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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 is /really/ 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, we didn't use the fact that $H$ was hermitian;
   1.114 +instead, we used the fact that we know how to represent $H$ and $\psi$
   1.115 +as functions of $x$: $H$ is a differential operator
   1.116 +\(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\) and $\psi$ is a quadratic
   1.117 +function of $x$. By applying $H$ to $\psi$, we took several
   1.118 +derivatives and arrived at our answer.
   1.119 +
   1.120 +These two methods gave different results. In the following sections,
   1.121 +I'll describe and analyze the source of this difference.
   1.122 +
   1.123 +** Physical operators only act on physical wavefunctions
   1.124 +   :PROPERTIES:
   1.125 +   :ORDERED:  t
   1.126 +   :END:
   1.127 +#In quantum mechanics, an operator is a function that takes in a
   1.128 +#physical state and produces another physical state as ouput. Some
   1.129 +#operators correspond to physical quantities such as energy,
   1.130 +#momentum, or position; the mathematical properties of these operators correspond to
   1.131 +#physical properties of the system.
   1.132 +
   1.133 +#Eigenstates are an example of this correspondence: an 
   1.134 +
   1.135 +Physical states are represented as wavefunctions in quantum
   1.136 +mechanics. Just as we disallow certain physically nonsensical states
   1.137 +in classical mechanics (for example, we consider it to be nonphysical
   1.138 +for an object to spontaneously disappear from one place and reappear
   1.139 +in another), we also disallow certain wavefunctions in quantum
   1.140 +mechanics.
   1.141 +
   1.142 +For example, since wavefunctions are supposed to correspond to
   1.143 +probability amplitudes, we require wavefunctions to be normalized
   1.144 +\((\langle \psi |\psi \rangle = 1)\). Generally, we disallow
   1.145 +wavefunctions that do not satisfy this property (although there are
   1.146 +some exceptions[fn:2]).
   1.147 +
   1.148 +As another example, we generally expect probability to vary smoothly\mdash{}if
   1.149 +a particle is very likely or very unlikely to be found at a particular
   1.150 +location, it should also be somewhat likely or somewhat unlikely to be
   1.151 +found /near/ that location. In more precise terms, we expect that for
   1.152 +physically meaningful wavefunctions, the probability 
   1.153 +\(\text{Pr}(x)=\int^x \psi^*\psi\) should be a continuous function of
   1.154 +$x$ and, again, we disallow wavefunctions that do not satisfy this
   1.155 +property because we consider them to be physically nonsensical.
   1.156 +
   1.157 +So, physical wavefunctions must satisfy certain properties
   1.158 +like the two just described. Wavefunctions that do not satisfy these properties are
   1.159 +rejected for being physically nonsensical: even though we can perform
   1.160 +calculations with them, the mathematical results we obtain do not mean
   1.161 +anything physically.
   1.162 +
   1.163 +Now, in quantum mechanics, an *operator* is a function that converts
   1.164 +states into other states. Some operators correspond to
   1.165 +physical quantities such as energy, momentum, or position, and as a
   1.166 +result, the mathematical properties of these operators correspond to
   1.167 +physical properties of the system. Physical operators are furthermore
   1.168 +subject to the following rule: they are only allowed to operate on 
   1.169 +#physical wavefunctions, and they are only allowed to produce
   1.170 +#physical wavefunctions[fn:why].
   1.171 +the wavefunctions we have deemed physical, and they only produce physical wavefunctions[fn::
   1.172 +
   1.173 + If you require a hermitian operator to have physical
   1.174 +  eigenstates, you get a very strong result: you guarantee that the
   1.175 +  operator will convert /every/ physical wavefunction into another
   1.176 +  physical wavefunction:
   1.177 +
   1.178 +  For any linear operator $\Omega$, the eigenvalue equation is
   1.179 +\(\Omega|\omega\rangle = \omega |\omega\rangle\). Notice that if an
   1.180 +eigenstate $|\omega\rangle$ is a physical wavefunction, the
   1.181 +eigenvalue equation equation forces $\Omega|\omega\rangle$ to be a
   1.182 +physical wavefunction as well. To elaborate, if the eigenstates of
   1.183 +$\Omega$ are physical functions, then $\Omega$ is guaranteed to
   1.184 +convert them into other physical functions.  Even more is true if the
   1.185 +operator $\Omega$ is also hermitian: there is a theorem which states
   1.186 +that \ldquo{}If \Omega is hermitian, then every physical wavefunction
   1.187 +can be written as a weighted sum of eigenstates of \Omega.\rdquo{} This
   1.188 +theorem implies that /if $\Omega$ is hermitian/, and /if the eigenstates
   1.189 +of \Omega are physically allowed/, then \Omega is guaranteed to
   1.190 +convert every physically allowed wavefunction into another physically
   1.191 +allowed wavefunction.].
