annotate bk/bk_quandary.org @ 6:b2f55bcf6853

fixed comments
author Robert McIntyre <rlm@mit.edu>
date Fri, 28 Oct 2011 04:56:15 -0700
parents 44d3dc936f6a
children
rev   line source
rlm@0 1 #+TITLE: Bugs in quantum mechanics
rlm@0 2 #+AUTHOR: Dylan Holmes
rlm@0 3 #+DESCRIPTION: I explain hermitian operators and use them to solve a confusing problem in quantum mechanics.
rlm@0 4 #+KEYWORDS: quantum mechanics, self-adjoint, hermitian, square well, momentum
rlm@0 5 #+SETUPFILE: ../../aurellem/org/setup.org
rlm@0 6 #+INCLUDE: ../../aurellem/org/level-0.org
rlm@0 7
rlm@0 8
rlm@0 9
rlm@0 10 #Bugs in Quantum Mechanics
rlm@0 11 #Bugs in the Quantum-Mechanical Momentum Operator
rlm@0 12
rlm@0 13
rlm@0 14 I studied quantum mechanics the same way I study most subjects\mdash{}
rlm@0 15 by collecting (and squashing) bugs in my understanding. One of these
rlm@0 16 bugs persisted throughout two semesters of
rlm@0 17 quantum mechanics coursework until I finally found
rlm@0 18 the paper
rlm@0 19 [[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
rlm@0 20 mechanics/]], which helped me stamp out the bug entirely. I decided to
rlm@0 21 write an article about the problem and its solution for a number of reasons:
rlm@0 22
rlm@0 23 - Although the paper was not unreasonably dense, it was written for
rlm@0 24 teachers. I wanted to write an article for students.
rlm@0 25 - I wanted to popularize the problem and its solution because other
rlm@0 26 explanations are currently too hard to find. (Even Shankar's
rlm@0 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.)
rlm@0 28 - Attempting an explanation is my way of making
rlm@0 29 sure that the bug is /really/ gone.
rlm@0 30 # entirely eradicated.
rlm@0 31
rlm@0 32 * COMMENT
rlm@0 33 I recommend the
rlm@0 34 paper not only for students who are learning
rlm@0 35 quantum mechanics, but especially for teachers interested in debugging
rlm@0 36 them.
rlm@0 37
rlm@0 38 * COMMENT
rlm@0 39 On my first exam in quantum mechanics, my professor asked us to
rlm@0 40 describe how certain measurements would affect a particle in a
rlm@0 41 box. Many of these measurement questions required routine application
rlm@0 42 of skills we had recently learned\mdash{}first, you recall (or
rlm@0 43 calculate) the eigenstates of the quantity
rlm@0 44 to be measured; second, you write the given state as a linear
rlm@0 45 sum of these eigenstates\mdash{} the coefficients on each term give
rlm@0 46 the probability amplitude.
rlm@0 47
rlm@0 48
rlm@0 49 * Two methods of calculation that give different results.
rlm@0 50
rlm@0 51 In the infinitely deep well, there is a particle in the the
rlm@0 52 normalized state
rlm@0 53
rlm@0 54 \(\psi(x) = \begin{cases}A\; x(x-a)&\text{for }0\lt{}x\lt{}a;\\0,&\text{for }x\lt{}0\text{ or }x>a.\end{cases}\)
rlm@0 55
rlm@0 56 This is apparently a perfectly respectable state: it is normalized ($A$ is a
rlm@0 57 normalization constant), it is zero
rlm@0 58 everywhere outside of the well, and it is moreover continuous.
rlm@0 59
rlm@0 60 Even so, we will find a problem if we attempt to calculate the average
rlm@0 61 energy-squared of this state (that is, the quantity \(\langle \psi | H^2 | \psi \rangle\)).
