view org/quandary.org @ 11:1f112b4f9e8f tip

Fixed what was baroque.
author Dylan Holmes <ocsenave@gmail.com>
date Tue, 01 Nov 2011 02:30:49 -0500
parents b2f55bcf6853
children
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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
15 I studied quantum mechanics the same way I study most subjects\mdash{}
16 by collecting (and squashing) bugs in my understanding. One of these
17 bugs persisted throughout two semesters of
18 quantum mechanics coursework until I finally found
19 the paper
20 [[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
21 mechanics/]], which helped me stamp out the bug entirely. I decided to
22 write an article about the problem and its solution for a number of reasons:
24 - Although the paper was not unreasonably dense, it was written for
25 teachers. I wanted to write an article for students.
26 - I wanted to popularize the problem and its solution because other
27 explanations are currently too hard to find. (Even Shankar's
28 excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it, as far as I know.)
29 - Attempting an explanation is my way of making
30 sure that the bug really /is/ gone.
31 # entirely eradicated.
33 * COMMENT
34 I recommend the
35 paper not only for students who are learning
36 quantum mechanics, but especially for teachers interested in debugging
37 them.
39 * COMMENT
40 On my first exam in quantum mechanics, my professor asked us to
41 describe how certain measurements would affect a particle in a
42 box. Many of these measurement questions required routine application
43 of skills we had recently learned\mdash{}first, you recall (or
44 calculate) the eigenstates of the quantity
45 to be measured; second, you write the given state as a linear
46 sum of these eigenstates\mdash{} the coefficients on each term give
47 the probability amplitude.
50 * Two methods of calculation that give different results.
52 In the infinitely deep well, there is a particle in the the
53 normalized state
55 \(\psi(x) = \begin{cases}A\; x(x-a)&\text{for }0\lt{}x\lt{}a;\\0,&\text{for }x\lt{}0\text{ or }x>a.\end{cases}\)
57 This is apparently a perfectly respectable state: it is normalized ($A$ is a
58 normalization constant), it is zero
59 everywhere outside of the well, and it is moreover continuous.
61 Even so, we will find a problem if we attempt to calculate the average
62 energy-squared of this state (that is, the quantity \(\langle \psi | H^2 | \psi \rangle\)).
64 ** First method
66 For short, define a new variable[fn:1] \(|\bar \psi\rangle \equiv
67 H|\psi\rangle\). We can express the state $|\bar\psi\rangle$ as a
68 function of $x$ because we know how to express $H$ and $\psi$ in terms
69 of $x$: $H$ is the differential operator $\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}$, and
70 $\psi(x)$ is the function defined above. So, we get \(\quad\bar\psi(x) = \frac{-A\hbar^2}{m}\). For future reference, observe that $\bar\psi(x)$
71 is constant.
73 Having introduced $|\bar\psi\rangle$, we can calculate the average energy-squared of $|\psi\rangle$ in the
74 following way.
76 \(\begin{eqnarray}
77 \langle \psi | H^2 |\psi\rangle &=& \langle \psi H^\dagger | H
78 \psi\rangle\\
79 &=& \langle \psi H | H\psi \rangle\\
80 &=& \langle \bar\psi | \bar\psi \rangle\\
81 &=& \int_0^a \left(\frac{A\hbar^2}{m}\right)^2\;dx\\
82 &=& \frac{A^2\hbar^4 a}{m^2}\\
83 \end{eqnarray}\)
85 For future reference, observe that this value is nonzero
86 (which makes sense).
88 ** Second method
89 We can also calculate the average energy-squared of $|\psi\rangle$ in the
90 following way.
92 \begin{eqnarray}
93 \langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\
94 &=& \langle \psi |H \bar\psi \rangle\\
95 &=&\int_0^a Ax(x-a)
96 \left(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\right)\frac{-A\hbar^2}{m}\;dx\\
97 &=& 0\quad (!)\\
98 \end{eqnarray}
100 The second-to-last term must be zero because the second derivative
101 of a constant \(\left(\frac{-A\hbar^2}{m}\right)\) is zero.
103 * What is the problem?
