### changeset 0:f743fd0f4d8b

initial commit of dylan's stuff
author Robert McIntyre Mon, 17 Oct 2011 23:17:55 -0700 8d8278e09888 org/bk.org org/bk2.org org/bk3.org org/bk4.org org/bk_quandary.org org/bkup.org org/quandary.org 7 files changed, 1997 insertions(+), 0 deletions(-) [+]
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     1.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
1.2 +++ b/org/bk.org	Mon Oct 17 23:17:55 2011 -0700
1.3 @@ -0,0 +1,88 @@
1.4 +#+TITLE: Bugs in Quantum Mechanics
1.5 +#+AUTHOR: Dylan Holmes
1.6 +#+SETUPFILE: ../../aurellem/org/setup.org
1.7 +#+INCLUDE:   ../../aurellem/org/level-0.org
1.8 +
1.9 +#Bugs in the Quantum-Mechanical Momentum Operator
1.10 +
1.11 +
1.12 +I studied quantum mechanics the same way I study most subjects\mdash{}
1.13 +by collecting (and squashing) bugs in my understanding. One of these
1.14 +bugs persisted throughout two semesters of
1.15 +quantum mechanics coursework until I finally found
1.16 +the paper
1.17 +[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
1.18 +mechanics/]], which helped me stamp out the bug entirely. I decided to
1.19 +write an article about the problem and its solution for a number of reasons:
1.20 +
1.21 +- Although the paper was not unreasonably dense, it was written for
1.22 +  teachers. I wanted to write an article for students.
1.23 +- I wanted to popularize the problem and its solution because other
1.24 +  explanations are currently too hard to find. (Even Shankar's
1.25 +  excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it.)
1.26 +- I wanted to check that the bug was indeed entirely
1.27 +  eradicated. Attempting an explanation is my way of making
1.28 +  sure.
1.29 +
1.30 +* COMMENT
1.31 + I recommend the
1.32 +paper not only for students who are learning
1.33 +quantum mechanics, but especially for teachers interested in debugging
1.34 +them.
1.35 +
1.36 +* COMMENT
1.37 +On my first exam in quantum mechanics, my professor asked us to
1.38 +describe how certain measurements would affect a particle in a
1.39 +box. Many of these measurement questions required routine application
1.40 +of skills we had recently learned\mdash{}first, you recall (or
1.41 +calculate) the eigenstates of the quantity
1.42 +to be measured; second, you write the given state as a linear
1.43 +sum of these eigenstates\mdash{} the coefficients on each term give
1.44 +the probability amplitude.
1.45 +
1.46 +* The infinite square well potential
1.47 +There is a particle in a one-dimensional potential well that has
1.48 +infinitely high walls and finite width $$a$$. This means that the
1.49 +particle exists in a potential[fn:coords][fn:infinity]
1.50 +
1.51 +
1.52 +$$V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for 1.53 +}\;x<0\text{ or }x>a.\end{cases}$$
1.54 +
1.55 +The Schr\ouml{}dinger equation describes how the particle's state
1.56 +$$|\psi\rangle$$ will change over time in this system.
1.57 +
1.58 +$$\begin{eqnarray} 1.59 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& 1.60 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}$$
1.61 +
1.62 +This is a differential equation whose solutions are the physically
1.63 +allowed states for the particle in this system. Like any differential
1.64 +equation,
1.65 +
1.66 +
1.67 +Like any differential equation, the Schr\ouml{}dinger equation
1.68 +#; physically allowed states are those that change in physically
1.69 +#allowed ways.
1.70 +
1.71 +
1.72 +** Boundary conditions
1.73 +Because the potential is infinite everywhere except within the well,
1.74 +a realistic particle must be confined to exist only within the
1.75 +well\mdash{}its wavefunction must be zero everywhere beyond the walls
1.76 +of the well.
1.77 +
1.78 +
1.79 +[fn:coords] I chose my coordinate system so that the well extends from
1.80 +$$0<x<a$$. Others choose a coordinate system so that the well extends from
1.81 +$$-\frac{a}{2}<x<\frac{a}{2}$$. Although both coordinate systems describe the same physical
1.82 +situation, they give different-looking answers.
1.83 +
1.84 +[fn:infinity] Of course, infinite potentials are not
1.85 +realistic. Instead, they are useful approximations to finite
1.86 +potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
1.87 +of the well\rdquo{} are close enough for your own practical
1.88 +purposes. Having introduced a physical impossibility into the problem
1.89 +already, we don't expect to get physically realistic solutions; we
1.90 +just expect to get mathematically consistent ones. The forthcoming
1.91 +trouble is that we don't.

     2.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
2.2 +++ b/org/bk2.org	Mon Oct 17 23:17:55 2011 -0700
2.3 @@ -0,0 +1,97 @@
2.4 +#+TITLE: Bugs in Quantum Mechanics
2.5 +#+AUTHOR: Dylan Holmes
2.6 +#+SETUPFILE: ../../aurellem/org/setup.org
2.7 +#+INCLUDE:   ../../aurellem/org/level-0.org
2.8 +
2.9 +
2.10 +#Bugs in the Quantum-Mechanical Momentum Operator
2.11 +
2.12 +
2.13 +I studied quantum mechanics the same way I study most subjects\mdash{}
2.14 +by collecting (and squashing) bugs in my understanding. One of these
2.15 +bugs persisted throughout two semesters of
2.16 +quantum mechanics coursework until I finally found
2.17 +the paper
2.18 +[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
2.19 +mechanics/]], which helped me stamp out the bug entirely. I decided to
2.20 +write an article about the problem and its solution for a number of reasons:
2.21 +
2.22 +- Although the paper was not unreasonably dense, it was written for
2.23 +  teachers. I wanted to write an article for students.
2.24 +- I wanted to popularize the problem and its solution because other
2.25 +  explanations are currently too hard to find. (Even Shankar's
2.26 +  excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it.)
2.27 +- I wanted to check that the bug was indeed entirely
2.28 +  eradicated. Attempting an explanation is my way of making
2.29 +  sure.
2.30 +
2.31 +* COMMENT
2.32 + I recommend the
2.33 +paper not only for students who are learning
2.34 +quantum mechanics, but especially for teachers interested in debugging
2.35 +them.
2.36 +
2.37 +* COMMENT
2.38 +On my first exam in quantum mechanics, my professor asked us to
2.39 +describe how certain measurements would affect a particle in a
2.40 +box. Many of these measurement questions required routine application
2.41 +of skills we had recently learned\mdash{}first, you recall (or
2.42 +calculate) the eigenstates of the quantity
2.43 +to be measured; second, you write the given state as a linear
2.44 +sum of these eigenstates\mdash{} the coefficients on each term give
2.45 +the probability amplitude.
2.46 +
2.47 +* The infinite square well potential
2.48 +
2.49 +There is a particle in a one-dimensional potential well that is
2.50 +infinite everywhere except for a well of length $$a$$. This means that the
2.51 +particle exists in a potential[fn:coords][fn:infinity]
2.52 +
2.53 +
2.54 +$$V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for 2.55 +}\;x<0\text{ or }x>a.\end{cases}$$
2.56 +
2.57 +The Schr\ouml{}dinger equation describes how the particle's state
2.58 +$$|\psi\rangle$$ will change over time in this system.
2.59 +
2.60 +$$\begin{eqnarray} 2.61 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& 2.62 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}$$
2.63 +
2.64 +This is a differential equation whose solutions are the physically
2.65 +allowed states for the particle in this system. Physically allowed
2.66 +states are those that change in physically allowed ways. Like any
2.67 +differential equation, the Schr\ouml{}dinger equation can be
2.68 +accompanied by /boundary conditions/\mdash{}conditions that
2.69 +further restrict which states qualify as physically allowed.
2.70 +
2.71 +Whenever possible, physicists impose these boundary conditions:
2.72 +- The state should be a /continuous function of/ $$x$$. This means
2.73 +  that if a particle is very likely to be /at/ a particular location,
2.74 +  it is also very likely to be /near/ that location.
2.75 +-
2.76 +
2.77 +#; physically allowed states are those that change in physically
2.78 +#allowed ways.
2.79 +
2.80 +
2.81 +** Boundary conditions
2.82 +Because the potential is infinite everywhere except within the well,
2.83 +a realistic particle must be confined to exist only within the
2.84 +well\mdash{}its wavefunction must be zero everywhere beyond the walls
2.85 +of the well.
2.86 +
2.87 +
2.88 +[fn:coords] I chose my coordinate system so that the well extends from
2.89 +$$0<x<a$$. Others choose a coordinate system so that the well extends from
2.90 +$$-\frac{a}{2}<x<\frac{a}{2}$$. Although both coordinate systems describe the same physical
2.91 +situation, they give different-looking answers.
2.92 +
2.93 +[fn:infinity] Of course, infinite potentials are not
2.94 +realistic. Instead, they are useful approximations to finite
2.95 +potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
2.96 +of the well\rdquo{} are close enough for your own practical
2.97 +purposes. Having introduced a physical impossibility into the problem
2.98 +already, we don't expect to get physically realistic solutions; we
2.99 +just expect to get mathematically consistent ones. The forthcoming
2.100 +trouble is that we don't.

     3.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
3.2 +++ b/org/bk3.org	Mon Oct 17 23:17:55 2011 -0700
3.3 @@ -0,0 +1,257 @@
3.4 +#+TITLE: Bugs in quantum mechanics
3.5 +#+AUTHOR: Dylan Holmes
3.6 +#+SETUPFILE: ../../aurellem/org/setup.org
3.7 +#+INCLUDE:   ../../aurellem/org/level-0.org
3.8 +
3.9 +#Bugs in Quantum Mechanics
3.10 +#Bugs in the Quantum-Mechanical Momentum Operator
3.11 +
3.12 +
3.13 +I studied quantum mechanics the same way I study most subjects\mdash{}
3.14 +by collecting (and squashing) bugs in my understanding. One of these
3.15 +bugs persisted throughout two semesters of
3.16 +quantum mechanics coursework until I finally found
3.17 +the paper
3.18 +[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
3.19 +mechanics/]], which helped me stamp out the bug entirely. I decided to
3.20 +write an article about the problem and its solution for a number of reasons:
3.21 +
3.22 +- Although the paper was not unreasonably dense, it was written for
3.23 +  teachers. I wanted to write an article for students.
3.24 +- I wanted to popularize the problem and its solution because other
3.25 +  explanations are currently too hard to find. (Even Shankar's
3.26 +  excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it.)
3.27 +- I wanted to check that the bug was indeed entirely
3.28 +  eradicated. Attempting an explanation is my way of making
3.29 +  sure.
3.30 +
3.31 +* COMMENT
3.32 + I recommend the
3.33 +paper not only for students who are learning
3.34 +quantum mechanics, but especially for teachers interested in debugging
3.35 +them.
3.36 +
3.37 +* COMMENT
3.38 +On my first exam in quantum mechanics, my professor asked us to
3.39 +describe how certain measurements would affect a particle in a
3.40 +box. Many of these measurement questions required routine application
3.41 +of skills we had recently learned\mdash{}first, you recall (or
3.42 +calculate) the eigenstates of the quantity
3.43 +to be measured; second, you write the given state as a linear
3.44 +sum of these eigenstates\mdash{} the coefficients on each term give
3.45 +the probability amplitude.
3.46 +
3.47 +
3.48 +* What I thought I knew
3.49 +
3.50 +The following is a list of things I thought were true of quantum
3.51 +mechanics; the catch is that the list contradicts itself.
3.52 +
3.53 +- For any hermitian operator: Eigenstates with different eigenvalues are orthogonal.
3.54 +- For any hermitian operator: Any physically allowed state can be
3.55 +  written as a linear sum of eigenstates of the operator.
3.56 +- The momentum operator and energy operator are hermitian, because
3.57 +  momentum and energy are measureable quantities.
3.58 +- In vacuum,
3.59 +  - the momentum operator has an eigenstate
3.60 +    $$p(x)=\exp{(ipx/\hbar)}$$ for each value of $p$.
