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     1.4 +#+TITLE: Bugs in quantum mechanics
     1.5 +#+AUTHOR: Dylan Holmes
     1.6 +#+DESCRIPTION: I explain hermitian operators and use them to solve a confusing problem in quantum mechanics.
     1.7 +#+KEYWORDS: quantum mechanics, self-adjoint, hermitian, square well, momentum
     1.8 +#+SETUPFILE: ../../aurellem/org/setup.org
     1.9 +#+INCLUDE:   ../../aurellem/org/level-0.org
    1.10 +
    1.11 +
    1.12 +
    1.13 +#Bugs in Quantum Mechanics
    1.14 +#Bugs in the Quantum-Mechanical Momentum Operator
    1.15 +
    1.16 +
    1.17 +I studied quantum mechanics the same way I study most subjects\mdash{}
    1.18 +by collecting (and squashing) bugs in my understanding. One of these
    1.19 +bugs persisted throughout two semesters of
    1.20 +quantum mechanics coursework until I finally found
    1.21 +the paper 
    1.22 +[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
    1.23 +mechanics/]], which helped me stamp out the bug entirely. I decided to
    1.24 +write an article about the problem and its solution for a number of reasons:
    1.25 +
    1.26 +- Although the paper was not unreasonably dense, it was written for
    1.27 +  teachers. I wanted to write an article for students.
    1.28 +- I wanted to popularize the problem and its solution because other
    1.29 +  explanations are currently too hard to find. (Even Shankar's
    1.30 +  excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it, as far as I know.)
    1.31 +- Attempting an explanation is my way of making
    1.32 +  sure that the bug really /is/ gone.
    1.33 +# entirely eradicated.
    1.34 +
    1.35 +* COMMENT
    1.36 + I recommend the
    1.37 +paper not only for students who are learning
    1.38 +quantum mechanics, but especially for teachers interested in debugging
    1.39 +them. 
    1.40 +
    1.41 +* COMMENT
    1.42 +On my first exam in quantum mechanics, my professor asked us to
    1.43 +describe how certain measurements would affect a particle in a
    1.44 +box. Many of these measurement questions required routine application
    1.45 +of skills we had recently learned\mdash{}first, you recall (or
    1.46 +calculate) the eigenstates of the quantity
    1.47 +to be measured; second, you write the given state as a linear
    1.48 +sum of these eigenstates\mdash{} the coefficients on each term give
    1.49 +the probability amplitude.
    1.50 +
    1.51 +
    1.52 +* Two methods of calculation that give different results.
    1.53 +
    1.54 +In the infinitely deep well, there is a particle in the the
    1.55 +normalized state
    1.56 +
    1.57 + \(\psi(x) = \begin{cases}A\; x(x-a)&\text{for }0\lt{}x\lt{}a;\\0,&\text{for }x\lt{}0\text{ or }x>a.\end{cases}\) 
    1.58 +
    1.59 +This is apparently a perfectly respectable state: it is normalized ($A$ is a
    1.60 +normalization constant), it is zero
    1.61 +everywhere outside of the well, and it is moreover continuous.
    1.62 +
    1.63 +Even so, we will find a problem if we attempt to calculate the average
    1.64 +energy-squared of this state (that is, the quantity \(\langle \psi | H^2 | \psi \rangle\)). 
    1.65 +
    1.66 +** First method
    1.67 +
    1.68 +For short, define a new variable[fn:1] \(|\bar \psi\rangle \equiv
    1.69 +H|\psi\rangle\). We can express the state $|\bar\psi\rangle$ as a
    1.70 +function of $x$ because we know how to express $H$ and $\psi$ in terms
    1.71 +of  $x$: $H$ is the differential operator $\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}$, and
    1.72 +$\psi(x)$ is the function defined above. So, we get \(\quad\bar\psi(x) = \frac{-A\hbar^2}{m}\). For future reference, observe that $\bar\psi(x)$
    1.73 +is constant.
    1.74 +
    1.75 +Having introduced $|\bar\psi\rangle$, we can calculate the average energy-squared of $|\psi\rangle$ in the
    1.76 +following way.