105 To recap: We used two different methods to calculate the average
106 energy-squared of a state $|\psi\rangle$. For the first method, we
107 used the fact that $H$ is a hermitian operator, replacing \(\langle
108 \psi H^\dagger | H \psi\rangle\) with \(\langle \psi H | H
109 \psi\rangle\). Using this substitution rule, we calculated the answer.
111 For the second method,
112 #we didn't use the fact that $H$ was hermitian;
113 we instead used the fact that we know how to represent $H$ and $\psi$
114 as functions of $x$: $H$ is a differential operator
115 \(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\) and $\psi$ is a quadratic
116 function of $x$. By applying $H$ to $\psi$, we took several
117 derivatives and arrived at our answer.
119 These two methods gave different results. In the following sections,
120 I'll describe and analyze the source of this difference.
122 ** Physical operators only act on physical wavefunctions
123 :PROPERTIES:
124 :ORDERED: t
125 :END:
126 #In quantum mechanics, an operator is a function that takes in a
127 #physical state and produces another physical state as ouput. Some
128 #operators correspond to physical quantities such as energy,
129 #momentum, or position; the mathematical properties of these operators correspond to
130 #physical properties of the system.
132 #Eigenstates are an example of this correspondence: an
134 Physical states are represented as wavefunctions in quantum
135 mechanics. Just as we disallow certain physically nonsensical states
136 in classical mechanics (for example, we consider it to be nonphysical
137 for an object to spontaneously disappear from one place and reappear
138 in another), we also disallow certain wavefunctions in quantum
139 mechanics.
141 For example, since wavefunctions are supposed to correspond to
142 probability amplitudes, we require wavefunctions to be normalized
143 \((\langle \psi |\psi \rangle = 1)\). Generally, we disallow
144 wavefunctions that do not satisfy this property (although there are
145 some exceptions[fn:2]).
147 As another example, we generally expect probability to vary smoothly\mdash{}if
148 a particle is very likely or very unlikely to be found at a particular
149 location, it should also be somewhat likely or somewhat unlikely to be
150 found /near/ that location. In more precise terms, we expect that for
151 physically meaningful wavefunctions, the probability
152 \(\text{Pr}(x)=\int^x \psi^*\psi\) should be a continuous function of
153 $x$ and, again, we disallow wavefunctions that do not satisfy this
154 property because we consider them to be physically nonsensical.
156 So, physical wavefunctions must satisfy certain properties
157 like the two just described. Wavefunctions that do not satisfy these properties are
158 rejected for being physically nonsensical: even though we can perform
159 calculations with them, the mathematical results we obtain do not mean
160 anything physically.
162 Now, in quantum mechanics, an *operator* is a function that converts
163 states into other states. Some operators correspond to
164 physical quantities such as energy, momentum, or position, and as a
165 result, the mathematical properties of these operators correspond to
166 physical properties of the system. Such operators are called
167 /hermitian operators/; one important property of hermitian operators
168 is this rule:
170 #+begin_quote
171 *Hermitian operator rule:* A hermitian operator must only operate on
172 the wavefunctions we have deemed physical, and must only produce
173 physical wavefunctions[fn:: If you require a hermitian operator to
174 have physical eigenstates (which is reasonable), you get a very strong result: you guarantee
175 that the operator will convert every physical wavefunction into
176 another physical wavefunction:
178 For any linear operator $\Omega$, the eigenvalue equation is
179 \(\Omega|\omega\rangle = \omega |\omega\rangle\). Notice that if an
180 eigenstate $|\omega\rangle$ is a physical wavefunction, the
181 eigenvalue equation forces $\Omega|\omega\rangle$ to be a
182 physical wavefunction as well. To elaborate, if the eigenstates of
183 $\Omega$ are physical functions, then $\Omega$ is guaranteed to
184 convert them into other physical functions. Even more is true if the
185 operator $\Omega$ is also hermitian: there is a theorem which states
186 that \ldquo{}If \Omega is hermitian, then every physical wavefunction
187 can be written as a weighted sum of eigenstates of \Omega.\rdquo{} This
188 theorem implies that /if $\Omega$ is hermitian/, and /if the eigenstates
189 of \Omega are physically allowed/, then \Omega is guaranteed to
190 convert every physically allowed wavefunction into another physically
191 allowed wavefunction.].
