
5.13 +#Bugs in Quantum Mechanics
5.14 +#Bugs in the Quantum-Mechanical Momentum Operator
5.15 +
5.16 +
5.17 +I studied quantum mechanics the same way I study most subjects\mdash{}
5.18 +by collecting (and squashing) bugs in my understanding. One of these
5.19 +bugs persisted throughout two semesters of
5.20 +quantum mechanics coursework until I finally found
5.21 +the paper
5.22 +[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
5.23 +mechanics/]], which helped me stamp out the bug entirely. I decided to
5.24 +write an article about the problem and its solution for a number of reasons:
5.25 +
5.26 +- Although the paper was not unreasonably dense, it was written for
5.27 +  teachers. I wanted to write an article for students.
5.28 +- I wanted to popularize the problem and its solution because other
5.29 +  explanations are currently too hard to find. (Even Shankar's
5.30 +  excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it, as far as I know.)
5.31 +- Attempting an explanation is my way of making
5.32 +  sure that the bug is /really/ gone.
5.33 +# entirely eradicated.
5.34 +
5.35 +* COMMENT
5.36 + I recommend the
5.37 +paper not only for students who are learning
5.38 +quantum mechanics, but especially for teachers interested in debugging
5.39 +them.
5.40 +
5.41 +* COMMENT
5.42 +On my first exam in quantum mechanics, my professor asked us to
5.43 +describe how certain measurements would affect a particle in a
5.44 +box. Many of these measurement questions required routine application
5.45 +of skills we had recently learned\mdash{}first, you recall (or
5.46 +calculate) the eigenstates of the quantity
5.47 +to be measured; second, you write the given state as a linear
5.48 +sum of these eigenstates\mdash{} the coefficients on each term give
5.49 +the probability amplitude.
5.50 +
5.51 +
5.52 +* Two methods of calculation that give different results.
5.53 +
5.54 +In the infinitely deep well, there is a particle in the the
5.55 +normalized state
5.56 +
5.57 + $$\psi(x) = \begin{cases}A\; x(x-a)&\text{for }0\lt{}x\lt{}a;\\0,&\text{for }x\lt{}0\text{ or }x>a.\end{cases}$$
5.58 +
5.59 +This is apparently a perfectly respectable state: it is normalized ($A$ is a
5.60 +normalization constant), it is zero
5.61 +everywhere outside of the well, and it is moreover continuous.
5.62 +
5.63 +Even so, we will find a problem if we attempt to calculate the average
5.64 +energy-squared of this state (that is, the quantity $$\langle \psi | H^2 | \psi \rangle$$).
5.65 +
5.66 +** First method
5.67 +
5.68 +For short, define a new variable[fn:1] $$|\bar \psi\rangle \equiv 5.69 +H|\psi\rangle$$. We can express the state $|\bar\psi\rangle$ as a
5.70 +function of $x$ because we know how to express $H$ and $\psi$ in terms
5.71 +of  $x$: $H$ is the differential operator $\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}$, and
5.72 +$\psi(x)$ is the function defined above. So, we get $$\quad\bar\psi(x) = \frac{-A\hbar^2}{m}$$. For future reference, observe that $\bar\psi(x)$
5.73 +is constant.
5.74 +
5.75 +Having introduced $|\bar\psi\rangle$, we can calculate the average energy-squared of $|\psi\rangle$ in the
5.76 +following way.
5.77 +
5.78 +$$\begin{eqnarray} 5.79 +\langle \psi | H^2 |\psi\rangle &=& \langle \psi H^\dagger | H 5.80 +\psi\rangle\\ 5.81 +&=& \langle \psi H | H\psi \rangle\\ 5.82 +&=& \langle \bar\psi | \bar\psi \rangle\\ 5.83 +&=& \int_0^a \left(\frac{A\hbar^2}{m}\right)^2\;dx\\ 5.84 +&=& \frac{A^2\hbar^4 a}{m^2}\\ 5.85 +\end{eqnarray}$$
5.86 +
5.87 +For future reference, observe that this value is  nonzero
5.88 +(which makes sense).
5.89 +
5.90 +** Second method
5.91 +We can also calculate the average energy-squared of $|\psi\rangle$ in the
5.92 +following way.
5.93 +
5.94 +\begin{eqnarray}
5.95 +\langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\
5.96 +&=& \langle \psi |H \bar\psi \rangle\\
5.97 +&=&\int_0^a Ax(x-a)
5.98 +\left(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\right)\frac{-A\hbar^2}{m}\;dx\\
5.99 +&=& 0\quad (!)\\
5.100 +\end{eqnarray}
5.101 +
5.102 +The second-to-last term must be zero because the second derivative
5.103 +of a constant $$\left(\frac{-A\hbar^2}{m}\right)$$ is zero.
5.104 +
5.105 +* What is the problem?
5.106 +
5.107 +To recap: We used two different methods to calculate the average
5.108 +energy-squared of a state $|\psi\rangle$. For the first method, we
5.109 +used the fact that $H$ is a hermitian operator, replacing $$\langle 5.110 +\psi H^\dagger | H \psi\rangle$$ with $$\langle \psi H | H 5.111 +\psi\rangle$$. Using this substitution rule, we calculated the answer.
5.112 +
5.113 +For the second method, we didn't use the fact that $H$ was hermitian;
5.114 +instead, we used the fact that we know how to represent $H$ and $\psi$
5.115 +as functions of $x$: $H$ is a differential operator
5.116 +$$\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}$$ and $\psi$ is a quadratic
5.117 +function of $x$. By applying $H$ to $\psi$, we took several
5.118 +derivatives and arrived at our answer.
5.119 +
5.120 +These two methods gave different results. In the following sections,
5.121 +I'll describe and analyze the source of this difference.
5.122 +
5.123 +** Physical operators only act on physical wavefunctions
5.124 +   :PROPERTIES:
5.125 +   :ORDERED:  t
5.126 +   :END:
5.127 +#In quantum mechanics, an operator is a function that takes in a
5.128 +#physical state and produces another physical state as ouput. Some
5.129 +#operators correspond to physical quantities such as energy,
5.130 +#momentum, or position; the mathematical properties of these operators correspond to
5.131 +#physical properties of the system.
5.132 +
5.133 +#Eigenstates are an example of this correspondence: an
5.134 +
5.135 +Physical states are represented as wavefunctions in quantum
5.136 +mechanics. Just as we disallow certain physically nonsensical states
5.137 +in classical mechanics (for example, we consider it to be nonphysical
5.138 +for an object to spontaneously disappear from one place and reappear
5.139 +in another), we also disallow certain wavefunctions in quantum
5.140 +mechanics.
5.141 +
5.142 +For example, since wavefunctions are supposed to correspond to
5.143 +probability amplitudes, we require wavefunctions to be normalized
5.144 +$$(\langle \psi |\psi \rangle = 1)$$. Generally, we disallow
5.145 +wavefunctions that do not satisfy this property (although there are
5.146 +some exceptions[fn:2]).
5.147 +
5.148 +As another example, we generally expect probability to vary smoothly\mdash{}if
5.149 +a particle is very likely or very unlikely to be found at a particular
5.150 +location, it should also be somewhat likely or somewhat unlikely to be
5.151 +found /near/ that location. In more precise terms, we expect that for
5.152 +physically meaningful wavefunctions, the probability
5.153 +$$\text{Pr}(x)=\int^x \psi^*\psi$$ should be a continuous function of
5.154 +$x$ and, again, we disallow wavefunctions that do not satisfy this
5.155 +property because we consider them to be physically nonsensical.
5.156 +
5.157 +So, physical wavefunctions must satisfy certain properties
5.158 +like the two just described. Wavefunctions that do not satisfy these properties are
5.159 +rejected for being physically nonsensical: even though we can perform
5.160 +calculations with them, the mathematical results we obtain do not mean
5.161 +anything physically.
5.162 +
5.163 +Now, in quantum mechanics, an *operator* is a function that converts
5.164 +states into other states. Some operators correspond to
5.165 +physical quantities such as energy, momentum, or position, and as a
5.166 +result, the mathematical properties of these operators correspond to
5.167 +physical properties of the system. Physical operators are furthermore
5.168 +subject to the following rule: they are only allowed to operate on
5.169 +#physical wavefunctions, and they are only allowed to produce
5.170 +#physical wavefunctions[fn:why].
5.171 +the wavefunctions we have deemed physical, and they only produce physical wavefunctions[fn::
5.172 +
5.173 + If you require a hermitian operator to have physical
5.174 +  eigenstates, you get a very strong result: you guarantee that the
5.175 +  operator will convert /every/ physical wavefunction into another
5.176 +  physical wavefunction:
5.177 +
5.178 +  For any linear operator $\Omega$, the eigenvalue equation is
5.179 +$$\Omega|\omega\rangle = \omega |\omega\rangle$$. Notice that if an
5.180 +eigenstate $|\omega\rangle$ is a physical wavefunction, the
5.181 +eigenvalue equation equation forces $\Omega|\omega\rangle$ to be a
5.182 +physical wavefunction as well. To elaborate, if the eigenstates of
5.183 +$\Omega$ are physical functions, then $\Omega$ is guaranteed to
5.184 +convert them into other physical functions.  Even more is true if the
5.185 +operator $\Omega$ is also hermitian: there is a theorem which states
5.186 +that \ldquo{}If \Omega is hermitian, then every physical wavefunction
5.187 +can be written as a weighted sum of eigenstates of \Omega.\rdquo{} This
5.188 +theorem implies that /if $\Omega$ is hermitian/, and /if the eigenstates
5.189 +of \Omega are physically allowed/, then \Omega is guaranteed to
5.190 +convert every physically allowed wavefunction into another physically
5.191 +allowed wavefunction.].
5.192 +
5.193 +In fact, this rule for physical operators is the source of our
5.194 +problem, as we unknowingly violated it when applying our second
5.195 +method!
5.196 +
5.197 +** The violation
5.198 +
5.199 +I'll start explaining this violation by being more specific about the
5.200 +infinitely deep well potential. We have said already that physicists
5.201 +require wavefunctions to satisfy certain properties in order to be
5.202 +deemed \ldquo{}physical\rdquo{}. To be specific, wavefunctions in the
5.203 +infinitely deep well
5.204 +- Must be *normalizable*, because they correspond to
5.205 +  probability amplitudes.
5.206 +- Must have *smoothly-varying probability*, because if a particle is very
5.207 +  likely to be at a location, it ought to be likely to be /near/
5.208 +  it as well.
5.209 +- Must *not exist outside the well*, because it
5.210 +  would take an infinite amount of energy to do so.
5.211 +
5.212 +Additionally, by combining the second and third conditions, some
5.213 +physicists reason that wavefunctions in the infinitely deep well
5.214 +
5.215 +- Must *become zero* towards the edges of the well.
5.216 +
5.217 +
5.218 +
5.219 +
5.220 +You'll remember we had
5.221 +
5.222 +$$5.223 +\begin{eqnarray} 5.224 +\psi(x) &=& A\;x(x-a)\\ 5.225 +\bar\psi(x)&=& \frac{-A\hbar^2}{m}\\ 5.226 +&&\text{for }0\lt{}x\lt{}a\\ 5.227 +\end{eqnarray} 5.228 +$$
5.229 +
5.230 +In our second method, we wrote
5.231 +
5.232 +
5.233 +$$\begin{eqnarray} 5.234 +\langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\ 5.235 +&=& \langle \psi |H \bar\psi \rangle\\ 5.236 +& \vdots&\\ 5.237 +&=& 0\\ 5.238 +\end{eqnarray}$$
5.239 +
5.240 +However, there are two issues here. First, we were not allowed to operate with $H$ on the wavefunction
5.241 +$|\bar \psi\rangle$, because $H$ is a physical operator and $|\bar 5.242 +\psi\rangle$ is a nonphysical state: in the infinite square well,
5.243 +physical wavefunctions must approach zero at the edges of the well,
5.244 +which the constant function $|\bar\psi\rangle$ does not do. By
5.245 +feeding $H$ a nonphysical wavefunction, we obtained nonsensical
5.246 +results.