    1.77 +
    1.78 +\(\begin{eqnarray}
    1.79 +\langle \psi | H^2 |\psi\rangle &=& \langle \psi H^\dagger | H
    1.80 +\psi\rangle\\
    1.81 +&=& \langle \psi H | H\psi \rangle\\
    1.82 +&=& \langle \bar\psi | \bar\psi \rangle\\
    1.83 +&=& \int_0^a \left(\frac{A\hbar^2}{m}\right)^2\;dx\\
    1.84 +&=& \frac{A^2\hbar^4 a}{m^2}\\
    1.85 +\end{eqnarray}\)
    1.86 + 
    1.87 +For future reference, observe that this value is  nonzero
    1.88 +(which makes sense).
    1.89 +
    1.90 +** Second method
    1.91 +We can also calculate the average energy-squared of $|\psi\rangle$ in the
    1.92 +following way.
    1.93 +
    1.94 +\begin{eqnarray}
    1.95 +\langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\
    1.96 +&=& \langle \psi |H \bar\psi \rangle\\
    1.97 +&=&\int_0^a Ax(x-a)
    1.98 +\left(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\right)\frac{-A\hbar^2}{m}\;dx\\
    1.99 +&=& 0\quad (!)\\
   1.100 +\end{eqnarray}
   1.101 +
   1.102 +The second-to-last term must be zero because the second derivative
   1.103 +of a constant \(\left(\frac{-A\hbar^2}{m}\right)\) is zero.
   1.104 +
   1.105 +* What is the problem?
   1.106 +
   1.107 +To recap: We used two different methods to calculate the average
   1.108 +energy-squared of a state $|\psi\rangle$. For the first method, we
   1.109 +used the fact that $H$ is a hermitian operator, replacing \(\langle
   1.110 +\psi H^\dagger | H \psi\rangle\) with \(\langle \psi H | H
   1.111 +\psi\rangle\). Using this substitution rule, we calculated the answer.
   1.112 +
   1.113 +For the second method, 
   1.114 +#we didn't use the fact that $H$ was hermitian;
   1.115 +we instead used the fact that we know how to represent $H$ and $\psi$
   1.116 +as functions of $x$: $H$ is a differential operator
   1.117 +\(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\) and $\psi$ is a quadratic
   1.118 +function of $x$. By applying $H$ to $\psi$, we took several
   1.119 +derivatives and arrived at our answer.
   1.120 +
   1.121 +These two methods gave different results. In the following sections,
   1.122 +I'll describe and analyze the source of this difference.
   1.123 +
   1.124 +** Physical operators only act on physical wavefunctions
   1.125 +   :PROPERTIES:
   1.126 +   :ORDERED:  t
   1.127 +   :END:
   1.128 +#In quantum mechanics, an operator is a function that takes in a
   1.129 +#physical state and produces another physical state as ouput. Some
   1.130 +#operators correspond to physical quantities such as energy,
   1.131 +#momentum, or position; the mathematical properties of these operators correspond to
   1.132 +#physical properties of the system.
   1.133 +
   1.134 +#Eigenstates are an example of this correspondence: an 
   1.135 +
   1.136 +Physical states are represented as wavefunctions in quantum
   1.137 +mechanics. Just as we disallow certain physically nonsensical states
   1.138 +in classical mechanics (for example, we consider it to be nonphysical
   1.139 +for an object to spontaneously disappear from one place and reappear
   1.140 +in another), we also disallow certain wavefunctions in quantum
   1.141 +mechanics.
   1.142 +
   1.143 +For example, since wavefunctions are supposed to correspond to
   1.144 +probability amplitudes, we require wavefunctions to be normalized
   1.145 +\((\langle \psi |\psi \rangle = 1)\). Generally, we disallow
   1.146 +wavefunctions that do not satisfy this property (although there are
   1.147 +some exceptions[fn:2]).
   1.148 +
   1.149 +As another example, we generally expect probability to vary smoothly\mdash{}if
   1.150 +a particle is very likely or very unlikely to be found at a particular
   1.151 +location, it should also be somewhat likely or somewhat unlikely to be
   1.152 +found /near/ that location. In more precise terms, we expect that for
   1.153 +physically meaningful wavefunctions, the probability 
   1.154 +\(\text{Pr}(x)=\int^x \psi^*\psi\) should be a continuous function of
   1.155 +$x$ and, again, we disallow wavefunctions that do not satisfy this
   1.156 +property because we consider them to be physically nonsensical.