192 #+end_quote
194 As you can see, this rule comes in two pieces. The first part is a
195 constraint on *you*, the physicist: you must never feed a nonphysical
196 state into a Hermitian operator, as it may produce nonsense. The
197 second part is a constraint on the *operator*: the operator is
198 guaranteed only to produce physical wavefunctions.
200 In fact, this rule for hermitian operators is the source of our
201 problem, as we unknowingly violated it when applying our second
202 method!
204 ** The Hamiltonian is nonphysical
205 You'll remember that in the second method we had wavefunctions within
206 the well
208 \(
209 \begin{eqnarray}
210 \psi(x) &=& A\;x(x-a)\\
211 \bar\psi(x)&\equiv& H|\psi\rangle = \frac{-A\hbar^2}{m}\\
212 \end{eqnarray}
213 \)
215 Using this, we wrote
218 \(\begin{eqnarray}
219 \langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\
220 &=& \langle \psi |H \bar\psi \rangle\\
221 & \vdots&\\
222 &=& 0\\
223 \end{eqnarray}\)
225 However, there are two issues here. First, we were not allowed to operate with $H$ on the wavefunction
226 $|\bar \psi\rangle$, because $H$ is supposed to be a physical operator and $|\bar
227 \psi\rangle$ is a nonphysical state. Indeed, the constant function $|\bar\psi\rangle$ does
228 not approach zero at the edges of the well. By
229 feeding $H$ a nonphysical wavefunction, we obtained nonsensical
230 results.
232 Second, and more importantly, we were wrong to claim that $H$ was a
233 physical operator\mdash{}that $H$ was hermitian. According to the
234 rule, a hermitian operator must convert physical states into other
235 physical states. But $|\psi\rangle$ is a physical state, as we said
236 when we first introduced it \mdash{}it is a normalized, continuous
237 function which approaches zero at the edges of the well and doesn't
238 exist outside it. On the other hand, $|\bar\psi\rangle$ is nonphysical
239 because it does not go to zero at the edges of the well. It is
240 therefore impermissible for $H$ to transform the physical state
241 $|\psi\rangle$ into the nonphysical state $|\bar\psi\rangle$. Because
242 $H$ converts some physical states into nonphysical states, it cannot
243 be a hermitian operator as we assumed.
245 # Boundary conditions affect hermiticity
246 ** Boundary conditions alter hermiticity
247 It may surprise you (and it certainly surprised me) to find that the
248 Hamiltonian is not hermitian. One of the fundamental principles of
249 quantum mechanics is that hermitian operators correspond to physically
250 observable quantities; for this reason, surely the
251 Hamiltonian\mdash{}the operator corresponding to energy\mdash{}ought to be hermitian?
253 But we must understand the correspondence between physically
254 observable quantities and hermitian operators: every hermitian
255 operator corresponds to a physically observable quantity, but not
256 every quantity that intuitively \ldquo{}ought\rdquo{} to be observable
257 will correspond to a hermitian operator[fn::For a simple example,
258 consider the differential operator \(D=\frac{d}{dx}\); although our
259 intuitions might suggest that $D$ is observable which leads us to
260 guess that $D$ is hermitian, it isn't. Still, the very closely related
261 operator $P=i\hbar\frac{d}{dx}$ /is/ hermitian. The point is that we
262 ought to validate our intuitions by checking the definitions.]. The
263 true definition of a hermitian operator imply that the Hamiltonian
264 stops being hermitian in the infinitely deep well. Here we arrive at a
265 crucial point:
267 Operators do not change /form/ between problems: the one-dimensional
268 Hamiltonian is always $H=\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}+V$, the
269 one-dimensional canonical momentum is always $P=i\hbar\frac{d}{dx}$,
270 and so on.
272 However, operators do change in this respect: hermitian operators must
273 only take in physical states, and must only produce physical states; because
274 in different problems we /do/ change the requirements for being a
275 physical state, we also change what it takes for
276 an operator to be called hermitian. As a result, an operator that
277 is hermitian in one setting may fail to be hermitian in another.
279 Having seen how boundary conditions can affect hermiticity, we
280 ought to be extra careful about which conditions we impose on our
281 wavefunctions.