3.61 +  - the energy operator has an eigenstate $$|E\rangle = 3.62 + \alpha|p\rangle + \beta|-p\rangle$$ for any $$\alpha,\beta$$ and
3.63 +    the particular choice of momentum $p=\sqrt{2mE}$.
3.64 +- In the infinitely deep potential well,
3.65 +  - the momentum operator has eigenstates with the same form $p(x) = 3.66 + \exp{(ipx/\hbar)}$, but because of the boundary conditions on the
3.67 +    well, the following modifications are required.
3.68 +    - The wavefunction must be zero everywhere outside the well. That
3.69 +      is, $$p(x) = \begin{cases}\exp{(ipx/\hbar)},& 0\lt{}x\lt{}a; 3.70 + \\0, & \text{for }x<0\text{ or }x>a \\ \end{cases}$$
3.71 +#0,&\text{for }x\lt{}0\text{ or }x\gt{}a\end{cases}\)
3.72 +    - no longer has an eigenstate for each value
3.73 +      of $p$. Instead, only values of $p$ that are integer multiples of
3.74 +      $\pi a/\hbar$ are physically realistic.
3.75 +
3.76 +
3.77 +
3.78 +* COMMENT:
3.79 +
3.80 +** Eigenstates with different eigenvalues are orthogonal
3.81 +
3.82 +#+begin_quote
3.83 +*Theorem:* Eigenstates with different eigenvalues are orthogonal.
3.84 +#+end_quote
3.85 +
3.86 +** COMMENT :
3.87 +I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$
3.88 +and $|b\rangle$ are eigenstates of $\Lambda$. This means that
3.89 +
3.90 +
3.91 +$$3.92 +\begin{eqnarray} 3.93 +\Lambda |a\rangle&=& a|a\rangle,\\ 3.94 +\Lambda|b\rangle&=& b|b\rangle.\\ 3.95 +\end{eqnarray} 3.96 +$$
3.97 +
3.98 +If we take the difference of these eigenstates, we find that
3.99 +
3.100 +$$3.101 +\begin{eqnarray} 3.102 +\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle 3.103 +\qquad \text{(because \Lambda is linear.)}\\ 3.104 +&=& a|a\rangle - b|b\rangle\qquad\text{(because |a\rangle and 3.105 +|b\rangle are eigenstates of \Lambda)} 3.106 +\end{eqnarray}$$
3.107 +
3.108 +
3.109 +which means that $a\neq b$.
3.110 +
3.111 +** Eigenvectors of hermitian operators span the space of solutions
3.112 +
3.113 +#+begin_quote
3.114 +*Theorem:* If $\Omega$ is a hermitian operator, then every physically
3.115 + allowed state can be written as a linear sum of eigenstates of
3.116 + $\Omega$.
3.117 +#+end_quote
3.118 +
3.119 +
3.120 +
3.121 +** Momentum and energy are hermitian operators
3.122 +This ought to be true because hermitian operators correspond to
3.123 +observable quantities. Since we expect momentum and energy to be
3.124 +measureable quantities, we expect that there are hermitian operators
3.125 +to represent them.
3.126 +
3.127 +
3.128 +** Momentum and energy eigenstates in vacuum
3.129 +An eigenstate of the momentum operator $P$ would be a state
3.130 +$$|p\rangle$$ such that $$P|p\rangle=p|p\rangle$$.
3.131 +
3.132 +** Momentum and energy eigenstates in the infinitely deep well
3.133 +
3.134 +
3.135 +
3.136 +* Can you measure momentum in the infinite square well?
3.137 +
3.138 +
3.139 +
3.140 +** COMMENT  Momentum eigenstates
3.141 +
3.142 +In free space, the Hamiltonian is $$H=\frac{1}{2m}P^2$$ and the
3.143 +momentum operator $P$ has eigenstates $$p(x) = \exp{(-ipx/\hbar)}$$.
3.144 +
3.145 +In the infinitely deep potential well, the Hamiltonian is the same but
3.146 +there is a new condition in order for states to qualify as physically
3.147 +allowed: the states must not exist anywhere outside of well, as it
3.148 +takes an infinite amount of energy to do so.
3.149 +
3.150 +Notice that the momentum eigenstates defined above do /not/ satisfy
3.151 +this condition.
3.152 +
3.153 +
3.154 +
3.155 +* COMMENT
3.156 +For each physical system, there is a Schr\ouml{}dinger equation that
3.157 +describes how a particle's state $|\psi\rangle$  will change over
3.158 +time.
3.159 +
3.160 +$$\begin{eqnarray} 3.161 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& 3.162 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}$$
3.163 +
3.164 +This is a differential equation; each solution to the
3.165 +Schr\ouml{}dinger equation is a state that is physically allowed for
3.166 +our particle. Here, physically allowed states are
3.167 +those that change in physically allowed ways. However, like any differential
3.168 +equation, the Schr\ouml{}dinger equation can be accompanied by
3.169 +/boundary conditions/\mdash{}conditions that further restrict which
3.170 +states qualify as physically allowed.
3.171 +
3.172 +
3.173 +
3.174 +
3.175 +** Eigenstates of momentum
3.176 +
3.177 +
3.178 +
3.179 +
3.180 +#In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger
3.181 +
3.182 +#$$i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle$$
3.183 +
3.184 +
3.185 +
3.186 +
3.187 +
3.188 +
3.189 +
3.190 +* COMMENT
3.191 +
3.192 +#* The infinite square well potential
3.193 +
3.194 +A particle exists in a potential that is
3.195 +infinite everywhere except for a region of length $$a$$, where the potential is zero. This means that the
3.196 +particle exists in a potential[fn:coords][fn:infinity]
3.197 +
3.198 +
3.199 +$$V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for 3.200 +}\;x<0\text{ or }x>a.\end{cases}$$
3.201 +
3.202 +The Schr\ouml{}dinger equation describes how the particle's state
3.203 +$$|\psi\rangle$$ will change over time in this system.
3.204 +
3.205 +$$\begin{eqnarray} 3.206 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& 3.207 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}$$
3.208 +
3.209 +This is a differential equation; each solution to the
3.210 +Schr\ouml{}dinger equation is a state that is physically allowed for
3.211 +our particle. Here, physically allowed states are
3.212 +those that change in physically allowed ways. However, like any differential
3.213 +equation, the Schr\ouml{}dinger equation can be accompanied by
3.214 +/boundary conditions/\mdash{}conditions that further restrict which
3.215 +states qualify as physically allowed.
3.216 +
3.217 +
3.218 +Whenever possible, physicists impose these boundary conditions:
3.219 +- A physically allowed state ought to be a /smoothly-varying function of position./ This means
3.220 +  that if a particle in the state  is likely to be /at/ a particular location,
3.221 +  it is also likely to be /near/ that location.
3.222 +
3.223 +These boundary conditions imply that for the square well potential in
3.224 +this problem,
3.225 +
3.226 +- Physically allowed states must be totally confined to the well,
3.227 +  because it takes an infinite amount of energy to exist anywhere
3.228 +  outside of the well (and physically allowed states ought to have
3.229 +  only finite energy).
3.230 +- Physically allowed states must be increasingly unlikely to find very
3.231 +  close to the walls of the well. This is because of two conditions: the above
3.232 +  condition says that the particle is /impossible/ to find
3.233 +  outside of the well, and the smoothly-varying condition says
3.234 +  that if a particle is impossible to find at a particular location,
3.235 +  it must be unlikely to be found nearby that location.
3.236 +
3.237 +#; physically allowed states are those that change in physically
3.238 +#allowed ways.
3.239 +
3.240 +
3.241 +#** Boundary conditions
3.242 +Because the potential is infinite everywhere except within the well,
3.243 +a realistic particle must be confined to exist only within the
3.244 +well\mdash{}its wavefunction must be zero everywhere beyond the walls
3.245 +of the well.
3.246 +
3.247 +
3.248 +[fn:coords] I chose my coordinate system so that the well extends from
3.249 +$$0<x<a$$. Others choose a coordinate system so that the well extends from
3.250 +$$-\frac{a}{2}<x<\frac{a}{2}$$. Although both coordinate systems describe the same physical
3.251 +situation, they give different-looking answers.
3.252 +
3.253 +[fn:infinity] Of course, infinite potentials are not
3.254 +realistic. Instead, they are useful approximations to finite
3.255 +potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
3.256 +of the well\rdquo{} are close enough for your own practical
3.257 +purposes. Having introduced a physical impossibility into the problem
3.258 +already, we don't expect to get physically realistic solutions; we
3.259 +just expect to get mathematically consistent ones. The forthcoming
3.260 +trouble is that we don't.

     4.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
4.2 +++ b/org/bk4.org	Mon Oct 17 23:17:55 2011 -0700
4.3 @@ -0,0 +1,309 @@
4.4 +#+TITLE: Bugs in quantum mechanics
4.5 +#+AUTHOR: Dylan Holmes
4.6 +#+SETUPFILE: ../../aurellem/org/setup.org
4.7 +#+INCLUDE:   ../../aurellem/org/level-0.org
4.8 +
4.9 +#Bugs in Quantum Mechanics
4.10 +#Bugs in the Quantum-Mechanical Momentum Operator
4.11 +
4.12 +
4.13 +I studied quantum mechanics the same way I study most subjects\mdash{}
4.14 +by collecting (and squashing) bugs in my understanding. One of these
4.15 +bugs persisted throughout two semesters of
4.16 +quantum mechanics coursework until I finally found
4.17 +the paper
4.18 +[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
4.19 +mechanics/]], which helped me stamp out the bug entirely. I decided to
4.20 +write an article about the problem and its solution for a number of reasons:
4.21 +
4.22 +- Although the paper was not unreasonably dense, it was written for
4.23 +  teachers. I wanted to write an article for students.
4.24 +- I wanted to popularize the problem and its solution because other
4.25 +  explanations are currently too hard to find. (Even Shankar's
4.26 +  excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it.)
4.27 +- I wanted to check that the bug was indeed entirely
4.28 +  eradicated. Attempting an explanation is my way of making
4.29 +  sure.
4.30 +
4.31 +* COMMENT
4.32 + I recommend the
4.33 +paper not only for students who are learning
4.34 +quantum mechanics, but especially for teachers interested in debugging
4.35 +them.
4.36 +
4.37 +* COMMENT
4.38 +On my first exam in quantum mechanics, my professor asked us to
4.39 +describe how certain measurements would affect a particle in a
4.40 +box. Many of these measurement questions required routine application
4.41 +of skills we had recently learned\mdash{}first, you recall (or
4.42 +calculate) the eigenstates of the quantity
4.43 +to be measured; second, you write the given state as a linear
4.44 +sum of these eigenstates\mdash{} the coefficients on each term give
4.45 +the probability amplitude.
4.46 +
4.47 +
4.48 +* What I thought I knew
4.49 +
4.50 +The following is a list of things I thought were true of quantum
4.51 +mechanics; the catch is that the list contradicts itself.
4.52 +
4.53 +1. For any hermitian operator: Eigenstates with different eigenvalues are orthogonal.
4.54 +2. For any hermitian operator: Any physically allowed state can be
4.55 +   written as a linear sum of eigenstates of the operator.
4.56 +3. The momentum operator and energy operator are hermitian, because
4.57 +   momentum and energy are measureable quantities.
4.58 +4. In the vacuum potential, the momentum and energy operators have these eigenstates:
4.59 +   - the momentum operator has an eigenstate
4.60 +     $$p(x)=\exp{(ipx/\hbar)}$$ for each value of $p$.
4.61 +   - the energy operator has an eigenstate $$|E\rangle = 4.62 + \alpha|p\rangle + \beta|-p\rangle$$ for any $$\alpha,\beta$$ and
4.63 +     the particular choice of momentum $p=\sqrt{2mE}$.
4.64 +5. In the infinitely deep potential well, the momentum and energy
4.65 +   operators have these eigenstates:
4.66 +   - The momentum eigenstates and energy eigenstates have the same form
4.67 +     as in the vacuum potential: $p(x) = 4.68 + \exp{(ipx/\hbar)}$ and $|E\rangle = \alpha|p\rangle + \beta|-p\rangle$.
