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author Robert McIntyre <rlm@mit.edu>
date Fri, 28 Oct 2011 04:56:15 -0700
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1 #+TITLE: Bugs in quantum mechanics
2 #+AUTHOR: Dylan Holmes
3 #+DESCRIPTION: I explain hermitian operators and use them to solve a confusing problem in quantum mechanics.
4 #+KEYWORDS: quantum mechanics, self-adjoint, hermitian, square well, momentum
5 #+SETUPFILE: ../../aurellem/org/setup.org
6 #+INCLUDE: ../../aurellem/org/level-0.org
10 #Bugs in Quantum Mechanics
11 #Bugs in the Quantum-Mechanical Momentum Operator
14 I studied quantum mechanics the same way I study most subjects\mdash{}
15 by collecting (and squashing) bugs in my understanding. One of these
16 bugs persisted throughout two semesters of
17 quantum mechanics coursework until I finally found
18 the paper
19 [[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
20 mechanics/]], which helped me stamp out the bug entirely. I decided to
21 write an article about the problem and its solution for a number of reasons:
23 - Although the paper was not unreasonably dense, it was written for
24 teachers. I wanted to write an article for students.
25 - I wanted to popularize the problem and its solution because other
26 explanations are currently too hard to find. (Even Shankar's
27 excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it, as far as I know.)
28 - Attempting an explanation is my way of making
29 sure that the bug is /really/ gone.
30 # entirely eradicated.
32 * COMMENT
33 I recommend the
34 paper not only for students who are learning
35 quantum mechanics, but especially for teachers interested in debugging
36 them.
38 * COMMENT
39 On my first exam in quantum mechanics, my professor asked us to
40 describe how certain measurements would affect a particle in a
41 box. Many of these measurement questions required routine application
42 of skills we had recently learned\mdash{}first, you recall (or
43 calculate) the eigenstates of the quantity
44 to be measured; second, you write the given state as a linear
45 sum of these eigenstates\mdash{} the coefficients on each term give
46 the probability amplitude.
49 * Two methods of calculation that give different results.
51 In the infinitely deep well, there is a particle in the the
52 normalized state
54 \(\psi(x) = \begin{cases}A\; x(x-a)&\text{for }0\lt{}x\lt{}a;\\0,&\text{for }x\lt{}0\text{ or }x>a.\end{cases}\)
56 This is apparently a perfectly respectable state: it is normalized ($A$ is a
57 normalization constant), it is zero
58 everywhere outside of the well, and it is moreover continuous.
60 Even so, we will find a problem if we attempt to calculate the average
61 energy-squared of this state (that is, the quantity \(\langle \psi | H^2 | \psi \rangle\)).
63 ** First method
65 For short, define a new variable[fn:1] \(|\bar \psi\rangle \equiv
66 H|\psi\rangle\). We can express the state $|\bar\psi\rangle$ as a
67 function of $x$ because we know how to express $H$ and $\psi$ in terms
68 of $x$: $H$ is the differential operator $\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}$, and
69 $\psi(x)$ is the function defined above. So, we get \(\quad\bar\psi(x) = \frac{-A\hbar^2}{m}\). For future reference, observe that $\bar\psi(x)$
70 is constant.
72 Having introduced $|\bar\psi\rangle$, we can calculate the average energy-squared of $|\psi\rangle$ in the
73 following way.
75 \(\begin{eqnarray}
76 \langle \psi | H^2 |\psi\rangle &=& \langle \psi H^\dagger | H
77 \psi\rangle\\
78 &=& \langle \psi H | H\psi \rangle\\
79 &=& \langle \bar\psi | \bar\psi \rangle\\
80 &=& \int_0^a \left(\frac{A\hbar^2}{m}\right)^2\;dx\\
81 &=& \frac{A^2\hbar^4 a}{m^2}\\
82 \end{eqnarray}\)
84 For future reference, observe that this value is nonzero
85 (which makes sense).
87 ** Second method
88 We can also calculate the average energy-squared of $|\psi\rangle$ in the
89 following way.
91 \begin{eqnarray}
92 \langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\
93 &=& \langle \psi |H \bar\psi \rangle\\
94 &=&\int_0^a Ax(x-a)
95 \left(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\right)\frac{-A\hbar^2}{m}\;dx\\
96 &=& 0\quad (!)\\
97 \end{eqnarray}
99 The second-to-last term must be zero because the second derivative
100 of a constant \(\left(\frac{-A\hbar^2}{m}\right)\) is zero.