5.247 +
5.248 +Second, we claimed that $H$ was a physical operator\mdash{}that $H$
5.249 +was hermitian. According to the rule, this means $H$ must convert physical states into other
5.250 +physical states. But $H$ converts the physical state $|\psi\rangle$
5.251 +into the nonphysical state $|\bar\psi\rangle = H|\psi\rangle$ is not. Because $H$ converts some
5.252 +physical states into nonphysical states, it cannot be a hermitian operator.
5.253 +
5.254 +** Boundary conditions affect hermiticity
5.255 +We have now discovered a flaw: when applied to the state
5.256 +$|\psi\rangle$, the second method violates the rule that physical
5.257 +operators must only take in physical states and must only produce
5.258 +physical states. This suggests that the problem was with the state
5.259 +$|\bar\psi\rangle$ that we wrongly fed into $H$. However, the problem
5.260 +is more serious still: the state $|\psi\rangle 5.261 + 5.262 +** COMMENT Re-examining physical constraints 5.263 + 5.264 +We have now discovered a flaw: when applied to the state 5.265 +$|\psi\rangle$, the second method violates the rule that physical 5.266 +operators must only take in physical states and must only produce 5.267 +physical states. Let's examine the problem more closely. 5.268 + 5.269 +We have said already that physicists require wavefunctions to satisfy 5.270 +certain properties in order to be deemed \ldquo{}physical\rdquo{}. To 5.271 +be specific, wavefunctions in the infinitely deep well 5.272 +- Must be *normalizable*, because they correspond to 5.273 + probability amplitudes. 5.274 +- Must have *smoothly-varying probability*, because if a particle is very 5.275 + likely to be at a location, it ought to be likely to be /near/ 5.276 + it as well. 5.277 +- Must *not exist outside the well*, because it 5.278 + would take an infinite amount of energy to do so. 5.279 + 5.280 +We now have discovered an important flaw in the second method: when 5.281 +applied to the state$|\bar\psi\rangle$, the second method violates 5.282 +the rule that physical operators must only take in 5.283 +physical states and must only produce physical states. The problem is 5.284 +even more serious, however 5.285 + 5.286 + 5.287 + 5.288 +[fn:1] I'm defining a new variable just to make certain expressions 5.289 + look shorter; this cannot affect the content of the answer we'll 5.290 + get. 5.291 + 5.292 +[fn:2] For example, in vaccuum (i.e., when the potential of the 5.293 + physical system is$V(x)=0$throughout all space), the momentum 5.294 + eigenstates are not normalizable\mdash{}the relevant integral blows 5.295 + up to infinity instead of converging to a number. Physicists modify 5.296 + the definition of normalization slightly so that 5.297 + \ldquo{}delta-normalizable \rdquo{} functions like these are included 5.298 + among the physical wavefunctions. 5.299 + 5.300 + 5.301 + 5.302 +* COMMENT: What I thought I knew 5.303 + 5.304 +The following is a list of things I thought were true of quantum 5.305 +mechanics; the catch is that the list contradicts itself. 5.306 + 5.307 +1. For any hermitian operator: Eigenstates with different eigenvalues are orthogonal. 5.308 +2. For any hermitian operator: Any physically allowed state can be 5.309 + written as a linear sum of eigenstates of the operator. 5.310 +3. The momentum operator and energy operator are hermitian, because 5.311 + momentum and energy are measureable quantities. 5.312 +4. In the vacuum potential, the momentum and energy operators have these eigenstates: 5.313 + - the momentum operator has an eigenstate 5.314 + $$p(x)=\exp{(ipx/\hbar)}$$ for each value of$p$. 5.315 + - the energy operator has an eigenstate $$|E\rangle = 5.316 + \alpha|p\rangle + \beta|-p\rangle$$ for any $$\alpha,\beta$$ and 5.317 + the particular choice of momentum$p=\sqrt{2mE}$. 5.318 +5. In the infinitely deep potential well, the momentum and energy 5.319 + operators have these eigenstates: 5.320 + - The momentum eigenstates and energy eigenstates have the same form 5.321 + as in the vacuum potential:$p(x) =
5.322 +     \exp{(ipx/\hbar)}$and$|E\rangle = \alpha|p\rangle + \beta|-p\rangle$. 5.323 + - Even so, because of the boundary conditions on the 5.324 + well, we must make the following modifications: 5.325 + + Physically realistic states must be impossible to find outside the well. (Only a state of infinite 5.326 + energy could exist outside the well, and infinite energy is not 5.327 + realistic.) This requirement means, for example, that momentum 5.328 + eigenstates in the infinitely deep well must be 5.329 + $$p(x) 5.330 + = \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a; 5.331 + \\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}$$ 5.332 + + Physically realistic states must vary smoothly throughout 5.333 + space. This means that if a particle in some state is very unlikely to be 5.334 + /at/ a particular location, it is also very unlikely be /near/ 5.335 + that location. Combining this requirement with the above 5.336 + requirement, we find that the momentum operator no longer has 5.337 + an eigenstate for each value of$p$; instead, only values of 5.338 +$p$that are integer multiples of$\pi \hbar/a$are physically 5.339 + realistic. Similarly, the energy operator no longer has an 5.340 + eigenstate for each value of$E$; instead, the only energy 5.341 + eigenstates in the infinitely deep well 5.342 + are$E_n(x)=\sin(n\pi x/ a)$for positive integers$n$. 5.343 + 5.344 +* COMMENT: 5.345 + 5.346 +** Eigenstates with different eigenvalues are orthogonal 5.347 + 5.348 +#+begin_quote 5.349 +*Theorem:* Eigenstates with different eigenvalues are orthogonal. 5.350 +#+end_quote 5.351 + 5.352 +** COMMENT : 5.353 +I can prove this: if$\Lambda$is any linear operator, suppose$|a\rangle$5.354 +and$|b\rangle$are eigenstates of$\Lambda$. This means that 5.355 + 5.356 + 5.357 +$$5.358 +\begin{eqnarray} 5.359 +\Lambda |a\rangle&=& a|a\rangle,\\ 5.360 +\Lambda|b\rangle&=& b|b\rangle.\\ 5.361 +\end{eqnarray} 5.362 +$$ 5.363 + 5.364 +If we take the difference of these eigenstates, we find that 5.365 + 5.366 +$$5.367 +\begin{eqnarray} 5.368 +\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle 5.369 +\qquad \text{(because \Lambda is linear.)}\\ 5.370 +&=& a|a\rangle - b|b\rangle\qquad\text{(because |a\rangle and 5.371 +|b\rangle are eigenstates of \Lambda)} 5.372 +\end{eqnarray}$$ 5.373 + 5.374 + 5.375 +which means that$a\neq b$. 5.376 + 5.377 +** Eigenvectors of hermitian operators span the space of solutions 5.378 + 5.379 +#+begin_quote 5.380 +*Theorem:* If$\Omega$is a hermitian operator, then every physically 5.381 + allowed state can be written as a linear sum of eigenstates of 5.382 +$\Omega$. 5.383 +#+end_quote 5.384 + 5.385 + 5.386 + 5.387 +** Momentum and energy are hermitian operators 5.388 +This ought to be true because hermitian operators correspond to 5.389 +observable quantities. Since we expect momentum and energy to be 5.390 +measureable quantities, we expect that there are hermitian operators 5.391 +to represent them. 5.392 + 5.393 + 5.394 +** Momentum and energy eigenstates in vacuum 5.395 +An eigenstate of the momentum operator$P$would be a state 5.396 +$$|p\rangle$$ such that $$P|p\rangle=p|p\rangle$$. 5.397 + 5.398 +** Momentum and energy eigenstates in the infinitely deep well 5.399 + 5.400 + 5.401 + 5.402 +* COMMENT Can you measure momentum in the infinitely deep well? 5.403 +In summary, I thought I knew: 5.404 +1. For any hermitian operator: eigenstates with different eigenvalues 5.405 + are orthogonal. 5.406 +2. For any hermitian operator: any physically realistic state can be 5.407 + written as a linear sum of eigenstates of the operator. 5.408 +3. The momentum operator and energy operator are hermitian, because 5.409 + momentum and energy are observable quantities. 5.410 +4. (The form of the momentum and energy eigenstates in the vacuum potential) 5.411 +5. (The form of the momentum and energy eigenstates in the infinitely deep well potential) 5.412 + 5.413 +Additionally, I understood that because the infinitely deep potential 5.414 +well is not realistic, states of such a system are not necessarily 5.415 +physically realistic. Instead, I understood 5.416 +\ldquo{}realistic states\rdquo{} to be those that satisfy the physically 5.417 +unrealistic Schr\ouml{}dinger equation and its boundary conditions. 5.418 + 5.419 +With that final caveat, here is the problem: 5.420 + 5.421 +According to (5), the momentum eigenstates in the well are 5.422 + 5.423 +$$p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}$$ 5.424 + 5.425 +(Additionally, we require that$p$be an integer multiple of$\pi\hbar/a$.) 5.426 + 5.427 +However, /these/ states are not orthogonal, which contradicts the 5.428 +assumption that (3) the momentum operator is hermitian and (2) 5.429 +eigenstates of a hermitian are orthogonal if they have different eigenvalues. 5.430 + 5.431 +#+begin_quote 5.432 +*Problem 1. The momentum eigenstates of the well are not orthogonal* 5.433 + 5.434 +/Proof./ If$p_1\neq p_2$, then 5.435 + 5.436 +$$\begin{eqnarray} 5.437 +\langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)\,dx\\ 5.438 +&=& \int_0^a p_1^*(x)p_2(x)\,dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{ 5.439 +outside the well.}\\ 5.440 +&=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)\,dx}}\\ 5.441 +&=& \int_0^a \exp{\left[i(p_1-p_2)x/\hbar\right]}\,dx\\ 5.442 +\end{eqnarray}$$ 5.443 +$\square$5.444 + 5.445 +#+end_quote 5.446 + 5.447 + 5.448 + 5.449 +** COMMENT Momentum eigenstates 5.450 + 5.451 +In free space, the Hamiltonian is $$H=\frac{1}{2m}P^2$$ and the 5.452 +momentum operator$P$has eigenstates $$p(x) = \exp{(-ipx/\hbar)}$$. 5.453 + 5.454 +In the infinitely deep potential well, the Hamiltonian is the same but 5.455 +there is a new condition in order for states to qualify as physically 5.456 +allowed: the states must not exist anywhere outside of well, as it 5.457 +takes an infinite amount of energy to do so. 5.458 + 5.459 +Notice that the momentum eigenstates defined above do /not/ satisfy 5.460 +this condition. 5.461 + 5.462 + 5.463 + 5.464 +* COMMENT 5.465 +For each physical system, there is a Schr\ouml{}dinger equation that 5.466 +describes how a particle's state$|\psi\rangle$will change over 5.467 +time. 5.468 + 5.469 +$$\begin{eqnarray} 5.470 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& 5.471 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}$$ 5.472 + 5.473 +This is a differential equation; each solution to the 5.474 +Schr\ouml{}dinger equation is a state that is physically allowed for 5.475 +our particle. Here, physically allowed states are 5.476 +those that change in physically allowed ways. However, like any differential 5.477 +equation, the Schr\ouml{}dinger equation can be accompanied by 5.478 +/boundary conditions/\mdash{}conditions that further restrict which 5.479 +states qualify as physically allowed. 