4.69 +   - Even so, because of the boundary conditions on the
4.70 +     well, we must make the following modifications:
4.71 +     + Physically realistic states must be impossible to find outside the well. (Only a state of infinite
4.72 +       energy could exist outside the well, and infinite energy is not
4.73 +       realistic.) This requirement means, for example, that momentum
4.74 +       eigenstates in the infinitely deep well must be
4.75 +       $$p(x) 4.76 + = \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a; 4.77 + \\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}$$
4.78 +     + Physically realistic states must vary smoothly throughout
4.79 +       space. This means that if a particle in some state is very unlikely to be
4.80 +       /at/ a particular location, it is also very unlikely be /near/
4.81 +       that location. Combining this requirement with the above
4.82 +       requirement, we find that the momentum operator no longer has
4.83 +       an eigenstate for each value of $p$; instead, only values of
4.84 +       $p$ that are integer multiples of $\pi a/\hbar$ are physically
4.85 +       realistic. Similarly, the energy operator no longer has an
4.86 +       eigenstate for each value of $E$; instead, the only energy
4.87 +       eigenstates in the infinitely deep well
4.88 +       are $E_n(x)=\sin(n\pi x/ a)$ for positive integers $n$.
4.89 +
4.90 +* COMMENT:
4.91 +
4.92 +** Eigenstates with different eigenvalues are orthogonal
4.93 +
4.94 +#+begin_quote
4.95 +*Theorem:* Eigenstates with different eigenvalues are orthogonal.
4.96 +#+end_quote
4.97 +
4.98 +** COMMENT :
4.99 +I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$
4.100 +and $|b\rangle$ are eigenstates of $\Lambda$. This means that
4.101 +
4.102 +
4.103 +$$4.104 +\begin{eqnarray} 4.105 +\Lambda |a\rangle&=& a|a\rangle,\\ 4.106 +\Lambda|b\rangle&=& b|b\rangle.\\ 4.107 +\end{eqnarray} 4.108 +$$
4.109 +
4.110 +If we take the difference of these eigenstates, we find that
4.111 +
4.112 +$$4.113 +\begin{eqnarray} 4.114 +\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle 4.115 +\qquad \text{(because \Lambda is linear.)}\\ 4.116 +&=& a|a\rangle - b|b\rangle\qquad\text{(because |a\rangle and 4.117 +|b\rangle are eigenstates of \Lambda)} 4.118 +\end{eqnarray}$$
4.119 +
4.120 +
4.121 +which means that $a\neq b$.
4.122 +
4.123 +** Eigenvectors of hermitian operators span the space of solutions
4.124 +
4.125 +#+begin_quote
4.126 +*Theorem:* If $\Omega$ is a hermitian operator, then every physically
4.127 + allowed state can be written as a linear sum of eigenstates of
4.128 + $\Omega$.
4.129 +#+end_quote
4.130 +
4.131 +
4.132 +
4.133 +** Momentum and energy are hermitian operators
4.134 +This ought to be true because hermitian operators correspond to
4.135 +observable quantities. Since we expect momentum and energy to be
4.136 +measureable quantities, we expect that there are hermitian operators
4.137 +to represent them.
4.138 +
4.139 +
4.140 +** Momentum and energy eigenstates in vacuum
4.141 +An eigenstate of the momentum operator $P$ would be a state
4.142 +$$|p\rangle$$ such that $$P|p\rangle=p|p\rangle$$.
4.143 +
4.144 +** Momentum and energy eigenstates in the infinitely deep well
4.145 +
4.146 +
4.147 +
4.148 +* Can you measure momentum in the infinitely deep well?
4.149 +In summary, I thought I knew:
4.150 +1. For any hermitian operator: eigenstates with different eigenvalues
4.151 +   are orthogonal.
4.152 +2. For any hermitian operator: any physically realistic state can be
4.153 +   written as a linear sum of eigenstates of the operator.
4.154 +3. The momentum operator and energy operator are hermitian, because
4.155 +   momentum and energy are observable quantities.
4.156 +4. (The form of the momentum and energy eigenstates in the vacuum potential)
4.157 +5. (The form of the momentum and energy eigenstates in the infinitely deep well potential)
4.158 +
4.159 +Additionally, I understood that because the infinitely deep potential
4.160 +well is not realistic, states of such a system  are not necessarily
4.161 +physically realistic. Instead, I understood
4.162 +\ldquo{}realistic states\rdquo{} to be those that satisfy the physically
4.163 +unrealistic Schr\ouml{}dinger equation and its boundary conditions.
4.164 +
4.165 +With that final caveat, here is the problem:
4.166 +
4.167 +According to (5), the momentum eigenstates in the well are
4.168 +
4.169 +$$p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}$$
4.170 +
4.171 +However, /these/ states are not orthogonal, which contradicts the
4.172 +assumption that (3) the momentum operator is hermitian and (2)
4.173 +eigenstates of a hermitian are orthogonal if they have different eigenvalues.
4.174 +
4.175 +#+begin_quote
4.176 +*Problem 1. The momentum eigenstates of the well are not orthogonal*
4.177 +
4.178 +/Proof./ If $p_1\neq p_2$, then
4.179 +
4.180 +$$\begin{eqnarray} 4.181 +\langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)dx\\ 4.182 +&=& \int_0^a p_1^*(x)p_2(x)dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{ 4.183 +outside the well.}\\ 4.184 +&=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)dx}} 4.185 +\end{eqnarray}$$
4.186 +$\square$
4.187 +
4.188 +#+end_quote
4.189 +
4.190 +
4.191 +
4.192 +** COMMENT  Momentum eigenstates
4.193 +
4.194 +In free space, the Hamiltonian is $$H=\frac{1}{2m}P^2$$ and the
4.195 +momentum operator $P$ has eigenstates $$p(x) = \exp{(-ipx/\hbar)}$$.
4.196 +
4.197 +In the infinitely deep potential well, the Hamiltonian is the same but
4.198 +there is a new condition in order for states to qualify as physically
4.199 +allowed: the states must not exist anywhere outside of well, as it
4.200 +takes an infinite amount of energy to do so.
4.201 +
4.202 +Notice that the momentum eigenstates defined above do /not/ satisfy
4.203 +this condition.
4.204 +
4.205 +
4.206 +
4.207 +* COMMENT
4.208 +For each physical system, there is a Schr\ouml{}dinger equation that
4.209 +describes how a particle's state $|\psi\rangle$  will change over
4.210 +time.
4.211 +
4.212 +$$\begin{eqnarray} 4.213 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& 4.214 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}$$
4.215 +
4.216 +This is a differential equation; each solution to the
4.217 +Schr\ouml{}dinger equation is a state that is physically allowed for
4.218 +our particle. Here, physically allowed states are
4.219 +those that change in physically allowed ways. However, like any differential
4.220 +equation, the Schr\ouml{}dinger equation can be accompanied by
4.221 +/boundary conditions/\mdash{}conditions that further restrict which
4.222 +states qualify as physically allowed.
4.223 +
4.224 +
4.225 +
4.226 +
4.227 +** Eigenstates of momentum
4.228 +
4.229 +
4.230 +
4.231 +
4.232 +#In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger
4.233 +
4.234 +#$$i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle$$
4.235 +
4.236 +
4.237 +
4.238 +
4.239 +
4.240 +
4.241 +
4.242 +* COMMENT
4.243 +
4.244 +#* The infinite square well potential
4.245 +
4.246 +A particle exists in a potential that is
4.247 +infinite everywhere except for a region of length $$a$$, where the potential is zero. This means that the
4.248 +particle exists in a potential[fn:coords][fn:infinity]
4.249 +
4.250 +
4.251 +$$V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for 4.252 +}\;x<0\text{ or }x>a.\end{cases}$$
4.253 +
4.254 +The Schr\ouml{}dinger equation describes how the particle's state
4.255 +$$|\psi\rangle$$ will change over time in this system.
4.256 +
4.257 +$$\begin{eqnarray} 4.258 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& 4.259 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}$$
4.260 +
4.261 +This is a differential equation; each solution to the
4.262 +Schr\ouml{}dinger equation is a state that is physically allowed for
4.263 +our particle. Here, physically allowed states are
4.264 +those that change in physically allowed ways. However, like any differential
4.265 +equation, the Schr\ouml{}dinger equation can be accompanied by
4.266 +/boundary conditions/\mdash{}conditions that further restrict which
4.267 +states qualify as physically allowed.
4.268 +
4.269 +
4.270 +Whenever possible, physicists impose these boundary conditions:
4.271 +- A physically allowed state ought to be a /smoothly-varying function of position./ This means
4.272 +  that if a particle in the state  is likely to be /at/ a particular location,
4.273 +  it is also likely to be /near/ that location.
4.274 +
4.275 +These boundary conditions imply that for the square well potential in
4.276 +this problem,
4.277 +
4.278 +- Physically allowed states must be totally confined to the well,
4.279 +  because it takes an infinite amount of energy to exist anywhere
4.280 +  outside of the well (and physically allowed states ought to have
4.281 +  only finite energy).
4.282 +- Physically allowed states must be increasingly unlikely to find very
4.283 +  close to the walls of the well. This is because of two conditions: the above
4.284 +  condition says that the particle is /impossible/ to find
4.285 +  outside of the well, and the smoothly-varying condition says
4.286 +  that if a particle is impossible to find at a particular location,
4.287 +  it must be unlikely to be found nearby that location.
4.288 +
4.289 +#; physically allowed states are those that change in physically
4.290 +#allowed ways.
4.291 +
4.292 +
4.293 +#** Boundary conditions
4.294 +Because the potential is infinite everywhere except within the well,
4.295 +a realistic particle must be confined to exist only within the
4.296 +well\mdash{}its wavefunction must be zero everywhere beyond the walls
4.297 +of the well.
4.298 +
4.299 +
4.300 +[fn:coords] I chose my coordinate system so that the well extends from
4.301 +$$0<x<a$$. Others choose a coordinate system so that the well extends from
4.302 +$$-\frac{a}{2}<x<\frac{a}{2}$$. Although both coordinate systems describe the same physical
4.303 +situation, they give different-looking answers.
4.304 +
4.305 +[fn:infinity] Of course, infinite potentials are not
4.306 +realistic. Instead, they are useful approximations to finite
4.307 +potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
4.308 +of the well\rdquo{} are close enough for your own practical
4.309 +purposes. Having introduced a physical impossibility into the problem
4.310 +already, we don't expect to get physically realistic solutions; we
4.311 +just expect to get mathematically consistent ones. The forthcoming
4.312 +trouble is that we don't.

     5.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
5.2 +++ b/org/bk_quandary.org	Mon Oct 17 23:17:55 2011 -0700
5.3 @@ -0,0 +1,566 @@
5.4 +#+TITLE: Bugs in quantum mechanics
5.5 +#+AUTHOR: Dylan Holmes
5.6 +#+DESCRIPTION: I explain hermitian operators and use them to solve a confusing problem in quantum mechanics.
5.7 +#+KEYWORDS: quantum mechanics, self-adjoint, hermitian, square well, momentum
5.8 +#+SETUPFILE: ../../aurellem/org/setup.org
5.9 +#+INCLUDE:   ../../aurellem/org/level-0.org
5.10 +
5.11 +
5.12 +
5.13 +#Bugs in Quantum Mechanics
5.14 +#Bugs in the Quantum-Mechanical Momentum Operator
5.15 +
5.16 +
5.17 +I studied quantum mechanics the same way I study most subjects\mdash{}
5.18 +by collecting (and squashing) bugs in my understanding. One of these
5.19 +bugs persisted throughout two semesters of
5.20 +quantum mechanics coursework until I finally found
5.21 +the paper
5.22 +[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
5.23 +mechanics/]], which helped me stamp out the bug entirely. I decided to
5.24 +write an article about the problem and its solution for a number of reasons:
5.25 +
5.26 +- Although the paper was not unreasonably dense, it was written for
5.27 +  teachers. I wanted to write an article for students.
