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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
7
8
9
10 #Bugs in Quantum Mechanics
11 #Bugs in the Quantum-Mechanical Momentum Operator
12
13
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:
22
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 really /is/ gone.
30 # entirely eradicated.
31
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.
37
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.
47
48
49 * Two methods of calculation that give different results.
50
51 In the infinitely deep well, there is a particle in the the
52 normalized state
53
54 \(\psi(x) = \begin{cases}A\; x(x-a)&\text{for }0\lt{}x\lt{}a;\\0,&\text{for }x\lt{}0\text{ or }x>a.\end{cases}\)
55
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.
59
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 \(\langle \psi | H^2 | \psi \rangle\)).
62
63 ** First method
64
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.
71
72 Having introduced $|\bar\psi\rangle$, we can calculate the average energy-squared of $|\psi\rangle$ in the
73 following way.
74
75 \(\begin{eqnarray}
76 \langle \psi | H^2 |\psi\rangle &=& \langle \psi H^\dagger | H
77 \psi\rangle\\
78 &=& \langle \psi H | H\psi \rangle\\
79 &=& \langle \bar\psi | \bar\psi \rangle\\
80 &=& \int_0^a \left(\frac{A\hbar^2}{m}\right)^2\;dx\\
81 &=& \frac{A^2\hbar^4 a}{m^2}\\
82 \end{eqnarray}\)
83
84 For future reference, observe that this value is nonzero
85 (which makes sense).
86
87 ** Second method
88 We can also calculate the average energy-squared of $|\psi\rangle$ in the
89 following way.
90
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}
98
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.
101
102 * What is the problem?
103
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.
109
110 For the second method,
111 #we didn't use the fact that $H$ was hermitian;
112 we instead used the fact that we know how to represent $H$ and $\psi$
113 as functions of $x$: $H$ is a differential operator
114 \(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\) and $\psi$ is a quadratic
115 function of $x$. By applying $H$ to $\psi$, we took several
116 derivatives and arrived at our answer.
117
118 These two methods gave different results. In the following sections,
119 I'll describe and analyze the source of this difference.
120
121 ** Physical operators only act on physical wavefunctions
122 :PROPERTIES:
123 :ORDERED: t
124 :END:
125 #In quantum mechanics, an operator is a function that takes in a
126 #physical state and produces another physical state as ouput. Some
127 #operators correspond to physical quantities such as energy,
128 #momentum, or position; the mathematical properties of these operators correspond to
129 #physical properties of the system.
130
131 #Eigenstates are an example of this correspondence: an
132
133 Physical states are represented as wavefunctions in quantum
134 mechanics. Just as we disallow certain physically nonsensical states
135 in classical mechanics (for example, we consider it to be nonphysical
136 for an object to spontaneously disappear from one place and reappear
137 in another), we also disallow certain wavefunctions in quantum
138 mechanics.
139
140 For example, since wavefunctions are supposed to correspond to
141 probability amplitudes, we require wavefunctions to be normalized
142 \((\langle \psi |\psi \rangle = 1)\). Generally, we disallow
143 wavefunctions that do not satisfy this property (although there are
144 some exceptions[fn:2]).
145
146 As another example, we generally expect probability to vary smoothly\mdash{}if
147 a particle is very likely or very unlikely to be found at a particular
148 location, it should also be somewhat likely or somewhat unlikely to be
149 found /near/ that location. In more precise terms, we expect that for
150 physically meaningful wavefunctions, the probability
151 \(\text{Pr}(x)=\int^x \psi^*\psi\) should be a continuous function of
152 $x$ and, again, we disallow wavefunctions that do not satisfy this
153 property because we consider them to be physically nonsensical.
154
155 So, physical wavefunctions must satisfy certain properties
156 like the two just described. Wavefunctions that do not satisfy these properties are
157 rejected for being physically nonsensical: even though we can perform
158 calculations with them, the mathematical results we obtain do not mean
159 anything physically.