5.480 + 5.481 + 5.482 + 5.483 + 5.484 +** Eigenstates of momentum 5.485 + 5.486 + 5.487 + 5.488 + 5.489 +#In the infinitely deep well potential$V(x)=0$, the Schr\ouml{}dinger 5.490 + 5.491 +#$$i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle$$ 5.492 + 5.493 + 5.494 + 5.495 + 5.496 + 5.497 + 5.498 + 5.499 +* COMMENT 5.500 + 5.501 +#* The infinite square well potential 5.502 + 5.503 +A particle exists in a potential that is 5.504 +infinite everywhere except for a region of length $$a$$, where the potential is zero. This means that the 5.505 +particle exists in a potential[fn:coords][fn:infinity] 5.506 + 5.507 + 5.508 +$$V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for 5.509 +}\;x<0\text{ or }x>a.\end{cases}$$ 5.510 + 5.511 +The Schr\ouml{}dinger equation describes how the particle's state 5.512 +$$|\psi\rangle$$ will change over time in this system. 5.513 + 5.514 +$$\begin{eqnarray} 5.515 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& 5.516 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}$$ 5.517 + 5.518 +This is a differential equation; each solution to the 5.519 +Schr\ouml{}dinger equation is a state that is physically allowed for 5.520 +our particle. Here, physically allowed states are 5.521 +those that change in physically allowed ways. However, like any differential 5.522 +equation, the Schr\ouml{}dinger equation can be accompanied by 5.523 +/boundary conditions/\mdash{}conditions that further restrict which 5.524 +states qualify as physically allowed. 5.525 + 5.526 + 5.527 +Whenever possible, physicists impose these boundary conditions: 5.528 +- A physically allowed state ought to be a /smoothly-varying function of position./ This means 5.529 + that if a particle in the state is likely to be /at/ a particular location, 5.530 + it is also likely to be /near/ that location. 5.531 + 5.532 +These boundary conditions imply that for the square well potential in 5.533 +this problem, 5.534 + 5.535 +- Physically allowed states must be totally confined to the well, 5.536 + because it takes an infinite amount of energy to exist anywhere 5.537 + outside of the well (and physically allowed states ought to have 5.538 + only finite energy). 5.539 +- Physically allowed states must be increasingly unlikely to find very 5.540 + close to the walls of the well. This is because of two conditions: the above 5.541 + condition says that the particle is /impossible/ to find 5.542 + outside of the well, and the smoothly-varying condition says 5.543 + that if a particle is impossible to find at a particular location, 5.544 + it must be unlikely to be found nearby that location. 5.545 + 5.546 +#; physically allowed states are those that change in physically 5.547 +#allowed ways. 5.548 + 5.549 + 5.550 +#** Boundary conditions 5.551 +Because the potential is infinite everywhere except within the well, 5.552 +a realistic particle must be confined to exist only within the 5.553 +well\mdash{}its wavefunction must be zero everywhere beyond the walls 5.554 +of the well. 5.555 + 5.556 + 5.557 +[fn:coords] I chose my coordinate system so that the well extends from 5.558 +$$0<x<a$$. Others choose a coordinate system so that the well extends from 5.559 +$$-\frac{a}{2}<x<\frac{a}{2}$$. Although both coordinate systems describe the same physical 5.560 +situation, they give different-looking answers. 5.561 + 5.562 +[fn:infinity] Of course, infinite potentials are not 5.563 +realistic. Instead, they are useful approximations to finite 5.564 +potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height 5.565 +of the well\rdquo{} are close enough for your own practical 5.566 +purposes. Having introduced a physical impossibility into the problem 5.567 +already, we don't expect to get physically realistic solutions; we 5.568 +just expect to get mathematically consistent ones. The forthcoming 5.569 +trouble is that we don't.   6.1 --- /dev/null Thu Jan 01 00:00:00 1970 +0000 6.2 +++ b/bk/bkup.org Fri Oct 28 00:06:37 2011 -0700 6.3 @@ -0,0 +1,49 @@ 6.4 +#+TITLE: Bugs in Quantum Mechanics 6.5 +#+AUTHOR: Dylan Holmes 6.6 +#+SETUPFILE: ../../aurellem/org/setup.org 6.7 +#+INCLUDE: ../../aurellem/org/level-0.org 6.8 + 6.9 + 6.10 +#Bugs in the Quantum-Mechanical Momentum Operator 6.11 + 6.12 + 6.13 +I studied quantum mechanics the same way I study most subjects\mdash{} 6.14 +by collecting (and squashing) bugs in my understanding. One of these 6.15 +bugs persisted throughout two semesters of 6.16 +quantum mechanics coursework until I finally found 6.17 +the paper 6.18 +[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum 6.19 +mechanics/]], which helped me stamp out the bug entirely. I decided to 6.20 +write an article about the problem and its solution for a number of reasons: 6.21 + 6.22 +- Although the paper was not unreasonably dense, it was written for 6.23 + teachers. I wanted to write an article for students. 6.24 +- I wanted to popularize the problem and its solution because 6.25 + other explanations are currently too hard to find. 6.26 +- I wanted to check that the bug was indeed entirely 6.27 + eradicated. Attempting an explanation is my way of making 6.28 + sure. 6.29 + 6.30 +* COMMENT 6.31 + I recommend the 6.32 +paper not only for students who are learning 6.33 +quantum mechanics, but especially for teachers interested in debugging 6.34 +them. 6.35 + 6.36 +* COMMENT 6.37 +On my first exam in quantum mechanics, my professor asked us to 6.38 +describe how certain measurements would affect a particle in a 6.39 +box. Many of these measurement questions required routine application 6.40 +of skills we had recently learned\mdash{}first, you recall (or 6.41 +calculate) the eigenstates of the quantity 6.42 +to be measured; second, you write the given state as a linear 6.43 +sum of these eigenstates\mdash{} the coefficients on each term give 6.44 +the probability amplitude. 6.45 + 6.46 +* Statement of the Problem 6.47 +A particle is 6.48 + 6.49 + 6.50 + 6.51 + 6.52 +* COMMENT [TABLE-OF-CONTENTS]   7.1 --- a/org/bk.org Fri Oct 28 00:03:05 2011 -0700 7.2 +++ /dev/null Thu Jan 01 00:00:00 1970 +0000 7.3 @@ -1,88 +0,0 @@ 7.4 -#+TITLE: Bugs in Quantum Mechanics 7.5 -#+AUTHOR: Dylan Holmes 7.6 -#+SETUPFILE: ../../aurellem/org/setup.org 7.7 -#+INCLUDE: ../../aurellem/org/level-0.org 7.8 - 7.9 -#Bugs in the Quantum-Mechanical Momentum Operator 7.10 - 7.11 - 7.12 -I studied quantum mechanics the same way I study most subjects\mdash{} 7.13 -by collecting (and squashing) bugs in my understanding. One of these 7.14 -bugs persisted throughout two semesters of 7.15 -quantum mechanics coursework until I finally found 7.16 -the paper 7.17 -[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum 7.18 -mechanics/]], which helped me stamp out the bug entirely. I decided to 7.19 -write an article about the problem and its solution for a number of reasons: 7.20 - 7.21 -- Although the paper was not unreasonably dense, it was written for 7.22 - teachers. I wanted to write an article for students. 7.23 -- I wanted to popularize the problem and its solution because other 7.24 - explanations are currently too hard to find. (Even Shankar's 7.25 - excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it.) 7.26 -- I wanted to check that the bug was indeed entirely 7.27 - eradicated. Attempting an explanation is my way of making 7.28 - sure. 7.29 - 7.30 -* COMMENT 7.31 - I recommend the 7.32 -paper not only for students who are learning 7.33 -quantum mechanics, but especially for teachers interested in debugging 7.34 -them. 7.35 - 7.36 -* COMMENT 7.37 -On my first exam in quantum mechanics, my professor asked us to 7.38 -describe how certain measurements would affect a particle in a 7.39 -box. Many of these measurement questions required routine application 7.40 -of skills we had recently learned\mdash{}first, you recall (or 7.41 -calculate) the eigenstates of the quantity 7.42 -to be measured; second, you write the given state as a linear 7.43 -sum of these eigenstates\mdash{} the coefficients on each term give 7.44 -the probability amplitude. 7.45 - 7.46 -* The infinite square well potential 7.47 -There is a particle in a one-dimensional potential well that has 7.48 -infinitely high walls and finite width $$a$$. This means that the 7.49 -particle exists in a potential[fn:coords][fn:infinity] 7.50 - 7.51 - 7.52 -$$V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for 7.53 -}\;x<0\text{ or }x>a.\end{cases}$$ 7.54 - 7.55 -The Schr\ouml{}dinger equation describes how the particle's state 7.56 -$$|\psi\rangle$$ will change over time in this system. 7.57 - 7.58 -$$\begin{eqnarray} 7.59 -i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& 7.60 -H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}$$ 7.61 - 7.62 -This is a differential equation whose solutions are the physically 7.63 -allowed states for the particle in this system. Like any differential 7.64 -equation, 7.65 - 7.66 - 7.67 -Like any differential equation, the Schr\ouml{}dinger equation 7.68 -#; physically allowed states are those that change in physically 7.69 -#allowed ways. 7.70 - 7.71 - 7.72 -** Boundary conditions 7.73 -Because the potential is infinite everywhere except within the well, 7.74 -a realistic particle must be confined to exist only within the 7.75 -well\mdash{}its wavefunction must be zero everywhere beyond the walls 7.76 -of the well. 7.77 - 7.78 - 7.79 -[fn:coords] I chose my coordinate system so that the well extends from 7.80 -$$0<x<a$$. Others choose a coordinate system so that the well extends from 7.81 -$$-\frac{a}{2}<x<\frac{a}{2}$$. Although both coordinate systems describe the same physical 7.82 -situation, they give different-looking answers. 7.83 - 7.84 -[fn:infinity] Of course, infinite potentials are not 7.85 -realistic. Instead, they are useful approximations to finite 7.86 -potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height 7.87 -of the well\rdquo{} are close enough for your own practical 7.88 -purposes. Having introduced a physical impossibility into the problem 7.89 -already, we don't expect to get physically realistic solutions; we 7.90 -just expect to get mathematically consistent ones. The forthcoming 7.91 -trouble is that we don't.   8.1 --- a/org/bk2.org Fri Oct 28 00:03:05 2011 -0700 8.2 +++ /dev/null Thu Jan 01 00:00:00 1970 +0000 8.3 @@ -1,97 +0,0 @@ 8.4 -#+TITLE: Bugs in Quantum Mechanics 8.5 -#+AUTHOR: Dylan Holmes 8.6 -#+SETUPFILE: ../../aurellem/org/setup.org 8.7 -#+INCLUDE: ../../aurellem/org/level-0.org 8.8 - 8.9 - 8.10 -#Bugs in the Quantum-Mechanical Momentum Operator 8.11 - 8.12 - 8.13 -I studied quantum mechanics the same way I study most subjects\mdash{} 8.14 -by collecting (and squashing) bugs in my understanding. One of these 8.15 -bugs persisted throughout two semesters of 8.16 -quantum mechanics coursework until I finally found 8.17 -the paper 8.18 -[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum 8.19 -mechanics/]], which helped me stamp out the bug entirely. I decided to 8.20 -write an article about the problem and its solution for a number of reasons: 8.21 - 8.22 -- Although the paper was not unreasonably dense, it was written for 8.23 - teachers. I wanted to write an article for students. 