Mercurial > dylan
diff org/bk3.org @ 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 |
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1.1 --- /dev/null Thu Jan 01 00:00:00 1970 +0000 1.2 +++ b/org/bk3.org Mon Oct 17 23:17:55 2011 -0700 1.3 @@ -0,0 +1,257 @@ 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 Quantum Mechanics 1.10 +#Bugs in the Quantum-Mechanical Momentum Operator 1.11 + 1.12 + 1.13 +I studied quantum mechanics the same way I study most subjects\mdash{} 1.14 +by collecting (and squashing) bugs in my understanding. One of these 1.15 +bugs persisted throughout two semesters of 1.16 +quantum mechanics coursework until I finally found 1.17 +the paper 1.18 +[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum 1.19 +mechanics/]], which helped me stamp out the bug entirely. I decided to 1.20 +write an article about the problem and its solution for a number of reasons: 1.21 + 1.22 +- Although the paper was not unreasonably dense, it was written for 1.23 + teachers. I wanted to write an article for students. 1.24 +- I wanted to popularize the problem and its solution because other 1.25 + explanations are currently too hard to find. (Even Shankar's 1.26 + excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it.) 1.27 +- I wanted to check that the bug was indeed entirely 1.28 + eradicated. Attempting an explanation is my way of making 1.29 + sure. 1.30 + 1.31 +* COMMENT 1.32 + I recommend the 1.33 +paper not only for students who are learning 1.34 +quantum mechanics, but especially for teachers interested in debugging 1.35 +them. 1.36 + 1.37 +* COMMENT 1.38 +On my first exam in quantum mechanics, my professor asked us to 1.39 +describe how certain measurements would affect a particle in a 1.40 +box. Many of these measurement questions required routine application 1.41 +of skills we had recently learned\mdash{}first, you recall (or 1.42 +calculate) the eigenstates of the quantity 1.43 +to be measured; second, you write the given state as a linear 1.44 +sum of these eigenstates\mdash{} the coefficients on each term give 1.45 +the probability amplitude. 1.46 + 1.47 + 1.48 +* What I thought I knew 1.49 + 1.50 +The following is a list of things I thought were true of quantum 1.51 +mechanics; the catch is that the list contradicts itself. 1.52 + 1.53 +- For any hermitian operator: Eigenstates with different eigenvalues are orthogonal. 1.54 +- For any hermitian operator: Any physically allowed state can be 1.55 + written as a linear sum of eigenstates of the operator. 1.56 +- The momentum operator and energy operator are hermitian, because 1.57 + momentum and energy are measureable quantities. 1.58 +- In vacuum, 1.59 + - the momentum operator has an eigenstate 1.60 + \(p(x)=\exp{(ipx/\hbar)}\) for each value of $p$. 1.61 + - the energy operator has an eigenstate \(|E\rangle = 1.62 + \alpha|p\rangle + \beta|-p\rangle\) for any \(\alpha,\beta\) and 1.63 + the particular choice of momentum $p=\sqrt{2mE}$. 1.64 +- In the infinitely deep potential well, 1.65 + - the momentum operator has eigenstates with the same form $p(x) = 1.66 + \exp{(ipx/\hbar)}$, but because of the boundary conditions on the 1.67 + well, the following modifications are required. 1.68 + - The wavefunction must be zero everywhere outside the well. That 1.69 + is, \(p(x) = \begin{cases}\exp{(ipx/\hbar)},& 0\lt{}x\lt{}a; 1.70 + \\0, & \text{for }x<0\text{ or }x>a \\ \end{cases}\) 1.71 +#0,&\text{for }x\lt{}0\text{ or }x\gt{}a\end{cases}\) 1.72 + - no longer has an eigenstate for each value 1.73 + of $p$. Instead, only values of $p$ that are integer multiples of 1.74 + $\pi a/\hbar$ are physically realistic. 1.75 + 1.76 + 1.77 + 1.78 +* COMMENT: 1.79 + 1.80 +** Eigenstates with different eigenvalues are orthogonal 1.81 + 1.82 +#+begin_quote 1.83 +*Theorem:* Eigenstates with different eigenvalues are orthogonal. 1.84 +#+end_quote 1.85 + 1.86 +** COMMENT : 1.87 +I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$ 1.88 +and $|b\rangle$ are eigenstates of $\Lambda$. This means that 1.89 + 1.90 + 1.91 +\( 1.92 +\begin{eqnarray} 1.93 +\Lambda |a\rangle&=& a|a\rangle,\\ 1.94 +\Lambda|b\rangle&=& b|b\rangle.\\ 1.95 +\end{eqnarray} 1.96 +\) 1.97 + 1.98 +If we take the difference of these eigenstates, we find that 1.99 + 1.100 +\( 1.101 +\begin{eqnarray} 1.102 +\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle 1.103 +\qquad \text{(because $\Lambda$ is linear.)}\\ 1.104 +&=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and 1.105 +$|b\rangle$ are eigenstates of $\Lambda$)} 1.106 +\end{eqnarray}\) 1.107 + 1.108 + 1.109 +which means that $a\neq b$. 1.110 + 1.111 +** Eigenvectors of hermitian operators span the space of solutions 1.112 + 1.113 +#+begin_quote 1.114 +*Theorem:* If $\Omega$ is a hermitian operator, then every physically 1.115 + allowed state can be written as a linear sum of eigenstates of 1.116 + $\Omega$. 1.117 +#+end_quote 1.118 + 1.119 + 1.120 + 1.121 +** Momentum and energy are hermitian operators 1.122 +This ought to be true because hermitian operators correspond to 1.123 +observable quantities. Since we expect momentum and energy to be 1.124 +measureable quantities, we expect that there are hermitian operators 1.125 +to represent them. 1.126 + 1.127 + 1.128 +** Momentum and energy eigenstates in vacuum 1.129 +An eigenstate of the momentum operator $P$ would be a state 1.130 +\(|p\rangle\) such that \(P|p\rangle=p|p\rangle\). 