annotate 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
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@4 14
rlm@0 15 I studied quantum mechanics the same way I study most subjects\mdash{}
rlm@0 16 by collecting (and squashing) bugs in my understanding. One of these
rlm@0 17 bugs persisted throughout two semesters of
rlm@0 18 quantum mechanics coursework until I finally found
rlm@0 19 the paper
rlm@0 20 [[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
rlm@0 21 mechanics/]], which helped me stamp out the bug entirely. I decided to
rlm@0 22 write an article about the problem and its solution for a number of reasons:
rlm@0 23
rlm@0 24 - Although the paper was not unreasonably dense, it was written for
rlm@0 25 teachers. I wanted to write an article for students.
rlm@0 26 - I wanted to popularize the problem and its solution because other
rlm@0 27 explanations are currently too hard to find. (Even Shankar's
rlm@0 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.)
rlm@0 29 - Attempting an explanation is my way of making
rlm@0 30 sure that the bug really /is/ gone.
rlm@0 31 # entirely eradicated.
rlm@0 32
rlm@0 33 * COMMENT
rlm@0 34 I recommend the
rlm@0 35 paper not only for students who are learning
rlm@0 36 quantum mechanics, but especially for teachers interested in debugging
rlm@0 37 them.
rlm@0 38
rlm@0 39 * COMMENT
rlm@0 40 On my first exam in quantum mechanics, my professor asked us to
rlm@0 41 describe how certain measurements would affect a particle in a
rlm@0 42 box. Many of these measurement questions required routine application
rlm@0 43 of skills we had recently learned\mdash{}first, you recall (or
rlm@0 44 calculate) the eigenstates of the quantity
rlm@0 45 to be measured; second, you write the given state as a linear
rlm@0 46 sum of these eigenstates\mdash{} the coefficients on each term give
rlm@0 47 the probability amplitude.
rlm@0 48
rlm@0 49
rlm@0 50 * Two methods of calculation that give different results.
rlm@0 51
rlm@0 52 In the infinitely deep well, there is a particle in the the
rlm@0 53 normalized state
rlm@0 54
rlm@0 55 \(\psi(x) = \begin{cases}A\; x(x-a)&\text{for }0\lt{}x\lt{}a;\\0,&\text{for }x\lt{}0\text{ or }x>a.\end{cases}\)
rlm@0 56
rlm@0 57 This is apparently a perfectly respectable state: it is normalized ($A$ is a
rlm@0 58 normalization constant), it is zero
rlm@0 59 everywhere outside of the well, and it is moreover continuous.
rlm@0 60
rlm@0 61 Even so, we will find a problem if we attempt to calculate the average
rlm@0 62 energy-squared of this state (that is, the quantity \(\langle \psi | H^2 | \psi \rangle\)).
rlm@0 63
rlm@0 64 ** First method
rlm@0 65
rlm@0 66 For short, define a new variable[fn:1] \(|\bar \psi\rangle \equiv
rlm@0 67 H|\psi\rangle\). We can express the state $|\bar\psi\rangle$ as a
rlm@0 68 function of $x$ because we know how to express $H$ and $\psi$ in terms
rlm@0 69 of $x$: $H$ is the differential operator $\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}$, and
rlm@0 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 $\bar\psi(x)$
rlm@0 71 is constant.
rlm@0 72
rlm@0 73 Having introduced $|\bar\psi\rangle$, we can calculate the average energy-squared of $|\psi\rangle$ in the
rlm@0 74 following way.
rlm@0 75
rlm@0 76 \(\begin{eqnarray}
rlm@0 77 \langle \psi | H^2 |\psi\rangle &=& \langle \psi H^\dagger | H
rlm@0 78 \psi\rangle\\
rlm@0 79 &=& \langle \psi H | H\psi \rangle\\
rlm@0 80 &=& \langle \bar\psi | \bar\psi \rangle\\
rlm@0 81 &=& \int_0^a \left(\frac{A\hbar^2}{m}\right)^2\;dx\\
rlm@0 82 &=& \frac{A^2\hbar^4 a}{m^2}\\
rlm@0 83 \end{eqnarray}\)
rlm@0 84
rlm@0 85 For future reference, observe that this value is nonzero
rlm@0 86 (which makes sense).
rlm@0 87
rlm@0 88 ** Second method
rlm@0 89 We can also calculate the average energy-squared of $|\psi\rangle$ in the
rlm@0 90 following way.
