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author Robert McIntyre <rlm@mit.edu>
date Mon, 17 Oct 2011 23:19:18 -0700
parents f743fd0f4d8b
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rlm@0 1 #+TITLE: Bugs in quantum mechanics
rlm@0 2 #+AUTHOR: Dylan Holmes
rlm@0 3 #+SETUPFILE: ../../aurellem/org/setup.org
rlm@0 4 #+INCLUDE: ../../aurellem/org/level-0.org
rlm@0 5
rlm@0 6 #Bugs in Quantum Mechanics
rlm@0 7 #Bugs in the Quantum-Mechanical Momentum Operator
rlm@0 8
rlm@0 9
rlm@0 10 I studied quantum mechanics the same way I study most subjects\mdash{}
rlm@0 11 by collecting (and squashing) bugs in my understanding. One of these
rlm@0 12 bugs persisted throughout two semesters of
rlm@0 13 quantum mechanics coursework until I finally found
rlm@0 14 the paper
rlm@0 15 [[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
rlm@0 16 mechanics/]], which helped me stamp out the bug entirely. I decided to
rlm@0 17 write an article about the problem and its solution for a number of reasons:
rlm@0 18
rlm@0 19 - Although the paper was not unreasonably dense, it was written for
rlm@0 20 teachers. I wanted to write an article for students.
rlm@0 21 - I wanted to popularize the problem and its solution because other
rlm@0 22 explanations are currently too hard to find. (Even Shankar's
rlm@0 23 excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it.)
rlm@0 24 - I wanted to check that the bug was indeed entirely
rlm@0 25 eradicated. Attempting an explanation is my way of making
rlm@0 26 sure.
rlm@0 27
rlm@0 28 * COMMENT
rlm@0 29 I recommend the
rlm@0 30 paper not only for students who are learning
rlm@0 31 quantum mechanics, but especially for teachers interested in debugging
rlm@0 32 them.
rlm@0 33
rlm@0 34 * COMMENT
rlm@0 35 On my first exam in quantum mechanics, my professor asked us to
rlm@0 36 describe how certain measurements would affect a particle in a
rlm@0 37 box. Many of these measurement questions required routine application
rlm@0 38 of skills we had recently learned\mdash{}first, you recall (or
rlm@0 39 calculate) the eigenstates of the quantity
rlm@0 40 to be measured; second, you write the given state as a linear
rlm@0 41 sum of these eigenstates\mdash{} the coefficients on each term give
rlm@0 42 the probability amplitude.
rlm@0 43
rlm@0 44
rlm@0 45 * What I thought I knew
rlm@0 46
rlm@0 47 The following is a list of things I thought were true of quantum
rlm@0 48 mechanics; the catch is that the list contradicts itself.
rlm@0 49
rlm@0 50 1. For any hermitian operator: Eigenstates with different eigenvalues are orthogonal.
rlm@0 51 2. For any hermitian operator: Any physically allowed state can be
rlm@0 52 written as a linear sum of eigenstates of the operator.
rlm@0 53 3. The momentum operator and energy operator are hermitian, because
rlm@0 54 momentum and energy are measureable quantities.
rlm@0 55 4. In the vacuum potential, the momentum and energy operators have these eigenstates:
rlm@0 56 - the momentum operator has an eigenstate
rlm@0 57 \(p(x)=\exp{(ipx/\hbar)}\) for each value of $p$.
rlm@0 58 - the energy operator has an eigenstate \(|E\rangle =
rlm@0 59 \alpha|p\rangle + \beta|-p\rangle\) for any \(\alpha,\beta\) and
rlm@0 60 the particular choice of momentum $p=\sqrt{2mE}$.
rlm@0 61 5. In the infinitely deep potential well, the momentum and energy
rlm@0 62 operators have these eigenstates:
rlm@0 63 - The momentum eigenstates and energy eigenstates have the same form
rlm@0 64 as in the vacuum potential: $p(x) =
rlm@0 65 \exp{(ipx/\hbar)}$ and $|E\rangle = \alpha|p\rangle + \beta|-p\rangle$.