160
161 Now, in quantum mechanics, an *operator* is a function that converts
162 states into other states. Some operators correspond to
163 physical quantities such as energy, momentum, or position, and as a
164 result, the mathematical properties of these operators correspond to
165 physical properties of the system. Such operators are called
166 /hermitian operators/; one important property of hermitian operators
167 is this rule:
168
169 #+begin_quote
170 *Hermitian operator rule:* A hermitian operator must only operate on
171 the wavefunctions we have deemed physical, and must only produce
172 physical wavefunctions[fn:: If you require a hermitian operator to
173 have physical eigenstates (which is reasonable), you get a very strong result: you guarantee
174 that the operator will convert every physical wavefunction into
175 another physical wavefunction:
176
177 For any linear operator $\Omega$, the eigenvalue equation is
178 \(\Omega|\omega\rangle = \omega |\omega\rangle\). Notice that if an
179 eigenstate $|\omega\rangle$ is a physical wavefunction, the
180 eigenvalue equation forces $\Omega|\omega\rangle$ to be a
181 physical wavefunction as well. To elaborate, if the eigenstates of
182 $\Omega$ are physical functions, then $\Omega$ is guaranteed to
183 convert them into other physical functions. Even more is true if the
184 operator $\Omega$ is also hermitian: there is a theorem which states
185 that \ldquo{}If \Omega is hermitian, then every physical wavefunction
186 can be written as a weighted sum of eigenstates of \Omega.\rdquo{} This
187 theorem implies that /if $\Omega$ is hermitian/, and /if the eigenstates
188 of \Omega are physically allowed/, then \Omega is guaranteed to
189 convert every physically allowed wavefunction into another physically
190 allowed wavefunction.].
191 #+end_quote
192
193 As you can see, this rule comes in two pieces. The first part is a
194 constraint on *you*, the physicist: you must never feed a nonphysical
195 state into a Hermitian operator, as it may produce nonsense. The
196 second part is a constraint on the *operator*: the operator is
197 guaranteed only to produce physical wavefunctions.
198
199 In fact, this rule for hermitian operators is the source of our
200 problem, as we unknowingly violated it when applying our second
201 method!
202
203 ** The Hamiltonian is nonphysical
204 You'll remember that in the second method we had wavefunctions within
205 the well
206
207 \(
208 \begin{eqnarray}
209 \psi(x) &=& A\;x(x-a)\\
210 \bar\psi(x)&\equiv& H|\psi\rangle = \frac{-A\hbar^2}{m}\\
211 \end{eqnarray}
212 \)
213
214 Using this, we wrote
215
216
217 \(\begin{eqnarray}
218 \langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\
219 &=& \langle \psi |H \bar\psi \rangle\\
220 & \vdots&\\
221 &=& 0\\
222 \end{eqnarray}\)
223
224 However, there are two issues here. First, we were not allowed to operate with $H$ on the wavefunction
225 $|\bar \psi\rangle$, because $H$ is supposed to be a physical operator and $|\bar
226 \psi\rangle$ is a nonphysical state. Indeed, the constant function $|\bar\psi\rangle$ does
227 not approach zero at the edges of the well. By
228 feeding $H$ a nonphysical wavefunction, we obtained nonsensical
229 results.
230
231 Second, and more importantly, we were wrong to claim that $H$ was a
232 physical operator\mdash{}that $H$ was hermitian. According to the
233 rule, a hermitian operator must convert physical states into other
234 physical states. But $|\psi\rangle$ is a physical state, as we said
235 when we first introduced it \mdash{}it is a normalized, continuous
236 function which approaches zero at the edges of the well and doesn't
237 exist outside it. On the other hand, $|\bar\psi\rangle$ is nonphysical
238 because it does not go to zero at the edges of the well. It is
239 therefore impermissible for $H$ to transform the physical state
240 $|\psi\rangle$ into the nonphysical state $|\bar\psi\rangle$. Because
241 $H$ converts some physical states into nonphysical states, it cannot
242 be a hermitian operator as we assumed.