rlm@0 91
rlm@0 92 \begin{eqnarray}
rlm@0 93 \langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\
rlm@0 94 &=& \langle \psi |H \bar\psi \rangle\\
rlm@0 95 &=&\int_0^a Ax(x-a)
rlm@0 96 \left(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\right)\frac{-A\hbar^2}{m}\;dx\\
rlm@0 97 &=& 0\quad (!)\\
rlm@0 98 \end{eqnarray}
rlm@0 99
rlm@0 100 The second-to-last term must be zero because the second derivative
rlm@0 101 of a constant \(\left(\frac{-A\hbar^2}{m}\right)\) is zero.
rlm@0 102
rlm@0 103 * What is the problem?
rlm@0 104
rlm@0 105 To recap: We used two different methods to calculate the average
rlm@0 106 energy-squared of a state $|\psi\rangle$. For the first method, we
rlm@0 107 used the fact that $H$ is a hermitian operator, replacing \(\langle
rlm@0 108 \psi H^\dagger | H \psi\rangle\) with \(\langle \psi H | H
rlm@0 109 \psi\rangle\). Using this substitution rule, we calculated the answer.
rlm@0 110
rlm@0 111 For the second method,
rlm@0 112 #we didn't use the fact that $H$ was hermitian;
rlm@0 113 we instead used the fact that we know how to represent $H$ and $\psi$
rlm@0 114 as functions of $x$: $H$ is a differential operator
rlm@0 115 \(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\) and $\psi$ is a quadratic
rlm@0 116 function of $x$. By applying $H$ to $\psi$, we took several
rlm@0 117 derivatives and arrived at our answer.
rlm@0 118
rlm@0 119 These two methods gave different results. In the following sections,
rlm@0 120 I'll describe and analyze the source of this difference.
rlm@0 121
rlm@0 122 ** Physical operators only act on physical wavefunctions
rlm@0 123 :PROPERTIES:
rlm@0 124 :ORDERED: t
rlm@0 125 :END:
rlm@0 126 #In quantum mechanics, an operator is a function that takes in a
rlm@0 127 #physical state and produces another physical state as ouput. Some
rlm@0 128 #operators correspond to physical quantities such as energy,
rlm@0 129 #momentum, or position; the mathematical properties of these operators correspond to
rlm@0 130 #physical properties of the system.
rlm@0 131
rlm@0 132 #Eigenstates are an example of this correspondence: an
rlm@0 133
rlm@0 134 Physical states are represented as wavefunctions in quantum
rlm@0 135 mechanics. Just as we disallow certain physically nonsensical states
rlm@0 136 in classical mechanics (for example, we consider it to be nonphysical
rlm@0 137 for an object to spontaneously disappear from one place and reappear
rlm@0 138 in another), we also disallow certain wavefunctions in quantum
rlm@0 139 mechanics.
rlm@0 140
rlm@0 141 For example, since wavefunctions are supposed to correspond to
rlm@0 142 probability amplitudes, we require wavefunctions to be normalized
rlm@0 143 \((\langle \psi |\psi \rangle = 1)\). Generally, we disallow
rlm@0 144 wavefunctions that do not satisfy this property (although there are
rlm@0 145 some exceptions[fn:2]).
rlm@0 146
rlm@0 147 As another example, we generally expect probability to vary smoothly\mdash{}if
rlm@0 148 a particle is very likely or very unlikely to be found at a particular
rlm@0 149 location, it should also be somewhat likely or somewhat unlikely to be
rlm@0 150 found /near/ that location. In more precise terms, we expect that for
rlm@0 151 physically meaningful wavefunctions, the probability
rlm@0 152 \(\text{Pr}(x)=\int^x \psi^*\psi\) should be a continuous function of
rlm@0 153 $x$ and, again, we disallow wavefunctions that do not satisfy this
rlm@0 154 property because we consider them to be physically nonsensical.
rlm@0 155
rlm@0 156 So, physical wavefunctions must satisfy certain properties
rlm@0 157 like the two just described. Wavefunctions that do not satisfy these properties are
rlm@0 158 rejected for being physically nonsensical: even though we can perform
rlm@0 159 calculations with them, the mathematical results we obtain do not mean
rlm@0 160 anything physically.