   1.157 +
   1.158 +So, physical wavefunctions must satisfy certain properties
   1.159 +like the two just described. Wavefunctions that do not satisfy these properties are
   1.160 +rejected for being physically nonsensical: even though we can perform
   1.161 +calculations with them, the mathematical results we obtain do not mean
   1.162 +anything physically.
   1.163 +
   1.164 +Now, in quantum mechanics, an *operator* is a function that converts
   1.165 +states into other states. Some operators correspond to
   1.166 +physical quantities such as energy, momentum, or position, and as a
   1.167 +result, the mathematical properties of these operators correspond to
   1.168 +physical properties of the system. Such operators are called
   1.169 +/hermitian operators/; one important property of hermitian operators
   1.170 +is this rule: 
   1.171 +
   1.172 +#+begin_quote 
   1.173 +*Hermitian operator rule:* A hermitian operator must only operate on
   1.174 +the wavefunctions we have deemed physical, and must only produce
   1.175 +physical wavefunctions[fn:: If you require a hermitian operator to
   1.176 +have physical eigenstates (which is reasonable), you get a very strong result: you guarantee
   1.177 +that the operator will convert every physical wavefunction into
   1.178 +another physical wavefunction: 
   1.179 +
   1.180 +  For any linear operator $\Omega$, the eigenvalue equation is
   1.181 +\(\Omega|\omega\rangle = \omega |\omega\rangle\). Notice that if an
   1.182 +eigenstate $|\omega\rangle$ is a physical wavefunction, the
   1.183 +eigenvalue equation forces $\Omega|\omega\rangle$ to be a
   1.184 +physical wavefunction as well. To elaborate, if the eigenstates of
   1.185 +$\Omega$ are physical functions, then $\Omega$ is guaranteed to
   1.186 +convert them into other physical functions.  Even more is true if the
   1.187 +operator $\Omega$ is also hermitian: there is a theorem which states
   1.188 +that \ldquo{}If \Omega is hermitian, then every physical wavefunction
   1.189 +can be written as a weighted sum of eigenstates of \Omega.\rdquo{} This
   1.190 +theorem implies that /if $\Omega$ is hermitian/, and /if the eigenstates
   1.191 +of \Omega are physically allowed/, then \Omega is guaranteed to
   1.192 +convert every physically allowed wavefunction into another physically
   1.193 +allowed wavefunction.].
   1.194 +#+end_quote
   1.195 +
   1.196 +As you can see, this rule comes in two pieces. The first part is a
   1.197 +constraint on *you*, the physicist: you must never feed a nonphysical
   1.198 +state into a Hermitian operator, as it may produce nonsense. The
   1.199 +second part is a constraint on the *operator*: the operator is
   1.200 +guaranteed only to produce physical wavefunctions.
   1.201 +
   1.202 +In fact, this rule for hermitian operators is the source of our
   1.203 +problem, as we unknowingly violated it when applying our second
   1.204 +method!
   1.205 +
   1.206 +** The Hamiltonian is nonphysical 
   1.207 +You'll remember that in the second method we had wavefunctions within
   1.208 +the well
   1.209 +
   1.210 +\(
   1.211 +\begin{eqnarray}
   1.212 +\psi(x) &=& A\;x(x-a)\\
   1.213 +\bar\psi(x)&\equiv& H|\psi\rangle = \frac{-A\hbar^2}{m}\\
   1.214 +\end{eqnarray}
   1.215 +\)
   1.216 +
   1.217 + Using this, we wrote
   1.218 +
   1.219 +
   1.220 +\(\begin{eqnarray}
   1.221 +\langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\
   1.222 +&=& \langle \psi |H \bar\psi \rangle\\
   1.223 +& \vdots&\\
   1.224 +&=& 0\\
   1.225 +\end{eqnarray}\)
   1.226 +
   1.227 +However, there are two issues here. First, we were not allowed to operate with $H$ on the wavefunction
   1.228 +$|\bar \psi\rangle$, because $H$ is supposed to be a physical operator and $|\bar
   1.229 +\psi\rangle$ is a nonphysical state. Indeed, the constant function $|\bar\psi\rangle$ does
   1.230 +not approach zero at the edges of the well. By
   1.231 +feeding $H$ a nonphysical wavefunction, we obtained nonsensical
   1.232 +results.