283 ** Choosing the right constraints
285 We have said already that physicists
286 require wavefunctions to satisfy certain properties in order to be
287 deemed \ldquo{}physical\rdquo{}. To be specific, wavefunctions in the
288 infinitely deep well
289 - Must be *normalizable*, because they correspond to
290 probability amplitudes.
291 - Must have *smoothly-varying probability*, because if a particle is very
292 likely to be at a location, it ought to be likely to be /near/
293 it as well.
294 - Must *not exist outside the well*, because it
295 would take an infinite amount of energy to do so.
297 These conditions are surely reasonable. However, physicists sometimes
298 assert that in order to satisfy the second and third conditions,
299 physical wavefunctions
301 - (?) Must *smoothly approach zero* towards the edges of the well.
303 This final constraint is our reason for rejecting $|\bar\psi\rangle$
304 as nonphysical and is consequently the reason why $H$ is not hermitian. If
305 we can convince ourselves that the final constraint is unnecessary,
306 $H$ may again be hermitian. This will satisfy our intuitions that the
307 energy operator /ought/ to be hermitian.
309 But in fact, we have the following mathematical observation to save
310 us: a function $f$ does not need to be continuous in order for the
311 integral \(\int^x f\) to be continuous. As a particularly relevant
312 example, you may now notice that the function $\bar\psi(x)$ is not
313 itself continuous, although the integral $\int_0^x \bar\psi$ /is/
314 continuous. Evidently, it doesn't matter that the wavefunction
315 $\bar\psi$ itself is not continuous; the probability corresponding to
316 $\bar\psi$ /does/ manage to vary continuously anyways. Because the
317 probability corresponding to $\bar\psi$ is the only aspect of
318 $\bar\psi$ which we can detect physically, we /can/ safely omit the
319 final constraint while keeping the other three.
321 ** Symmetric operators look like hermitian operators, but sometimes aren't.
324 #+end_quote
325 ** COMMENT Re-examining physical constraints
327 We have now discovered a flaw: when applied to the state
328 $|\psi\rangle$, the second method violates the rule that physical
329 operators must only take in physical states and must only produce
330 physical states. Let's examine the problem more closely.
332 We have said already that physicists require wavefunctions to satisfy
333 certain properties in order to be deemed \ldquo{}physical\rdquo{}. To
334 be specific, wavefunctions in the infinitely deep well
335 - Must be *normalizable*, because they correspond to
336 probability amplitudes.
337 - Must have *smoothly-varying probability*, because if a particle is very
338 likely to be at a location, it ought to be likely to be /near/
339 it as well.
340 - Must *not exist outside the well*, because it
341 would take an infinite amount of energy to do so.
343 We now have discovered an important flaw in the second method: when
344 applied to the state $|\bar\psi\rangle$, the second method violates
345 the rule that physical operators must only take in
346 physical states and must only produce physical states. The problem is
347 even more serious, however
351 [fn:1] I'm defining a new variable just to make certain expressions
352 look shorter; this cannot affect the content of the answer we'll
353 get.
355 [fn:2] For example, in vaccuum (i.e., when the potential of the
356 physical system is $V(x)=0$ throughout all space), the momentum
357 eigenstates are not normalizable\mdash{}the relevant integral blows
358 up to infinity instead of converging to a number. Physicists modify
359 the definition of normalization slightly so that
360 \ldquo{}delta-normalizable \rdquo{} functions like these are included
361 among the physical wavefunctions.
365 * COMMENT : What I thought I knew
367 The following is a list of things I thought were true of quantum
368 mechanics; the catch is that the list contradicts itself.
370 1. For any hermitian operator: Eigenstates with different eigenvalues are orthogonal.
371 2. For any hermitian operator: Any physically allowed state can be
372 written as a linear sum of eigenstates of the operator.
373 3. The momentum operator and energy operator are hermitian, because
374 momentum and energy are measureable quantities.
375 4. In the vacuum potential, the momentum and energy operators have these eigenstates:
376 - the momentum operator has an eigenstate
377 \(p(x)=\exp{(ipx/\hbar)}\) for each value of $p$.
378 - the energy operator has an eigenstate \(|E\rangle =
379 \alpha|p\rangle + \beta|-p\rangle\) for any \(\alpha,\beta\) and
380 the particular choice of momentum $p=\sqrt{2mE}$.