rlm@0 161
rlm@0 162 Now, in quantum mechanics, an *operator* is a function that converts
rlm@0 163 states into other states. Some operators correspond to
rlm@0 164 physical quantities such as energy, momentum, or position, and as a
rlm@0 165 result, the mathematical properties of these operators correspond to
rlm@0 166 physical properties of the system. Such operators are called
rlm@0 167 /hermitian operators/; one important property of hermitian operators
rlm@0 168 is this rule:
rlm@0 169
rlm@0 170 #+begin_quote
rlm@0 171 *Hermitian operator rule:* A hermitian operator must only operate on
rlm@0 172 the wavefunctions we have deemed physical, and must only produce
rlm@0 173 physical wavefunctions[fn:: If you require a hermitian operator to
rlm@0 174 have physical eigenstates (which is reasonable), you get a very strong result: you guarantee
rlm@0 175 that the operator will convert every physical wavefunction into
rlm@0 176 another physical wavefunction:
rlm@0 177
rlm@0 178 For any linear operator $\Omega$, the eigenvalue equation is
rlm@0 179 \(\Omega|\omega\rangle = \omega |\omega\rangle\). Notice that if an
rlm@0 180 eigenstate $|\omega\rangle$ is a physical wavefunction, the
rlm@0 181 eigenvalue equation forces $\Omega|\omega\rangle$ to be a
rlm@0 182 physical wavefunction as well. To elaborate, if the eigenstates of
rlm@0 183 $\Omega$ are physical functions, then $\Omega$ is guaranteed to
rlm@0 184 convert them into other physical functions. Even more is true if the
rlm@0 185 operator $\Omega$ is also hermitian: there is a theorem which states
rlm@0 186 that \ldquo{}If \Omega is hermitian, then every physical wavefunction
rlm@0 187 can be written as a weighted sum of eigenstates of \Omega.\rdquo{} This
rlm@0 188 theorem implies that /if $\Omega$ is hermitian/, and /if the eigenstates
rlm@0 189 of \Omega are physically allowed/, then \Omega is guaranteed to
rlm@0 190 convert every physically allowed wavefunction into another physically
rlm@0 191 allowed wavefunction.].
rlm@0 192 #+end_quote
rlm@0 193
rlm@0 194 As you can see, this rule comes in two pieces. The first part is a
rlm@0 195 constraint on *you*, the physicist: you must never feed a nonphysical
rlm@0 196 state into a Hermitian operator, as it may produce nonsense. The
rlm@0 197 second part is a constraint on the *operator*: the operator is
rlm@0 198 guaranteed only to produce physical wavefunctions.
rlm@0 199
rlm@0 200 In fact, this rule for hermitian operators is the source of our
rlm@0 201 problem, as we unknowingly violated it when applying our second
rlm@0 202 method!
rlm@0 203
rlm@0 204 ** The Hamiltonian is nonphysical
rlm@0 205 You'll remember that in the second method we had wavefunctions within
rlm@0 206 the well
rlm@0 207
rlm@0 208 \(
rlm@0 209 \begin{eqnarray}
rlm@0 210 \psi(x) &=& A\;x(x-a)\\
rlm@0 211 \bar\psi(x)&\equiv& H|\psi\rangle = \frac{-A\hbar^2}{m}\\
rlm@0 212 \end{eqnarray}
rlm@0 213 \)
rlm@0 214
rlm@0 215 Using this, we wrote
rlm@0 216
rlm@0 217
rlm@0 218 \(\begin{eqnarray}
rlm@0 219 \langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\
rlm@0 220 &=& \langle \psi |H \bar\psi \rangle\\
rlm@0 221 & \vdots&\\
rlm@0 222 &=& 0\\
rlm@0 223 \end{eqnarray}\)
rlm@0 224
rlm@0 225 However, there are two issues here. First, we were not allowed to operate with $H$ on the wavefunction
rlm@0 226 $|\bar \psi\rangle$, because $H$ is supposed to be a physical operator and $|\bar
rlm@0 227 \psi\rangle$ is a nonphysical state. Indeed, the constant function $|\bar\psi\rangle$ does
rlm@0 228 not approach zero at the edges of the well. By
rlm@0 229 feeding $H$ a nonphysical wavefunction, we obtained nonsensical
rlm@0 230 results.