102 * What is the problem?
104 To recap: We used two different methods to calculate the average
105 energy-squared of a state $|\psi\rangle$. For the first method, we
106 used the fact that $H$ is a hermitian operator, replacing \(\langle
107 \psi H^\dagger | H \psi\rangle\) with \(\langle \psi H | H
108 \psi\rangle\). Using this substitution rule, we calculated the answer.
110 For the second method, we didn't use the fact that $H$ was hermitian;
111 instead, we used the fact that we know how to represent $H$ and $\psi$
112 as functions of $x$: $H$ is a differential operator
113 \(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\) and $\psi$ is a quadratic
114 function of $x$. By applying $H$ to $\psi$, we took several
115 derivatives and arrived at our answer.
117 These two methods gave different results. In the following sections,
118 I'll describe and analyze the source of this difference.
120 ** Physical operators only act on physical wavefunctions
121 :PROPERTIES:
122 :ORDERED: t
123 :END:
124 #In quantum mechanics, an operator is a function that takes in a
125 #physical state and produces another physical state as ouput. Some
126 #operators correspond to physical quantities such as energy,
127 #momentum, or position; the mathematical properties of these operators correspond to
128 #physical properties of the system.
130 #Eigenstates are an example of this correspondence: an
132 Physical states are represented as wavefunctions in quantum
133 mechanics. Just as we disallow certain physically nonsensical states
134 in classical mechanics (for example, we consider it to be nonphysical
135 for an object to spontaneously disappear from one place and reappear
136 in another), we also disallow certain wavefunctions in quantum
137 mechanics.
139 For example, since wavefunctions are supposed to correspond to
140 probability amplitudes, we require wavefunctions to be normalized
141 \((\langle \psi |\psi \rangle = 1)\). Generally, we disallow
142 wavefunctions that do not satisfy this property (although there are
143 some exceptions[fn:2]).
145 As another example, we generally expect probability to vary smoothly\mdash{}if
146 a particle is very likely or very unlikely to be found at a particular
147 location, it should also be somewhat likely or somewhat unlikely to be
148 found /near/ that location. In more precise terms, we expect that for
149 physically meaningful wavefunctions, the probability
150 \(\text{Pr}(x)=\int^x \psi^*\psi\) should be a continuous function of
151 $x$ and, again, we disallow wavefunctions that do not satisfy this
152 property because we consider them to be physically nonsensical.
154 So, physical wavefunctions must satisfy certain properties
155 like the two just described. Wavefunctions that do not satisfy these properties are
156 rejected for being physically nonsensical: even though we can perform
157 calculations with them, the mathematical results we obtain do not mean
158 anything physically.
160 Now, in quantum mechanics, an *operator* is a function that converts
161 states into other states. Some operators correspond to
162 physical quantities such as energy, momentum, or position, and as a
163 result, the mathematical properties of these operators correspond to
164 physical properties of the system. Physical operators are furthermore
165 subject to the following rule: they are only allowed to operate on
166 #physical wavefunctions, and they are only allowed to produce
167 #physical wavefunctions[fn:why].
168 the wavefunctions we have deemed physical, and they only produce physical wavefunctions[fn::
170 If you require a hermitian operator to have physical
171 eigenstates, you get a very strong result: you guarantee that the
172 operator will convert /every/ physical wavefunction into another
173 physical wavefunction:
175 For any linear operator $\Omega$, the eigenvalue equation is
176 \(\Omega|\omega\rangle = \omega |\omega\rangle\). Notice that if an
177 eigenstate $|\omega\rangle$ is a physical wavefunction, the
178 eigenvalue equation equation forces $\Omega|\omega\rangle$ to be a
179 physical wavefunction as well. To elaborate, if the eigenstates of
180 $\Omega$ are physical functions, then $\Omega$ is guaranteed to
181 convert them into other physical functions. Even more is true if the
182 operator $\Omega$ is also hermitian: there is a theorem which states
183 that \ldquo{}If \Omega is hermitian, then every physical wavefunction
184 can be written as a weighted sum of eigenstates of \Omega.\rdquo{} This
185 theorem implies that /if $\Omega$ is hermitian/, and /if the eigenstates
186 of \Omega are physically allowed/, then \Omega is guaranteed to
187 convert every physically allowed wavefunction into another physically
188 allowed wavefunction.].
190 In fact, this rule for physical operators is the source of our
191 problem, as we unknowingly violated it when applying our second
192 method!