   1.233 +
   1.234 +Second, and more importantly, we were wrong to claim that $H$ was a
   1.235 +physical operator\mdash{}that $H$ was hermitian. According to the
   1.236 +rule, a hermitian operator must convert physical states into other
   1.237 +physical states. But $|\psi\rangle$ is a physical state, as we said
   1.238 +when we first introduced it \mdash{}it is a normalized, continuous
   1.239 +function which approaches zero at the edges of the well and doesn't
   1.240 +exist outside it. On the other hand, $|\bar\psi\rangle$ is nonphysical
   1.241 +because it does not go to zero at the edges of the well. It is
   1.242 +therefore impermissible for $H$ to transform the physical state
   1.243 +$|\psi\rangle$ into the nonphysical state $|\bar\psi\rangle$.  Because
   1.244 +$H$ converts some physical states into nonphysical states, it cannot
   1.245 +be a hermitian operator as we assumed.
   1.246 +
   1.247 +# Boundary conditions affect hermiticity
   1.248 +** Boundary conditions alter hermiticity
   1.249 +It may surprise you (and it certainly surprised me) to find that the
   1.250 +Hamiltonian is not hermitian. One of the fundamental principles of
   1.251 +quantum mechanics is that hermitian operators correspond to physically
   1.252 +observable quantities; for this reason, surely the
   1.253 +Hamiltonian\mdash{}the operator corresponding to energy\mdash{}ought to be hermitian?
   1.254 +
   1.255 +But we must understand the correspondence between physically
   1.256 +observable quantities and hermitian operators: every hermitian
   1.257 +operator corresponds to a physically observable quantity, but not
   1.258 +every quantity that intuitively \ldquo{}ought\rdquo{} to be observable
   1.259 +will correspond to a hermitian operator[fn::For a simple example,
   1.260 +consider the differential operator \(D=\frac{d}{dx}\); although our
   1.261 +intuitions might suggest that $D$ is observable which leads us to
   1.262 +guess that $D$ is hermitian, it isn't. Still, the very closely related
   1.263 +operator $P=i\hbar\frac{d}{dx}$ /is/ hermitian. The point is that we
   1.264 +ought to validate our intuitions by checking the definitions.]. The
   1.265 +true definition of a hermitian operator imply that the Hamiltonian
   1.266 +stops being hermitian in the infinitely deep well. Here we arrive at a
   1.267 +crucial point:
   1.268 +
   1.269 +Operators do not change /form/ between problems: the one-dimensional
   1.270 +Hamiltonian is always $H=\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}+V$, the
   1.271 +one-dimensional canonical momentum is always $P=i\hbar\frac{d}{dx}$,
   1.272 +and so on.
   1.273 +
   1.274 +However, operators do change in this respect: hermitian operators must
   1.275 +only take in physical states, and must only produce physical states; because
   1.276 +in different problems we /do/ change the requirements for being a
   1.277 +physical state, we also change what it takes for
   1.278 +an operator to be called hermitian.  As a result, an operator that
   1.279 +is hermitian in one setting may fail to be hermitian in another.
   1.280 +
   1.281 +Having seen how boundary conditions can affect hermiticity, we
   1.282 +ought to be extra careful about which conditions we impose on our
   1.283 +wavefunctions.
   1.284 +
   1.285 +** Choosing the right constraints 
   1.286 +
   1.287 + We have said already that physicists
   1.288 +require wavefunctions to satisfy certain properties in order to be
   1.289 +deemed \ldquo{}physical\rdquo{}. To be specific, wavefunctions in the
   1.290 +infinitely deep well
   1.291 +- Must be *normalizable*, because they correspond to
   1.292 +  probability amplitudes.
   1.293 +- Must have *smoothly-varying probability*, because if a particle is very
   1.294 +  likely to be at a location, it ought to be likely to be /near/
   1.295 +  it as well.