rlm@0 231
rlm@0 232 Second, and more importantly, we were wrong to claim that $H$ was a
rlm@0 233 physical operator\mdash{}that $H$ was hermitian. According to the
rlm@0 234 rule, a hermitian operator must convert physical states into other
rlm@0 235 physical states. But $|\psi\rangle$ is a physical state, as we said
rlm@0 236 when we first introduced it \mdash{}it is a normalized, continuous
rlm@0 237 function which approaches zero at the edges of the well and doesn't
rlm@0 238 exist outside it. On the other hand, $|\bar\psi\rangle$ is nonphysical
rlm@0 239 because it does not go to zero at the edges of the well. It is
rlm@0 240 therefore impermissible for $H$ to transform the physical state
rlm@0 241 $|\psi\rangle$ into the nonphysical state $|\bar\psi\rangle$. Because
rlm@0 242 $H$ converts some physical states into nonphysical states, it cannot
rlm@0 243 be a hermitian operator as we assumed.
rlm@0 244
rlm@0 245 # Boundary conditions affect hermiticity
rlm@0 246 ** Boundary conditions alter hermiticity
rlm@0 247 It may surprise you (and it certainly surprised me) to find that the
rlm@0 248 Hamiltonian is not hermitian. One of the fundamental principles of
rlm@0 249 quantum mechanics is that hermitian operators correspond to physically
rlm@0 250 observable quantities; for this reason, surely the
rlm@0 251 Hamiltonian\mdash{}the operator corresponding to energy\mdash{}ought to be hermitian?
rlm@0 252
rlm@0 253 But we must understand the correspondence between physically
rlm@0 254 observable quantities and hermitian operators: every hermitian
rlm@0 255 operator corresponds to a physically observable quantity, but not
rlm@0 256 every quantity that intuitively \ldquo{}ought\rdquo{} to be observable
rlm@0 257 will correspond to a hermitian operator[fn::For a simple example,
rlm@0 258 consider the differential operator \(D=\frac{d}{dx}\); although our
rlm@0 259 intuitions might suggest that $D$ is observable which leads us to
rlm@0 260 guess that $D$ is hermitian, it isn't. Still, the very closely related
rlm@0 261 operator $P=i\hbar\frac{d}{dx}$ /is/ hermitian. The point is that we
rlm@0 262 ought to validate our intuitions by checking the definitions.]. The
rlm@0 263 true definition of a hermitian operator imply that the Hamiltonian
rlm@0 264 stops being hermitian in the infinitely deep well. Here we arrive at a
rlm@0 265 crucial point:
rlm@0 266
rlm@0 267 Operators do not change /form/ between problems: the one-dimensional
rlm@0 268 Hamiltonian is always $H=\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}+V$, the
rlm@0 269 one-dimensional canonical momentum is always $P=i\hbar\frac{d}{dx}$,
rlm@0 270 and so on.
rlm@0 271
rlm@0 272 However, operators do change in this respect: hermitian operators must
rlm@0 273 only take in physical states, and must only produce physical states; because
rlm@0 274 in different problems we /do/ change the requirements for being a
rlm@0 275 physical state, we also change what it takes for
rlm@0 276 an operator to be called hermitian. As a result, an operator that
rlm@0 277 is hermitian in one setting may fail to be hermitian in another.
rlm@0 278
rlm@0 279 Having seen how boundary conditions can affect hermiticity, we
rlm@0 280 ought to be extra careful about which conditions we impose on our
rlm@0 281 wavefunctions.
rlm@0 282
rlm@0 283 ** Choosing the right constraints
rlm@0 284
rlm@0 285 We have said already that physicists
rlm@0 286 require wavefunctions to satisfy certain properties in order to be
rlm@0 287 deemed \ldquo{}physical\rdquo{}. To be specific, wavefunctions in the
rlm@0 288 infinitely deep well
rlm@0 289 - Must be *normalizable*, because they correspond to
rlm@0 290 probability amplitudes.
rlm@0 291 - Must have *smoothly-varying probability*, because if a particle is very
rlm@0 292 likely to be at a location, it ought to be likely to be /near/
rlm@0 293 it as well.
rlm@0 294 - Must *not exist outside the well*, because it
rlm@0 295 would take an infinite amount of energy to do so.
rlm@0 296
rlm@0 297 These conditions are surely reasonable. However, physicists sometimes
rlm@0 298 assert that in order to satisfy the second and third conditions,
rlm@0 299 physical wavefunctions
rlm@0 300
rlm@0 301 - (?) Must *smoothly approach zero* towards the edges of the well.