1.131 + 1.132 +** Momentum and energy eigenstates in the infinitely deep well 1.133 + 1.134 + 1.135 + 1.136 +* Can you measure momentum in the infinite square well? 1.137 + 1.138 + 1.139 + 1.140 +** COMMENT Momentum eigenstates 1.141 + 1.142 +In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the 1.143 +momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\). 1.144 + 1.145 +In the infinitely deep potential well, the Hamiltonian is the same but 1.146 +there is a new condition in order for states to qualify as physically 1.147 +allowed: the states must not exist anywhere outside of well, as it 1.148 +takes an infinite amount of energy to do so. 1.149 + 1.150 +Notice that the momentum eigenstates defined above do /not/ satisfy 1.151 +this condition. 1.152 + 1.153 + 1.154 + 1.155 +* COMMENT 1.156 +For each physical system, there is a Schr\ouml{}dinger equation that 1.157 +describes how a particle's state $|\psi\rangle$ will change over 1.158 +time. 1.159 + 1.160 +\(\begin{eqnarray} 1.161 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& 1.162 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\) 1.163 + 1.164 +This is a differential equation; each solution to the 1.165 +Schr\ouml{}dinger equation is a state that is physically allowed for 1.166 +our particle. Here, physically allowed states are 1.167 +those that change in physically allowed ways. However, like any differential 1.168 +equation, the Schr\ouml{}dinger equation can be accompanied by 1.169 +/boundary conditions/\mdash{}conditions that further restrict which 1.170 +states qualify as physically allowed. 1.171 + 1.172 + 1.173 + 1.174 + 1.175 +** Eigenstates of momentum 1.176 + 1.177 + 1.178 + 1.179 + 1.180 +#In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger 1.181 + 1.182 +#\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\) 1.183 + 1.184 + 1.185 + 1.186 + 1.187 + 1.188 + 1.189 + 1.190 +* COMMENT 1.191 + 1.192 +#* The infinite square well potential 1.193 + 1.194 +A particle exists in a potential that is 1.195 +infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the 1.196 +particle exists in a potential[fn:coords][fn:infinity] 1.197 + 1.198 + 1.199 +\(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for 1.200 +}\;x<0\text{ or }x>a.\end{cases}\) 1.201 + 1.202 +The Schr\ouml{}dinger equation describes how the particle's state 1.203 +\(|\psi\rangle\) will change over time in this system. 1.204 + 1.205 +\(\begin{eqnarray} 1.206 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& 1.207 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\) 1.208 + 1.209 +This is a differential equation; each solution to the 1.210 +Schr\ouml{}dinger equation is a state that is physically allowed for 1.211 +our particle. Here, physically allowed states are 1.212 +those that change in physically allowed ways. However, like any differential 1.213 +equation, the Schr\ouml{}dinger equation can be accompanied by 1.214 +/boundary conditions/\mdash{}conditions that further restrict which 1.215 +states qualify as physically allowed. 1.216 + 1.217 + 1.218 +Whenever possible, physicists impose these boundary conditions: 1.219 +- A physically allowed state ought to be a /smoothly-varying function of position./ This means 1.220 + that if a particle in the state is likely to be /at/ a particular location, 1.221 + it is also likely to be /near/ that location. 1.222 + 1.223 +These boundary conditions imply that for the square well potential in 1.224 +this problem, 1.225 + 1.226 +- Physically allowed states must be totally confined to the well, 1.227 + because it takes an infinite amount of energy to exist anywhere 1.228 + outside of the well (and physically allowed states ought to have 1.229 + only finite energy). 1.230 +- Physically allowed states must be increasingly unlikely to find very 1.231 + close to the walls of the well. This is because of two conditions: the above 1.232 + condition says that the particle is /impossible/ to find 1.233 + outside of the well, and the smoothly-varying condition says 1.234 + that if a particle is impossible to find at a particular location, 1.235 + it must be unlikely to be found nearby that location. 1.236 + 1.237 +#; physically allowed states are those that change in physically 1.238 +#allowed ways. 1.239 + 1.240 + 1.241 +#** Boundary conditions 1.242 +Because the potential is infinite everywhere except within the well, 1.243 +a realistic particle must be confined to exist only within the 1.244 +well\mdash{}its wavefunction must be zero everywhere beyond the walls 1.245 +of the well. 1.246 + 1.247 + 1.248 +[fn:coords] I chose my coordinate system so that the well extends from 1.249 +\(0<x<a\). Others choose a coordinate system so that the well extends from 1.250 +\(-\frac{a}{2}<x<\frac{a}{2}\). Although both coordinate systems describe the same physical 1.251 +situation, they give different-looking answers. 1.252 + 1.253 +[fn:infinity] Of course, infinite potentials are not 1.254 +realistic. Instead, they are useful approximations to finite 1.255 +potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height 1.256 +of the well\rdquo{} are close enough for your own practical 1.257 +purposes. Having introduced a physical impossibility into the problem 1.258 +already, we don't expect to get physically realistic solutions; we 1.259 +just expect to get mathematically consistent ones. The forthcoming 1.260 +trouble is that we don't.