194 ** The violation
196 I'll start explaining this violation by being more specific about the
197 infinitely deep well potential. We have said already that physicists
198 require wavefunctions to satisfy certain properties in order to be
199 deemed \ldquo{}physical\rdquo{}. To be specific, wavefunctions in the
200 infinitely deep well
201 - Must be *normalizable*, because they correspond to
202 probability amplitudes.
203 - Must have *smoothly-varying probability*, because if a particle is very
204 likely to be at a location, it ought to be likely to be /near/
205 it as well.
206 - Must *not exist outside the well*, because it
207 would take an infinite amount of energy to do so.
209 Additionally, by combining the second and third conditions, some
210 physicists reason that wavefunctions in the infinitely deep well
212 - Must *become zero* towards the edges of the well.
217 You'll remember we had
219 \(
220 \begin{eqnarray}
221 \psi(x) &=& A\;x(x-a)\\
222 \bar\psi(x)&=& \frac{-A\hbar^2}{m}\\
223 &&\text{for }0\lt{}x\lt{}a\\
224 \end{eqnarray}
225 \)
227 In our second method, we wrote
230 \(\begin{eqnarray}
231 \langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\
232 &=& \langle \psi |H \bar\psi \rangle\\
233 & \vdots&\\
234 &=& 0\\
235 \end{eqnarray}\)
237 However, there are two issues here. First, we were not allowed to operate with $H$ on the wavefunction
238 $|\bar \psi\rangle$, because $H$ is a physical operator and $|\bar
239 \psi\rangle$ is a nonphysical state: in the infinite square well,
240 physical wavefunctions must approach zero at the edges of the well,
241 which the constant function $|\bar\psi\rangle$ does not do. By
242 feeding $H$ a nonphysical wavefunction, we obtained nonsensical
243 results.
245 Second, we claimed that $H$ was a physical operator\mdash{}that $H$
246 was hermitian. According to the rule, this means $H$ must convert physical states into other
247 physical states. But $H$ converts the physical state $|\psi\rangle$
248 into the nonphysical state $|\bar\psi\rangle = H|\psi\rangle$ is not. Because $H$ converts some
249 physical states into nonphysical states, it cannot be a hermitian operator.
251 ** Boundary conditions affect hermiticity
252 We have now discovered a flaw: when applied to the state
253 $|\psi\rangle$, the second method violates the rule that physical
254 operators must only take in physical states and must only produce
255 physical states. This suggests that the problem was with the state
256 $|\bar\psi\rangle$ that we wrongly fed into $H$. However, the problem
257 is more serious still: the state $|\psi\rangle
259 ** COMMENT Re-examining physical constraints
261 We have now discovered a flaw: when applied to the state
262 $|\psi\rangle$, the second method violates the rule that physical
263 operators must only take in physical states and must only produce
264 physical states. Let's examine the problem more closely.
266 We have said already that physicists require wavefunctions to satisfy
267 certain properties in order to be deemed \ldquo{}physical\rdquo{}. To
268 be specific, wavefunctions in the infinitely deep well
269 - Must be *normalizable*, because they correspond to
270 probability amplitudes.
271 - Must have *smoothly-varying probability*, because if a particle is very
272 likely to be at a location, it ought to be likely to be /near/
273 it as well.
274 - Must *not exist outside the well*, because it
275 would take an infinite amount of energy to do so.
277 We now have discovered an important flaw in the second method: when
278 applied to the state $|\bar\psi\rangle$, the second method violates
279 the rule that physical operators must only take in
280 physical states and must only produce physical states. The problem is
281 even more serious, however
285 [fn:1] I'm defining a new variable just to make certain expressions
286 look shorter; this cannot affect the content of the answer we'll
287 get.
289 [fn:2] For example, in vaccuum (i.e., when the potential of the
290 physical system is $V(x)=0$ throughout all space), the momentum
291 eigenstates are not normalizable\mdash{}the relevant integral blows
292 up to infinity instead of converging to a number. Physicists modify
293 the definition of normalization slightly so that
294 \ldquo{}delta-normalizable \rdquo{} functions like these are included
295 among the physical wavefunctions.
299 * COMMENT: What I thought I knew
301 The following is a list of things I thought were true of quantum
302 mechanics; the catch is that the list contradicts itself.
304 1. For any hermitian operator: Eigenstates with different eigenvalues are orthogonal.
305 2. For any hermitian operator: Any physically allowed state can be
306 written as a linear sum of eigenstates of the operator.
307 3. The momentum operator and energy operator are hermitian, because
308 momentum and energy are measureable quantities.