rlm@0 62
rlm@0 63 ** First method
rlm@0 64
rlm@0 65 For short, define a new variable[fn:1] \(|\bar \psi\rangle \equiv
rlm@0 66 H|\psi\rangle\). We can express the state $|\bar\psi\rangle$ as a
rlm@0 67 function of $x$ because we know how to express $H$ and $\psi$ in terms
rlm@0 68 of $x$: $H$ is the differential operator $\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}$, and
rlm@0 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 $\bar\psi(x)$
rlm@0 70 is constant.
rlm@0 71
rlm@0 72 Having introduced $|\bar\psi\rangle$, we can calculate the average energy-squared of $|\psi\rangle$ in the
rlm@0 73 following way.
rlm@0 74
rlm@0 75 \(\begin{eqnarray}
rlm@0 76 \langle \psi | H^2 |\psi\rangle &=& \langle \psi H^\dagger | H
rlm@0 77 \psi\rangle\\
rlm@0 78 &=& \langle \psi H | H\psi \rangle\\
rlm@0 79 &=& \langle \bar\psi | \bar\psi \rangle\\
rlm@0 80 &=& \int_0^a \left(\frac{A\hbar^2}{m}\right)^2\;dx\\
rlm@0 81 &=& \frac{A^2\hbar^4 a}{m^2}\\
rlm@0 82 \end{eqnarray}\)
rlm@0 83
rlm@0 84 For future reference, observe that this value is nonzero
rlm@0 85 (which makes sense).
rlm@0 86
rlm@0 87 ** Second method
rlm@0 88 We can also calculate the average energy-squared of $|\psi\rangle$ in the
rlm@0 89 following way.
rlm@0 90
rlm@0 91 \begin{eqnarray}
rlm@0 92 \langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\
rlm@0 93 &=& \langle \psi |H \bar\psi \rangle\\
rlm@0 94 &=&\int_0^a Ax(x-a)
rlm@0 95 \left(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\right)\frac{-A\hbar^2}{m}\;dx\\
rlm@0 96 &=& 0\quad (!)\\
rlm@0 97 \end{eqnarray}
rlm@0 98
rlm@0 99 The second-to-last term must be zero because the second derivative
rlm@0 100 of a constant \(\left(\frac{-A\hbar^2}{m}\right)\) is zero.
rlm@0 101
rlm@0 102 * What is the problem?
rlm@0 103
rlm@0 104 To recap: We used two different methods to calculate the average
rlm@0 105 energy-squared of a state $|\psi\rangle$. For the first method, we
rlm@0 106 used the fact that $H$ is a hermitian operator, replacing \(\langle
rlm@0 107 \psi H^\dagger | H \psi\rangle\) with \(\langle \psi H | H
rlm@0 108 \psi\rangle\). Using this substitution rule, we calculated the answer.
rlm@0 109
rlm@0 110 For the second method, we didn't use the fact that $H$ was hermitian;
rlm@0 111 instead, we used the fact that we know how to represent $H$ and $\psi$
rlm@0 112 as functions of $x$: $H$ is a differential operator
rlm@0 113 \(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\) and $\psi$ is a quadratic
rlm@0 114 function of $x$. By applying $H$ to $\psi$, we took several
rlm@0 115 derivatives and arrived at our answer.
rlm@0 116
rlm@0 117 These two methods gave different results. In the following sections,
rlm@0 118 I'll describe and analyze the source of this difference.
rlm@0 119
rlm@0 120 ** Physical operators only act on physical wavefunctions
rlm@0 121 :PROPERTIES:
rlm@0 122 :ORDERED: t
rlm@0 123 :END:
rlm@0 124 #In quantum mechanics, an operator is a function that takes in a
rlm@0 125 #physical state and produces another physical state as ouput. Some
rlm@0 126 #operators correspond to physical quantities such as energy,
rlm@0 127 #momentum, or position; the mathematical properties of these operators correspond to
rlm@0 128 #physical properties of the system.