5.28 +- I wanted to popularize the problem and its solution because other
5.29 +  explanations are currently too hard to find. (Even Shankar's
5.30 +  excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it, as far as I know.)
5.31 +- Attempting an explanation is my way of making
5.32 +  sure that the bug is /really/ gone.
5.33 +# entirely eradicated.
5.34 +
5.35 +* COMMENT
5.36 + I recommend the
5.37 +paper not only for students who are learning
5.38 +quantum mechanics, but especially for teachers interested in debugging
5.39 +them.
5.40 +
5.41 +* COMMENT
5.42 +On my first exam in quantum mechanics, my professor asked us to
5.43 +describe how certain measurements would affect a particle in a
5.44 +box. Many of these measurement questions required routine application
5.45 +of skills we had recently learned\mdash{}first, you recall (or
5.46 +calculate) the eigenstates of the quantity
5.47 +to be measured; second, you write the given state as a linear
5.48 +sum of these eigenstates\mdash{} the coefficients on each term give
5.49 +the probability amplitude.
5.50 +
5.51 +
5.52 +* Two methods of calculation that give different results.
5.53 +
5.54 +In the infinitely deep well, there is a particle in the the
5.55 +normalized state
5.56 +
5.57 + $$\psi(x) = \begin{cases}A\; x(x-a)&\text{for }0\lt{}x\lt{}a;\\0,&\text{for }x\lt{}0\text{ or }x>a.\end{cases}$$
5.58 +
5.59 +This is apparently a perfectly respectable state: it is normalized ($A$ is a
5.60 +normalization constant), it is zero
5.61 +everywhere outside of the well, and it is moreover continuous.
5.62 +
5.63 +Even so, we will find a problem if we attempt to calculate the average
5.64 +energy-squared of this state (that is, the quantity $$\langle \psi | H^2 | \psi \rangle$$).
5.65 +
5.66 +** First method
5.67 +
5.68 +For short, define a new variable[fn:1] $$|\bar \psi\rangle \equiv 5.69 +H|\psi\rangle$$. We can express the state $|\bar\psi\rangle$ as a
5.70 +function of $x$ because we know how to express $H$ and $\psi$ in terms
5.71 +of  $x$: $H$ is the differential operator $\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}$, and
5.72 +$\psi(x)$ is the function defined above. So, we get $$\quad\bar\psi(x) = \frac{-A\hbar^2}{m}$$. For future reference, observe that $\bar\psi(x)$
5.73 +is constant.
5.74 +
5.75 +Having introduced $|\bar\psi\rangle$, we can calculate the average energy-squared of $|\psi\rangle$ in the
5.76 +following way.
5.77 +
5.78 +$$\begin{eqnarray} 5.79 +\langle \psi | H^2 |\psi\rangle &=& \langle \psi H^\dagger | H 5.80 +\psi\rangle\\ 5.81 +&=& \langle \psi H | H\psi \rangle\\ 5.82 +&=& \langle \bar\psi | \bar\psi \rangle\\ 5.83 +&=& \int_0^a \left(\frac{A\hbar^2}{m}\right)^2\;dx\\ 5.84 +&=& \frac{A^2\hbar^4 a}{m^2}\\ 5.85 +\end{eqnarray}$$
5.86 +
5.87 +For future reference, observe that this value is  nonzero
5.88 +(which makes sense).
5.89 +
5.90 +** Second method
5.91 +We can also calculate the average energy-squared of $|\psi\rangle$ in the
5.92 +following way.
5.93 +
5.94 +\begin{eqnarray}
5.95 +\langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\
5.96 +&=& \langle \psi |H \bar\psi \rangle\\
5.97 +&=&\int_0^a Ax(x-a)
5.98 +\left(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\right)\frac{-A\hbar^2}{m}\;dx\\
5.99 +&=& 0\quad (!)\\
5.100 +\end{eqnarray}
5.101 +
5.102 +The second-to-last term must be zero because the second derivative
5.103 +of a constant $$\left(\frac{-A\hbar^2}{m}\right)$$ is zero.
5.104 +
5.105 +* What is the problem?
5.106 +
5.107 +To recap: We used two different methods to calculate the average
5.108 +energy-squared of a state $|\psi\rangle$. For the first method, we
5.109 +used the fact that $H$ is a hermitian operator, replacing $$\langle 5.110 +\psi H^\dagger | H \psi\rangle$$ with $$\langle \psi H | H 5.111 +\psi\rangle$$. Using this substitution rule, we calculated the answer.
5.112 +
5.113 +For the second method, we didn't use the fact that $H$ was hermitian;
5.114 +instead, we used the fact that we know how to represent $H$ and $\psi$
5.115 +as functions of $x$: $H$ is a differential operator
5.116 +$$\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}$$ and $\psi$ is a quadratic
5.117 +function of $x$. By applying $H$ to $\psi$, we took several
5.118 +derivatives and arrived at our answer.
5.119 +
5.120 +These two methods gave different results. In the following sections,
5.121 +I'll describe and analyze the source of this difference.
5.122 +
5.123 +** Physical operators only act on physical wavefunctions
5.124 +   :PROPERTIES:
5.125 +   :ORDERED:  t
5.126 +   :END:
5.127 +#In quantum mechanics, an operator is a function that takes in a
5.128 +#physical state and produces another physical state as ouput. Some
5.129 +#operators correspond to physical quantities such as energy,
5.130 +#momentum, or position; the mathematical properties of these operators correspond to
5.131 +#physical properties of the system.
5.132 +
5.133 +#Eigenstates are an example of this correspondence: an
5.134 +
5.135 +Physical states are represented as wavefunctions in quantum
5.136 +mechanics. Just as we disallow certain physically nonsensical states
5.137 +in classical mechanics (for example, we consider it to be nonphysical
5.138 +for an object to spontaneously disappear from one place and reappear
5.139 +in another), we also disallow certain wavefunctions in quantum
5.140 +mechanics.
5.141 +
5.142 +For example, since wavefunctions are supposed to correspond to
5.143 +probability amplitudes, we require wavefunctions to be normalized
5.144 +$$(\langle \psi |\psi \rangle = 1)$$. Generally, we disallow
5.145 +wavefunctions that do not satisfy this property (although there are
5.146 +some exceptions[fn:2]).
5.147 +
5.148 +As another example, we generally expect probability to vary smoothly\mdash{}if
5.149 +a particle is very likely or very unlikely to be found at a particular
5.150 +location, it should also be somewhat likely or somewhat unlikely to be
5.151 +found /near/ that location. In more precise terms, we expect that for
5.152 +physically meaningful wavefunctions, the probability
5.153 +$$\text{Pr}(x)=\int^x \psi^*\psi$$ should be a continuous function of
5.154 +$x$ and, again, we disallow wavefunctions that do not satisfy this
5.155 +property because we consider them to be physically nonsensical.
5.156 +
5.157 +So, physical wavefunctions must satisfy certain properties
5.158 +like the two just described. Wavefunctions that do not satisfy these properties are
5.159 +rejected for being physically nonsensical: even though we can perform
5.160 +calculations with them, the mathematical results we obtain do not mean
5.161 +anything physically.
5.162 +
5.163 +Now, in quantum mechanics, an *operator* is a function that converts
5.164 +states into other states. Some operators correspond to
5.165 +physical quantities such as energy, momentum, or position, and as a
5.166 +result, the mathematical properties of these operators correspond to
5.167 +physical properties of the system. Physical operators are furthermore
5.168 +subject to the following rule: they are only allowed to operate on
5.169 +#physical wavefunctions, and they are only allowed to produce
5.170 +#physical wavefunctions[fn:why].
5.171 +the wavefunctions we have deemed physical, and they only produce physical wavefunctions[fn::
5.172 +
5.173 + If you require a hermitian operator to have physical
5.174 +  eigenstates, you get a very strong result: you guarantee that the
5.175 +  operator will convert /every/ physical wavefunction into another
5.176 +  physical wavefunction:
5.177 +
5.178 +  For any linear operator $\Omega$, the eigenvalue equation is
5.179 +$$\Omega|\omega\rangle = \omega |\omega\rangle$$. Notice that if an
5.180 +eigenstate $|\omega\rangle$ is a physical wavefunction, the
5.181 +eigenvalue equation equation forces $\Omega|\omega\rangle$ to be a
5.182 +physical wavefunction as well. To elaborate, if the eigenstates of
5.183 +$\Omega$ are physical functions, then $\Omega$ is guaranteed to
5.184 +convert them into other physical functions.  Even more is true if the
5.185 +operator $\Omega$ is also hermitian: there is a theorem which states
5.186 +that \ldquo{}If \Omega is hermitian, then every physical wavefunction
5.187 +can be written as a weighted sum of eigenstates of \Omega.\rdquo{} This
5.188 +theorem implies that /if $\Omega$ is hermitian/, and /if the eigenstates
5.189 +of \Omega are physically allowed/, then \Omega is guaranteed to
5.190 +convert every physically allowed wavefunction into another physically
5.191 +allowed wavefunction.].
5.192 +
5.193 +In fact, this rule for physical operators is the source of our
5.194 +problem, as we unknowingly violated it when applying our second
5.195 +method!
5.196 +
5.197 +** The violation
5.198 +
5.199 +I'll start explaining this violation by being more specific about the
5.200 +infinitely deep well potential. We have said already that physicists
5.201 +require wavefunctions to satisfy certain properties in order to be
5.202 +deemed \ldquo{}physical\rdquo{}. To be specific, wavefunctions in the
5.203 +infinitely deep well
5.204 +- Must be *normalizable*, because they correspond to
5.205 +  probability amplitudes.
5.206 +- Must have *smoothly-varying probability*, because if a particle is very
5.207 +  likely to be at a location, it ought to be likely to be /near/
5.208 +  it as well.
5.209 +- Must *not exist outside the well*, because it
5.210 +  would take an infinite amount of energy to do so.
5.211 +
5.212 +Additionally, by combining the second and third conditions, some
5.213 +physicists reason that wavefunctions in the infinitely deep well
5.214 +
5.215 +- Must *become zero* towards the edges of the well.
5.216 +
5.217 +
5.218 +
5.219 +
5.220 +You'll remember we had
5.221 +
5.222 +$$5.223 +\begin{eqnarray} 5.224 +\psi(x) &=& A\;x(x-a)\\ 5.225 +\bar\psi(x)&=& \frac{-A\hbar^2}{m}\\ 5.226 +&&\text{for }0\lt{}x\lt{}a\\ 5.227 +\end{eqnarray} 5.228 +$$
5.229 +
5.230 +In our second method, we wrote
5.231 +
5.232 +
5.233 +$$\begin{eqnarray} 5.234 +\langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\ 5.235 +&=& \langle \psi |H \bar\psi \rangle\\ 5.236 +& \vdots&\\ 5.237 +&=& 0\\ 5.238 +\end{eqnarray}$$
5.239 +
5.240 +However, there are two issues here. First, we were not allowed to operate with $H$ on the wavefunction
5.241 +$|\bar \psi\rangle$, because $H$ is a physical operator and $|\bar 5.242 +\psi\rangle$ is a nonphysical state: in the infinite square well,
5.243 +physical wavefunctions must approach zero at the edges of the well,
5.244 +which the constant function $|\bar\psi\rangle$ does not do. By
5.245 +feeding $H$ a nonphysical wavefunction, we obtained nonsensical
5.246 +results.
5.247 +
5.248 +Second, we claimed that $H$ was a physical operator\mdash{}that $H$
5.249 +was hermitian. According to the rule, this means $H$ must convert physical states into other
5.250 +physical states. But $H$ converts the physical state $|\psi\rangle$
5.251 +into the nonphysical state $|\bar\psi\rangle = H|\psi\rangle$ is not. Because $H$ converts some
5.252 +physical states into nonphysical states, it cannot be a hermitian operator.