8.24 -- I wanted to popularize the problem and its solution because other 8.25 - explanations are currently too hard to find. (Even Shankar's 8.26 - excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it.) 8.27 -- I wanted to check that the bug was indeed entirely 8.28 - eradicated. Attempting an explanation is my way of making 8.29 - sure. 8.30 - 8.31 -* COMMENT 8.32 - I recommend the 8.33 -paper not only for students who are learning 8.34 -quantum mechanics, but especially for teachers interested in debugging 8.35 -them. 8.36 - 8.37 -* COMMENT 8.38 -On my first exam in quantum mechanics, my professor asked us to 8.39 -describe how certain measurements would affect a particle in a 8.40 -box. Many of these measurement questions required routine application 8.41 -of skills we had recently learned\mdash{}first, you recall (or 8.42 -calculate) the eigenstates of the quantity 8.43 -to be measured; second, you write the given state as a linear 8.44 -sum of these eigenstates\mdash{} the coefficients on each term give 8.45 -the probability amplitude. 8.46 - 8.47 -* The infinite square well potential 8.48 - 8.49 -There is a particle in a one-dimensional potential well that is 8.50 -infinite everywhere except for a well of length $$a$$. This means that the 8.51 -particle exists in a potential[fn:coords][fn:infinity] 8.52 - 8.53 - 8.54 -$$V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for 8.55 -}\;x<0\text{ or }x>a.\end{cases}$$ 8.56 - 8.57 -The Schr\ouml{}dinger equation describes how the particle's state 8.58 -$$|\psi\rangle$$ will change over time in this system. 8.59 - 8.60 -$$\begin{eqnarray} 8.61 -i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& 8.62 -H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}$$ 8.63 - 8.64 -This is a differential equation whose solutions are the physically 8.65 -allowed states for the particle in this system. Physically allowed 8.66 -states are those that change in physically allowed ways. Like any 8.67 -differential equation, the Schr\ouml{}dinger equation can be 8.68 -accompanied by /boundary conditions/\mdash{}conditions that 8.69 -further restrict which states qualify as physically allowed. 8.70 - 8.71 -Whenever possible, physicists impose these boundary conditions: 8.72 -- The state should be a /continuous function of/ $$x$$. This means 8.73 - that if a particle is very likely to be /at/ a particular location, 8.74 - it is also very likely to be /near/ that location. 8.75 -- 8.76 - 8.77 -#; physically allowed states are those that change in physically 8.78 -#allowed ways. 8.79 - 8.80 - 8.81 -** Boundary conditions 8.82 -Because the potential is infinite everywhere except within the well, 8.83 -a realistic particle must be confined to exist only within the 8.84 -well\mdash{}its wavefunction must be zero everywhere beyond the walls 8.85 -of the well. 8.86 - 8.87 - 8.88 -[fn:coords] I chose my coordinate system so that the well extends from 8.89 -$$0<x<a$$. Others choose a coordinate system so that the well extends from 8.90 -$$-\frac{a}{2}<x<\frac{a}{2}$$. Although both coordinate systems describe the same physical 8.91 -situation, they give different-looking answers. 8.92 - 8.93 -[fn:infinity] Of course, infinite potentials are not 8.94 -realistic. Instead, they are useful approximations to finite 8.95 -potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height 8.96 -of the well\rdquo{} are close enough for your own practical 8.97 -purposes. Having introduced a physical impossibility into the problem 8.98 -already, we don't expect to get physically realistic solutions; we 8.99 -just expect to get mathematically consistent ones. The forthcoming 8.100 -trouble is that we don't.   9.1 --- a/org/bk3.org Fri Oct 28 00:03:05 2011 -0700 9.2 +++ /dev/null Thu Jan 01 00:00:00 1970 +0000 9.3 @@ -1,257 +0,0 @@ 9.4 -#+TITLE: Bugs in quantum mechanics 9.5 -#+AUTHOR: Dylan Holmes 9.6 -#+SETUPFILE: ../../aurellem/org/setup.org 9.7 -#+INCLUDE: ../../aurellem/org/level-0.org 9.8 - 9.9 -#Bugs in Quantum Mechanics 9.10 -#Bugs in the Quantum-Mechanical Momentum Operator 9.11 - 9.12 - 9.13 -I studied quantum mechanics the same way I study most subjects\mdash{} 9.14 -by collecting (and squashing) bugs in my understanding. One of these 9.15 -bugs persisted throughout two semesters of 9.16 -quantum mechanics coursework until I finally found 9.17 -the paper 9.18 -[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum 9.19 -mechanics/]], which helped me stamp out the bug entirely. I decided to 9.20 -write an article about the problem and its solution for a number of reasons: 9.21 - 9.22 -- Although the paper was not unreasonably dense, it was written for 9.23 - teachers. I wanted to write an article for students. 9.24 -- I wanted to popularize the problem and its solution because other 9.25 - explanations are currently too hard to find. (Even Shankar's 9.26 - excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it.) 9.27 -- I wanted to check that the bug was indeed entirely 9.28 - eradicated. Attempting an explanation is my way of making 9.29 - sure. 9.30 - 9.31 -* COMMENT 9.32 - I recommend the 9.33 -paper not only for students who are learning 9.34 -quantum mechanics, but especially for teachers interested in debugging 9.35 -them. 9.36 - 9.37 -* COMMENT 9.38 -On my first exam in quantum mechanics, my professor asked us to 9.39 -describe how certain measurements would affect a particle in a 9.40 -box. Many of these measurement questions required routine application 9.41 -of skills we had recently learned\mdash{}first, you recall (or 9.42 -calculate) the eigenstates of the quantity 9.43 -to be measured; second, you write the given state as a linear 9.44 -sum of these eigenstates\mdash{} the coefficients on each term give 9.45 -the probability amplitude. 9.46 - 9.47 - 9.48 -* What I thought I knew 9.49 - 9.50 -The following is a list of things I thought were true of quantum 9.51 -mechanics; the catch is that the list contradicts itself. 9.52 - 9.53 -- For any hermitian operator: Eigenstates with different eigenvalues are orthogonal. 9.54 -- For any hermitian operator: Any physically allowed state can be 9.55 - written as a linear sum of eigenstates of the operator. 9.56 -- The momentum operator and energy operator are hermitian, because 9.57 - momentum and energy are measureable quantities. 9.58 -- In vacuum, 9.59 - - the momentum operator has an eigenstate 9.60 - $$p(x)=\exp{(ipx/\hbar)}$$ for each value of$p$. 9.61 - - the energy operator has an eigenstate $$|E\rangle = 9.62 - \alpha|p\rangle + \beta|-p\rangle$$ for any $$\alpha,\beta$$ and 9.63 - the particular choice of momentum$p=\sqrt{2mE}$. 9.64 -- In the infinitely deep potential well, 9.65 - - the momentum operator has eigenstates with the same form$p(x) =
9.66 -    \exp{(ipx/\hbar)}$, but because of the boundary conditions on the 9.67 - well, the following modifications are required. 9.68 - - The wavefunction must be zero everywhere outside the well. That 9.69 - is, $$p(x) = \begin{cases}\exp{(ipx/\hbar)},& 0\lt{}x\lt{}a; 9.70 - \\0, & \text{for }x<0\text{ or }x>a \\ \end{cases}$$ 9.71 -#0,&\text{for }x\lt{}0\text{ or }x\gt{}a\end{cases}\) 9.72 - - no longer has an eigenstate for each value 9.73 - of$p$. Instead, only values of$p$that are integer multiples of 9.74 -$\pi a/\hbar$are physically realistic. 9.75 - 9.76 - 9.77 - 9.78 -* COMMENT: 9.79 - 9.80 -** Eigenstates with different eigenvalues are orthogonal 9.81 - 9.82 -#+begin_quote 9.83 -*Theorem:* Eigenstates with different eigenvalues are orthogonal. 9.84 -#+end_quote 9.85 - 9.86 -** COMMENT : 9.87 -I can prove this: if$\Lambda$is any linear operator, suppose$|a\rangle$9.88 -and$|b\rangle$are eigenstates of$\Lambda$. This means that 9.89 - 9.90 - 9.91 -$$9.92 -\begin{eqnarray} 9.93 -\Lambda |a\rangle&=& a|a\rangle,\\ 9.94 -\Lambda|b\rangle&=& b|b\rangle.\\ 9.95 -\end{eqnarray} 9.96 -$$ 9.97 - 9.98 -If we take the difference of these eigenstates, we find that 9.99 - 9.100 -$$9.101 -\begin{eqnarray} 9.102 -\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle 9.103 -\qquad \text{(because \Lambda is linear.)}\\ 9.104 -&=& a|a\rangle - b|b\rangle\qquad\text{(because |a\rangle and 9.105 -|b\rangle are eigenstates of \Lambda)} 9.106 -\end{eqnarray}$$ 9.107 - 9.108 - 9.109 -which means that$a\neq b$. 9.110 - 9.111 -** Eigenvectors of hermitian operators span the space of solutions 9.112 - 9.113 -#+begin_quote 9.114 -*Theorem:* If$\Omega$is a hermitian operator, then every physically 9.115 - allowed state can be written as a linear sum of eigenstates of 9.116 -$\Omega$. 9.117 -#+end_quote 9.118 - 9.119 - 9.120 - 9.121 -** Momentum and energy are hermitian operators 9.122 -This ought to be true because hermitian operators correspond to 9.123 -observable quantities. Since we expect momentum and energy to be 9.124 -measureable quantities, we expect that there are hermitian operators 9.125 -to represent them. 9.126 - 9.127 - 9.128 -** Momentum and energy eigenstates in vacuum 9.129 -An eigenstate of the momentum operator$P$would be a state 9.130 -$$|p\rangle$$ such that $$P|p\rangle=p|p\rangle$$. 9.131 - 9.132 -** Momentum and energy eigenstates in the infinitely deep well 9.133 - 9.134 - 9.135 - 9.136 -* Can you measure momentum in the infinite square well? 9.137 - 9.138 - 9.139 - 9.140 -** COMMENT Momentum eigenstates 9.141 - 9.142 -In free space, the Hamiltonian is $$H=\frac{1}{2m}P^2$$ and the 9.143 -momentum operator$P$has eigenstates $$p(x) = \exp{(-ipx/\hbar)}$$. 9.144 - 9.145 -In the infinitely deep potential well, the Hamiltonian is the same but 9.146 -there is a new condition in order for states to qualify as physically 9.147 -allowed: the states must not exist anywhere outside of well, as it 9.148 -takes an infinite amount of energy to do so. 9.149 - 9.150 -Notice that the momentum eigenstates defined above do /not/ satisfy 9.151 -this condition. 9.152 - 9.153 - 9.154 - 9.155 -* COMMENT 9.156 -For each physical system, there is a Schr\ouml{}dinger equation that 9.157 -describes how a particle's state$|\psi\rangle$will change over 9.158 -time. 9.159 - 9.160 -$$\begin{eqnarray} 9.161 -i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& 9.162 -H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}$$ 9.163 - 9.164 -This is a differential equation; each solution to the 9.165 -Schr\ouml{}dinger equation is a state that is physically allowed for 9.166 -our particle. Here, physically allowed states are 9.167 -those that change in physically allowed ways. However, like any differential 9.168 -equation, the Schr\ouml{}dinger equation can be accompanied by 9.169 -/boundary conditions/\mdash{}conditions that further restrict which 9.170 -states qualify as physically allowed. 9.171 - 9.172 - 9.173 - 9.174 - 9.175 -** Eigenstates of momentum 9.176 - 9.177 - 9.178 - 9.179 - 9.180 -#In the infinitely deep well potential$V(x)=0$, the Schr\ouml{}dinger 9.181 - 9.182 -#$$i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle$$ 9.183 - 9.184 - 9.185 - 9.186 - 9.187 - 9.188 - 9.189 - 9.190 -* COMMENT 9.191 - 9.192 -#* The infinite square well potential 9.193 - 9.194 -A particle exists in a potential that is 9.195 -infinite everywhere except for a region of length $$a$$, where the potential is zero. This means that the 9.196 -particle exists in a potential[fn:coords][fn:infinity] 9.197 - 9.198 - 9.199 -$$V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for 9.200 -}\;x<0\text{ or }x>a.