rlm@0 302
rlm@0 303 This final constraint is our reason for rejecting $|\bar\psi\rangle$
rlm@0 304 as nonphysical and is consequently the reason why $H$ is not hermitian. If
rlm@0 305 we can convince ourselves that the final constraint is unnecessary,
rlm@0 306 $H$ may again be hermitian. This will satisfy our intuitions that the
rlm@0 307 energy operator /ought/ to be hermitian.
rlm@0 308
rlm@0 309 But in fact, we have the following mathematical observation to save
rlm@0 310 us: a function $f$ does not need to be continuous in order for the
rlm@0 311 integral \(\int^x f\) to be continuous. As a particularly relevant
rlm@0 312 example, you may now notice that the function $\bar\psi(x)$ is not
rlm@0 313 itself continuous, although the integral $\int_0^x \bar\psi$ /is/
rlm@0 314 continuous. Evidently, it doesn't matter that the wavefunction
rlm@0 315 $\bar\psi$ itself is not continuous; the probability corresponding to
rlm@0 316 $\bar\psi$ /does/ manage to vary continuously anyways. Because the
rlm@0 317 probability corresponding to $\bar\psi$ is the only aspect of
rlm@0 318 $\bar\psi$ which we can detect physically, we /can/ safely omit the
rlm@0 319 final constraint while keeping the other three.
rlm@0 320
rlm@0 321 ** Symmetric operators look like hermitian operators, but sometimes aren't.
rlm@0 322
rlm@0 323
rlm@0 324 #+end_quote
rlm@0 325 ** COMMENT Re-examining physical constraints
rlm@0 326
rlm@0 327 We have now discovered a flaw: when applied to the state
rlm@0 328 $|\psi\rangle$, the second method violates the rule that physical
rlm@0 329 operators must only take in physical states and must only produce
rlm@0 330 physical states. Let's examine the problem more closely.
rlm@0 331
rlm@0 332 We have said already that physicists require wavefunctions to satisfy
rlm@0 333 certain properties in order to be deemed \ldquo{}physical\rdquo{}. To
rlm@0 334 be specific, wavefunctions in the infinitely deep well
rlm@0 335 - Must be *normalizable*, because they correspond to
rlm@0 336 probability amplitudes.
rlm@0 337 - Must have *smoothly-varying probability*, because if a particle is very
rlm@0 338 likely to be at a location, it ought to be likely to be /near/
rlm@0 339 it as well.
rlm@0 340 - Must *not exist outside the well*, because it
rlm@0 341 would take an infinite amount of energy to do so.
rlm@0 342
rlm@0 343 We now have discovered an important flaw in the second method: when
rlm@0 344 applied to the state $|\bar\psi\rangle$, the second method violates
rlm@0 345 the rule that physical operators must only take in
rlm@0 346 physical states and must only produce physical states. The problem is
rlm@0 347 even more serious, however
rlm@0 348
rlm@0 349
rlm@0 350
rlm@0 351 [fn:1] I'm defining a new variable just to make certain expressions
rlm@0 352 look shorter; this cannot affect the content of the answer we'll
rlm@0 353 get.
rlm@0 354
rlm@0 355 [fn:2] For example, in vaccuum (i.e., when the potential of the
rlm@0 356 physical system is $V(x)=0$ throughout all space), the momentum
rlm@0 357 eigenstates are not normalizable\mdash{}the relevant integral blows
rlm@0 358 up to infinity instead of converging to a number. Physicists modify
rlm@0 359 the definition of normalization slightly so that
rlm@0 360 \ldquo{}delta-normalizable \rdquo{} functions like these are included
rlm@0 361 among the physical wavefunctions.
rlm@0 362
rlm@0 363
rlm@0 364
rlm@6 365 * COMMENT : What I thought I knew
rlm@0 366
rlm@0 367 The following is a list of things I thought were true of quantum
rlm@0 368 mechanics; the catch is that the list contradicts itself.
rlm@0 369
rlm@0 370 1. For any hermitian operator: Eigenstates with different eigenvalues are orthogonal.
rlm@0 371 2. For any hermitian operator: Any physically allowed state can be
rlm@0 372 written as a linear sum of eigenstates of the operator.
rlm@0 373 3. The momentum operator and energy operator are hermitian, because
rlm@0 374 momentum and energy are measureable quantities.