rlm@0 129
rlm@0 130 #Eigenstates are an example of this correspondence: an
rlm@0 131
rlm@0 132 Physical states are represented as wavefunctions in quantum
rlm@0 133 mechanics. Just as we disallow certain physically nonsensical states
rlm@0 134 in classical mechanics (for example, we consider it to be nonphysical
rlm@0 135 for an object to spontaneously disappear from one place and reappear
rlm@0 136 in another), we also disallow certain wavefunctions in quantum
rlm@0 137 mechanics.
rlm@0 138
rlm@0 139 For example, since wavefunctions are supposed to correspond to
rlm@0 140 probability amplitudes, we require wavefunctions to be normalized
rlm@0 141 \((\langle \psi |\psi \rangle = 1)\). Generally, we disallow
rlm@0 142 wavefunctions that do not satisfy this property (although there are
rlm@0 143 some exceptions[fn:2]).
rlm@0 144
rlm@0 145 As another example, we generally expect probability to vary smoothly\mdash{}if
rlm@0 146 a particle is very likely or very unlikely to be found at a particular
rlm@0 147 location, it should also be somewhat likely or somewhat unlikely to be
rlm@0 148 found /near/ that location. In more precise terms, we expect that for
rlm@0 149 physically meaningful wavefunctions, the probability
rlm@0 150 \(\text{Pr}(x)=\int^x \psi^*\psi\) should be a continuous function of
rlm@0 151 $x$ and, again, we disallow wavefunctions that do not satisfy this
rlm@0 152 property because we consider them to be physically nonsensical.
rlm@0 153
rlm@0 154 So, physical wavefunctions must satisfy certain properties
rlm@0 155 like the two just described. Wavefunctions that do not satisfy these properties are
rlm@0 156 rejected for being physically nonsensical: even though we can perform
rlm@0 157 calculations with them, the mathematical results we obtain do not mean
rlm@0 158 anything physically.
rlm@0 159
rlm@0 160 Now, in quantum mechanics, an *operator* is a function that converts
rlm@0 161 states into other states. Some operators correspond to
rlm@0 162 physical quantities such as energy, momentum, or position, and as a
rlm@0 163 result, the mathematical properties of these operators correspond to
rlm@0 164 physical properties of the system. Physical operators are furthermore
rlm@0 165 subject to the following rule: they are only allowed to operate on
rlm@0 166 #physical wavefunctions, and they are only allowed to produce
rlm@0 167 #physical wavefunctions[fn:why].
rlm@0 168 the wavefunctions we have deemed physical, and they only produce physical wavefunctions[fn::
rlm@0 169
rlm@0 170 If you require a hermitian operator to have physical
rlm@0 171 eigenstates, you get a very strong result: you guarantee that the
rlm@0 172 operator will convert /every/ physical wavefunction into another
rlm@0 173 physical wavefunction:
rlm@0 174
rlm@0 175 For any linear operator $\Omega$, the eigenvalue equation is
rlm@0 176 \(\Omega|\omega\rangle = \omega |\omega\rangle\). Notice that if an
rlm@0 177 eigenstate $|\omega\rangle$ is a physical wavefunction, the
rlm@0 178 eigenvalue equation equation forces $\Omega|\omega\rangle$ to be a
rlm@0 179 physical wavefunction as well. To elaborate, if the eigenstates of
rlm@0 180 $\Omega$ are physical functions, then $\Omega$ is guaranteed to
rlm@0 181 convert them into other physical functions. Even more is true if the
rlm@0 182 operator $\Omega$ is also hermitian: there is a theorem which states
rlm@0 183 that \ldquo{}If \Omega is hermitian, then every physical wavefunction
rlm@0 184 can be written as a weighted sum of eigenstates of \Omega.\rdquo{} This
rlm@0 185 theorem implies that /if $\Omega$ is hermitian/, and /if the eigenstates
rlm@0 186 of \Omega are physically allowed/, then \Omega is guaranteed to
rlm@0 187 convert every physically allowed wavefunction into another physically
rlm@0 188 allowed wavefunction.].
rlm@0 189
rlm@0 190 In fact, this rule for physical operators is the source of our
rlm@0 191 problem, as we unknowingly violated it when applying our second
rlm@0 192 method!