rlm@0 66 - Even so, because of the boundary conditions on the
rlm@0 67 well, we must make the following modifications:
rlm@0 68 + Physically realistic states must be impossible to find outside the well. (Only a state of infinite
rlm@0 69 energy could exist outside the well, and infinite energy is not
rlm@0 70 realistic.) This requirement means, for example, that momentum
rlm@0 71 eigenstates in the infinitely deep well must be
rlm@0 72 \(p(x)
rlm@0 73 = \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;
rlm@0 74 \\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
rlm@0 75 + Physically realistic states must vary smoothly throughout
rlm@0 76 space. This means that if a particle in some state is very unlikely to be
rlm@0 77 /at/ a particular location, it is also very unlikely be /near/
rlm@0 78 that location. Combining this requirement with the above
rlm@0 79 requirement, we find that the momentum operator no longer has
rlm@0 80 an eigenstate for each value of $p$; instead, only values of
rlm@0 81 $p$ that are integer multiples of $\pi a/\hbar$ are physically
rlm@0 82 realistic. Similarly, the energy operator no longer has an
rlm@0 83 eigenstate for each value of $E$; instead, the only energy
rlm@0 84 eigenstates in the infinitely deep well
rlm@0 85 are $E_n(x)=\sin(n\pi x/ a)$ for positive integers $n$.
rlm@0 86
rlm@0 87 * COMMENT:
rlm@0 88
rlm@0 89 ** Eigenstates with different eigenvalues are orthogonal
rlm@0 90
rlm@0 91 #+begin_quote
rlm@0 92 *Theorem:* Eigenstates with different eigenvalues are orthogonal.
rlm@0 93 #+end_quote
rlm@0 94
rlm@0 95 ** COMMENT :
rlm@0 96 I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$
rlm@0 97 and $|b\rangle$ are eigenstates of $\Lambda$. This means that
rlm@0 98
rlm@0 99
rlm@0 100 \(
rlm@0 101 \begin{eqnarray}
rlm@0 102 \Lambda |a\rangle&=& a|a\rangle,\\
rlm@0 103 \Lambda|b\rangle&=& b|b\rangle.\\
rlm@0 104 \end{eqnarray}
rlm@0 105 \)
rlm@0 106
rlm@0 107 If we take the difference of these eigenstates, we find that
rlm@0 108
rlm@0 109 \(
rlm@0 110 \begin{eqnarray}
rlm@0 111 \Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle
rlm@0 112 \qquad \text{(because $\Lambda$ is linear.)}\\
rlm@0 113 &=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and
rlm@0 114 $|b\rangle$ are eigenstates of $\Lambda$)}
rlm@0 115 \end{eqnarray}\)
rlm@0 116
rlm@0 117
rlm@0 118 which means that $a\neq b$.
rlm@0 119
rlm@0 120 ** Eigenvectors of hermitian operators span the space of solutions
rlm@0 121
rlm@0 122 #+begin_quote
rlm@0 123 *Theorem:* If $\Omega$ is a hermitian operator, then every physically
rlm@0 124 allowed state can be written as a linear sum of eigenstates of
rlm@0 125 $\Omega$.
rlm@0 126 #+end_quote
rlm@0 127
rlm@0 128
rlm@0 129
rlm@0 130 ** Momentum and energy are hermitian operators
rlm@0 131 This ought to be true because hermitian operators correspond to
rlm@0 132 observable quantities. Since we expect momentum and energy to be
rlm@0 133 measureable quantities, we expect that there are hermitian operators
rlm@0 134 to represent them.
rlm@0 135
rlm@0 136
rlm@0 137 ** Momentum and energy eigenstates in vacuum
rlm@0 138 An eigenstate of the momentum operator $P$ would be a state
rlm@0 139 \(|p\rangle\) such that \(P|p\rangle=p|p\rangle\).
rlm@0 140
rlm@0 141 ** Momentum and energy eigenstates in the infinitely deep well
rlm@0 142
rlm@0 143
rlm@0 144
rlm@0 145 * Can you measure momentum in the infinitely deep well?