309 4. In the vacuum potential, the momentum and energy operators have these eigenstates:
310 - the momentum operator has an eigenstate
311 \(p(x)=\exp{(ipx/\hbar)}\) for each value of $p$.
312 - the energy operator has an eigenstate \(|E\rangle =
313 \alpha|p\rangle + \beta|-p\rangle\) for any \(\alpha,\beta\) and
314 the particular choice of momentum $p=\sqrt{2mE}$.
315 5. In the infinitely deep potential well, the momentum and energy
316 operators have these eigenstates:
317 - The momentum eigenstates and energy eigenstates have the same form
318 as in the vacuum potential: $p(x) =
319 \exp{(ipx/\hbar)}$ and $|E\rangle = \alpha|p\rangle + \beta|-p\rangle$.
320 - Even so, because of the boundary conditions on the
321 well, we must make the following modifications:
322 + Physically realistic states must be impossible to find outside the well. (Only a state of infinite
323 energy could exist outside the well, and infinite energy is not
324 realistic.) This requirement means, for example, that momentum
325 eigenstates in the infinitely deep well must be
326 \(p(x)
327 = \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;
328 \\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
329 + Physically realistic states must vary smoothly throughout
330 space. This means that if a particle in some state is very unlikely to be
331 /at/ a particular location, it is also very unlikely be /near/
332 that location. Combining this requirement with the above
333 requirement, we find that the momentum operator no longer has
334 an eigenstate for each value of $p$; instead, only values of
335 $p$ that are integer multiples of $\pi \hbar/a$ are physically
336 realistic. Similarly, the energy operator no longer has an
337 eigenstate for each value of $E$; instead, the only energy
338 eigenstates in the infinitely deep well
339 are $E_n(x)=\sin(n\pi x/ a)$ for positive integers $n$.
341 * COMMENT:
343 ** Eigenstates with different eigenvalues are orthogonal
345 #+begin_quote
346 *Theorem:* Eigenstates with different eigenvalues are orthogonal.
347 #+end_quote
349 ** COMMENT :
350 I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$
351 and $|b\rangle$ are eigenstates of $\Lambda$. This means that
354 \(
355 \begin{eqnarray}
356 \Lambda |a\rangle&=& a|a\rangle,\\
357 \Lambda|b\rangle&=& b|b\rangle.\\
358 \end{eqnarray}
359 \)
361 If we take the difference of these eigenstates, we find that
363 \(
364 \begin{eqnarray}
365 \Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle
366 \qquad \text{(because $\Lambda$ is linear.)}\\
367 &=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and
368 $|b\rangle$ are eigenstates of $\Lambda$)}
369 \end{eqnarray}\)
372 which means that $a\neq b$.
374 ** Eigenvectors of hermitian operators span the space of solutions
376 #+begin_quote
377 *Theorem:* If $\Omega$ is a hermitian operator, then every physically
378 allowed state can be written as a linear sum of eigenstates of
379 $\Omega$.
380 #+end_quote
384 ** Momentum and energy are hermitian operators
385 This ought to be true because hermitian operators correspond to
386 observable quantities. Since we expect momentum and energy to be
387 measureable quantities, we expect that there are hermitian operators
388 to represent them.
391 ** Momentum and energy eigenstates in vacuum
392 An eigenstate of the momentum operator $P$ would be a state
393 \(|p\rangle\) such that \(P|p\rangle=p|p\rangle\).
395 ** Momentum and energy eigenstates in the infinitely deep well
399 * COMMENT Can you measure momentum in the infinitely deep well?
400 In summary, I thought I knew:
401 1. For any hermitian operator: eigenstates with different eigenvalues
402 are orthogonal.
403 2. For any hermitian operator: any physically realistic state can be
404 written as a linear sum of eigenstates of the operator.
405 3. The momentum operator and energy operator are hermitian, because
406 momentum and energy are observable quantities.
407 4. (The form of the momentum and energy eigenstates in the vacuum potential)
408 5. (The form of the momentum and energy eigenstates in the infinitely deep well potential)
410 Additionally, I understood that because the infinitely deep potential
411 well is not realistic, states of such a system are not necessarily
412 physically realistic. Instead, I understood
413 \ldquo{}realistic states\rdquo{} to be those that satisfy the physically
414 unrealistic Schr\ouml{}dinger equation and its boundary conditions.
416 With that final caveat, here is the problem:
418 According to (5), the momentum eigenstates in the well are
420 \(p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
422 (Additionally, we require that $p$ be an integer multiple of $\pi\hbar/a$.)