rlm@0 193
rlm@0 194 ** The violation
rlm@0 195
rlm@0 196 I'll start explaining this violation by being more specific about the
rlm@0 197 infinitely deep well potential. We have said already that physicists
rlm@0 198 require wavefunctions to satisfy certain properties in order to be
rlm@0 199 deemed \ldquo{}physical\rdquo{}. To be specific, wavefunctions in the
rlm@0 200 infinitely deep well
rlm@0 201 - Must be *normalizable*, because they correspond to
rlm@0 202 probability amplitudes.
rlm@0 203 - Must have *smoothly-varying probability*, because if a particle is very
rlm@0 204 likely to be at a location, it ought to be likely to be /near/
rlm@0 205 it as well.
rlm@0 206 - Must *not exist outside the well*, because it
rlm@0 207 would take an infinite amount of energy to do so.
rlm@0 208
rlm@0 209 Additionally, by combining the second and third conditions, some
rlm@0 210 physicists reason that wavefunctions in the infinitely deep well
rlm@0 211
rlm@0 212 - Must *become zero* towards the edges of the well.
rlm@0 213
rlm@0 214
rlm@0 215
rlm@0 216
rlm@0 217 You'll remember we had
rlm@0 218
rlm@0 219 \(
rlm@0 220 \begin{eqnarray}
rlm@0 221 \psi(x) &=& A\;x(x-a)\\
rlm@0 222 \bar\psi(x)&=& \frac{-A\hbar^2}{m}\\
rlm@0 223 &&\text{for }0\lt{}x\lt{}a\\
rlm@0 224 \end{eqnarray}
rlm@0 225 \)
rlm@0 226
rlm@0 227 In our second method, we wrote
rlm@0 228
rlm@0 229
rlm@0 230 \(\begin{eqnarray}
rlm@0 231 \langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\
rlm@0 232 &=& \langle \psi |H \bar\psi \rangle\\
rlm@0 233 & \vdots&\\
rlm@0 234 &=& 0\\
rlm@0 235 \end{eqnarray}\)
rlm@0 236
rlm@0 237 However, there are two issues here. First, we were not allowed to operate with $H$ on the wavefunction
rlm@0 238 $|\bar \psi\rangle$, because $H$ is a physical operator and $|\bar
rlm@0 239 \psi\rangle$ is a nonphysical state: in the infinite square well,
rlm@0 240 physical wavefunctions must approach zero at the edges of the well,
rlm@0 241 which the constant function $|\bar\psi\rangle$ does not do. By
rlm@0 242 feeding $H$ a nonphysical wavefunction, we obtained nonsensical
rlm@0 243 results.
rlm@0 244
rlm@0 245 Second, we claimed that $H$ was a physical operator\mdash{}that $H$
rlm@0 246 was hermitian. According to the rule, this means $H$ must convert physical states into other
rlm@0 247 physical states. But $H$ converts the physical state $|\psi\rangle$
rlm@0 248 into the nonphysical state $|\bar\psi\rangle = H|\psi\rangle$ is not. Because $H$ converts some
rlm@0 249 physical states into nonphysical states, it cannot be a hermitian operator.
rlm@0 250
rlm@0 251 ** Boundary conditions affect hermiticity
rlm@0 252 We have now discovered a flaw: when applied to the state
rlm@0 253 $|\psi\rangle$, the second method violates the rule that physical
rlm@0 254 operators must only take in physical states and must only produce
rlm@0 255 physical states. This suggests that the problem was with the state
rlm@0 256 $|\bar\psi\rangle$ that we wrongly fed into $H$. However, the problem
rlm@0 257 is more serious still: the state $|\psi\rangle
rlm@0 258
rlm@0 259 ** COMMENT Re-examining physical constraints
rlm@0 260
rlm@0 261 We have now discovered a flaw: when applied to the state
rlm@0 262 $|\psi\rangle$, the second method violates the rule that physical
rlm@0 263 operators must only take in physical states and must only produce
rlm@0 264 physical states. Let's examine the problem more closely.