243
244 # Boundary conditions affect hermiticity
245 ** Boundary conditions alter hermiticity
246 It may surprise you (and it certainly surprised me) to find that the
247 Hamiltonian is not hermitian. One of the fundamental principles of
248 quantum mechanics is that hermitian operators correspond to physically
249 observable quantities; for this reason, surely the
250 Hamiltonian\mdash{}the operator corresponding to energy\mdash{}ought to be hermitian?
251
252 But we must understand the correspondence between physically
253 observable quantities and hermitian operators: every hermitian
254 operator corresponds to a physically observable quantity, but not
255 every quantity that intuitively \ldquo{}ought\rdquo{} to be observable
256 will correspond to a hermitian operator[fn::For a simple example,
257 consider the differential operator \(D=\frac{d}{dx}\); although our
258 intuitions might suggest that $D$ is observable which leads us to
259 guess that $D$ is hermitian, it isn't. Still, the very closely related
260 operator $P=i\hbar\frac{d}{dx}$ /is/ hermitian. The point is that we
261 ought to validate our intuitions by checking the definitions.]. The
262 true definition of a hermitian operator imply that the Hamiltonian
263 stops being hermitian in the infinitely deep well. Here we arrive at a
264 crucial point:
265
266 Operators do not change /form/ between problems: the one-dimensional
267 Hamiltonian is always $H=\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}+V$, the
268 one-dimensional canonical momentum is always $P=i\hbar\frac{d}{dx}$,
269 and so on.
270
271 However, operators do change in this respect: hermitian operators must
272 only take in physical states, and must only produce physical states; because
273 in different problems we /do/ change the requirements for being a
274 physical state, we also change what it takes for
275 an operator to be called hermitian. As a result, an operator that
276 is hermitian in one setting may fail to be hermitian in another.
277
278 Having seen how boundary conditions can affect hermiticity, we
279 ought to be extra careful about which conditions we impose on our
280 wavefunctions.
281
282 ** Choosing the right constraints
283
284 We have said already that physicists
285 require wavefunctions to satisfy certain properties in order to be
286 deemed \ldquo{}physical\rdquo{}. To be specific, wavefunctions in the
287 infinitely deep well
288 - Must be *normalizable*, because they correspond to
289 probability amplitudes.
290 - Must have *smoothly-varying probability*, because if a particle is very
291 likely to be at a location, it ought to be likely to be /near/
292 it as well.
293 - Must *not exist outside the well*, because it
294 would take an infinite amount of energy to do so.
295
296 These conditions are surely reasonable. However, physicists sometimes
297 assert that in order to satisfy the second and third conditions,
298 physical wavefunctions
299
300 - (?) Must *smoothly approach zero* towards the edges of the well.
301
302 This final constraint is our reason for rejecting $|\bar\psi\rangle$
303 as nonphysical and is consequently the reason why $H$ is not hermitian. If
304 we can convince ourselves that the final constraint is unnecessary,
305 $H$ may again be hermitian. This will satisfy our intuitions that the
306 energy operator /ought/ to be hermitian.
307
308 But in fact, we have the following mathematical observation to save
309 us: a function $f$ does not need to be continuous in order for the
310 integral \(\int^x f\) to be continuous. As a particularly relevant
311 example, you may now notice that the function $\bar\psi(x)$ is not
312 itself continuous, although the integral $\int_0^x \bar\psi$ /is/
313 continuous. Evidently, it doesn't matter that the wavefunction
314 $\bar\psi$ itself is not continuous; the probability corresponding to
315 $\bar\psi$ /does/ manage to vary continuously anyways. Because the
316 probability corresponding to $\bar\psi$ is the only aspect of
317 $\bar\psi$ which we can detect physically, we /can/ safely omit the
318 final constraint while keeping the other three.