rlm@0 375 4. In the vacuum potential, the momentum and energy operators have these eigenstates:
rlm@0 376 - the momentum operator has an eigenstate
rlm@0 377 \(p(x)=\exp{(ipx/\hbar)}\) for each value of $p$.
rlm@0 378 - the energy operator has an eigenstate \(|E\rangle =
rlm@0 379 \alpha|p\rangle + \beta|-p\rangle\) for any \(\alpha,\beta\) and
rlm@0 380 the particular choice of momentum $p=\sqrt{2mE}$.
rlm@0 381 5. In the infinitely deep potential well, the momentum and energy
rlm@0 382 operators have these eigenstates:
rlm@0 383 - The momentum eigenstates and energy eigenstates have the same form
rlm@0 384 as in the vacuum potential: $p(x) =
rlm@0 385 \exp{(ipx/\hbar)}$ and $|E\rangle = \alpha|p\rangle + \beta|-p\rangle$.
rlm@0 386 - Even so, because of the boundary conditions on the
rlm@0 387 well, we must make the following modifications:
rlm@0 388 + Physically realistic states must be impossible to find outside the well. (Only a state of infinite
rlm@0 389 energy could exist outside the well, and infinite energy is not
rlm@0 390 realistic.) This requirement means, for example, that momentum
rlm@0 391 eigenstates in the infinitely deep well must be
rlm@0 392 \(p(x)
rlm@0 393 = \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;
rlm@0 394 \\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
rlm@0 395 + Physically realistic states must vary smoothly throughout
rlm@0 396 space. This means that if a particle in some state is very unlikely to be
rlm@0 397 /at/ a particular location, it is also very unlikely be /near/
rlm@0 398 that location. Combining this requirement with the above
rlm@0 399 requirement, we find that the momentum operator no longer has
rlm@0 400 an eigenstate for each value of $p$; instead, only values of
rlm@0 401 $p$ that are integer multiples of $\pi \hbar/a$ are physically
rlm@0 402 realistic. Similarly, the energy operator no longer has an
rlm@0 403 eigenstate for each value of $E$; instead, the only energy
rlm@0 404 eigenstates in the infinitely deep well
rlm@0 405 are $E_n(x)=\sin(n\pi x/ a)$ for positive integers $n$.
rlm@0 406
rlm@6 407 * COMMENT :
rlm@0 408
rlm@0 409 ** Eigenstates with different eigenvalues are orthogonal
rlm@0 410
rlm@0 411 #+begin_quote
rlm@0 412 *Theorem:* Eigenstates with different eigenvalues are orthogonal.
rlm@0 413 #+end_quote
rlm@0 414
rlm@0 415 ** COMMENT :
rlm@0 416 I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$
rlm@0 417 and $|b\rangle$ are eigenstates of $\Lambda$. This means that
rlm@0 418
rlm@0 419
rlm@0 420 \(
rlm@0 421 \begin{eqnarray}
rlm@0 422 \Lambda |a\rangle&=& a|a\rangle,\\
rlm@0 423 \Lambda|b\rangle&=& b|b\rangle.\\
rlm@0 424 \end{eqnarray}
rlm@0 425 \)
rlm@0 426
rlm@0 427 If we take the difference of these eigenstates, we find that
rlm@0 428
rlm@0 429 \(
rlm@0 430 \begin{eqnarray}
rlm@0 431 \Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle
rlm@0 432 \qquad \text{(because $\Lambda$ is linear.)}\\
rlm@0 433 &=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and
rlm@0 434 $|b\rangle$ are eigenstates of $\Lambda$)}
rlm@0 435 \end{eqnarray}\)
rlm@0 436
rlm@0 437
rlm@0 438 which means that $a\neq b$.
rlm@0 439
rlm@0 440 ** Eigenvectors of hermitian operators span the space of solutions
rlm@0 441
rlm@0 442 #+begin_quote
rlm@0 443 *Theorem:* If $\Omega$ is a hermitian operator, then every physically
rlm@0 444 allowed state can be written as a linear sum of eigenstates of
rlm@0 445 $\Omega$.
rlm@0 446 #+end_quote
rlm@0 447
rlm@0 448
rlm@0 449
rlm@0 450 ** Momentum and energy are hermitian operators
rlm@0 451 This ought to be true because hermitian operators correspond to
rlm@0 452 observable quantities. Since we expect momentum and energy to be
rlm@0 453 measureable quantities, we expect that there are hermitian operators
rlm@0 454 to represent them.
