Mercurial > dylan
changeset 0:f743fd0f4d8b
initial commit of dylan's stuff
author | Robert McIntyre <rlm@mit.edu> |
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date | Mon, 17 Oct 2011 23:17:55 -0700 |
parents | |
children | 8d8278e09888 |
files | org/bk.org org/bk2.org org/bk3.org org/bk4.org org/bk_quandary.org org/bkup.org org/quandary.org |
diffstat | 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.