\end{cases}$$ 9.201 - 9.202 -The Schr\ouml{}dinger equation describes how the particle's state 9.203 -$$|\psi\rangle$$ will change over time in this system. 9.204 - 9.205 -$$\begin{eqnarray} 9.206 -i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& 9.207 -H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}$$ 9.208 - 9.209 -This is a differential equation; each solution to the 9.210 -Schr\ouml{}dinger equation is a state that is physically allowed for 9.211 -our particle. Here, physically allowed states are 9.212 -those that change in physically allowed ways. However, like any differential 9.213 -equation, the Schr\ouml{}dinger equation can be accompanied by 9.214 -/boundary conditions/\mdash{}conditions that further restrict which 9.215 -states qualify as physically allowed. 9.216 - 9.217 - 9.218 -Whenever possible, physicists impose these boundary conditions: 9.219 -- A physically allowed state ought to be a /smoothly-varying function of position./ This means 9.220 - that if a particle in the state is likely to be /at/ a particular location, 9.221 - it is also likely to be /near/ that location. 9.222 - 9.223 -These boundary conditions imply that for the square well potential in 9.224 -this problem, 9.225 - 9.226 -- Physically allowed states must be totally confined to the well, 9.227 - because it takes an infinite amount of energy to exist anywhere 9.228 - outside of the well (and physically allowed states ought to have 9.229 - only finite energy). 9.230 -- Physically allowed states must be increasingly unlikely to find very 9.231 - close to the walls of the well. This is because of two conditions: the above 9.232 - condition says that the particle is /impossible/ to find 9.233 - outside of the well, and the smoothly-varying condition says 9.234 - that if a particle is impossible to find at a particular location, 9.235 - it must be unlikely to be found nearby that location. 9.236 - 9.237 -#; physically allowed states are those that change in physically 9.238 -#allowed ways. 9.239 - 9.240 - 9.241 -#** Boundary conditions 9.242 -Because the potential is infinite everywhere except within the well, 9.243 -a realistic particle must be confined to exist only within the 9.244 -well\mdash{}its wavefunction must be zero everywhere beyond the walls 9.245 -of the well. 9.246 - 9.247 - 9.248 -[fn:coords] I chose my coordinate system so that the well extends from 9.249 -$$0<x<a$$. Others choose a coordinate system so that the well extends from 9.250 -$$-\frac{a}{2}<x<\frac{a}{2}$$. Although both coordinate systems describe the same physical 9.251 -situation, they give different-looking answers. 9.252 - 9.253 -[fn:infinity] Of course, infinite potentials are not 9.254 -realistic. Instead, they are useful approximations to finite 9.255 -potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height 9.256 -of the well\rdquo{} are close enough for your own practical 9.257 -purposes. Having introduced a physical impossibility into the problem 9.258 -already, we don't expect to get physically realistic solutions; we 9.259 -just expect to get mathematically consistent ones. The forthcoming 9.260 -trouble is that we don't.   10.1 --- a/org/bk4.org Fri Oct 28 00:03:05 2011 -0700 10.2 +++ /dev/null Thu Jan 01 00:00:00 1970 +0000 10.3 @@ -1,309 +0,0 @@ 10.4 -#+TITLE: Bugs in quantum mechanics 10.5 -#+AUTHOR: Dylan Holmes 10.6 -#+SETUPFILE: ../../aurellem/org/setup.org 10.7 -#+INCLUDE: ../../aurellem/org/level-0.org 10.8 - 10.9 -#Bugs in Quantum Mechanics 10.10 -#Bugs in the Quantum-Mechanical Momentum Operator 10.11 - 10.12 - 10.13 -I studied quantum mechanics the same way I study most subjects\mdash{} 10.14 -by collecting (and squashing) bugs in my understanding. One of these 10.15 -bugs persisted throughout two semesters of 10.16 -quantum mechanics coursework until I finally found 10.17 -the paper 10.18 -[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum 10.19 -mechanics/]], which helped me stamp out the bug entirely. I decided to 10.20 -write an article about the problem and its solution for a number of reasons: 10.21 - 10.22 -- Although the paper was not unreasonably dense, it was written for 10.23 - teachers. I wanted to write an article for students. 10.24 -- I wanted to popularize the problem and its solution because other 10.25 - explanations are currently too hard to find. (Even Shankar's 10.26 - excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it.) 10.27 -- I wanted to check that the bug was indeed entirely 10.28 - eradicated. Attempting an explanation is my way of making 10.29 - sure. 10.30 - 10.31 -* COMMENT 10.32 - I recommend the 10.33 -paper not only for students who are learning 10.34 -quantum mechanics, but especially for teachers interested in debugging 10.35 -them. 10.36 - 10.37 -* COMMENT 10.38 -On my first exam in quantum mechanics, my professor asked us to 10.39 -describe how certain measurements would affect a particle in a 10.40 -box. Many of these measurement questions required routine application 10.41 -of skills we had recently learned\mdash{}first, you recall (or 10.42 -calculate) the eigenstates of the quantity 10.43 -to be measured; second, you write the given state as a linear 10.44 -sum of these eigenstates\mdash{} the coefficients on each term give 10.45 -the probability amplitude. 10.46 - 10.47 - 10.48 -* What I thought I knew 10.49 - 10.50 -The following is a list of things I thought were true of quantum 10.51 -mechanics; the catch is that the list contradicts itself. 10.52 - 10.53 -1. For any hermitian operator: Eigenstates with different eigenvalues are orthogonal. 10.54 -2. For any hermitian operator: Any physically allowed state can be 10.55 - written as a linear sum of eigenstates of the operator. 10.56 -3. The momentum operator and energy operator are hermitian, because 10.57 - momentum and energy are measureable quantities. 10.58 -4. In the vacuum potential, the momentum and energy operators have these eigenstates: 10.59 - - the momentum operator has an eigenstate 10.60 - $$p(x)=\exp{(ipx/\hbar)}$$ for each value of$p$. 10.61 - - the energy operator has an eigenstate $$|E\rangle = 10.62 - \alpha|p\rangle + \beta|-p\rangle$$ for any $$\alpha,\beta$$ and 10.63 - the particular choice of momentum$p=\sqrt{2mE}$. 10.64 -5. In the infinitely deep potential well, the momentum and energy 10.65 - operators have these eigenstates: 10.66 - - The momentum eigenstates and energy eigenstates have the same form 10.67 - as in the vacuum potential:$p(x) =
10.68 -     \exp{(ipx/\hbar)}$and$|E\rangle = \alpha|p\rangle + \beta|-p\rangle$. 10.69 - - Even so, because of the boundary conditions on the 10.70 - well, we must make the following modifications: 10.71 - + Physically realistic states must be impossible to find outside the well. (Only a state of infinite 10.72 - energy could exist outside the well, and infinite energy is not 10.73 - realistic.) This requirement means, for example, that momentum 10.74 - eigenstates in the infinitely deep well must be 10.75 - $$p(x) 10.76 - = \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a; 10.77 - \\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}$$ 10.78 - + Physically realistic states must vary smoothly throughout 10.79 - space. This means that if a particle in some state is very unlikely to be 10.80 - /at/ a particular location, it is also very unlikely be /near/ 10.81 - that location. Combining this requirement with the above 10.82 - requirement, we find that the momentum operator no longer has 10.83 - an eigenstate for each value of$p$; instead, only values of 10.84 -$p$that are integer multiples of$\pi a/\hbar$are physically 10.85 - realistic. Similarly, the energy operator no longer has an 10.86 - eigenstate for each value of$E$; instead, the only energy 10.87 - eigenstates in the infinitely deep well 10.88 - are$E_n(x)=\sin(n\pi x/ a)$for positive integers$n$. 10.89 - 10.90 -* COMMENT: 10.91 - 10.92 -** Eigenstates with different eigenvalues are orthogonal 10.93 - 10.94 -#+begin_quote 10.95 -*Theorem:* Eigenstates with different eigenvalues are orthogonal. 10.96 -#+end_quote 10.97 - 10.98 -** COMMENT : 10.99 -I can prove this: if$\Lambda$is any linear operator, suppose$|a\rangle$10.100 -and$|b\rangle$are eigenstates of$\Lambda$. This means that 10.101 - 10.102 - 10.103 -$$10.104 -\begin{eqnarray} 10.105 -\Lambda |a\rangle&=& a|a\rangle,\\ 10.106 -\Lambda|b\rangle&=& b|b\rangle.\\ 10.107 -\end{eqnarray} 10.108 -$$ 10.109 - 10.110 -If we take the difference of these eigenstates, we find that 10.111 - 10.112 -$$10.113 -\begin{eqnarray} 10.114 -\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle 10.115 -\qquad \text{(because \Lambda is linear.)}\\ 10.116 -&=& a|a\rangle - b|b\rangle\qquad\text{(because |a\rangle and 10.117 -|b\rangle are eigenstates of \Lambda)} 10.118 -\end{eqnarray}$$ 10.119 - 10.120 - 10.121 -which means that$a\neq b$. 10.122 - 10.123 -** Eigenvectors of hermitian operators span the space of solutions 10.124 - 10.125 -#+begin_quote 10.126 -*Theorem:* If$\Omega$is a hermitian operator, then every physically 10.127 - allowed state can be written as a linear sum of eigenstates of 10.128 -$\Omega$. 10.129 -#+end_quote 10.130 - 10.131 - 10.132 - 10.133 -** Momentum and energy are hermitian operators 10.134 -This ought to be true because hermitian operators correspond to 10.135 -observable quantities. Since we expect momentum and energy to be 10.136 -measureable quantities, we expect that there are hermitian operators 10.137 -to represent them. 10.138 - 10.139 - 10.140 -** Momentum and energy eigenstates in vacuum 10.141 -An eigenstate of the momentum operator$P$would be a state 10.142 -$$|p\rangle$$ such that $$P|p\rangle=p|p\rangle$$. 10.143 - 10.144 -** Momentum and energy eigenstates in the infinitely deep well 10.145 - 10.146 - 10.147 - 10.148 -* Can you measure momentum in the infinitely deep well? 10.149 -In summary, I thought I knew: 10.150 -1. For any hermitian operator: eigenstates with different eigenvalues 10.151 - are orthogonal. 10.152 -2. For any hermitian operator: any physically realistic state can be 10.153 - written as a linear sum of eigenstates of the operator. 10.154 -3. The momentum operator and energy operator are hermitian, because 10.155 - momentum and energy are observable quantities. 10.156 -4. (The form of the momentum and energy eigenstates in the vacuum potential) 10.157 -5. (The form of the momentum and energy eigenstates in the infinitely deep well potential) 10.158 - 10.159 -Additionally, I understood that because the infinitely deep potential 10.160 -well is not realistic, states of such a system are not necessarily 10.161 -physically realistic. Instead, I understood 10.162 -\ldquo{}realistic states\rdquo{} to be those that satisfy the physically 10.163 -unrealistic Schr\ouml{}dinger equation and its boundary conditions. 