   1.296 +- Must *not exist outside the well*, because it
   1.297 +  would take an infinite amount of energy to do so.
   1.298 +
   1.299 +These conditions are surely reasonable. However, physicists sometimes
   1.300 +assert that in order to satisfy the second and third conditions,
   1.301 +physical wavefunctions
   1.302 +
   1.303 +- (?) Must *smoothly approach zero* towards the edges of the well.
   1.304 +
   1.305 +This final constraint is our reason for rejecting $|\bar\psi\rangle$
   1.306 +as nonphysical and is consequently the reason why $H$ is not hermitian. If
   1.307 +we can convince ourselves that the final constraint is unnecessary,
   1.308 +$H$ may again be hermitian. This will satisfy our intuitions that the
   1.309 +energy operator /ought/ to be hermitian.
   1.310 +
   1.311 +But in fact, we have the following mathematical observation to save
   1.312 +us: a function $f$ does not need to be continuous in order for the
   1.313 +integral \(\int^x f\) to be continuous. As a particularly relevant
   1.314 +example, you may now notice that the function $\bar\psi(x)$ is not
   1.315 +itself continuous, although the integral $\int_0^x \bar\psi$ /is/
   1.316 +continuous. Evidently, it doesn't matter that the wavefunction
   1.317 +$\bar\psi$ itself is not continuous; the probability corresponding to
   1.318 +$\bar\psi$ /does/ manage to vary continuously anyways. Because the
   1.319 +probability corresponding to $\bar\psi$ is the only aspect of
   1.320 +$\bar\psi$ which we can detect physically, we /can/ safely omit the
   1.321 +final constraint while keeping the other three.
   1.322 +
   1.323 +** Symmetric operators look like hermitian operators, but sometimes aren't.
   1.324 +
   1.325 +
   1.326 +#+end_quote
   1.327 +** COMMENT Re-examining physical constraints
   1.328 +
   1.329 +We have now discovered a flaw: when applied to the state
   1.330 +$|\psi\rangle$, the second method violates the rule that physical
   1.331 +operators must only take in physical states and must only produce
   1.332 +physical states. Let's examine the problem more closely.
   1.333 +
   1.334 +We have said already that physicists require wavefunctions to satisfy
   1.335 +certain properties in order to be deemed \ldquo{}physical\rdquo{}. To
   1.336 +be specific, wavefunctions in the infinitely deep well
   1.337 +- Must be *normalizable*, because they correspond to
   1.338 +  probability amplitudes.
   1.339 +- Must have *smoothly-varying probability*, because if a particle is very
   1.340 +  likely to be at a location, it ought to be likely to be /near/
   1.341 +  it as well.
   1.342 +- Must *not exist outside the well*, because it
   1.343 +  would take an infinite amount of energy to do so.
   1.344 +
   1.345 +We now have discovered an important flaw in the second method: when
   1.346 +applied to the state $|\bar\psi\rangle$, the second method violates
   1.347 +the rule that physical operators must only take in
   1.348 +physical states and must only produce physical states. The problem is
   1.349 +even more serious, however
   1.350 +
   1.351 +
   1.352 +
   1.353 +[fn:1] I'm defining a new variable just to make certain expressions
   1.354 +  look shorter; this cannot affect the content of the answer we'll
   1.355 +  get. 
   1.356 +
   1.357 +[fn:2] For example, in vaccuum (i.e., when the potential of the
   1.358 +  physical system is $V(x)=0$ throughout all space), the momentum
   1.359 +  eigenstates are not normalizable\mdash{}the relevant integral blows
   1.360 +  up to infinity instead of converging to a number. Physicists modify
   1.361 +  the definition of normalization slightly so that
   1.362 +  \ldquo{}delta-normalizable \rdquo{} functions like these are included
   1.363 +  among the physical wavefunctions.
   1.364 +
   1.365 +
   1.366 +
   1.367 +* COMMENT: What I thought I knew
   1.368 +
   1.369 +The following is a list of things I thought were true of quantum
   1.370 +mechanics; the catch is that the list contradicts itself.
   1.371 +
   1.372 +1. For any hermitian operator: Eigenstates with different eigenvalues are orthogonal.