5.253 +
5.254 +** Boundary conditions affect hermiticity
5.255 +We have now discovered a flaw: when applied to the state
5.256 +$|\psi\rangle$, the second method violates the rule that physical
5.257 +operators must only take in physical states and must only produce
5.258 +physical states. This suggests that the problem was with the state
5.259 +$|\bar\psi\rangle$ that we wrongly fed into $H$. However, the problem
5.260 +is more serious still: the state $|\psi\rangle 5.261 + 5.262 +** COMMENT Re-examining physical constraints 5.263 + 5.264 +We have now discovered a flaw: when applied to the state 5.265 +$|\psi\rangle$, the second method violates the rule that physical 5.266 +operators must only take in physical states and must only produce 5.267 +physical states. Let's examine the problem more closely. 5.268 + 5.269 +We have said already that physicists require wavefunctions to satisfy 5.270 +certain properties in order to be deemed \ldquo{}physical\rdquo{}. To 5.271 +be specific, wavefunctions in the infinitely deep well 5.272 +- Must be *normalizable*, because they correspond to 5.273 + probability amplitudes. 5.274 +- Must have *smoothly-varying probability*, because if a particle is very 5.275 + likely to be at a location, it ought to be likely to be /near/ 5.276 + it as well. 5.277 +- Must *not exist outside the well*, because it 5.278 + would take an infinite amount of energy to do so. 5.279 + 5.280 +We now have discovered an important flaw in the second method: when 5.281 +applied to the state$|\bar\psi\rangle$, the second method violates 5.282 +the rule that physical operators must only take in 5.283 +physical states and must only produce physical states. The problem is 5.284 +even more serious, however 5.285 + 5.286 + 5.287 + 5.288 +[fn:1] I'm defining a new variable just to make certain expressions 5.289 + look shorter; this cannot affect the content of the answer we'll 5.290 + get. 5.291 + 5.292 +[fn:2] For example, in vaccuum (i.e., when the potential of the 5.293 + physical system is$V(x)=0$throughout all space), the momentum 5.294 + eigenstates are not normalizable\mdash{}the relevant integral blows 5.295 + up to infinity instead of converging to a number. Physicists modify 5.296 + the definition of normalization slightly so that 5.297 + \ldquo{}delta-normalizable \rdquo{} functions like these are included 5.298 + among the physical wavefunctions. 5.299 + 5.300 + 5.301 + 5.302 +* COMMENT: What I thought I knew 5.303 + 5.304 +The following is a list of things I thought were true of quantum 5.305 +mechanics; the catch is that the list contradicts itself. 5.306 + 5.307 +1. For any hermitian operator: Eigenstates with different eigenvalues are orthogonal. 5.308 +2. For any hermitian operator: Any physically allowed state can be 5.309 + written as a linear sum of eigenstates of the operator. 5.310 +3. The momentum operator and energy operator are hermitian, because 5.311 + momentum and energy are measureable quantities. 5.312 +4. In the vacuum potential, the momentum and energy operators have these eigenstates: 5.313 + - the momentum operator has an eigenstate 5.314 + $$p(x)=\exp{(ipx/\hbar)}$$ for each value of$p$. 5.315 + - the energy operator has an eigenstate $$|E\rangle = 5.316 + \alpha|p\rangle + \beta|-p\rangle$$ for any $$\alpha,\beta$$ and 5.317 + the particular choice of momentum$p=\sqrt{2mE}$. 5.318 +5. In the infinitely deep potential well, the momentum and energy 5.319 + operators have these eigenstates: 5.320 + - The momentum eigenstates and energy eigenstates have the same form 5.321 + as in the vacuum potential:$p(x) =
5.322 +     \exp{(ipx/\hbar)}$and$|E\rangle = \alpha|p\rangle + \beta|-p\rangle$. 5.323 + - Even so, because of the boundary conditions on the 5.324 + well, we must make the following modifications: 5.325 + + Physically realistic states must be impossible to find outside the well. (Only a state of infinite 5.326 + energy could exist outside the well, and infinite energy is not 5.327 + realistic.) This requirement means, for example, that momentum 5.328 + eigenstates in the infinitely deep well must be 5.329 + $$p(x) 5.330 + = \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a; 5.331 + \\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}$$ 5.332 + + Physically realistic states must vary smoothly throughout 5.333 + space. This means that if a particle in some state is very unlikely to be 5.334 + /at/ a particular location, it is also very unlikely be /near/ 5.335 + that location. Combining this requirement with the above 5.336 + requirement, we find that the momentum operator no longer has 5.337 + an eigenstate for each value of$p$; instead, only values of 5.338 +$p$that are integer multiples of$\pi \hbar/a$are physically 5.339 + realistic. Similarly, the energy operator no longer has an 5.340 + eigenstate for each value of$E$; instead, the only energy 5.341 + eigenstates in the infinitely deep well 5.342 + are$E_n(x)=\sin(n\pi x/ a)$for positive integers$n$. 5.343 + 5.344 +* COMMENT: 5.345 + 5.346 +** Eigenstates with different eigenvalues are orthogonal 5.347 + 5.348 +#+begin_quote 5.349 +*Theorem:* Eigenstates with different eigenvalues are orthogonal. 5.350 +#+end_quote 5.351 + 5.352 +** COMMENT : 5.353 +I can prove this: if$\Lambda$is any linear operator, suppose$|a\rangle$5.354 +and$|b\rangle$are eigenstates of$\Lambda$. This means that 5.355 + 5.356 + 5.357 +$$5.358 +\begin{eqnarray} 5.359 +\Lambda |a\rangle&=& a|a\rangle,\\ 5.360 +\Lambda|b\rangle&=& b|b\rangle.\\ 5.361 +\end{eqnarray} 5.362 +$$ 5.363 + 5.364 +If we take the difference of these eigenstates, we find that 5.365 + 5.366 +$$5.367 +\begin{eqnarray} 5.368 +\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle 5.369 +\qquad \text{(because \Lambda is linear.)}\\ 5.370 +&=& a|a\rangle - b|b\rangle\qquad\text{(because |a\rangle and 5.371 +|b\rangle are eigenstates of \Lambda)} 5.372 +\end{eqnarray}$$ 5.373 + 5.374 + 5.375 +which means that$a\neq b$. 5.376 + 5.377 +** Eigenvectors of hermitian operators span the space of solutions 5.378 + 5.379 +#+begin_quote 5.380 +*Theorem:* If$\Omega$is a hermitian operator, then every physically 5.381 + allowed state can be written as a linear sum of eigenstates of 5.382 +$\Omega$. 5.383 +#+end_quote 5.384 + 5.385 + 5.386 + 5.387 +** Momentum and energy are hermitian operators 5.388 +This ought to be true because hermitian operators correspond to 5.389 +observable quantities. Since we expect momentum and energy to be 5.390 +measureable quantities, we expect that there are hermitian operators 5.391 +to represent them. 5.392 + 5.393 + 5.394 +** Momentum and energy eigenstates in vacuum 5.395 +An eigenstate of the momentum operator$P$would be a state 5.396 +$$|p\rangle$$ such that $$P|p\rangle=p|p\rangle$$. 5.397 + 5.398 +** Momentum and energy eigenstates in the infinitely deep well 5.399 + 5.400 + 5.401 + 5.402 +* COMMENT Can you measure momentum in the infinitely deep well? 5.403 +In summary, I thought I knew: 5.404 +1. For any hermitian operator: eigenstates with different eigenvalues 5.405 + are orthogonal. 5.406 +2. For any hermitian operator: any physically realistic state can be 5.407 + written as a linear sum of eigenstates of the operator. 5.408 +3. The momentum operator and energy operator are hermitian, because 5.409 + momentum and energy are observable quantities. 5.410 +4. (The form of the momentum and energy eigenstates in the vacuum potential) 5.411 +5. (The form of the momentum and energy eigenstates in the infinitely deep well potential) 5.412 + 5.413 +Additionally, I understood that because the infinitely deep potential 5.414 +well is not realistic, states of such a system are not necessarily 5.415 +physically realistic. Instead, I understood 5.416 +\ldquo{}realistic states\rdquo{} to be those that satisfy the physically 5.417 +unrealistic Schr\ouml{}dinger equation and its boundary conditions. 5.418 + 5.419 +With that final caveat, here is the problem: 5.420 + 5.421 +According to (5), the momentum eigenstates in the well are 5.422 + 5.423 +$$p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}$$ 5.424 + 5.425 +(Additionally, we require that$p$be an integer multiple of$\pi\hbar/a$.) 5.426 + 5.427 +However, /these/ states are not orthogonal, which contradicts the 5.428 +assumption that (3) the momentum operator is hermitian and (2) 5.429 +eigenstates of a hermitian are orthogonal if they have different eigenvalues. 5.430 + 5.431 +#+begin_quote 5.432 +*Problem 1. The momentum eigenstates of the well are not orthogonal* 5.433 + 5.434 +/Proof./ If$p_1\neq p_2$, then 5.435 + 5.436 +$$\begin{eqnarray} 5.437 +\langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)\,dx\\ 5.438 +&=& \int_0^a p_1^*(x)p_2(x)\,dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{ 5.439 +outside the well.}\\ 5.440 +&=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)\,dx}}\\ 5.441 +&=& \int_0^a \exp{\left[i(p_1-p_2)x/\hbar\right]}\,dx\\ 5.442 +\end{eqnarray}$$ 5.443 +$\square$5.444 + 5.445 +#+end_quote 5.446 + 5.447 + 5.448 + 5.449 +** COMMENT Momentum eigenstates 5.450 + 5.451 +In free space, the Hamiltonian is $$H=\frac{1}{2m}P^2$$ and the 5.452 +momentum operator$P$has eigenstates $$p(x) = \exp{(-ipx/\hbar)}$$. 5.453 + 5.454 +In the infinitely deep potential well, the Hamiltonian is the same but 5.455 +there is a new condition in order for states to qualify as physically 5.456 +allowed: the states must not exist anywhere outside of well, as it 5.457 +takes an infinite amount of energy to do so. 5.458 + 5.459 +Notice that the momentum eigenstates defined above do /not/ satisfy 5.460 +this condition. 5.461 + 5.462 + 5.463 + 5.464 +* COMMENT 5.465 +For each physical system, there is a Schr\ouml{}dinger equation that 5.466 +describes how a particle's state$|\psi\rangle$will change over 5.467 +time. 5.468 + 5.469 +$$\begin{eqnarray} 5.470 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& 5.471 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}$$ 5.472 + 5.473 +This is a differential equation; each solution to the 5.474 +Schr\ouml{}dinger equation is a state that is physically allowed for 5.475 +our particle. Here, physically allowed states are 5.476 +those that change in physically allowed ways. However, like any differential 5.477 +equation, the Schr\ouml{}dinger equation can be accompanied by 5.478 +/boundary conditions/\mdash{}conditions that further restrict which 5.479 +states qualify as physically allowed. 5.480 + 5.481 + 5.482 + 5.483 + 5.484 +** Eigenstates of momentum 5.485 + 5.486 + 5.487 + 5.488 + 5.489 +#In the infinitely deep well potential$V(x)=0$, the Schr\ouml{}dinger 5.490 + 5.491 +#$$i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle$$ 5.492 + 5.493 + 5.494 + 5.495 + 5.496 + 5.497 + 5.498 + 5.499 +* COMMENT 5.500 + 5.501 +#* The infinite square well potential 5.502 + 5.503 +A particle exists in a potential that is 5.504 +infinite everywhere except for a region of length $$a$$, where the potential is zero. This means that the 5.505 +particle exists in a potential[fn:coords][fn:infinity] 5.506 + 5.507 + 5.508 +$$V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for 5.509 +}\;x<0\text{ or }x>a.\end{cases}$$ 5.510 + 5.511 +The Schr\ouml{}dinger equation describes how the particle's state 5.512 +$$|\psi\rangle$$ will change over time in this system. 5.513 + 5.514 +$$\begin{eqnarray} 5.515 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& 5.516 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}$$ 5.517 + 5.518 +This is a differential equation; each solution to the 5.519 +Schr\ouml{}dinger equation is a state that is physically allowed for 5.520 +our particle. Here, physically allowed states are 5.521 +those that change in physically allowed ways. However, like any differential 5.522 +equation, the Schr\ouml{}dinger equation can be accompanied by 5.523 +/boundary conditions/\mdash{}conditions that further restrict which 5.524 +states qualify as physically allowed. 