rlm@0 265
rlm@0 266 We have said already that physicists require wavefunctions to satisfy
rlm@0 267 certain properties in order to be deemed \ldquo{}physical\rdquo{}. To
rlm@0 268 be specific, wavefunctions in the infinitely deep well
rlm@0 269 - Must be *normalizable*, because they correspond to
rlm@0 270 probability amplitudes.
rlm@0 271 - Must have *smoothly-varying probability*, because if a particle is very
rlm@0 272 likely to be at a location, it ought to be likely to be /near/
rlm@0 273 it as well.
rlm@0 274 - Must *not exist outside the well*, because it
rlm@0 275 would take an infinite amount of energy to do so.
rlm@0 276
rlm@0 277 We now have discovered an important flaw in the second method: when
rlm@0 278 applied to the state $|\bar\psi\rangle$, the second method violates
rlm@0 279 the rule that physical operators must only take in
rlm@0 280 physical states and must only produce physical states. The problem is
rlm@0 281 even more serious, however
rlm@0 282
rlm@0 283
rlm@0 284
rlm@0 285 [fn:1] I'm defining a new variable just to make certain expressions
rlm@0 286 look shorter; this cannot affect the content of the answer we'll
rlm@0 287 get.
rlm@0 288
rlm@0 289 [fn:2] For example, in vaccuum (i.e., when the potential of the
rlm@0 290 physical system is $V(x)=0$ throughout all space), the momentum
rlm@0 291 eigenstates are not normalizable\mdash{}the relevant integral blows
rlm@0 292 up to infinity instead of converging to a number. Physicists modify
rlm@0 293 the definition of normalization slightly so that
rlm@0 294 \ldquo{}delta-normalizable \rdquo{} functions like these are included
rlm@0 295 among the physical wavefunctions.
rlm@0 296
rlm@0 297
rlm@0 298
rlm@0 299 * COMMENT: What I thought I knew
rlm@0 300
rlm@0 301 The following is a list of things I thought were true of quantum
rlm@0 302 mechanics; the catch is that the list contradicts itself.
rlm@0 303
rlm@0 304 1. For any hermitian operator: Eigenstates with different eigenvalues are orthogonal.
rlm@0 305 2. For any hermitian operator: Any physically allowed state can be
rlm@0 306 written as a linear sum of eigenstates of the operator.
rlm@0 307 3. The momentum operator and energy operator are hermitian, because
rlm@0 308 momentum and energy are measureable quantities.
rlm@0 309 4. In the vacuum potential, the momentum and energy operators have these eigenstates:
rlm@0 310 - the momentum operator has an eigenstate
rlm@0 311 \(p(x)=\exp{(ipx/\hbar)}\) for each value of $p$.
rlm@0 312 - the energy operator has an eigenstate \(|E\rangle =
rlm@0 313 \alpha|p\rangle + \beta|-p\rangle\) for any \(\alpha,\beta\) and
rlm@0 314 the particular choice of momentum $p=\sqrt{2mE}$.
rlm@0 315 5. In the infinitely deep potential well, the momentum and energy
rlm@0 316 operators have these eigenstates:
rlm@0 317 - The momentum eigenstates and energy eigenstates have the same form
rlm@0 318 as in the vacuum potential: $p(x) =
rlm@0 319 \exp{(ipx/\hbar)}$ and $|E\rangle = \alpha|p\rangle + \beta|-p\rangle$.