   1.373 +2. For any hermitian operator: Any physically allowed state can be
   1.374 +   written as a linear sum of eigenstates of the operator.
   1.375 +3. The momentum operator and energy operator are hermitian, because
   1.376 +   momentum and energy are measureable quantities.
   1.377 +4. In the vacuum potential, the momentum and energy operators have these eigenstates:
   1.378 +   - the momentum operator has an eigenstate
   1.379 +     \(p(x)=\exp{(ipx/\hbar)}\) for each value of $p$.
   1.380 +   - the energy operator has an eigenstate \(|E\rangle =
   1.381 +     \alpha|p\rangle + \beta|-p\rangle\) for any \(\alpha,\beta\) and
   1.382 +     the particular choice of momentum $p=\sqrt{2mE}$.
   1.383 +5. In the infinitely deep potential well, the momentum and energy
   1.384 +   operators have these eigenstates:
   1.385 +   - The momentum eigenstates and energy eigenstates have the same form
   1.386 +     as in the vacuum potential: $p(x) =
   1.387 +     \exp{(ipx/\hbar)}$ and $|E\rangle = \alpha|p\rangle + \beta|-p\rangle$.
   1.388 +   - Even so, because of the boundary conditions on the
   1.389 +     well, we must make the following modifications:
   1.390 +     + Physically realistic states must be impossible to find outside the well. (Only a state of infinite
   1.391 +       energy could exist outside the well, and infinite energy is not
   1.392 +       realistic.) This requirement means, for example, that momentum
   1.393 +       eigenstates in the infinitely deep well must be
   1.394 +       \(p(x)
   1.395 +       = \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;
   1.396 +       \\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
   1.397 +     + Physically realistic states must vary smoothly throughout
   1.398 +       space. This means that if a particle in some state is very unlikely to be
   1.399 +       /at/ a particular location, it is also very unlikely be /near/
   1.400 +       that location. Combining this requirement with the above
   1.401 +       requirement, we find that the momentum operator no longer has
   1.402 +       an eigenstate for each value of $p$; instead, only values of
   1.403 +       $p$ that are integer multiples of $\pi \hbar/a$ are physically
   1.404 +       realistic. Similarly, the energy operator no longer has an
   1.405 +       eigenstate for each value of $E$; instead, the only energy
   1.406 +       eigenstates in the infinitely deep well
   1.407 +       are $E_n(x)=\sin(n\pi x/ a)$ for positive integers $n$.
   1.408 +
   1.409 +* COMMENT: 
   1.410 +
   1.411 +** Eigenstates with different eigenvalues are orthogonal
   1.412 +
   1.413 +#+begin_quote
   1.414 +*Theorem:* Eigenstates with different eigenvalues are orthogonal.
   1.415 +#+end_quote
   1.416 +
   1.417 +** COMMENT :
   1.418 +I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$
   1.419 +and $|b\rangle$ are eigenstates of $\Lambda$. This means that
   1.420 +
   1.421 +
   1.422 +\(
   1.423 +\begin{eqnarray}
   1.424 +\Lambda |a\rangle&=& a|a\rangle,\\
   1.425 +\Lambda|b\rangle&=& b|b\rangle.\\
   1.426 +\end{eqnarray}
   1.427 +\)
   1.428 +
   1.429 +If we take the difference of these eigenstates, we find that
   1.430 +
   1.431 +\(
   1.432 +\begin{eqnarray}
   1.433 +\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle
   1.434 +\qquad \text{(because $\Lambda$ is linear.)}\\
   1.435 +&=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and
   1.436 +$|b\rangle$ are eigenstates of $\Lambda$)}
   1.437 +\end{eqnarray}\)
   1.438 +
   1.439 +
   1.440 +which means that $a\neq b$.
   1.441 +
   1.442 +** Eigenvectors of hermitian operators span the space of solutions
   1.443 +
   1.444 +#+begin_quote
   1.445 +*Theorem:* If $\Omega$ is a hermitian operator, then every physically
   1.446 + allowed state can be written as a linear sum of eigenstates of
   1.447 + $\Omega$.