319
320 ** Symmetric operators look like hermitian operators, but sometimes aren't.
321
322
323 #+end_quote
324 ** COMMENT Re-examining physical constraints
325
326 We have now discovered a flaw: when applied to the state
327 $|\psi\rangle$, the second method violates the rule that physical
328 operators must only take in physical states and must only produce
329 physical states. Let's examine the problem more closely.
330
331 We have said already that physicists require wavefunctions to satisfy
332 certain properties in order to be deemed \ldquo{}physical\rdquo{}. To
333 be specific, wavefunctions in the infinitely deep well
334 - Must be *normalizable*, because they correspond to
335 probability amplitudes.
336 - Must have *smoothly-varying probability*, because if a particle is very
337 likely to be at a location, it ought to be likely to be /near/
338 it as well.
339 - Must *not exist outside the well*, because it
340 would take an infinite amount of energy to do so.
341
342 We now have discovered an important flaw in the second method: when
343 applied to the state $|\bar\psi\rangle$, the second method violates
344 the rule that physical operators must only take in
345 physical states and must only produce physical states. The problem is
346 even more serious, however
347
348
349
350 [fn:1] I'm defining a new variable just to make certain expressions
351 look shorter; this cannot affect the content of the answer we'll
352 get.
353
354 [fn:2] For example, in vaccuum (i.e., when the potential of the
355 physical system is $V(x)=0$ throughout all space), the momentum
356 eigenstates are not normalizable\mdash{}the relevant integral blows
357 up to infinity instead of converging to a number. Physicists modify
358 the definition of normalization slightly so that
359 \ldquo{}delta-normalizable \rdquo{} functions like these are included
360 among the physical wavefunctions.
361
362
363
364 * COMMENT: What I thought I knew
365
366 The following is a list of things I thought were true of quantum
367 mechanics; the catch is that the list contradicts itself.
368
369 1. For any hermitian operator: Eigenstates with different eigenvalues are orthogonal.
370 2. For any hermitian operator: Any physically allowed state can be
371 written as a linear sum of eigenstates of the operator.
372 3. The momentum operator and energy operator are hermitian, because
373 momentum and energy are measureable quantities.
374 4. In the vacuum potential, the momentum and energy operators have these eigenstates:
375 - the momentum operator has an eigenstate
376 \(p(x)=\exp{(ipx/\hbar)}\) for each value of $p$.
377 - the energy operator has an eigenstate \(|E\rangle =
378 \alpha|p\rangle + \beta|-p\rangle\) for any \(\alpha,\beta\) and
379 the particular choice of momentum $p=\sqrt{2mE}$.
380 5. In the infinitely deep potential well, the momentum and energy
381 operators have these eigenstates:
382 - The momentum eigenstates and energy eigenstates have the same form
383 as in the vacuum potential: $p(x) =
384 \exp{(ipx/\hbar)}$ and $|E\rangle = \alpha|p\rangle + \beta|-p\rangle$.
385 - Even so, because of the boundary conditions on the
386 well, we must make the following modifications:
387 + Physically realistic states must be impossible to find outside the well. (Only a state of infinite
388 energy could exist outside the well, and infinite energy is not
389 realistic.) This requirement means, for example, that momentum
390 eigenstates in the infinitely deep well must be
391 \(p(x)
392 = \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;
393 \\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
394 + Physically realistic states must vary smoothly throughout
395 space. This means that if a particle in some state is very unlikely to be
396 /at/ a particular location, it is also very unlikely be /near/
397 that location. Combining this requirement with the above
398 requirement, we find that the momentum operator no longer has
399 an eigenstate for each value of $p$; instead, only values of
400 $p$ that are integer multiples of $\pi \hbar/a$ are physically
401 realistic. Similarly, the energy operator no longer has an
402 eigenstate for each value of $E$; instead, the only energy
403 eigenstates in the infinitely deep well
404 are $E_n(x)=\sin(n\pi x/ a)$ for positive integers $n$.