rlm@0 320 - Even so, because of the boundary conditions on the
rlm@0 321 well, we must make the following modifications:
rlm@0 322 + Physically realistic states must be impossible to find outside the well. (Only a state of infinite
rlm@0 323 energy could exist outside the well, and infinite energy is not
rlm@0 324 realistic.) This requirement means, for example, that momentum
rlm@0 325 eigenstates in the infinitely deep well must be
rlm@0 326 \(p(x)
rlm@0 327 = \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;
rlm@0 328 \\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
rlm@0 329 + Physically realistic states must vary smoothly throughout
rlm@0 330 space. This means that if a particle in some state is very unlikely to be
rlm@0 331 /at/ a particular location, it is also very unlikely be /near/
rlm@0 332 that location. Combining this requirement with the above
rlm@0 333 requirement, we find that the momentum operator no longer has
rlm@0 334 an eigenstate for each value of $p$; instead, only values of
rlm@0 335 $p$ that are integer multiples of $\pi \hbar/a$ are physically
rlm@0 336 realistic. Similarly, the energy operator no longer has an
rlm@0 337 eigenstate for each value of $E$; instead, the only energy
rlm@0 338 eigenstates in the infinitely deep well
rlm@0 339 are $E_n(x)=\sin(n\pi x/ a)$ for positive integers $n$.
rlm@0 340
rlm@0 341 * COMMENT:
rlm@0 342
rlm@0 343 ** Eigenstates with different eigenvalues are orthogonal
rlm@0 344
rlm@0 345 #+begin_quote
rlm@0 346 *Theorem:* Eigenstates with different eigenvalues are orthogonal.
rlm@0 347 #+end_quote
rlm@0 348
rlm@0 349 ** COMMENT :
rlm@0 350 I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$
rlm@0 351 and $|b\rangle$ are eigenstates of $\Lambda$. This means that
rlm@0 352
rlm@0 353
rlm@0 354 \(
rlm@0 355 \begin{eqnarray}
rlm@0 356 \Lambda |a\rangle&=& a|a\rangle,\\
rlm@0 357 \Lambda|b\rangle&=& b|b\rangle.\\
rlm@0 358 \end{eqnarray}
rlm@0 359 \)
rlm@0 360
rlm@0 361 If we take the difference of these eigenstates, we find that
rlm@0 362
rlm@0 363 \(
rlm@0 364 \begin{eqnarray}
rlm@0 365 \Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle
rlm@0 366 \qquad \text{(because $\Lambda$ is linear.)}\\
rlm@0 367 &=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and
rlm@0 368 $|b\rangle$ are eigenstates of $\Lambda$)}
rlm@0 369 \end{eqnarray}\)
rlm@0 370
rlm@0 371
rlm@0 372 which means that $a\neq b$.
rlm@0 373
rlm@0 374 ** Eigenvectors of hermitian operators span the space of solutions
rlm@0 375
rlm@0 376 #+begin_quote
rlm@0 377 *Theorem:* If $\Omega$ is a hermitian operator, then every physically
rlm@0 378 allowed state can be written as a linear sum of eigenstates of
rlm@0 379 $\Omega$.
rlm@0 380 #+end_quote
rlm@0 381
rlm@0 382
rlm@0 383
rlm@0 384 ** Momentum and energy are hermitian operators
rlm@0 385 This ought to be true because hermitian operators correspond to
rlm@0 386 observable quantities. Since we expect momentum and energy to be
rlm@0 387 measureable quantities, we expect that there are hermitian operators
rlm@0 388 to represent them.
rlm@0 389
rlm@0 390
rlm@0 391 ** Momentum and energy eigenstates in vacuum
rlm@0 392 An eigenstate of the momentum operator $P$ would be a state
rlm@0 393 \(|p\rangle\) such that \(P|p\rangle=p|p\rangle\).
rlm@0 394
rlm@0 395 ** Momentum and energy eigenstates in the infinitely deep well
rlm@0 396
rlm@0 397
rlm@0 398
rlm@0 399 * COMMENT Can you measure momentum in the infinitely deep well?
rlm@0 400 In summary, I thought I knew:
rlm@0 401 1. For any hermitian operator: eigenstates with different eigenvalues
rlm@0 402 are orthogonal.
rlm@0 403 2. For any hermitian operator: any physically realistic state can be
rlm@0 404 written as a linear sum of eigenstates of the operator.