381 5. In the infinitely deep potential well, the momentum and energy
382 operators have these eigenstates:
383 - The momentum eigenstates and energy eigenstates have the same form
384 as in the vacuum potential: $p(x) =
385 \exp{(ipx/\hbar)}$ and $|E\rangle = \alpha|p\rangle + \beta|-p\rangle$.
386 - Even so, because of the boundary conditions on the
387 well, we must make the following modifications:
388 + Physically realistic states must be impossible to find outside the well. (Only a state of infinite
389 energy could exist outside the well, and infinite energy is not
390 realistic.) This requirement means, for example, that momentum
391 eigenstates in the infinitely deep well must be
392 \(p(x)
393 = \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;
394 \\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
395 + Physically realistic states must vary smoothly throughout
396 space. This means that if a particle in some state is very unlikely to be
397 /at/ a particular location, it is also very unlikely be /near/
398 that location. Combining this requirement with the above
399 requirement, we find that the momentum operator no longer has
400 an eigenstate for each value of $p$; instead, only values of
401 $p$ that are integer multiples of $\pi \hbar/a$ are physically
402 realistic. Similarly, the energy operator no longer has an
403 eigenstate for each value of $E$; instead, the only energy
404 eigenstates in the infinitely deep well
405 are $E_n(x)=\sin(n\pi x/ a)$ for positive integers $n$.
407 * COMMENT :
409 ** Eigenstates with different eigenvalues are orthogonal
411 #+begin_quote
412 *Theorem:* Eigenstates with different eigenvalues are orthogonal.
413 #+end_quote
415 ** COMMENT :
416 I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$
417 and $|b\rangle$ are eigenstates of $\Lambda$. This means that
420 \(
421 \begin{eqnarray}
422 \Lambda |a\rangle&=& a|a\rangle,\\
423 \Lambda|b\rangle&=& b|b\rangle.\\
424 \end{eqnarray}
425 \)
427 If we take the difference of these eigenstates, we find that
429 \(
430 \begin{eqnarray}
431 \Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle
432 \qquad \text{(because $\Lambda$ is linear.)}\\
433 &=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and
434 $|b\rangle$ are eigenstates of $\Lambda$)}
435 \end{eqnarray}\)
438 which means that $a\neq b$.
440 ** Eigenvectors of hermitian operators span the space of solutions
442 #+begin_quote
443 *Theorem:* If $\Omega$ is a hermitian operator, then every physically
444 allowed state can be written as a linear sum of eigenstates of
445 $\Omega$.
446 #+end_quote
450 ** Momentum and energy are hermitian operators
451 This ought to be true because hermitian operators correspond to
452 observable quantities. Since we expect momentum and energy to be
453 measureable quantities, we expect that there are hermitian operators
454 to represent them.
457 ** Momentum and energy eigenstates in vacuum
458 An eigenstate of the momentum operator $P$ would be a state
459 \(|p\rangle\) such that \(P|p\rangle=p|p\rangle\).
461 ** Momentum and energy eigenstates in the infinitely deep well
465 * COMMENT Can you measure momentum in the infinitely deep well?
466 In summary, I thought I knew:
467 1. For any hermitian operator: eigenstates with different eigenvalues
468 are orthogonal.
469 2. For any hermitian operator: any physically realistic state can be
470 written as a linear sum of eigenstates of the operator.
471 3. The momentum operator and energy operator are hermitian, because
472 momentum and energy are observable quantities.
473 4. (The form of the momentum and energy eigenstates in the vacuum potential)
474 5. (The form of the momentum and energy eigenstates in the infinitely deep well potential)
476 Additionally, I understood that because the infinitely deep potential
477 well is not realistic, states of such a system are not necessarily
478 physically realistic. Instead, I understood
479 \ldquo{}realistic states\rdquo{} to be those that satisfy the physically
480 unrealistic Schr\ouml{}dinger equation and its boundary conditions.
482 With that final caveat, here is the problem:
484 According to (5), the momentum eigenstates in the well are
486 \(p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
488 (Additionally, we require that $p$ be an integer multiple of $\pi\hbar/a$.)
490 However, /these/ states are not orthogonal, which contradicts the
491 assumption that (3) the momentum operator is hermitian and (2)
492 eigenstates of a hermitian are orthogonal if they have different eigenvalues.