10.164 - 10.165 -With that final caveat, here is the problem: 10.166 - 10.167 -According to (5), the momentum eigenstates in the well are 10.168 - 10.169 -$$p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}$$ 10.170 - 10.171 -However, /these/ states are not orthogonal, which contradicts the 10.172 -assumption that (3) the momentum operator is hermitian and (2) 10.173 -eigenstates of a hermitian are orthogonal if they have different eigenvalues. 10.174 - 10.175 -#+begin_quote 10.176 -*Problem 1. The momentum eigenstates of the well are not orthogonal* 10.177 - 10.178 -/Proof./ If$p_1\neq p_2$, then 10.179 - 10.180 -$$\begin{eqnarray} 10.181 -\langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)dx\\ 10.182 -&=& \int_0^a p_1^*(x)p_2(x)dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{ 10.183 -outside the well.}\\ 10.184 -&=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)dx}} 10.185 -\end{eqnarray}$$ 10.186 -$\square$10.187 - 10.188 -#+end_quote 10.189 - 10.190 - 10.191 - 10.192 -** COMMENT Momentum eigenstates 10.193 - 10.194 -In free space, the Hamiltonian is $$H=\frac{1}{2m}P^2$$ and the 10.195 -momentum operator$P$has eigenstates $$p(x) = \exp{(-ipx/\hbar)}$$. 10.196 - 10.197 -In the infinitely deep potential well, the Hamiltonian is the same but 10.198 -there is a new condition in order for states to qualify as physically 10.199 -allowed: the states must not exist anywhere outside of well, as it 10.200 -takes an infinite amount of energy to do so. 10.201 - 10.202 -Notice that the momentum eigenstates defined above do /not/ satisfy 10.203 -this condition. 10.204 - 10.205 - 10.206 - 10.207 -* COMMENT 10.208 -For each physical system, there is a Schr\ouml{}dinger equation that 10.209 -describes how a particle's state$|\psi\rangle$will change over 10.210 -time. 10.211 - 10.212 -$$\begin{eqnarray} 10.213 -i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& 10.214 -H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}$$ 10.215 - 10.216 -This is a differential equation; each solution to the 10.217 -Schr\ouml{}dinger equation is a state that is physically allowed for 10.218 -our particle. Here, physically allowed states are 10.219 -those that change in physically allowed ways. However, like any differential 10.220 -equation, the Schr\ouml{}dinger equation can be accompanied by 10.221 -/boundary conditions/\mdash{}conditions that further restrict which 10.222 -states qualify as physically allowed. 10.223 - 10.224 - 10.225 - 10.226 - 10.227 -** Eigenstates of momentum 10.228 - 10.229 - 10.230 - 10.231 - 10.232 -#In the infinitely deep well potential$V(x)=0$, the Schr\ouml{}dinger 10.233 - 10.234 -#$$i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle$$ 10.235 - 10.236 - 10.237 - 10.238 - 10.239 - 10.240 - 10.241 - 10.242 -* COMMENT 10.243 - 10.244 -#* The infinite square well potential 10.245 - 10.246 -A particle exists in a potential that is 10.247 -infinite everywhere except for a region of length $$a$$, where the potential is zero. This means that the 10.248 -particle exists in a potential[fn:coords][fn:infinity] 10.249 - 10.250 - 10.251 -$$V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for 10.252 -}\;x<0\text{ or }x>a.\end{cases}$$ 10.253 - 10.254 -The Schr\ouml{}dinger equation describes how the particle's state 10.255 -$$|\psi\rangle$$ will change over time in this system. 10.256 - 10.257 -$$\begin{eqnarray} 10.258 -i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& 10.259 -H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}$$ 10.260 - 10.261 -This is a differential equation; each solution to the 10.262 -Schr\ouml{}dinger equation is a state that is physically allowed for 10.263 -our particle. Here, physically allowed states are 10.264 -those that change in physically allowed ways. However, like any differential 10.265 -equation, the Schr\ouml{}dinger equation can be accompanied by 10.266 -/boundary conditions/\mdash{}conditions that further restrict which 10.267 -states qualify as physically allowed. 10.268 - 10.269 - 10.270 -Whenever possible, physicists impose these boundary conditions: 10.271 -- A physically allowed state ought to be a /smoothly-varying function of position./ This means 10.272 - that if a particle in the state is likely to be /at/ a particular location, 10.273 - it is also likely to be /near/ that location. 10.274 - 10.275 -These boundary conditions imply that for the square well potential in 10.276 -this problem, 10.277 - 10.278 -- Physically allowed states must be totally confined to the well, 10.279 - because it takes an infinite amount of energy to exist anywhere 10.280 - outside of the well (and physically allowed states ought to have 10.281 - only finite energy). 10.282 -- Physically allowed states must be increasingly unlikely to find very 10.283 - close to the walls of the well. This is because of two conditions: the above 10.284 - condition says that the particle is /impossible/ to find 10.285 - outside of the well, and the smoothly-varying condition says 10.286 - that if a particle is impossible to find at a particular location, 10.287 - it must be unlikely to be found nearby that location. 10.288 - 10.289 -#; physically allowed states are those that change in physically 10.290 -#allowed ways. 10.291 - 10.292 - 10.293 -#** Boundary conditions 10.294 -Because the potential is infinite everywhere except within the well, 10.295 -a realistic particle must be confined to exist only within the 10.296 -well\mdash{}its wavefunction must be zero everywhere beyond the walls 10.297 -of the well. 10.298 - 10.299 - 10.300 -[fn:coords] I chose my coordinate system so that the well extends from 10.301 -$$0<x<a$$. Others choose a coordinate system so that the well extends from 10.302 -$$-\frac{a}{2}<x<\frac{a}{2}$$. Although both coordinate systems describe the same physical 10.303 -situation, they give different-looking answers. 10.304 - 10.305 -[fn:infinity] Of course, infinite potentials are not 10.306 -realistic. Instead, they are useful approximations to finite 10.307 -potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height 10.308 -of the well\rdquo{} are close enough for your own practical 10.309 -purposes. Having introduced a physical impossibility into the problem 10.310 -already, we don't expect to get physically realistic solutions; we 10.311 -just expect to get mathematically consistent ones. The forthcoming 10.312 -trouble is that we don't.   11.1 --- a/org/bk_quandary.org Fri Oct 28 00:03:05 2011 -0700 11.2 +++ /dev/null Thu Jan 01 00:00:00 1970 +0000 11.3 @@ -1,566 +0,0 @@ 11.4 -#+TITLE: Bugs in quantum mechanics 11.5 -#+AUTHOR: Dylan Holmes 11.6 -#+DESCRIPTION: I explain hermitian operators and use them to solve a confusing problem in quantum mechanics. 11.7 -#+KEYWORDS: quantum mechanics, self-adjoint, hermitian, square well, momentum 11.8 -#+SETUPFILE: ../../aurellem/org/setup.org 11.9 -#+INCLUDE: ../../aurellem/org/level-0.org 11.10 - 11.11 - 11.12 - 11.13 -#Bugs in Quantum Mechanics 11.14 -#Bugs in the Quantum-Mechanical Momentum Operator 11.15 - 11.16 - 11.17 -I studied quantum mechanics the same way I study most subjects\mdash{} 11.18 -by collecting (and squashing) bugs in my understanding. One of these 11.19 -bugs persisted throughout two semesters of 11.20 -quantum mechanics coursework until I finally found 11.21 -the paper 11.22 -[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum 11.23 -mechanics/]], which helped me stamp out the bug entirely. I decided to 11.24 -write an article about the problem and its solution for a number of reasons: 11.25 - 11.26 -- Although the paper was not unreasonably dense, it was written for 11.27 - teachers. I wanted to write an article for students. 11.28 -- I wanted to popularize the problem and its solution because other 11.29 - explanations are currently too hard to find. (Even Shankar's 11.30 - excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it, as far as I know.) 11.31 -- Attempting an explanation is my way of making 11.32 - sure that the bug is /really/ gone. 11.33 -# entirely eradicated. 11.34 - 11.35 -* COMMENT 11.36 - I recommend the 11.37 -paper not only for students who are learning 11.38 -quantum mechanics, but especially for teachers interested in debugging 11.39 -them. 11.40 - 11.41 -* COMMENT 11.42 -On my first exam in quantum mechanics, my professor asked us to 11.43 -describe how certain measurements would affect a particle in a 11.44 -box. Many of these measurement questions required routine application 11.45 -of skills we had recently learned\mdash{}first, you recall (or 11.46 -calculate) the eigenstates of the quantity 11.47 -to be measured; second, you write the given state as a linear 11.48 -sum of these eigenstates\mdash{} the coefficients on each term give 11.49 -the probability amplitude. 11.50 - 11.51 - 11.52 -* Two methods of calculation that give different results. 11.53 - 11.54 -In the infinitely deep well, there is a particle in the the 11.55 -normalized state 11.56 - 11.57 - $$\psi(x) = \begin{cases}A\; x(x-a)&\text{for }0\lt{}x\lt{}a;\\0,&\text{for }x\lt{}0\text{ or }x>a.\end{cases}$$ 11.58 - 11.59 -This is apparently a perfectly respectable state: it is normalized ($A$is a 11.60 -normalization constant), it is zero 11.61 -everywhere outside of the well, and it is moreover continuous. 11.62 - 11.63 -Even so, we will find a problem if we attempt to calculate the average 11.64 -energy-squared of this state (that is, the quantity $$\langle \psi | H^2 | \psi \rangle$$). 11.65 - 11.66 -** First method 11.67 - 11.68 -For short, define a new variable[fn:1] $$|\bar \psi\rangle \equiv 11.69 -H|\psi\rangle$$. We can express the state$|\bar\psi\rangle$as a 11.70 -function of$x$because we know how to express$H$and$\psi$in terms 11.71 -of$x$:$H$is the differential operator$\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}$, and 11.72 -$\psi(x)$is the function defined above. So, we get $$\quad\bar\psi(x) = \frac{-A\hbar^2}{m}$$. For future reference, observe that$\bar\psi(x)$11.73 -is constant. 11.74 - 11.75 -Having introduced$|\bar\psi\rangle$, we can calculate the average energy-squared of$|\psi\rangle$in the 11.76 -following way. 11.77 - 11.78 -$$\begin{eqnarray} 11.79 -\langle \psi | H^2 |\psi\rangle &=& \langle \psi H^\dagger | H 11.80 -\psi\rangle\\ 11.81 -&=& \langle \psi H | H\psi \rangle\\ 11.82 -&=& \langle \bar\psi | \bar\psi \rangle\\ 11.83 -&=& \int_0^a \left(\frac{A\hbar^2}{m}\right)^2\;dx\\ 11.84 -&=& \frac{A^2\hbar^4 a}{m^2}\\ 11.85 -\end{eqnarray}$$ 11.86 - 11.87 -For future reference, observe that this value is nonzero 11.88 -(which makes sense). 11.89 - 11.90 -** Second method 11.91 -We can also calculate the average energy-squared of$|\psi\rangle$in the 11.92 -following way. 11.93 - 11.94 -\begin{eqnarray} 11.95 -\langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\ 11.96 -&=& \langle \psi |H \bar\psi \rangle\\ 11.97 -&=&\int_0^a Ax(x-a) 11.98 -\left(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\right)\frac{-A\hbar^2}{m}\;dx\\ 11.99 -&=& 0\quad (!)\\ 11.100 -\end{eqnarray} 11.101 - 11.102 -The second-to-last term must be zero because the second derivative 11.103 -of a constant $$\left(\frac{-A\hbar^2}{m}\right)$$ is zero. 11.104 - 11.105 -* What is the problem? 