rlm@0 146 In summary, I thought I knew:
rlm@0 147 1. For any hermitian operator: eigenstates with different eigenvalues
rlm@0 148 are orthogonal.
rlm@0 149 2. For any hermitian operator: any physically realistic state can be
rlm@0 150 written as a linear sum of eigenstates of the operator.
rlm@0 151 3. The momentum operator and energy operator are hermitian, because
rlm@0 152 momentum and energy are observable quantities.
rlm@0 153 4. (The form of the momentum and energy eigenstates in the vacuum potential)
rlm@0 154 5. (The form of the momentum and energy eigenstates in the infinitely deep well potential)
rlm@0 155
rlm@0 156 Additionally, I understood that because the infinitely deep potential
rlm@0 157 well is not realistic, states of such a system are not necessarily
rlm@0 158 physically realistic. Instead, I understood
rlm@0 159 \ldquo{}realistic states\rdquo{} to be those that satisfy the physically
rlm@0 160 unrealistic Schr\ouml{}dinger equation and its boundary conditions.
rlm@0 161
rlm@0 162 With that final caveat, here is the problem:
rlm@0 163
rlm@0 164 According to (5), the momentum eigenstates in the well are
rlm@0 165
rlm@0 166 \(p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
rlm@0 167
rlm@0 168 However, /these/ states are not orthogonal, which contradicts the
rlm@0 169 assumption that (3) the momentum operator is hermitian and (2)
rlm@0 170 eigenstates of a hermitian are orthogonal if they have different eigenvalues.
rlm@0 171
rlm@0 172 #+begin_quote
rlm@0 173 *Problem 1. The momentum eigenstates of the well are not orthogonal*
rlm@0 174
rlm@0 175 /Proof./ If $p_1\neq p_2$, then
rlm@0 176
rlm@0 177 \(\begin{eqnarray}
rlm@0 178 \langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)dx\\
rlm@0 179 &=& \int_0^a p_1^*(x)p_2(x)dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{
rlm@0 180 outside the well.}\\
rlm@0 181 &=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)dx}}
rlm@0 182 \end{eqnarray}\)
rlm@0 183 $\square$
rlm@0 184
rlm@0 185 #+end_quote
rlm@0 186
rlm@0 187
rlm@0 188
rlm@0 189 ** COMMENT Momentum eigenstates
rlm@0 190
rlm@0 191 In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the
rlm@0 192 momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\).
rlm@0 193
rlm@0 194 In the infinitely deep potential well, the Hamiltonian is the same but
rlm@0 195 there is a new condition in order for states to qualify as physically
rlm@0 196 allowed: the states must not exist anywhere outside of well, as it
rlm@0 197 takes an infinite amount of energy to do so.
rlm@0 198
rlm@0 199 Notice that the momentum eigenstates defined above do /not/ satisfy
rlm@0 200 this condition.
rlm@0 201
rlm@0 202
rlm@0 203
rlm@0 204 * COMMENT
rlm@0 205 For each physical system, there is a Schr\ouml{}dinger equation that
rlm@0 206 describes how a particle's state $|\psi\rangle$ will change over
rlm@0 207 time.
rlm@0 208
rlm@0 209 \(\begin{eqnarray}
rlm@0 210 i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
rlm@0 211 H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
rlm@0 212
rlm@0 213 This is a differential equation; each solution to the
rlm@0 214 Schr\ouml{}dinger equation is a state that is physically allowed for
rlm@0 215 our particle. Here, physically allowed states are
rlm@0 216 those that change in physically allowed ways. However, like any differential
rlm@0 217 equation, the Schr\ouml{}dinger equation can be accompanied by
rlm@0 218 /boundary conditions/\mdash{}conditions that further restrict which
rlm@0 219 states qualify as physically allowed.