5.525 + 5.526 + 5.527 +Whenever possible, physicists impose these boundary conditions: 5.528 +- A physically allowed state ought to be a /smoothly-varying function of position./ This means 5.529 + that if a particle in the state is likely to be /at/ a particular location, 5.530 + it is also likely to be /near/ that location. 5.531 + 5.532 +These boundary conditions imply that for the square well potential in 5.533 +this problem, 5.534 + 5.535 +- Physically allowed states must be totally confined to the well, 5.536 + because it takes an infinite amount of energy to exist anywhere 5.537 + outside of the well (and physically allowed states ought to have 5.538 + only finite energy). 5.539 +- Physically allowed states must be increasingly unlikely to find very 5.540 + close to the walls of the well. This is because of two conditions: the above 5.541 + condition says that the particle is /impossible/ to find 5.542 + outside of the well, and the smoothly-varying condition says 5.543 + that if a particle is impossible to find at a particular location, 5.544 + it must be unlikely to be found nearby that location. 5.545 + 5.546 +#; physically allowed states are those that change in physically 5.547 +#allowed ways. 5.548 + 5.549 + 5.550 +#** Boundary conditions 5.551 +Because the potential is infinite everywhere except within the well, 5.552 +a realistic particle must be confined to exist only within the 5.553 +well\mdash{}its wavefunction must be zero everywhere beyond the walls 5.554 +of the well. 5.555 + 5.556 + 5.557 +[fn:coords] I chose my coordinate system so that the well extends from 5.558 +$$0<x<a$$. Others choose a coordinate system so that the well extends from 5.559 +$$-\frac{a}{2}<x<\frac{a}{2}$$. Although both coordinate systems describe the same physical 5.560 +situation, they give different-looking answers. 5.561 + 5.562 +[fn:infinity] Of course, infinite potentials are not 5.563 +realistic. Instead, they are useful approximations to finite 5.564 +potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height 5.565 +of the well\rdquo{} are close enough for your own practical 5.566 +purposes. Having introduced a physical impossibility into the problem 5.567 +already, we don't expect to get physically realistic solutions; we 5.568 +just expect to get mathematically consistent ones. The forthcoming 5.569 +trouble is that we don't.   6.1 --- /dev/null Thu Jan 01 00:00:00 1970 +0000 6.2 +++ b/org/bkup.org Mon Oct 17 23:17:55 2011 -0700 6.3 @@ -0,0 +1,49 @@ 6.4 +#+TITLE: Bugs in Quantum Mechanics 6.5 +#+AUTHOR: Dylan Holmes 6.6 +#+SETUPFILE: ../../aurellem/org/setup.org 6.7 +#+INCLUDE: ../../aurellem/org/level-0.org 6.8 + 6.9 + 6.10 +#Bugs in the Quantum-Mechanical Momentum Operator 6.11 + 6.12 + 6.13 +I studied quantum mechanics the same way I study most subjects\mdash{} 6.14 +by collecting (and squashing) bugs in my understanding. One of these 6.15 +bugs persisted throughout two semesters of 6.16 +quantum mechanics coursework until I finally found 6.17 +the paper 6.18 +[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum 6.19 +mechanics/]], which helped me stamp out the bug entirely. I decided to 6.20 +write an article about the problem and its solution for a number of reasons: 6.21 + 6.22 +- Although the paper was not unreasonably dense, it was written for 6.23 + teachers. I wanted to write an article for students. 6.24 +- I wanted to popularize the problem and its solution because 6.25 + other explanations are currently too hard to find. 6.26 +- I wanted to check that the bug was indeed entirely 6.27 + eradicated. Attempting an explanation is my way of making 6.28 + sure. 6.29 + 6.30 +* COMMENT 6.31 + I recommend the 6.32 +paper not only for students who are learning 6.33 +quantum mechanics, but especially for teachers interested in debugging 6.34 +them. 6.35 + 6.36 +* COMMENT 6.37 +On my first exam in quantum mechanics, my professor asked us to 6.38 +describe how certain measurements would affect a particle in a 6.39 +box. Many of these measurement questions required routine application 6.40 +of skills we had recently learned\mdash{}first, you recall (or 6.41 +calculate) the eigenstates of the quantity 6.42 +to be measured; second, you write the given state as a linear 6.43 +sum of these eigenstates\mdash{} the coefficients on each term give 6.44 +the probability amplitude. 6.45 + 6.46 +* Statement of the Problem 6.47 +A particle is 6.48 + 6.49 + 6.50 + 6.51 + 6.52 +* COMMENT [TABLE-OF-CONTENTS]   7.1 --- /dev/null Thu Jan 01 00:00:00 1970 +0000 7.2 +++ b/org/quandary.org Mon Oct 17 23:17:55 2011 -0700 7.3 @@ -0,0 +1,631 @@ 7.4 +#+TITLE: Bugs in quantum mechanics 7.5 +#+AUTHOR: Dylan Holmes 7.6 +#+DESCRIPTION: I explain hermitian operators and use them to solve a confusing problem in quantum mechanics. 7.7 +#+KEYWORDS: quantum mechanics, self-adjoint, hermitian, square well, momentum 7.8 +#+SETUPFILE: ../../aurellem/org/setup.org 7.9 +#+INCLUDE: ../../aurellem/org/level-0.org 7.10 + 7.11 + 7.12 + 7.13 +#Bugs in Quantum Mechanics 7.14 +#Bugs in the Quantum-Mechanical Momentum Operator 7.15 + 7.16 + 7.17 +I studied quantum mechanics the same way I study most subjects\mdash{} 7.18 +by collecting (and squashing) bugs in my understanding. One of these 7.19 +bugs persisted throughout two semesters of 7.20 +quantum mechanics coursework until I finally found 7.21 +the paper 7.22 +[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum 7.23 +mechanics/]], which helped me stamp out the bug entirely. I decided to 7.24 +write an article about the problem and its solution for a number of reasons: 7.25 + 7.26 +- Although the paper was not unreasonably dense, it was written for 7.27 + teachers. I wanted to write an article for students. 7.28 +- I wanted to popularize the problem and its solution because other 7.29 + explanations are currently too hard to find. (Even Shankar's 7.30 + excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it, as far as I know.) 7.31 +- Attempting an explanation is my way of making 7.32 + sure that the bug really /is/ gone. 7.33 +# entirely eradicated. 7.34 + 7.35 +* COMMENT 7.36 + I recommend the 7.37 +paper not only for students who are learning 7.38 +quantum mechanics, but especially for teachers interested in debugging 7.39 +them. 7.40 + 7.41 +* COMMENT 7.42 +On my first exam in quantum mechanics, my professor asked us to 7.43 +describe how certain measurements would affect a particle in a 7.44 +box. Many of these measurement questions required routine application 7.45 +of skills we had recently learned\mdash{}first, you recall (or 7.46 +calculate) the eigenstates of the quantity 7.47 +to be measured; second, you write the given state as a linear 7.48 +sum of these eigenstates\mdash{} the coefficients on each term give 7.49 +the probability amplitude. 7.50 + 7.51 + 7.52 +* Two methods of calculation that give different results. 7.53 + 7.54 +In the infinitely deep well, there is a particle in the the 7.55 +normalized state 7.56 + 7.57 + $$\psi(x) = \begin{cases}A\; x(x-a)&\text{for }0\lt{}x\lt{}a;\\0,&\text{for }x\lt{}0\text{ or }x>a.\end{cases}$$ 7.58 + 7.59 +This is apparently a perfectly respectable state: it is normalized ($A$is a 7.60 +normalization constant), it is zero 7.61 +everywhere outside of the well, and it is moreover continuous. 7.62 + 7.63 +Even so, we will find a problem if we attempt to calculate the average 7.64 +energy-squared of this state (that is, the quantity $$\langle \psi | H^2 | \psi \rangle$$). 7.65 + 7.66 +** First method 7.67 + 7.68 +For short, define a new variable[fn:1] $$|\bar \psi\rangle \equiv 7.69 +H|\psi\rangle$$. We can express the state$|\bar\psi\rangle$as a 7.70 +function of$x$because we know how to express$H$and$\psi$in terms 7.71 +of$x$:$H$is the differential operator$\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}$, and 7.72 +$\psi(x)$is the function defined above. So, we get $$\quad\bar\psi(x) = \frac{-A\hbar^2}{m}$$. For future reference, observe that$\bar\psi(x)$7.73 +is constant. 7.74 + 7.75 +Having introduced$|\bar\psi\rangle$, we can calculate the average energy-squared of$|\psi\rangle$in the 7.76 +following way. 7.77 + 7.78 +$$\begin{eqnarray} 7.79 +\langle \psi | H^2 |\psi\rangle &=& \langle \psi H^\dagger | H 7.80 +\psi\rangle\\ 7.81 +&=& \langle \psi H | H\psi \rangle\\ 7.82 +&=& \langle \bar\psi | \bar\psi \rangle\\ 7.83 +&=& \int_0^a \left(\frac{A\hbar^2}{m}\right)^2\;dx\\ 7.84 +&=& \frac{A^2\hbar^4 a}{m^2}\\ 7.85 +\end{eqnarray}$$ 7.86 + 7.87 +For future reference, observe that this value is nonzero 7.88 +(which makes sense). 7.89 + 7.90 +** Second method 7.91 +We can also calculate the average energy-squared of$|\psi\rangle$in the 7.92 +following way. 7.93 + 7.94 +\begin{eqnarray} 7.95 +\langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\ 7.96 +&=& \langle \psi |H \bar\psi \rangle\\ 7.97 +&=&\int_0^a Ax(x-a) 7.98 +\left(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\right)\frac{-A\hbar^2}{m}\;dx\\ 7.99 +&=& 0\quad (!)\\ 7.100 +\end{eqnarray} 7.101 + 7.102 +The second-to-last term must be zero because the second derivative 7.103 +of a constant $$\left(\frac{-A\hbar^2}{m}\right)$$ is zero. 7.104 + 7.105 +* What is the problem? 7.106 + 7.107 +To recap: We used two different methods to calculate the average 7.108 +energy-squared of a state$|\psi\rangle$. For the first method, we 7.109 +used the fact that$H$is a hermitian operator, replacing $$\langle 7.110 +\psi H^\dagger | H \psi\rangle$$ with $$\langle \psi H | H 7.111 +\psi\rangle$$. Using this substitution rule, we calculated the answer. 7.112 + 7.113 +For the second method, 7.114 +#we didn't use the fact that$H$was hermitian; 7.115 +we instead used the fact that we know how to represent$H$and$\psi$7.116 +as functions of$x$:$H$is a differential operator 7.117 +$$\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}$$ and$\psi$is a quadratic 7.118 +function of$x$. By applying$H$to$\psi$, we took several 7.119 +derivatives and arrived at our answer. 7.120 + 7.121 +These two methods gave different results. In the following sections, 7.122 +I'll describe and analyze the source of this difference. 7.123 + 7.124 +** Physical operators only act on physical wavefunctions 7.125 + :PROPERTIES: 7.126 + :ORDERED: t 7.127 + :END: 7.128 +#In quantum mechanics, an operator is a function that takes in a 7.129 +#physical state and produces another physical state as ouput. Some 7.130 +#operators correspond to physical quantities such as energy, 7.131 +#momentum, or position; the mathematical properties of these operators correspond to 7.132 +#physical properties of the system. 7.133 + 7.134 +#Eigenstates are an example of this correspondence: an 7.135 + 7.136 +Physical states are represented as wavefunctions in quantum 7.137 +mechanics. Just as we disallow certain physically nonsensical states 7.138 +in classical mechanics (for example, we consider it to be nonphysical 7.139 +for an object to spontaneously disappear from one place and reappear 7.140 +in another), we also disallow certain wavefunctions in quantum 7.141 +mechanics. 7.142 + 7.143 +For example, since wavefunctions are supposed to correspond to 7.144 +probability amplitudes, we require wavefunctions to be normalized 7.145 +$$(\langle \psi |\psi \rangle = 1)$$. Generally, we disallow 7.146 +wavefunctions that do not satisfy this property (although there are 7.147 +some exceptions[fn:2]). 