405
406 * COMMENT:
407
408 ** Eigenstates with different eigenvalues are orthogonal
409
410 #+begin_quote
411 *Theorem:* Eigenstates with different eigenvalues are orthogonal.
412 #+end_quote
413
414 ** COMMENT :
415 I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$
416 and $|b\rangle$ are eigenstates of $\Lambda$. This means that
417
418
419 \(
420 \begin{eqnarray}
421 \Lambda |a\rangle&=& a|a\rangle,\\
422 \Lambda|b\rangle&=& b|b\rangle.\\
423 \end{eqnarray}
424 \)
425
426 If we take the difference of these eigenstates, we find that
427
428 \(
429 \begin{eqnarray}
430 \Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle
431 \qquad \text{(because $\Lambda$ is linear.)}\\
432 &=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and
433 $|b\rangle$ are eigenstates of $\Lambda$)}
434 \end{eqnarray}\)
435
436
437 which means that $a\neq b$.
438
439 ** Eigenvectors of hermitian operators span the space of solutions
440
441 #+begin_quote
442 *Theorem:* If $\Omega$ is a hermitian operator, then every physically
443 allowed state can be written as a linear sum of eigenstates of
444 $\Omega$.
445 #+end_quote
446
447
448
449 ** Momentum and energy are hermitian operators
450 This ought to be true because hermitian operators correspond to
451 observable quantities. Since we expect momentum and energy to be
452 measureable quantities, we expect that there are hermitian operators
453 to represent them.
454
455
456 ** Momentum and energy eigenstates in vacuum
457 An eigenstate of the momentum operator $P$ would be a state
458 \(|p\rangle\) such that \(P|p\rangle=p|p\rangle\).
459
460 ** Momentum and energy eigenstates in the infinitely deep well
461
462
463
464 * COMMENT Can you measure momentum in the infinitely deep well?
465 In summary, I thought I knew:
466 1. For any hermitian operator: eigenstates with different eigenvalues
467 are orthogonal.
468 2. For any hermitian operator: any physically realistic state can be
469 written as a linear sum of eigenstates of the operator.
470 3. The momentum operator and energy operator are hermitian, because
471 momentum and energy are observable quantities.
472 4. (The form of the momentum and energy eigenstates in the vacuum potential)
473 5. (The form of the momentum and energy eigenstates in the infinitely deep well potential)
474
475 Additionally, I understood that because the infinitely deep potential
476 well is not realistic, states of such a system are not necessarily
477 physically realistic. Instead, I understood
478 \ldquo{}realistic states\rdquo{} to be those that satisfy the physically
479 unrealistic Schr\ouml{}dinger equation and its boundary conditions.
480
481 With that final caveat, here is the problem:
482
483 According to (5), the momentum eigenstates in the well are
484
485 \(p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
486
487 (Additionally, we require that $p$ be an integer multiple of $\pi\hbar/a$.)
488
489 However, /these/ states are not orthogonal, which contradicts the
490 assumption that (3) the momentum operator is hermitian and (2)
491 eigenstates of a hermitian are orthogonal if they have different eigenvalues.
492
493 #+begin_quote
494 *Problem 1. The momentum eigenstates of the well are not orthogonal*
495
496 /Proof./ If $p_1\neq p_2$, then
497
498 \(\begin{eqnarray}
499 \langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)\,dx\\
500 &=& \int_0^a p_1^*(x)p_2(x)\,dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{
501 outside the well.}\\
502 &=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)\,dx}}\\
503 &=& \int_0^a \exp{\left[i(p_1-p_2)x/\hbar\right]}\,dx\\
504 \end{eqnarray}\)
505 $\square$
506
507 #+end_quote
508
509
510
511 ** COMMENT Momentum eigenstates
512
513 In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the
514 momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\).