rlm@0 455
rlm@0 456
rlm@0 457 ** Momentum and energy eigenstates in vacuum
rlm@0 458 An eigenstate of the momentum operator $P$ would be a state
rlm@0 459 \(|p\rangle\) such that \(P|p\rangle=p|p\rangle\).
rlm@0 460
rlm@0 461 ** Momentum and energy eigenstates in the infinitely deep well
rlm@0 462
rlm@0 463
rlm@0 464
rlm@0 465 * COMMENT Can you measure momentum in the infinitely deep well?
rlm@0 466 In summary, I thought I knew:
rlm@0 467 1. For any hermitian operator: eigenstates with different eigenvalues
rlm@0 468 are orthogonal.
rlm@0 469 2. For any hermitian operator: any physically realistic state can be
rlm@0 470 written as a linear sum of eigenstates of the operator.
rlm@0 471 3. The momentum operator and energy operator are hermitian, because
rlm@0 472 momentum and energy are observable quantities.
rlm@0 473 4. (The form of the momentum and energy eigenstates in the vacuum potential)
rlm@0 474 5. (The form of the momentum and energy eigenstates in the infinitely deep well potential)
rlm@0 475
rlm@0 476 Additionally, I understood that because the infinitely deep potential
rlm@0 477 well is not realistic, states of such a system are not necessarily
rlm@0 478 physically realistic. Instead, I understood
rlm@0 479 \ldquo{}realistic states\rdquo{} to be those that satisfy the physically
rlm@0 480 unrealistic Schr\ouml{}dinger equation and its boundary conditions.
rlm@0 481
rlm@0 482 With that final caveat, here is the problem:
rlm@0 483
rlm@0 484 According to (5), the momentum eigenstates in the well are
rlm@0 485
rlm@0 486 \(p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
rlm@0 487
rlm@0 488 (Additionally, we require that $p$ be an integer multiple of $\pi\hbar/a$.)
rlm@0 489
rlm@0 490 However, /these/ states are not orthogonal, which contradicts the
rlm@0 491 assumption that (3) the momentum operator is hermitian and (2)
rlm@0 492 eigenstates of a hermitian are orthogonal if they have different eigenvalues.
rlm@0 493
rlm@0 494 #+begin_quote
rlm@0 495 *Problem 1. The momentum eigenstates of the well are not orthogonal*
rlm@0 496
rlm@0 497 /Proof./ If $p_1\neq p_2$, then
rlm@0 498
rlm@0 499 \(\begin{eqnarray}
rlm@0 500 \langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)\,dx\\
rlm@0 501 &=& \int_0^a p_1^*(x)p_2(x)\,dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{
rlm@0 502 outside the well.}\\
rlm@0 503 &=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)\,dx}}\\
rlm@0 504 &=& \int_0^a \exp{\left[i(p_1-p_2)x/\hbar\right]}\,dx\\
rlm@0 505 \end{eqnarray}\)
rlm@0 506 $\square$
rlm@0 507
rlm@0 508 #+end_quote
rlm@0 509
rlm@0 510
rlm@0 511
rlm@0 512 ** COMMENT Momentum eigenstates
rlm@0 513
rlm@0 514 In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the
rlm@0 515 momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\).
rlm@0 516
rlm@0 517 In the infinitely deep potential well, the Hamiltonian is the same but
rlm@0 518 there is a new condition in order for states to qualify as physically
rlm@0 519 allowed: the states must not exist anywhere outside of well, as it
rlm@0 520 takes an infinite amount of energy to do so.
rlm@0 521
rlm@0 522 Notice that the momentum eigenstates defined above do /not/ satisfy
rlm@0 523 this condition.
rlm@0 524
rlm@0 525
rlm@0 526
rlm@0 527 * COMMENT
rlm@0 528 For each physical system, there is a Schr\ouml{}dinger equation that
rlm@0 529 describes how a particle's state $|\psi\rangle$ will change over
rlm@0 530 time.
rlm@0 531
rlm@0 532 \(\begin{eqnarray}
rlm@0 533 i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
rlm@0 534 H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
rlm@0 535
rlm@0 536 This is a differential equation; each solution to the
rlm@0 537 Schr\ouml{}dinger equation is a state that is physically allowed for
rlm@0 538 our particle. Here, physically allowed states are
rlm@0 539 those that change in physically allowed ways. However, like any differential
rlm@0 540 equation, the Schr\ouml{}dinger equation can be accompanied by
rlm@0 541 /boundary conditions/\mdash{}conditions that further restrict which
rlm@0 542 states qualify as physically allowed.