7.148 + 7.149 +As another example, we generally expect probability to vary smoothly\mdash{}if 7.150 +a particle is very likely or very unlikely to be found at a particular 7.151 +location, it should also be somewhat likely or somewhat unlikely to be 7.152 +found /near/ that location. In more precise terms, we expect that for 7.153 +physically meaningful wavefunctions, the probability 7.154 +$$\text{Pr}(x)=\int^x \psi^*\psi$$ should be a continuous function of 7.155 +$x$and, again, we disallow wavefunctions that do not satisfy this 7.156 +property because we consider them to be physically nonsensical. 7.157 + 7.158 +So, physical wavefunctions must satisfy certain properties 7.159 +like the two just described. Wavefunctions that do not satisfy these properties are 7.160 +rejected for being physically nonsensical: even though we can perform 7.161 +calculations with them, the mathematical results we obtain do not mean 7.162 +anything physically. 7.163 + 7.164 +Now, in quantum mechanics, an *operator* is a function that converts 7.165 +states into other states. Some operators correspond to 7.166 +physical quantities such as energy, momentum, or position, and as a 7.167 +result, the mathematical properties of these operators correspond to 7.168 +physical properties of the system. Such operators are called 7.169 +/hermitian operators/; one important property of hermitian operators 7.170 +is this rule: 7.171 + 7.172 +#+begin_quote 7.173 +*Hermitian operator rule:* A hermitian operator must only operate on 7.174 +the wavefunctions we have deemed physical, and must only produce 7.175 +physical wavefunctions[fn:: If you require a hermitian operator to 7.176 +have physical eigenstates (which is reasonable), you get a very strong result: you guarantee 7.177 +that the operator will convert every physical wavefunction into 7.178 +another physical wavefunction: 7.179 + 7.180 + For any linear operator$\Omega$, the eigenvalue equation is 7.181 +$$\Omega|\omega\rangle = \omega |\omega\rangle$$. Notice that if an 7.182 +eigenstate$|\omega\rangle$is a physical wavefunction, the 7.183 +eigenvalue equation forces$\Omega|\omega\rangle$to be a 7.184 +physical wavefunction as well. To elaborate, if the eigenstates of 7.185 +$\Omega$are physical functions, then$\Omega$is guaranteed to 7.186 +convert them into other physical functions. Even more is true if the 7.187 +operator$\Omega$is also hermitian: there is a theorem which states 7.188 +that \ldquo{}If \Omega is hermitian, then every physical wavefunction 7.189 +can be written as a weighted sum of eigenstates of \Omega.\rdquo{} This 7.190 +theorem implies that /if$\Omega$is hermitian/, and /if the eigenstates 7.191 +of \Omega are physically allowed/, then \Omega is guaranteed to 7.192 +convert every physically allowed wavefunction into another physically 7.193 +allowed wavefunction.]. 7.194 +#+end_quote 7.195 + 7.196 +As you can see, this rule comes in two pieces. The first part is a 7.197 +constraint on *you*, the physicist: you must never feed a nonphysical 7.198 +state into a Hermitian operator, as it may produce nonsense. The 7.199 +second part is a constraint on the *operator*: the operator is 7.200 +guaranteed only to produce physical wavefunctions. 7.201 + 7.202 +In fact, this rule for hermitian operators is the source of our 7.203 +problem, as we unknowingly violated it when applying our second 7.204 +method! 7.205 + 7.206 +** The Hamiltonian is nonphysical 7.207 +You'll remember that in the second method we had wavefunctions within 7.208 +the well 7.209 + 7.210 +$$7.211 +\begin{eqnarray} 7.212 +\psi(x) &=& A\;x(x-a)\\ 7.213 +\bar\psi(x)&\equiv& H|\psi\rangle = \frac{-A\hbar^2}{m}\\ 7.214 +\end{eqnarray} 7.215 +$$ 7.216 + 7.217 + Using this, we wrote 7.218 + 7.219 + 7.220 +$$\begin{eqnarray} 7.221 +\langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\ 7.222 +&=& \langle \psi |H \bar\psi \rangle\\ 7.223 +& \vdots&\\ 7.224 +&=& 0\\ 7.225 +\end{eqnarray}$$ 7.226 + 7.227 +However, there are two issues here. First, we were not allowed to operate with$H$on the wavefunction 7.228 +$|\bar \psi\rangle$, because$H$is supposed to be a physical operator and$|\bar
7.229 +\psi\rangle$is a nonphysical state. Indeed, the constant function$|\bar\psi\rangle$does 7.230 +not approach zero at the edges of the well. By 7.231 +feeding$H$a nonphysical wavefunction, we obtained nonsensical 7.232 +results. 7.233 + 7.234 +Second, and more importantly, we were wrong to claim that$H$was a 7.235 +physical operator\mdash{}that$H$was hermitian. According to the 7.236 +rule, a hermitian operator must convert physical states into other 7.237 +physical states. But$|\psi\rangle$is a physical state, as we said 7.238 +when we first introduced it \mdash{}it is a normalized, continuous 7.239 +function which approaches zero at the edges of the well and doesn't 7.240 +exist outside it. On the other hand,$|\bar\psi\rangle$is nonphysical 7.241 +because it does not go to zero at the edges of the well. It is 7.242 +therefore impermissible for$H$to transform the physical state 7.243 +$|\psi\rangle$into the nonphysical state$|\bar\psi\rangle$. Because 7.244 +$H$converts some physical states into nonphysical states, it cannot 7.245 +be a hermitian operator as we assumed. 7.246 + 7.247 +# Boundary conditions affect hermiticity 7.248 +** Boundary conditions alter hermiticity 7.249 +It may surprise you (and it certainly surprised me) to find that the 7.250 +Hamiltonian is not hermitian. One of the fundamental principles of 7.251 +quantum mechanics is that hermitian operators correspond to physically 7.252 +observable quantities; for this reason, surely the 7.253 +Hamiltonian\mdash{}the operator corresponding to energy\mdash{}ought to be hermitian? 7.254 + 7.255 +But we must understand the correspondence between physically 7.256 +observable quantities and hermitian operators: every hermitian 7.257 +operator corresponds to a physically observable quantity, but not 7.258 +every quantity that intuitively \ldquo{}ought\rdquo{} to be observable 7.259 +will correspond to a hermitian operator[fn::For a simple example, 7.260 +consider the differential operator $$D=\frac{d}{dx}$$; although our 7.261 +intuitions might suggest that$D$is observable which leads us to 7.262 +guess that$D$is hermitian, it isn't. Still, the very closely related 7.263 +operator$P=i\hbar\frac{d}{dx}$/is/ hermitian. The point is that we 7.264 +ought to validate our intuitions by checking the definitions.]. The 7.265 +true definition of a hermitian operator imply that the Hamiltonian 7.266 +stops being hermitian in the infinitely deep well. Here we arrive at a 7.267 +crucial point: 7.268 + 7.269 +Operators do not change /form/ between problems: the one-dimensional 7.270 +Hamiltonian is always$H=\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}+V$, the 7.271 +one-dimensional canonical momentum is always$P=i\hbar\frac{d}{dx}$, 7.272 +and so on. 7.273 + 7.274 +However, operators do change in this respect: hermitian operators must 7.275 +only take in physical states, and must only produce physical states; because 7.276 +in different problems we /do/ change the requirements for being a 7.277 +physical state, we also change what it takes for 7.278 +an operator to be called hermitian. As a result, an operator that 7.279 +is hermitian in one setting may fail to be hermitian in another. 7.280 + 7.281 +Having seen how boundary conditions can affect hermiticity, we 7.282 +ought to be extra careful about which conditions we impose on our 7.283 +wavefunctions. 7.284 + 7.285 +** Choosing the right constraints 7.286 + 7.287 + We have said already that physicists 7.288 +require wavefunctions to satisfy certain properties in order to be 7.289 +deemed \ldquo{}physical\rdquo{}. To be specific, wavefunctions in the 7.290 +infinitely deep well 7.291 +- Must be *normalizable*, because they correspond to 7.292 + probability amplitudes. 7.293 +- Must have *smoothly-varying probability*, because if a particle is very 7.294 + likely to be at a location, it ought to be likely to be /near/ 7.295 + it as well. 7.296 +- Must *not exist outside the well*, because it 7.297 + would take an infinite amount of energy to do so. 7.298 + 7.299 +These conditions are surely reasonable. However, physicists sometimes 7.300 +assert that in order to satisfy the second and third conditions, 7.301 +physical wavefunctions 7.302 + 7.303 +- (?) Must *smoothly approach zero* towards the edges of the well. 7.304 + 7.305 +This final constraint is our reason for rejecting$|\bar\psi\rangle$7.306 +as nonphysical and is consequently the reason why$H$is not hermitian. If 7.307 +we can convince ourselves that the final constraint is unnecessary, 7.308 +$H$may again be hermitian. This will satisfy our intuitions that the 7.309 +energy operator /ought/ to be hermitian. 7.310 + 7.311 +But in fact, we have the following mathematical observation to save 7.312 +us: a function$f$does not need to be continuous in order for the 7.313 +integral $$\int^x f$$ to be continuous. As a particularly relevant 7.314 +example, you may now notice that the function$\bar\psi(x)$is not 7.315 +itself continuous, although the integral$\int_0^x \bar\psi$/is/ 7.316 +continuous. Evidently, it doesn't matter that the wavefunction 7.317 +$\bar\psi$itself is not continuous; the probability corresponding to 7.318 +$\bar\psi$/does/ manage to vary continuously anyways. Because the 7.319 +probability corresponding to$\bar\psi$is the only aspect of 7.320 +$\bar\psi$which we can detect physically, we /can/ safely omit the 7.321 +final constraint while keeping the other three. 7.322 + 7.323 +** Symmetric operators look like hermitian operators, but sometimes aren't. 7.324 + 7.325 + 7.326 +#+end_quote 7.327 +** COMMENT Re-examining physical constraints 7.328 + 7.329 +We have now discovered a flaw: when applied to the state 7.330 +$|\psi\rangle$, the second method violates the rule that physical 7.331 +operators must only take in physical states and must only produce 7.332 +physical states. Let's examine the problem more closely. 7.333 + 7.334 +We have said already that physicists require wavefunctions to satisfy 7.335 +certain properties in order to be deemed \ldquo{}physical\rdquo{}. To 7.336 +be specific, wavefunctions in the infinitely deep well 7.337 +- Must be *normalizable*, because they correspond to 7.338 + probability amplitudes. 7.339 +- Must have *smoothly-varying probability*, because if a particle is very 7.340 + likely to be at a location, it ought to be likely to be /near/ 7.341 + it as well. 7.342 +- Must *not exist outside the well*, because it 7.343 + would take an infinite amount of energy to do so. 7.344 + 7.345 +We now have discovered an important flaw in the second method: when 7.346 +applied to the state$|\bar\psi\rangle$, the second method violates 7.347 +the rule that physical operators must only take in 7.348 +physical states and must only produce physical states. The problem is 7.349 +even more serious, however 7.350 + 7.351 + 7.352 + 7.353 +[fn:1] I'm defining a new variable just to make certain expressions 7.354 + look shorter; this cannot affect the content of the answer we'll 7.355 + get. 7.356 + 7.357 +[fn:2] For example, in vaccuum (i.e., when the potential of the 7.358 + physical system is$V(x)=0$throughout all space), the momentum 7.359 + eigenstates are not normalizable\mdash{}the relevant integral blows 7.360 + up to infinity instead of converging to a number. Physicists modify 7.361 + the definition of normalization slightly so that 7.362 + \ldquo{}delta-normalizable \rdquo{} functions like these are included 7.363 + among the physical wavefunctions. 7.364 + 7.365 + 7.366 + 7.367 +* COMMENT: What I thought I knew 7.368 + 7.369 +The following is a list of things I thought were true of quantum 7.370 +mechanics; the catch is that the list contradicts itself. 7.371 + 7.372 +1. For any hermitian operator: Eigenstates with different eigenvalues are orthogonal. 7.373 +2. For any hermitian operator: Any physically allowed state can be 7.374 + written as a linear sum of eigenstates of the operator. 7.375 +3. The momentum operator and energy operator are hermitian, because 7.376 + momentum and energy are measureable quantities. 7.377 +4. In the vacuum potential, the momentum and energy operators have these eigenstates: 7.378 + - the momentum operator has an eigenstate 7.379 + $$p(x)=\exp{(ipx/\hbar)}$$ for each value of$p$. 7.380 + - the energy operator has an eigenstate $$|E\rangle = 7.381 + \alpha|p\rangle + \beta|-p\rangle$$ for any $$\alpha,\beta$$ and 7.382 + the particular choice of momentum$p=\sqrt{2mE}$. 7.383 +5. In the infinitely deep potential well, the momentum and energy 7.384 + operators have these eigenstates: 7.385 + - The momentum eigenstates and energy eigenstates have the same form 7.386 + as in the vacuum potential:$p(x) =