515
516 In the infinitely deep potential well, the Hamiltonian is the same but
517 there is a new condition in order for states to qualify as physically
518 allowed: the states must not exist anywhere outside of well, as it
519 takes an infinite amount of energy to do so.
520
521 Notice that the momentum eigenstates defined above do /not/ satisfy
522 this condition.
523
524
525
526 * COMMENT
527 For each physical system, there is a Schr\ouml{}dinger equation that
528 describes how a particle's state $|\psi\rangle$ will change over
529 time.
530
531 \(\begin{eqnarray}
532 i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
533 H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
534
535 This is a differential equation; each solution to the
536 Schr\ouml{}dinger equation is a state that is physically allowed for
537 our particle. Here, physically allowed states are
538 those that change in physically allowed ways. However, like any differential
539 equation, the Schr\ouml{}dinger equation can be accompanied by
540 /boundary conditions/\mdash{}conditions that further restrict which
541 states qualify as physically allowed.
542
543
544
545
546 ** Eigenstates of momentum
547
548
549
550
551 #In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger
552
553 #\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\)
554
555
556
557
558
559
560
561 * COMMENT
562
563 #* The infinite square well potential
564
565 A particle exists in a potential that is
566 infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the
567 particle exists in a potential[fn:coords][fn:infinity]
568
569
570 \(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for
571 }\;x<0\text{ or }x>a.\end{cases}\)
572
573 The Schr\ouml{}dinger equation describes how the particle's state
574 \(|\psi\rangle\) will change over time in this system.
575
576 \(\begin{eqnarray}
577 i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
578 H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
579
580 This is a differential equation; each solution to the
581 Schr\ouml{}dinger equation is a state that is physically allowed for
582 our particle. Here, physically allowed states are
583 those that change in physically allowed ways. However, like any differential
584 equation, the Schr\ouml{}dinger equation can be accompanied by
585 /boundary conditions/\mdash{}conditions that further restrict which
586 states qualify as physically allowed.
587
588
589 Whenever possible, physicists impose these boundary conditions:
590 - A physically allowed state ought to be a /smoothly-varying function of position./ This means
591 that if a particle in the state is likely to be /at/ a particular location,
592 it is also likely to be /near/ that location.
593
594 These boundary conditions imply that for the square well potential in
595 this problem,
596
597 - Physically allowed states must be totally confined to the well,
598 because it takes an infinite amount of energy to exist anywhere
599 outside of the well (and physically allowed states ought to have
600 only finite energy).
601 - Physically allowed states must be increasingly unlikely to find very
602 close to the walls of the well. This is because of two conditions: the above
603 condition says that the particle is /impossible/ to find
604 outside of the well, and the smoothly-varying condition says
605 that if a particle is impossible to find at a particular location,
606 it must be unlikely to be found nearby that location.
607
608 #; physically allowed states are those that change in physically
609 #allowed ways.
610
611
612 #** Boundary conditions
613 Because the potential is infinite everywhere except within the well,
614 a realistic particle must be confined to exist only within the
615 well\mdash{}its wavefunction must be zero everywhere beyond the walls
616 of the well.
617
618
619 [fn:coords] I chose my coordinate system so that the well extends from
620 \(0<x<a\). Others choose a coordinate system so that the well extends from
621 \(-\frac{a}{2}<x<\frac{a}{2}\). Although both coordinate systems describe the same physical
622 situation, they give different-looking answers.
623
624 [fn:infinity] Of course, infinite potentials are not
625 realistic. Instead, they are useful approximations to finite
626 potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
627 of the well\rdquo{} are close enough for your own practical
628 purposes. Having introduced a physical impossibility into the problem
629 already, we don't expect to get physically realistic solutions; we
630 just expect to get mathematically consistent ones. The forthcoming
631 trouble is that we don't.