424 However, /these/ states are not orthogonal, which contradicts the
425 assumption that (3) the momentum operator is hermitian and (2)
426 eigenstates of a hermitian are orthogonal if they have different eigenvalues.
428 #+begin_quote
429 *Problem 1. The momentum eigenstates of the well are not orthogonal*
431 /Proof./ If $p_1\neq p_2$, then
433 \(\begin{eqnarray}
434 \langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)\,dx\\
435 &=& \int_0^a p_1^*(x)p_2(x)\,dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{
436 outside the well.}\\
437 &=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)\,dx}}\\
438 &=& \int_0^a \exp{\left[i(p_1-p_2)x/\hbar\right]}\,dx\\
439 \end{eqnarray}\)
440 $\square$
442 #+end_quote
446 ** COMMENT Momentum eigenstates
448 In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the
449 momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\).
451 In the infinitely deep potential well, the Hamiltonian is the same but
452 there is a new condition in order for states to qualify as physically
453 allowed: the states must not exist anywhere outside of well, as it
454 takes an infinite amount of energy to do so.
456 Notice that the momentum eigenstates defined above do /not/ satisfy
457 this condition.
461 * COMMENT
462 For each physical system, there is a Schr\ouml{}dinger equation that
463 describes how a particle's state $|\psi\rangle$ will change over
464 time.
466 \(\begin{eqnarray}
467 i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
468 H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
470 This is a differential equation; each solution to the
471 Schr\ouml{}dinger equation is a state that is physically allowed for
472 our particle. Here, physically allowed states are
473 those that change in physically allowed ways. However, like any differential
474 equation, the Schr\ouml{}dinger equation can be accompanied by
475 /boundary conditions/\mdash{}conditions that further restrict which
476 states qualify as physically allowed.
481 ** Eigenstates of momentum
486 #In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger
488 #\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\)
496 * COMMENT
498 #* The infinite square well potential
500 A particle exists in a potential that is
501 infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the
502 particle exists in a potential[fn:coords][fn:infinity]
505 \(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for
506 }\;x<0\text{ or }x>a.\end{cases}\)
508 The Schr\ouml{}dinger equation describes how the particle's state
509 \(|\psi\rangle\) will change over time in this system.
511 \(\begin{eqnarray}
512 i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
513 H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
515 This is a differential equation; each solution to the
516 Schr\ouml{}dinger equation is a state that is physically allowed for
517 our particle. Here, physically allowed states are
518 those that change in physically allowed ways. However, like any differential
519 equation, the Schr\ouml{}dinger equation can be accompanied by
520 /boundary conditions/\mdash{}conditions that further restrict which
521 states qualify as physically allowed.
524 Whenever possible, physicists impose these boundary conditions:
525 - A physically allowed state ought to be a /smoothly-varying function of position./ This means
526 that if a particle in the state is likely to be /at/ a particular location,
527 it is also likely to be /near/ that location.
529 These boundary conditions imply that for the square well potential in
530 this problem,
532 - Physically allowed states must be totally confined to the well,
533 because it takes an infinite amount of energy to exist anywhere
534 outside of the well (and physically allowed states ought to have
535 only finite energy).
536 - Physically allowed states must be increasingly unlikely to find very
537 close to the walls of the well. This is because of two conditions: the above
538 condition says that the particle is /impossible/ to find
539 outside of the well, and the smoothly-varying condition says
540 that if a particle is impossible to find at a particular location,
541 it must be unlikely to be found nearby that location.
543 #; physically allowed states are those that change in physically
544 #allowed ways.
547 #** Boundary conditions
548 Because the potential is infinite everywhere except within the well,
549 a realistic particle must be confined to exist only within the
550 well\mdash{}its wavefunction must be zero everywhere beyond the walls
551 of the well.
554 [fn:coords] I chose my coordinate system so that the well extends from
555 \(0<x<a\). Others choose a coordinate system so that the well extends from
556 \(-\frac{a}{2}<x<\frac{a}{2}\). Although both coordinate systems describe the same physical
557 situation, they give different-looking answers.
559 [fn:infinity] Of course, infinite potentials are not
560 realistic. Instead, they are useful approximations to finite
561 potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
562 of the well\rdquo{} are close enough for your own practical
563 purposes. Having introduced a physical impossibility into the problem
564 already, we don't expect to get physically realistic solutions; we
565 just expect to get mathematically consistent ones. The forthcoming
566 trouble is that we don't.