7.387 +     \exp{(ipx/\hbar)}$and$|E\rangle = \alpha|p\rangle + \beta|-p\rangle$. 7.388 + - Even so, because of the boundary conditions on the 7.389 + well, we must make the following modifications: 7.390 + + Physically realistic states must be impossible to find outside the well. (Only a state of infinite 7.391 + energy could exist outside the well, and infinite energy is not 7.392 + realistic.) This requirement means, for example, that momentum 7.393 + eigenstates in the infinitely deep well must be 7.394 + $$p(x) 7.395 + = \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a; 7.396 + \\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}$$ 7.397 + + Physically realistic states must vary smoothly throughout 7.398 + space. This means that if a particle in some state is very unlikely to be 7.399 + /at/ a particular location, it is also very unlikely be /near/ 7.400 + that location. Combining this requirement with the above 7.401 + requirement, we find that the momentum operator no longer has 7.402 + an eigenstate for each value of$p$; instead, only values of 7.403 +$p$that are integer multiples of$\pi \hbar/a$are physically 7.404 + realistic. Similarly, the energy operator no longer has an 7.405 + eigenstate for each value of$E$; instead, the only energy 7.406 + eigenstates in the infinitely deep well 7.407 + are$E_n(x)=\sin(n\pi x/ a)$for positive integers$n$. 7.408 + 7.409 +* COMMENT: 7.410 + 7.411 +** Eigenstates with different eigenvalues are orthogonal 7.412 + 7.413 +#+begin_quote 7.414 +*Theorem:* Eigenstates with different eigenvalues are orthogonal. 7.415 +#+end_quote 7.416 + 7.417 +** COMMENT : 7.418 +I can prove this: if$\Lambda$is any linear operator, suppose$|a\rangle$7.419 +and$|b\rangle$are eigenstates of$\Lambda$. This means that 7.420 + 7.421 + 7.422 +$$7.423 +\begin{eqnarray} 7.424 +\Lambda |a\rangle&=& a|a\rangle,\\ 7.425 +\Lambda|b\rangle&=& b|b\rangle.\\ 7.426 +\end{eqnarray} 7.427 +$$ 7.428 + 7.429 +If we take the difference of these eigenstates, we find that 7.430 + 7.431 +$$7.432 +\begin{eqnarray} 7.433 +\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle 7.434 +\qquad \text{(because \Lambda is linear.)}\\ 7.435 +&=& a|a\rangle - b|b\rangle\qquad\text{(because |a\rangle and 7.436 +|b\rangle are eigenstates of \Lambda)} 7.437 +\end{eqnarray}$$ 7.438 + 7.439 + 7.440 +which means that$a\neq b$. 7.441 + 7.442 +** Eigenvectors of hermitian operators span the space of solutions 7.443 + 7.444 +#+begin_quote 7.445 +*Theorem:* If$\Omega$is a hermitian operator, then every physically 7.446 + allowed state can be written as a linear sum of eigenstates of 7.447 +$\Omega$. 7.448 +#+end_quote 7.449 + 7.450 + 7.451 + 7.452 +** Momentum and energy are hermitian operators 7.453 +This ought to be true because hermitian operators correspond to 7.454 +observable quantities. Since we expect momentum and energy to be 7.455 +measureable quantities, we expect that there are hermitian operators 7.456 +to represent them. 7.457 + 7.458 + 7.459 +** Momentum and energy eigenstates in vacuum 7.460 +An eigenstate of the momentum operator$P$would be a state 7.461 +$$|p\rangle$$ such that $$P|p\rangle=p|p\rangle$$. 7.462 + 7.463 +** Momentum and energy eigenstates in the infinitely deep well 7.464 + 7.465 + 7.466 + 7.467 +* COMMENT Can you measure momentum in the infinitely deep well? 7.468 +In summary, I thought I knew: 7.469 +1. For any hermitian operator: eigenstates with different eigenvalues 7.470 + are orthogonal. 7.471 +2. For any hermitian operator: any physically realistic state can be 7.472 + written as a linear sum of eigenstates of the operator. 7.473 +3. The momentum operator and energy operator are hermitian, because 7.474 + momentum and energy are observable quantities. 7.475 +4. (The form of the momentum and energy eigenstates in the vacuum potential) 7.476 +5. (The form of the momentum and energy eigenstates in the infinitely deep well potential) 7.477 + 7.478 +Additionally, I understood that because the infinitely deep potential 7.479 +well is not realistic, states of such a system are not necessarily 7.480 +physically realistic. Instead, I understood 7.481 +\ldquo{}realistic states\rdquo{} to be those that satisfy the physically 7.482 +unrealistic Schr\ouml{}dinger equation and its boundary conditions. 7.483 + 7.484 +With that final caveat, here is the problem: 7.485 + 7.486 +According to (5), the momentum eigenstates in the well are 7.487 + 7.488 +$$p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}$$ 7.489 + 7.490 +(Additionally, we require that$p$be an integer multiple of$\pi\hbar/a$.) 7.491 + 7.492 +However, /these/ states are not orthogonal, which contradicts the 7.493 +assumption that (3) the momentum operator is hermitian and (2) 7.494 +eigenstates of a hermitian are orthogonal if they have different eigenvalues. 7.495 + 7.496 +#+begin_quote 7.497 +*Problem 1. The momentum eigenstates of the well are not orthogonal* 7.498 + 7.499 +/Proof./ If$p_1\neq p_2$, then 7.500 + 7.501 +$$\begin{eqnarray} 7.502 +\langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)\,dx\\ 7.503 +&=& \int_0^a p_1^*(x)p_2(x)\,dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{ 7.504 +outside the well.}\\ 7.505 +&=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)\,dx}}\\ 7.506 +&=& \int_0^a \exp{\left[i(p_1-p_2)x/\hbar\right]}\,dx\\ 7.507 +\end{eqnarray}$$ 7.508 +$\square$7.509 + 7.510 +#+end_quote 7.511 + 7.512 + 7.513 + 7.514 +** COMMENT Momentum eigenstates 7.515 + 7.516 +In free space, the Hamiltonian is $$H=\frac{1}{2m}P^2$$ and the 7.517 +momentum operator$P$has eigenstates $$p(x) = \exp{(-ipx/\hbar)}$$. 7.518 + 7.519 +In the infinitely deep potential well, the Hamiltonian is the same but 7.520 +there is a new condition in order for states to qualify as physically 7.521 +allowed: the states must not exist anywhere outside of well, as it 7.522 +takes an infinite amount of energy to do so. 7.523 + 7.524 +Notice that the momentum eigenstates defined above do /not/ satisfy 7.525 +this condition. 7.526 + 7.527 + 7.528 + 7.529 +* COMMENT 7.530 +For each physical system, there is a Schr\ouml{}dinger equation that 7.531 +describes how a particle's state$|\psi\rangle$will change over 7.532 +time. 7.533 + 7.534 +$$\begin{eqnarray} 7.535 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& 7.536 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}$$ 7.537 + 7.538 +This is a differential equation; each solution to the 7.539 +Schr\ouml{}dinger equation is a state that is physically allowed for 7.540 +our particle. Here, physically allowed states are 7.541 +those that change in physically allowed ways. However, like any differential 7.542 +equation, the Schr\ouml{}dinger equation can be accompanied by 7.543 +/boundary conditions/\mdash{}conditions that further restrict which 7.544 +states qualify as physically allowed. 7.545 + 7.546 + 7.547 + 7.548 + 7.549 +** Eigenstates of momentum 7.550 + 7.551 + 7.552 + 7.553 + 7.554 +#In the infinitely deep well potential$V(x)=0\$, the Schr\ouml{}dinger
7.555 +
7.556 +#$$i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle$$
7.557 +
7.558 +
7.559 +
7.560 +
7.561 +
7.562 +
7.563 +
7.564 +* COMMENT
7.565 +
7.566 +#* The infinite square well potential
7.567 +
7.568 +A particle exists in a potential that is
7.569 +infinite everywhere except for a region of length $$a$$, where the potential is zero. This means that the
7.570 +particle exists in a potential[fn:coords][fn:infinity]
7.571 +
7.572 +
7.573 +$$V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for 7.574 +}\;x<0\text{ or }x>a.\end{cases}$$
7.575 +
7.576 +The Schr\ouml{}dinger equation describes how the particle's state
7.577 +$$|\psi\rangle$$ will change over time in this system.
7.578 +
7.579 +$$\begin{eqnarray} 7.580 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& 7.581 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}$$
7.582 +
7.583 +This is a differential equation; each solution to the
7.584 +Schr\ouml{}dinger equation is a state that is physically allowed for
7.585 +our particle. Here, physically allowed states are
7.586 +those that change in physically allowed ways. However, like any differential
7.587 +equation, the Schr\ouml{}dinger equation can be accompanied by
7.588 +/boundary conditions/\mdash{}conditions that further restrict which
7.589 +states qualify as physically allowed.
7.590 +
7.591 +
7.592 +Whenever possible, physicists impose these boundary conditions:
7.593 +- A physically allowed state ought to be a /smoothly-varying function of position./ This means
7.594 +  that if a particle in the state  is likely to be /at/ a particular location,
7.595 +  it is also likely to be /near/ that location.
7.596 +
7.597 +These boundary conditions imply that for the square well potential in
7.598 +this problem,
7.599 +
7.600 +- Physically allowed states must be totally confined to the well,
7.601 +  because it takes an infinite amount of energy to exist anywhere
7.602 +  outside of the well (and physically allowed states ought to have
7.603 +  only finite energy).
7.604 +- Physically allowed states must be increasingly unlikely to find very
7.605 +  close to the walls of the well. This is because of two conditions: the above
7.606 +  condition says that the particle is /impossible/ to find
7.607 +  outside of the well, and the smoothly-varying condition says
7.608 +  that if a particle is impossible to find at a particular location,
7.609 +  it must be unlikely to be found nearby that location.
7.610 +
7.611 +#; physically allowed states are those that change in physically
7.612 +#allowed ways.
7.613 +
7.614 +
7.615 +#** Boundary conditions
7.616 +Because the potential is infinite everywhere except within the well,
7.617 +a realistic particle must be confined to exist only within the
7.618 +well\mdash{}its wavefunction must be zero everywhere beyond the walls
7.619 +of the well.
7.620 +
7.621 +
7.622 +[fn:coords] I chose my coordinate system so that the well extends from
7.623 +$$0<x<a$$. Others choose a coordinate system so that the well extends from
7.624 +$$-\frac{a}{2}<x<\frac{a}{2}$$. Although both coordinate systems describe the same physical
7.625 +situation, they give different-looking answers.
7.626 +
7.627 +[fn:infinity] Of course, infinite potentials are not
7.628 +realistic. Instead, they are useful approximations to finite
7.629 +potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
7.630 +of the well\rdquo{} are close enough for your own practical
7.631 +purposes. Having introduced a physical impossibility into the problem
7.632 +already, we don't expect to get physically realistic solutions; we
7.633 +just expect to get mathematically consistent ones. The forthcoming
7.634 +trouble is that we don't.