   1.448 +#+end_quote
   1.449 +
   1.450 +
   1.451 +
   1.452 +** Momentum and energy are hermitian operators
   1.453 +This ought to be true because hermitian operators correspond to
   1.454 +observable quantities. Since we expect momentum and energy to be
   1.455 +measureable quantities, we expect that there are hermitian operators
   1.456 +to represent them.
   1.457 +
   1.458 +
   1.459 +** Momentum and energy eigenstates in vacuum
   1.460 +An eigenstate of the momentum operator $P$ would be a state
   1.461 +\(|p\rangle\) such that \(P|p\rangle=p|p\rangle\).
   1.462 +
   1.463 +** Momentum and energy eigenstates in the infinitely deep well
   1.464 +
   1.465 +
   1.466 +
   1.467 +* COMMENT Can you measure momentum in the infinitely deep well?
   1.468 +In summary, I thought I knew:
   1.469 +1. For any hermitian operator: eigenstates with different eigenvalues
   1.470 +   are orthogonal.
   1.471 +2. For any hermitian operator: any physically realistic state can be
   1.472 +   written as a linear sum of eigenstates of the operator.
   1.473 +3. The momentum operator and energy operator are hermitian, because
   1.474 +   momentum and energy are observable quantities. 
   1.475 +4. (The form of the momentum and energy eigenstates in the vacuum potential)
   1.476 +5. (The form of the momentum and energy eigenstates in the infinitely deep well potential)
   1.477 +
   1.478 +Additionally, I understood that because the infinitely deep potential
   1.479 +well is not realistic, states of such a system  are not necessarily
   1.480 +physically realistic. Instead, I understood
   1.481 +\ldquo{}realistic states\rdquo{} to be those that satisfy the physically
   1.482 +unrealistic Schr\ouml{}dinger equation and its boundary conditions.
   1.483 +
   1.484 +With that final caveat, here is the problem:
   1.485 +
   1.486 +According to (5), the momentum eigenstates in the well are 
   1.487 +
   1.488 +\(p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
   1.489 +
   1.490 +(Additionally, we require that $p$ be an integer multiple of $\pi\hbar/a$.) 
   1.491 +
   1.492 +However, /these/ states are not orthogonal, which contradicts the
   1.493 +assumption that (3) the momentum operator is hermitian and (2)
   1.494 +eigenstates of a hermitian are orthogonal if they have different eigenvalues.
   1.495 +
   1.496 +#+begin_quote 
   1.497 +*Problem 1. The momentum eigenstates of the well are not orthogonal*
   1.498 +
   1.499 +/Proof./ If $p_1\neq p_2$, then 
   1.500 +
   1.501 +\(\begin{eqnarray}
   1.502 +\langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)\,dx\\
   1.503 +&=& \int_0^a p_1^*(x)p_2(x)\,dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{
   1.504 +outside the well.}\\
   1.505 +&=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)\,dx}}\\
   1.506 +&=& \int_0^a \exp{\left[i(p_1-p_2)x/\hbar\right]}\,dx\\
   1.507 +\end{eqnarray}\)
   1.508 +$\square$
   1.509 +
   1.510 +#+end_quote
   1.511 +
   1.512 +
   1.513 +
   1.514 +** COMMENT  Momentum eigenstates
   1.515 +
   1.516 +In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the
   1.517 +momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\).
   1.518 +
   1.519 +In the infinitely deep potential well, the Hamiltonian is the same but
   1.520 +there is a new condition in order for states to qualify as physically
   1.521 +allowed: the states must not exist anywhere outside of well, as it
   1.522 +takes an infinite amount of energy to do so. 
   1.523 +
   1.524 +Notice that the momentum eigenstates defined above do /not/ satisfy
   1.525 +this condition.
   1.526 +
   1.527 +
   1.528 +
   1.529 +* COMMENT
   1.530 +For each physical system, there is a Schr\ouml{}dinger equation that
   1.531 +describes how a particle's state $|\psi\rangle$  will change over
   1.532 +time.