rlm@0 220
rlm@0 221
rlm@0 222
rlm@0 223
rlm@0 224 ** Eigenstates of momentum
rlm@0 225
rlm@0 226
rlm@0 227
rlm@0 228
rlm@0 229 #In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger
rlm@0 230
rlm@0 231 #\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\)
rlm@0 232
rlm@0 233
rlm@0 234
rlm@0 235
rlm@0 236
rlm@0 237
rlm@0 238
rlm@0 239 * COMMENT
rlm@0 240
rlm@0 241 #* The infinite square well potential
rlm@0 242
rlm@0 243 A particle exists in a potential that is
rlm@0 244 infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the
rlm@0 245 particle exists in a potential[fn:coords][fn:infinity]
rlm@0 246
rlm@0 247
rlm@0 248 \(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for
rlm@0 249 }\;x<0\text{ or }x>a.\end{cases}\)
rlm@0 250
rlm@0 251 The Schr\ouml{}dinger equation describes how the particle's state
rlm@0 252 \(|\psi\rangle\) will change over time in this system.
rlm@0 253
rlm@0 254 \(\begin{eqnarray}
rlm@0 255 i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
rlm@0 256 H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
rlm@0 257
rlm@0 258 This is a differential equation; each solution to the
rlm@0 259 Schr\ouml{}dinger equation is a state that is physically allowed for
rlm@0 260 our particle. Here, physically allowed states are
rlm@0 261 those that change in physically allowed ways. However, like any differential
rlm@0 262 equation, the Schr\ouml{}dinger equation can be accompanied by
rlm@0 263 /boundary conditions/\mdash{}conditions that further restrict which
rlm@0 264 states qualify as physically allowed.
rlm@0 265
rlm@0 266
rlm@0 267 Whenever possible, physicists impose these boundary conditions:
rlm@0 268 - A physically allowed state ought to be a /smoothly-varying function of position./ This means
rlm@0 269 that if a particle in the state is likely to be /at/ a particular location,
rlm@0 270 it is also likely to be /near/ that location.
rlm@0 271
rlm@0 272 These boundary conditions imply that for the square well potential in
rlm@0 273 this problem,
rlm@0 274
rlm@0 275 - Physically allowed states must be totally confined to the well,
rlm@0 276 because it takes an infinite amount of energy to exist anywhere
rlm@0 277 outside of the well (and physically allowed states ought to have
rlm@0 278 only finite energy).
rlm@0 279 - Physically allowed states must be increasingly unlikely to find very
rlm@0 280 close to the walls of the well. This is because of two conditions: the above
rlm@0 281 condition says that the particle is /impossible/ to find
rlm@0 282 outside of the well, and the smoothly-varying condition says
rlm@0 283 that if a particle is impossible to find at a particular location,
rlm@0 284 it must be unlikely to be found nearby that location.
rlm@0 285
rlm@0 286 #; physically allowed states are those that change in physically
rlm@0 287 #allowed ways.
rlm@0 288
rlm@0 289
rlm@0 290 #** Boundary conditions
rlm@0 291 Because the potential is infinite everywhere except within the well,
rlm@0 292 a realistic particle must be confined to exist only within the
rlm@0 293 well\mdash{}its wavefunction must be zero everywhere beyond the walls
rlm@0 294 of the well.
rlm@0 295
rlm@0 296
rlm@0 297 [fn:coords] I chose my coordinate system so that the well extends from
rlm@0 298 \(0<x<a\). Others choose a coordinate system so that the well extends from
rlm@0 299 \(-\frac{a}{2}<x<\frac{a}{2}\). Although both coordinate systems describe the same physical
rlm@0 300 situation, they give different-looking answers.
rlm@0 301
rlm@0 302 [fn:infinity] Of course, infinite potentials are not
rlm@0 303 realistic. Instead, they are useful approximations to finite
rlm@0 304 potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
rlm@0 305 of the well\rdquo{} are close enough for your own practical
rlm@0 306 purposes. Having introduced a physical impossibility into the problem
rlm@0 307 already, we don't expect to get physically realistic solutions; we
rlm@0 308 just expect to get mathematically consistent ones. The forthcoming
rlm@0 309 trouble is that we don't.