rlm@0 405 3. The momentum operator and energy operator are hermitian, because
rlm@0 406 momentum and energy are observable quantities.
rlm@0 407 4. (The form of the momentum and energy eigenstates in the vacuum potential)
rlm@0 408 5. (The form of the momentum and energy eigenstates in the infinitely deep well potential)
rlm@0 409
rlm@0 410 Additionally, I understood that because the infinitely deep potential
rlm@0 411 well is not realistic, states of such a system are not necessarily
rlm@0 412 physically realistic. Instead, I understood
rlm@0 413 \ldquo{}realistic states\rdquo{} to be those that satisfy the physically
rlm@0 414 unrealistic Schr\ouml{}dinger equation and its boundary conditions.
rlm@0 415
rlm@0 416 With that final caveat, here is the problem:
rlm@0 417
rlm@0 418 According to (5), the momentum eigenstates in the well are
rlm@0 419
rlm@0 420 \(p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
rlm@0 421
rlm@0 422 (Additionally, we require that $p$ be an integer multiple of $\pi\hbar/a$.)
rlm@0 423
rlm@0 424 However, /these/ states are not orthogonal, which contradicts the
rlm@0 425 assumption that (3) the momentum operator is hermitian and (2)
rlm@0 426 eigenstates of a hermitian are orthogonal if they have different eigenvalues.
rlm@0 427
rlm@0 428 #+begin_quote
rlm@0 429 *Problem 1. The momentum eigenstates of the well are not orthogonal*
rlm@0 430
rlm@0 431 /Proof./ If $p_1\neq p_2$, then
rlm@0 432
rlm@0 433 \(\begin{eqnarray}
rlm@0 434 \langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)\,dx\\
rlm@0 435 &=& \int_0^a p_1^*(x)p_2(x)\,dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{
rlm@0 436 outside the well.}\\
rlm@0 437 &=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)\,dx}}\\
rlm@0 438 &=& \int_0^a \exp{\left[i(p_1-p_2)x/\hbar\right]}\,dx\\
rlm@0 439 \end{eqnarray}\)
rlm@0 440 $\square$
rlm@0 441
rlm@0 442 #+end_quote
rlm@0 443
rlm@0 444
rlm@0 445
rlm@0 446 ** COMMENT Momentum eigenstates
rlm@0 447
rlm@0 448 In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the
rlm@0 449 momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\).
rlm@0 450
rlm@0 451 In the infinitely deep potential well, the Hamiltonian is the same but
rlm@0 452 there is a new condition in order for states to qualify as physically
rlm@0 453 allowed: the states must not exist anywhere outside of well, as it
rlm@0 454 takes an infinite amount of energy to do so.
rlm@0 455
rlm@0 456 Notice that the momentum eigenstates defined above do /not/ satisfy
rlm@0 457 this condition.
rlm@0 458
rlm@0 459
rlm@0 460
rlm@0 461 * COMMENT
rlm@0 462 For each physical system, there is a Schr\ouml{}dinger equation that
rlm@0 463 describes how a particle's state $|\psi\rangle$ will change over
rlm@0 464 time.
rlm@0 465
rlm@0 466 \(\begin{eqnarray}
rlm@0 467 i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
rlm@0 468 H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
rlm@0 469
rlm@0 470 This is a differential equation; each solution to the
rlm@0 471 Schr\ouml{}dinger equation is a state that is physically allowed for
rlm@0 472 our particle. Here, physically allowed states are
rlm@0 473 those that change in physically allowed ways. However, like any differential
rlm@0 474 equation, the Schr\ouml{}dinger equation can be accompanied by
rlm@0 475 /boundary conditions/\mdash{}conditions that further restrict which
rlm@0 476 states qualify as physically allowed.