11.106 - 11.107 -To recap: We used two different methods to calculate the average 11.108 -energy-squared of a state$|\psi\rangle$. For the first method, we 11.109 -used the fact that$H$is a hermitian operator, replacing $$\langle 11.110 -\psi H^\dagger | H \psi\rangle$$ with $$\langle \psi H | H 11.111 -\psi\rangle$$. Using this substitution rule, we calculated the answer. 11.112 - 11.113 -For the second method, we didn't use the fact that$H$was hermitian; 11.114 -instead, we used the fact that we know how to represent$H$and$\psi$11.115 -as functions of$x$:$H$is a differential operator 11.116 -$$\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}$$ and$\psi$is a quadratic 11.117 -function of$x$. By applying$H$to$\psi$, we took several 11.118 -derivatives and arrived at our answer. 11.119 - 11.120 -These two methods gave different results. In the following sections, 11.121 -I'll describe and analyze the source of this difference. 11.122 - 11.123 -** Physical operators only act on physical wavefunctions 11.124 - :PROPERTIES: 11.125 - :ORDERED: t 11.126 - :END: 11.127 -#In quantum mechanics, an operator is a function that takes in a 11.128 -#physical state and produces another physical state as ouput. Some 11.129 -#operators correspond to physical quantities such as energy, 11.130 -#momentum, or position; the mathematical properties of these operators correspond to 11.131 -#physical properties of the system. 11.132 - 11.133 -#Eigenstates are an example of this correspondence: an 11.134 - 11.135 -Physical states are represented as wavefunctions in quantum 11.136 -mechanics. Just as we disallow certain physically nonsensical states 11.137 -in classical mechanics (for example, we consider it to be nonphysical 11.138 -for an object to spontaneously disappear from one place and reappear 11.139 -in another), we also disallow certain wavefunctions in quantum 11.140 -mechanics. 11.141 - 11.142 -For example, since wavefunctions are supposed to correspond to 11.143 -probability amplitudes, we require wavefunctions to be normalized 11.144 -$$(\langle \psi |\psi \rangle = 1)$$. Generally, we disallow 11.145 -wavefunctions that do not satisfy this property (although there are 11.146 -some exceptions[fn:2]). 11.147 - 11.148 -As another example, we generally expect probability to vary smoothly\mdash{}if 11.149 -a particle is very likely or very unlikely to be found at a particular 11.150 -location, it should also be somewhat likely or somewhat unlikely to be 11.151 -found /near/ that location. In more precise terms, we expect that for 11.152 -physically meaningful wavefunctions, the probability 11.153 -$$\text{Pr}(x)=\int^x \psi^*\psi$$ should be a continuous function of 11.154 -$x$and, again, we disallow wavefunctions that do not satisfy this 11.155 -property because we consider them to be physically nonsensical. 11.156 - 11.157 -So, physical wavefunctions must satisfy certain properties 11.158 -like the two just described. Wavefunctions that do not satisfy these properties are 11.159 -rejected for being physically nonsensical: even though we can perform 11.160 -calculations with them, the mathematical results we obtain do not mean 11.161 -anything physically. 11.162 - 11.163 -Now, in quantum mechanics, an *operator* is a function that converts 11.164 -states into other states. Some operators correspond to 11.165 -physical quantities such as energy, momentum, or position, and as a 11.166 -result, the mathematical properties of these operators correspond to 11.167 -physical properties of the system. Physical operators are furthermore 11.168 -subject to the following rule: they are only allowed to operate on 11.169 -#physical wavefunctions, and they are only allowed to produce 11.170 -#physical wavefunctions[fn:why]. 11.171 -the wavefunctions we have deemed physical, and they only produce physical wavefunctions[fn:: 11.172 - 11.173 - If you require a hermitian operator to have physical 11.174 - eigenstates, you get a very strong result: you guarantee that the 11.175 - operator will convert /every/ physical wavefunction into another 11.176 - physical wavefunction: 11.177 - 11.178 - For any linear operator$\Omega$, the eigenvalue equation is 11.179 -$$\Omega|\omega\rangle = \omega |\omega\rangle$$. Notice that if an 11.180 -eigenstate$|\omega\rangle$is a physical wavefunction, the 11.181 -eigenvalue equation equation forces$\Omega|\omega\rangle$to be a 11.182 -physical wavefunction as well. To elaborate, if the eigenstates of 11.183 -$\Omega$are physical functions, then$\Omega$is guaranteed to 11.184 -convert them into other physical functions. Even more is true if the 11.185 -operator$\Omega$is also hermitian: there is a theorem which states 11.186 -that \ldquo{}If \Omega is hermitian, then every physical wavefunction 11.187 -can be written as a weighted sum of eigenstates of \Omega.\rdquo{} This 11.188 -theorem implies that /if$\Omega$is hermitian/, and /if the eigenstates 11.189 -of \Omega are physically allowed/, then \Omega is guaranteed to 11.190 -convert every physically allowed wavefunction into another physically 11.191 -allowed wavefunction.]. 11.192 - 11.193 -In fact, this rule for physical operators is the source of our 11.194 -problem, as we unknowingly violated it when applying our second 11.195 -method! 11.196 - 11.197 -** The violation 11.198 - 11.199 -I'll start explaining this violation by being more specific about the 11.200 -infinitely deep well potential. We have said already that physicists 11.201 -require wavefunctions to satisfy certain properties in order to be 11.202 -deemed \ldquo{}physical\rdquo{}. To be specific, wavefunctions in the 11.203 -infinitely deep well 11.204 -- Must be *normalizable*, because they correspond to 11.205 - probability amplitudes. 11.206 -- Must have *smoothly-varying probability*, because if a particle is very 11.207 - likely to be at a location, it ought to be likely to be /near/ 11.208 - it as well. 11.209 -- Must *not exist outside the well*, because it 11.210 - would take an infinite amount of energy to do so. 11.211 - 11.212 -Additionally, by combining the second and third conditions, some 11.213 -physicists reason that wavefunctions in the infinitely deep well 11.214 - 11.215 -- Must *become zero* towards the edges of the well. 11.216 - 11.217 - 11.218 - 11.219 - 11.220 -You'll remember we had 11.221 - 11.222 -$$11.223 -\begin{eqnarray} 11.224 -\psi(x) &=& A\;x(x-a)\\ 11.225 -\bar\psi(x)&=& \frac{-A\hbar^2}{m}\\ 11.226 -&&\text{for }0\lt{}x\lt{}a\\ 11.227 -\end{eqnarray} 11.228 -$$ 11.229 - 11.230 -In our second method, we wrote 11.231 - 11.232 - 11.233 -$$\begin{eqnarray} 11.234 -\langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\ 11.235 -&=& \langle \psi |H \bar\psi \rangle\\ 11.236 -& \vdots&\\ 11.237 -&=& 0\\ 11.238 -\end{eqnarray}$$ 11.239 - 11.240 -However, there are two issues here. First, we were not allowed to operate with$H$on the wavefunction 11.241 -$|\bar \psi\rangle$, because$H$is a physical operator and$|\bar
11.242 -\psi\rangle$is a nonphysical state: in the infinite square well, 11.243 -physical wavefunctions must approach zero at the edges of the well, 11.244 -which the constant function$|\bar\psi\rangle$does not do. By 11.245 -feeding$H$a nonphysical wavefunction, we obtained nonsensical 11.246 -results. 11.247 - 11.248 -Second, we claimed that$H$was a physical operator\mdash{}that$H$11.249 -was hermitian. According to the rule, this means$H$must convert physical states into other 11.250 -physical states. But$H$converts the physical state$|\psi\rangle$11.251 -into the nonphysical state$|\bar\psi\rangle = H|\psi\rangle$is not. Because$H$converts some 11.252 -physical states into nonphysical states, it cannot be a hermitian operator. 11.253 - 11.254 -** Boundary conditions affect hermiticity 11.255 -We have now discovered a flaw: when applied to the state 11.256 -$|\psi\rangle$, the second method violates the rule that physical 11.257 -operators must only take in physical states and must only produce 11.258 -physical states. This suggests that the problem was with the state 11.259 -$|\bar\psi\rangle$that we wrongly fed into$H$. However, the problem 11.260 -is more serious still: the state$|\psi\rangle
11.261 -
11.262 -** COMMENT Re-examining physical constraints
11.263 -
11.264 -We have now discovered a flaw: when applied to the state
11.265 -$|\psi\rangle$, the second method violates the rule that physical
11.266 -operators must only take in physical states and must only produce
11.267 -physical states. Let's examine the problem more closely.
11.268 -
11.269 -We have said already that physicists require wavefunctions to satisfy
11.270 -certain properties in order to be deemed \ldquo{}physical\rdquo{}. To
11.271 -be specific, wavefunctions in the infinitely deep well
11.272 -- Must be *normalizable*, because they correspond to
11.273 -  probability amplitudes.
11.274 -- Must have *smoothly-varying probability*, because if a particle is very
11.275 -  likely to be at a location, it ought to be likely to be /near/
11.276 -  it as well.
11.277 -- Must *not exist outside the well*, because it
11.278 -  would take an infinite amount of energy to do so.
11.279 -
11.280 -We now have discovered an important flaw in the second method: when
11.281 -applied to the state $|\bar\psi\rangle$, the second method violates
11.282 -the rule that physical operators must only take in
11.283 -physical states and must only produce physical states. The problem is
11.284 -even more serious, however
11.285 -
11.286 -
11.287 -
11.288 -[fn:1] I'm defining a new variable just to make certain expressions
11.289 -  look shorter; this cannot affect the content of the answer we'll
11.290 -  get.
11.291 -
11.292 -[fn:2] For example, in vaccuum (i.e., when the potential of the
11.293 -  physical system is $V(x)=0$ throughout all space), the momentum
11.294 -  eigenstates are not normalizable\mdash{}the relevant integral blows
11.295 -  up to infinity instead of converging to a number. Physicists modify
11.296 -  the definition of normalization slightly so that
11.297 -  \ldquo{}delta-normalizable \rdquo{} functions like these are included
11.298 -  among the physical wavefunctions.
11.299 -
11.300 -
11.301 -
11.302 -* COMMENT: What I thought I knew
11.303 -
11.304 -The following is a list of things I thought were true of quantum
11.305 -mechanics; the catch is that the list contradicts itself.
11.306 -
11.307 -1. For any hermitian operator: Eigenstates with different eigenvalues are orthogonal.
11.308 -2. For any hermitian operator: Any physically allowed state can be
11.309 -   written as a linear sum of eigenstates of the operator.
11.310 -3. The momentum operator and energy operator are hermitian, because
11.311 -   momentum and energy are measureable quantities.
11.312 -4. In the vacuum potential, the momentum and energy operators have these eigenstates:
11.313 -   - the momentum operator has an eigenstate
11.314 -     $$p(x)=\exp{(ipx/\hbar)}$$ for each value of $p$.