   1.533 +
   1.534 +\(\begin{eqnarray}
   1.535 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
   1.536 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
   1.537 +
   1.538 +This is a differential equation; each solution to the
   1.539 +Schr\ouml{}dinger equation is a state that is physically allowed for
   1.540 +our particle. Here, physically allowed states are
   1.541 +those that change in physically allowed ways. However, like any differential
   1.542 +equation, the Schr\ouml{}dinger equation can be accompanied by
   1.543 +/boundary conditions/\mdash{}conditions that further restrict which
   1.544 +states qualify as physically allowed.
   1.545 +
   1.546 +
   1.547 +
   1.548 +
   1.549 +** Eigenstates of momentum
   1.550 +
   1.551 +
   1.552 +
   1.553 +
   1.554 +#In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger
   1.555 +
   1.556 +#\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\)
   1.557 +
   1.558 +
   1.559 +
   1.560 +
   1.561 +
   1.562 +
   1.563 +
   1.564 +* COMMENT
   1.565 +
   1.566 +#* The infinite square well potential
   1.567 +
   1.568 +A particle exists in a potential that is
   1.569 +infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the
   1.570 +particle exists in a potential[fn:coords][fn:infinity]
   1.571 +
   1.572 +
   1.573 +\(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for
   1.574 +}\;x<0\text{ or }x>a.\end{cases}\)
   1.575 +
   1.576 +The Schr\ouml{}dinger equation describes how the particle's state 
   1.577 +\(|\psi\rangle\) will change over time in this system.
   1.578 +
   1.579 +\(\begin{eqnarray}
   1.580 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
   1.581 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
   1.582 +
   1.583 +This is a differential equation; each solution to the
   1.584 +Schr\ouml{}dinger equation is a state that is physically allowed for
   1.585 +our particle. Here, physically allowed states are
   1.586 +those that change in physically allowed ways. However, like any differential
   1.587 +equation, the Schr\ouml{}dinger equation can be accompanied by
   1.588 +/boundary conditions/\mdash{}conditions that further restrict which
   1.589 +states qualify as physically allowed.
   1.590 +
   1.591 +
   1.592 +Whenever possible, physicists impose these boundary conditions:
   1.593 +- A physically allowed state ought to be a /smoothly-varying function of position./ This means
   1.594 +  that if a particle in the state  is likely to be /at/ a particular location,
   1.595 +  it is also likely to be /near/ that location.
   1.596 +
   1.597 +These boundary conditions imply that for the square well potential in
   1.598 +this problem,
   1.599 +
   1.600 +- Physically allowed states must be totally confined to the well,
   1.601 +  because it takes an infinite amount of energy to exist anywhere
   1.602 +  outside of the well (and physically allowed states ought to have
   1.603 +  only finite energy).
   1.604 +- Physically allowed states must be increasingly unlikely to find very
   1.605 +  close to the walls of the well. This is because of two conditions: the above
   1.606 +  condition says that the particle is /impossible/ to find
   1.607 +  outside of the well, and the smoothly-varying condition says
   1.608 +  that if a particle is impossible to find at a particular location,
   1.609 +  it must be unlikely to be found nearby that location.
   1.610 +
   1.611 +#; physically allowed states are those that change in physically
   1.612 +#allowed ways.
   1.613 +
   1.614 +
   1.615 +#** Boundary conditions
   1.616 +Because the potential is infinite everywhere except within the well,
   1.617 +a realistic particle must be confined to exist only within the
   1.618 +well\mdash{}its wavefunction must be zero everywhere beyond the walls
   1.619 +of the well.
   1.620 +
   1.621 +
   1.622 +[fn:coords] I chose my coordinate system so that the well extends from
   1.623 +\(0<x<a\). Others choose a coordinate system so that the well extends from
   1.624 +\(-\frac{a}{2}<x<\frac{a}{2}\). Although both coordinate systems describe the same physical
   1.625 +situation, they give different-looking answers.
   1.626 +
   1.627 +[fn:infinity] Of course, infinite potentials are not
   1.628 +realistic. Instead, they are useful approximations to finite
   1.629 +potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
   1.630 +of the well\rdquo{} are close enough for your own practical
   1.631 +purposes. Having introduced a physical impossibility into the problem
   1.632 +already, we don't expect to get physically realistic solutions; we
   1.633 +just expect to get mathematically consistent ones. The forthcoming
   1.634 +trouble is that we don't.