rlm@0 477
rlm@0 478
rlm@0 479
rlm@0 480
rlm@0 481 ** Eigenstates of momentum
rlm@0 482
rlm@0 483
rlm@0 484
rlm@0 485
rlm@0 486 #In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger
rlm@0 487
rlm@0 488 #\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\)
rlm@0 489
rlm@0 490
rlm@0 491
rlm@0 492
rlm@0 493
rlm@0 494
rlm@0 495
rlm@0 496 * COMMENT
rlm@0 497
rlm@0 498 #* The infinite square well potential
rlm@0 499
rlm@0 500 A particle exists in a potential that is
rlm@0 501 infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the
rlm@0 502 particle exists in a potential[fn:coords][fn:infinity]
rlm@0 503
rlm@0 504
rlm@0 505 \(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for
rlm@0 506 }\;x<0\text{ or }x>a.\end{cases}\)
rlm@0 507
rlm@0 508 The Schr\ouml{}dinger equation describes how the particle's state
rlm@0 509 \(|\psi\rangle\) will change over time in this system.
rlm@0 510
rlm@0 511 \(\begin{eqnarray}
rlm@0 512 i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
rlm@0 513 H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
rlm@0 514
rlm@0 515 This is a differential equation; each solution to the
rlm@0 516 Schr\ouml{}dinger equation is a state that is physically allowed for
rlm@0 517 our particle. Here, physically allowed states are
rlm@0 518 those that change in physically allowed ways. However, like any differential
rlm@0 519 equation, the Schr\ouml{}dinger equation can be accompanied by
rlm@0 520 /boundary conditions/\mdash{}conditions that further restrict which
rlm@0 521 states qualify as physically allowed.
rlm@0 522
rlm@0 523
rlm@0 524 Whenever possible, physicists impose these boundary conditions:
rlm@0 525 - A physically allowed state ought to be a /smoothly-varying function of position./ This means
rlm@0 526 that if a particle in the state is likely to be /at/ a particular location,
rlm@0 527 it is also likely to be /near/ that location.
rlm@0 528
rlm@0 529 These boundary conditions imply that for the square well potential in
rlm@0 530 this problem,
rlm@0 531
rlm@0 532 - Physically allowed states must be totally confined to the well,
rlm@0 533 because it takes an infinite amount of energy to exist anywhere
rlm@0 534 outside of the well (and physically allowed states ought to have
rlm@0 535 only finite energy).
rlm@0 536 - Physically allowed states must be increasingly unlikely to find very
rlm@0 537 close to the walls of the well. This is because of two conditions: the above
rlm@0 538 condition says that the particle is /impossible/ to find
rlm@0 539 outside of the well, and the smoothly-varying condition says
rlm@0 540 that if a particle is impossible to find at a particular location,
rlm@0 541 it must be unlikely to be found nearby that location.
rlm@0 542
rlm@0 543 #; physically allowed states are those that change in physically
rlm@0 544 #allowed ways.
rlm@0 545
rlm@0 546
rlm@0 547 #** Boundary conditions
rlm@0 548 Because the potential is infinite everywhere except within the well,
rlm@0 549 a realistic particle must be confined to exist only within the
rlm@0 550 well\mdash{}its wavefunction must be zero everywhere beyond the walls
rlm@0 551 of the well.
rlm@0 552
rlm@0 553
rlm@0 554 [fn:coords] I chose my coordinate system so that the well extends from
rlm@0 555 \(0<x<a\). Others choose a coordinate system so that the well extends from
rlm@0 556 \(-\frac{a}{2}<x<\frac{a}{2}\). Although both coordinate systems describe the same physical
rlm@0 557 situation, they give different-looking answers.
rlm@0 558
rlm@0 559 [fn:infinity] Of course, infinite potentials are not
rlm@0 560 realistic. Instead, they are useful approximations to finite
rlm@0 561 potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
rlm@0 562 of the well\rdquo{} are close enough for your own practical
rlm@0 563 purposes. Having introduced a physical impossibility into the problem
rlm@0 564 already, we don't expect to get physically realistic solutions; we
rlm@0 565 just expect to get mathematically consistent ones. The forthcoming
rlm@0 566 trouble is that we don't.