494 #+begin_quote
495 *Problem 1. The momentum eigenstates of the well are not orthogonal*
497 /Proof./ If $p_1\neq p_2$, then
499 \(\begin{eqnarray}
500 \langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)\,dx\\
501 &=& \int_0^a p_1^*(x)p_2(x)\,dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{
502 outside the well.}\\
503 &=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)\,dx}}\\
504 &=& \int_0^a \exp{\left[i(p_1-p_2)x/\hbar\right]}\,dx\\
505 \end{eqnarray}\)
506 $\square$
508 #+end_quote
512 ** COMMENT Momentum eigenstates
514 In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the
515 momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\).
517 In the infinitely deep potential well, the Hamiltonian is the same but
518 there is a new condition in order for states to qualify as physically
519 allowed: the states must not exist anywhere outside of well, as it
520 takes an infinite amount of energy to do so.
522 Notice that the momentum eigenstates defined above do /not/ satisfy
523 this condition.
527 * COMMENT
528 For each physical system, there is a Schr\ouml{}dinger equation that
529 describes how a particle's state $|\psi\rangle$ will change over
530 time.
532 \(\begin{eqnarray}
533 i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
534 H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
536 This is a differential equation; each solution to the
537 Schr\ouml{}dinger equation is a state that is physically allowed for
538 our particle. Here, physically allowed states are
539 those that change in physically allowed ways. However, like any differential
540 equation, the Schr\ouml{}dinger equation can be accompanied by
541 /boundary conditions/\mdash{}conditions that further restrict which
542 states qualify as physically allowed.
547 ** Eigenstates of momentum
552 #In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger
554 #\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\)
562 * COMMENT
564 #* The infinite square well potential
566 A particle exists in a potential that is
567 infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the
568 particle exists in a potential[fn:coords][fn:infinity]
571 \(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for
572 }\;x<0\text{ or }x>a.\end{cases}\)
574 The Schr\ouml{}dinger equation describes how the particle's state
575 \(|\psi\rangle\) will change over time in this system.
577 \(\begin{eqnarray}
578 i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
579 H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
581 This is a differential equation; each solution to the
582 Schr\ouml{}dinger equation is a state that is physically allowed for
583 our particle. Here, physically allowed states are
584 those that change in physically allowed ways. However, like any differential
585 equation, the Schr\ouml{}dinger equation can be accompanied by
586 /boundary conditions/\mdash{}conditions that further restrict which
587 states qualify as physically allowed.
590 Whenever possible, physicists impose these boundary conditions:
591 - A physically allowed state ought to be a /smoothly-varying function of position./ This means
592 that if a particle in the state is likely to be /at/ a particular location,
593 it is also likely to be /near/ that location.
595 These boundary conditions imply that for the square well potential in
596 this problem,
598 - Physically allowed states must be totally confined to the well,
599 because it takes an infinite amount of energy to exist anywhere
600 outside of the well (and physically allowed states ought to have
601 only finite energy).
602 - Physically allowed states must be increasingly unlikely to find very
603 close to the walls of the well. This is because of two conditions: the above
604 condition says that the particle is /impossible/ to find
605 outside of the well, and the smoothly-varying condition says
606 that if a particle is impossible to find at a particular location,
607 it must be unlikely to be found nearby that location.
609 #; physically allowed states are those that change in physically
610 #allowed ways.
613 #** Boundary conditions
614 Because the potential is infinite everywhere except within the well,
615 a realistic particle must be confined to exist only within the
616 well\mdash{}its wavefunction must be zero everywhere beyond the walls
617 of the well.
620 [fn:coords] I chose my coordinate system so that the well extends from
621 \(0<x<a\). Others choose a coordinate system so that the well extends from
622 \(-\frac{a}{2}<x<\frac{a}{2}\). Although both coordinate systems describe the same physical
623 situation, they give different-looking answers.
625 [fn:infinity] Of course, infinite potentials are not
626 realistic. Instead, they are useful approximations to finite
627 potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
628 of the well\rdquo{} are close enough for your own practical
629 purposes. Having introduced a physical impossibility into the problem
630 already, we don't expect to get physically realistic solutions; we
631 just expect to get mathematically consistent ones. The forthcoming
632 trouble is that we don't.