11.315 -   - the energy operator has an eigenstate $$|E\rangle = 11.316 - \alpha|p\rangle + \beta|-p\rangle$$ for any $$\alpha,\beta$$ and
11.317 -     the particular choice of momentum $p=\sqrt{2mE}$.
11.318 -5. In the infinitely deep potential well, the momentum and energy
11.319 -   operators have these eigenstates:
11.320 -   - The momentum eigenstates and energy eigenstates have the same form
11.321 -     as in the vacuum potential: $p(x) = 11.322 - \exp{(ipx/\hbar)}$ and $|E\rangle = \alpha|p\rangle + \beta|-p\rangle$.
11.323 -   - Even so, because of the boundary conditions on the
11.324 -     well, we must make the following modifications:
11.325 -     + Physically realistic states must be impossible to find outside the well. (Only a state of infinite
11.326 -       energy could exist outside the well, and infinite energy is not
11.327 -       realistic.) This requirement means, for example, that momentum
11.328 -       eigenstates in the infinitely deep well must be
11.329 -       $$p(x) 11.330 - = \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a; 11.331 - \\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}$$
11.332 -     + Physically realistic states must vary smoothly throughout
11.333 -       space. This means that if a particle in some state is very unlikely to be
11.334 -       /at/ a particular location, it is also very unlikely be /near/
11.335 -       that location. Combining this requirement with the above
11.336 -       requirement, we find that the momentum operator no longer has
11.337 -       an eigenstate for each value of $p$; instead, only values of
11.338 -       $p$ that are integer multiples of $\pi \hbar/a$ are physically
11.339 -       realistic. Similarly, the energy operator no longer has an
11.340 -       eigenstate for each value of $E$; instead, the only energy
11.341 -       eigenstates in the infinitely deep well
11.342 -       are $E_n(x)=\sin(n\pi x/ a)$ for positive integers $n$.
11.343 -
11.344 -* COMMENT:
11.345 -
11.346 -** Eigenstates with different eigenvalues are orthogonal
11.347 -
11.348 -#+begin_quote
11.349 -*Theorem:* Eigenstates with different eigenvalues are orthogonal.
11.350 -#+end_quote
11.351 -
11.352 -** COMMENT :
11.353 -I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$
11.354 -and $|b\rangle$ are eigenstates of $\Lambda$. This means that
11.355 -
11.356 -
11.357 -$$11.358 -\begin{eqnarray} 11.359 -\Lambda |a\rangle&=& a|a\rangle,\\ 11.360 -\Lambda|b\rangle&=& b|b\rangle.\\ 11.361 -\end{eqnarray} 11.362 -$$
11.363 -
11.364 -If we take the difference of these eigenstates, we find that
11.365 -
11.366 -$$11.367 -\begin{eqnarray} 11.368 -\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle 11.369 -\qquad \text{(because \Lambda is linear.)}\\ 11.370 -&=& a|a\rangle - b|b\rangle\qquad\text{(because |a\rangle and 11.371 -|b\rangle are eigenstates of \Lambda)} 11.372 -\end{eqnarray}$$
11.373 -
11.374 -
11.375 -which means that $a\neq b$.
11.376 -
11.377 -** Eigenvectors of hermitian operators span the space of solutions
11.378 -
11.379 -#+begin_quote
11.380 -*Theorem:* If $\Omega$ is a hermitian operator, then every physically
11.381 - allowed state can be written as a linear sum of eigenstates of
11.382 - $\Omega$.
11.383 -#+end_quote
11.384 -
11.385 -
11.386 -
11.387 -** Momentum and energy are hermitian operators
11.388 -This ought to be true because hermitian operators correspond to
11.389 -observable quantities. Since we expect momentum and energy to be
11.390 -measureable quantities, we expect that there are hermitian operators
11.391 -to represent them.
11.392 -
11.393 -
11.394 -** Momentum and energy eigenstates in vacuum
11.395 -An eigenstate of the momentum operator $P$ would be a state
11.396 -$$|p\rangle$$ such that $$P|p\rangle=p|p\rangle$$.
11.397 -
11.398 -** Momentum and energy eigenstates in the infinitely deep well
11.399 -
11.400 -
11.401 -
11.402 -* COMMENT Can you measure momentum in the infinitely deep well?
11.403 -In summary, I thought I knew:
11.404 -1. For any hermitian operator: eigenstates with different eigenvalues
11.405 -   are orthogonal.
11.406 -2. For any hermitian operator: any physically realistic state can be
11.407 -   written as a linear sum of eigenstates of the operator.
11.408 -3. The momentum operator and energy operator are hermitian, because
11.409 -   momentum and energy are observable quantities.
11.410 -4. (The form of the momentum and energy eigenstates in the vacuum potential)
11.411 -5. (The form of the momentum and energy eigenstates in the infinitely deep well potential)
11.412 -
11.413 -Additionally, I understood that because the infinitely deep potential
11.414 -well is not realistic, states of such a system  are not necessarily
11.415 -physically realistic. Instead, I understood
11.416 -\ldquo{}realistic states\rdquo{} to be those that satisfy the physically
11.417 -unrealistic Schr\ouml{}dinger equation and its boundary conditions.
11.418 -
11.419 -With that final caveat, here is the problem:
11.420 -
11.421 -According to (5), the momentum eigenstates in the well are
11.422 -
11.423 -$$p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}$$
11.424 -
11.425 -(Additionally, we require that $p$ be an integer multiple of $\pi\hbar/a$.)
11.426 -
11.427 -However, /these/ states are not orthogonal, which contradicts the
11.428 -assumption that (3) the momentum operator is hermitian and (2)
11.429 -eigenstates of a hermitian are orthogonal if they have different eigenvalues.
11.430 -
11.431 -#+begin_quote
11.432 -*Problem 1. The momentum eigenstates of the well are not orthogonal*
11.433 -
11.434 -/Proof./ If $p_1\neq p_2$, then
11.435 -
11.436 -$$\begin{eqnarray} 11.437 -\langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)\,dx\\ 11.438 -&=& \int_0^a p_1^*(x)p_2(x)\,dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{ 11.439 -outside the well.}\\ 11.440 -&=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)\,dx}}\\ 11.441 -&=& \int_0^a \exp{\left[i(p_1-p_2)x/\hbar\right]}\,dx\\ 11.442 -\end{eqnarray}$$
11.443 -$\square$
11.444 -
11.445 -#+end_quote
11.446 -
11.447 -
11.448 -
11.449 -** COMMENT  Momentum eigenstates
11.450 -
11.451 -In free space, the Hamiltonian is $$H=\frac{1}{2m}P^2$$ and the
11.452 -momentum operator $P$ has eigenstates $$p(x) = \exp{(-ipx/\hbar)}$$.
11.453 -
11.454 -In the infinitely deep potential well, the Hamiltonian is the same but
11.455 -there is a new condition in order for states to qualify as physically
11.456 -allowed: the states must not exist anywhere outside of well, as it
11.457 -takes an infinite amount of energy to do so.
11.458 -
11.459 -Notice that the momentum eigenstates defined above do /not/ satisfy
11.460 -this condition.
11.461 -
11.462 -
11.463 -
11.464 -* COMMENT
11.465 -For each physical system, there is a Schr\ouml{}dinger equation that
11.466 -describes how a particle's state $|\psi\rangle$  will change over
11.467 -time.
11.468 -
11.469 -$$\begin{eqnarray} 11.470 -i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& 11.471 -H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}$$
11.472 -
11.473 -This is a differential equation; each solution to the
11.474 -Schr\ouml{}dinger equation is a state that is physically allowed for
11.475 -our particle. Here, physically allowed states are
11.476 -those that change in physically allowed ways. However, like any differential
11.477 -equation, the Schr\ouml{}dinger equation can be accompanied by
11.478 -/boundary conditions/\mdash{}conditions that further restrict which
11.479 -states qualify as physically allowed.
11.480 -
11.481 -
11.482 -
11.483 -
11.484 -** Eigenstates of momentum
11.485 -
11.486 -
11.487 -
11.488 -
11.489 -#In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger
11.490 -
11.491 -#$$i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle$$
11.492 -
11.493 -
11.494 -
11.495 -
11.496 -
11.497 -
11.498 -
11.499 -* COMMENT
11.500 -
11.501 -#* The infinite square well potential
11.502 -
11.503 -A particle exists in a potential that is
11.504 -infinite everywhere except for a region of length $$a$$, where the potential is zero. This means that the
11.505 -particle exists in a potential[fn:coords][fn:infinity]
11.506 -
11.507 -
11.508 -$$V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for 11.509 -}\;x<0\text{ or }x>a.\end{cases}$$
11.510 -
11.511 -The Schr\ouml{}dinger equation describes how the particle's state
11.512 -$$|\psi\rangle$$ will change over time in this system.
11.513 -
11.514 -$$\begin{eqnarray} 11.515 -i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& 11.516 -H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}$$
11.517 -
11.518 -This is a differential equation; each solution to the
11.519 -Schr\ouml{}dinger equation is a state that is physically allowed for
11.520 -our particle. Here, physically allowed states are
11.521 -those that change in physically allowed ways. However, like any differential
11.522 -equation, the Schr\ouml{}dinger equation can be accompanied by
11.523 -/boundary conditions/\mdash{}conditions that further restrict which
11.524 -states qualify as physically allowed.
11.525 -
11.526 -
11.527 -Whenever possible, physicists impose these boundary conditions:
11.528 -- A physically allowed state ought to be a /smoothly-varying function of position./ This means
11.529 -  that if a particle in the state  is likely to be /at/ a particular location,
11.530 -  it is also likely to be /near/ that location.
11.531 -
11.532 -These boundary conditions imply that for the square well potential in
11.533 -this problem,
11.534 -
11.535 -- Physically allowed states must be totally confined to the well,
11.536 -  because it takes an infinite amount of energy to exist anywhere
11.537 -  outside of the well (and physically allowed states ought to have
11.538 -  only finite energy).
11.539 -- Physically allowed states must be increasingly unlikely to find very
11.540 -  close to the walls of the well. This is because of two conditions: the above
11.541 -  condition says that the particle is /impossible/ to find
11.542 -  outside of the well, and the smoothly-varying condition says
11.543 -  that if a particle is impossible to find at a particular location,
11.544 -  it must be unlikely to be found nearby that location.
11.545 -
11.546 -#; physically allowed states are those that change in physically
11.547 -#allowed ways.
11.548 -
11.549 -
11.550 -#** Boundary conditions
11.551 -Because the potential is infinite everywhere except within the well,
11.552 -a realistic particle must be confined to exist only within the
11.553 -well\mdash{}its wavefunction must be zero everywhere beyond the walls
11.554 -of the well.
11.555 -
11.556 -
11.557 -[fn:coords] I chose my coordinate system so that the well extends from
11.558 -$$0<x<a$$. Others choose a coordinate system so that the well extends from
11.559 -$$-\frac{a}{2}<x<\frac{a}{2}$$. Although both coordinate systems describe the same physical
11.560 -situation, they give different-looking answers.
11.561 -
11.562 -[fn:infinity] Of course, infinite potentials are not
11.563 -realistic. Instead, they are useful approximations to finite
11.564 -potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
11.565 -of the well\rdquo{} are close enough for your own practical
11.566 -purposes. Having introduced a physical impossibility into the problem
11.567 -already, we don't expect to get physically realistic solutions; we
11.568 -just expect to get mathematically consistent ones. The forthcoming
11.569 -trouble is that we don't.

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