rlm@0 543
rlm@0 544
rlm@0 545
rlm@0 546
rlm@0 547 ** Eigenstates of momentum
rlm@0 548
rlm@0 549
rlm@0 550
rlm@0 551
rlm@0 552 #In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger
rlm@0 553
rlm@0 554 #\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\)
rlm@0 555
rlm@0 556
rlm@0 557
rlm@0 558
rlm@0 559
rlm@0 560
rlm@0 561
rlm@0 562 * COMMENT
rlm@0 563
rlm@0 564 #* The infinite square well potential
rlm@0 565
rlm@0 566 A particle exists in a potential that is
rlm@0 567 infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the
rlm@0 568 particle exists in a potential[fn:coords][fn:infinity]
rlm@0 569
rlm@0 570
rlm@0 571 \(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for
rlm@0 572 }\;x<0\text{ or }x>a.\end{cases}\)
rlm@0 573
rlm@0 574 The Schr\ouml{}dinger equation describes how the particle's state
rlm@0 575 \(|\psi\rangle\) will change over time in this system.
rlm@0 576
rlm@0 577 \(\begin{eqnarray}
rlm@0 578 i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
rlm@0 579 H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
rlm@0 580
rlm@0 581 This is a differential equation; each solution to the
rlm@0 582 Schr\ouml{}dinger equation is a state that is physically allowed for
rlm@0 583 our particle. Here, physically allowed states are
rlm@0 584 those that change in physically allowed ways. However, like any differential
rlm@0 585 equation, the Schr\ouml{}dinger equation can be accompanied by
rlm@0 586 /boundary conditions/\mdash{}conditions that further restrict which
rlm@0 587 states qualify as physically allowed.
rlm@0 588
rlm@0 589
rlm@0 590 Whenever possible, physicists impose these boundary conditions:
rlm@0 591 - A physically allowed state ought to be a /smoothly-varying function of position./ This means
rlm@0 592 that if a particle in the state is likely to be /at/ a particular location,
rlm@0 593 it is also likely to be /near/ that location.
rlm@0 594
rlm@0 595 These boundary conditions imply that for the square well potential in
rlm@0 596 this problem,
rlm@0 597
rlm@0 598 - Physically allowed states must be totally confined to the well,
rlm@0 599 because it takes an infinite amount of energy to exist anywhere
rlm@0 600 outside of the well (and physically allowed states ought to have
rlm@0 601 only finite energy).
rlm@0 602 - Physically allowed states must be increasingly unlikely to find very
rlm@0 603 close to the walls of the well. This is because of two conditions: the above
rlm@0 604 condition says that the particle is /impossible/ to find
rlm@0 605 outside of the well, and the smoothly-varying condition says
rlm@0 606 that if a particle is impossible to find at a particular location,
rlm@0 607 it must be unlikely to be found nearby that location.
rlm@0 608
rlm@0 609 #; physically allowed states are those that change in physically
rlm@0 610 #allowed ways.
rlm@0 611
rlm@0 612
rlm@0 613 #** Boundary conditions
rlm@0 614 Because the potential is infinite everywhere except within the well,
rlm@0 615 a realistic particle must be confined to exist only within the
rlm@0 616 well\mdash{}its wavefunction must be zero everywhere beyond the walls
rlm@0 617 of the well.
rlm@0 618
rlm@0 619
rlm@0 620 [fn:coords] I chose my coordinate system so that the well extends from
rlm@0 621 \(0<x<a\). Others choose a coordinate system so that the well extends from
rlm@0 622 \(-\frac{a}{2}<x<\frac{a}{2}\). Although both coordinate systems describe the same physical
rlm@0 623 situation, they give different-looking answers.
rlm@0 624
rlm@0 625 [fn:infinity] Of course, infinite potentials are not
rlm@0 626 realistic. Instead, they are useful approximations to finite
rlm@0 627 potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
rlm@0 628 of the well\rdquo{} are close enough for your own practical
rlm@0 629 purposes. Having introduced a physical impossibility into the problem
rlm@0 630 already, we don't expect to get physically realistic solutions; we
rlm@0 631 just expect to get mathematically consistent ones. The forthcoming
rlm@0 632 trouble is that we don't.