   1.192 +
   1.193 +In fact, this rule for physical operators is the source of our
   1.194 +problem, as we unknowingly violated it when applying our second
   1.195 +method!
   1.196 +
   1.197 +** The violation
   1.198 +
   1.199 +I'll start explaining this violation by being more specific about the
   1.200 +infinitely deep well potential. We have said already that physicists
   1.201 +require wavefunctions to satisfy certain properties in order to be
   1.202 +deemed \ldquo{}physical\rdquo{}. To be specific, wavefunctions in the
   1.203 +infinitely deep well
   1.204 +- Must be *normalizable*, because they correspond to
   1.205 +  probability amplitudes.
   1.206 +- Must have *smoothly-varying probability*, because if a particle is very
   1.207 +  likely to be at a location, it ought to be likely to be /near/
   1.208 +  it as well.
   1.209 +- Must *not exist outside the well*, because it
   1.210 +  would take an infinite amount of energy to do so.
   1.211 +
   1.212 +Additionally, by combining the second and third conditions, some
   1.213 +physicists reason that wavefunctions in the infinitely deep well
   1.214 +
   1.215 +- Must *become zero* towards the edges of the well.
   1.216 +
   1.217 +
   1.218 +
   1.219 +
   1.220 +You'll remember we had
   1.221 +
   1.222 +\(
   1.223 +\begin{eqnarray}
   1.224 +\psi(x) &=& A\;x(x-a)\\
   1.225 +\bar\psi(x)&=& \frac{-A\hbar^2}{m}\\
   1.226 +&&\text{for }0\lt{}x\lt{}a\\
   1.227 +\end{eqnarray}
   1.228 +\)
   1.229 +
   1.230 +In our second method, we wrote 
   1.231 +
   1.232 +
   1.233 +\(\begin{eqnarray}
   1.234 +\langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\
   1.235 +&=& \langle \psi |H \bar\psi \rangle\\
   1.236 +& \vdots&\\
   1.237 +&=& 0\\
   1.238 +\end{eqnarray}\)
   1.239 +
   1.240 +However, there are two issues here. First, we were not allowed to operate with $H$ on the wavefunction
   1.241 +$|\bar \psi\rangle$, because $H$ is a physical operator and $|\bar
   1.242 +\psi\rangle$ is a nonphysical state: in the infinite square well,
   1.243 +physical wavefunctions must approach zero at the edges of the well,
   1.244 +which the constant function $|\bar\psi\rangle$ does not do. By
   1.245 +feeding $H$ a nonphysical wavefunction, we obtained nonsensical
   1.246 +results.
   1.247 +
   1.248 +Second, we claimed that $H$ was a physical operator\mdash{}that $H$
   1.249 +was hermitian. According to the rule, this means $H$ must convert physical states into other
   1.250 +physical states. But $H$ converts the physical state $|\psi\rangle$
   1.251 +into the nonphysical state $|\bar\psi\rangle = H|\psi\rangle$ is not. Because $H$ converts some
   1.252 +physical states into nonphysical states, it cannot be a hermitian operator.
   1.253 +
   1.254 +** Boundary conditions affect hermiticity
   1.255 +We have now discovered a flaw: when applied to the state
   1.256 +$|\psi\rangle$, the second method violates the rule that physical
   1.257 +operators must only take in physical states and must only produce
   1.258 +physical states. This suggests that the problem was with the state
   1.259 +$|\bar\psi\rangle$ that we wrongly fed into $H$. However, the problem
   1.260 +is more serious still: the state $|\psi\rangle 
   1.261 +
   1.262 +** COMMENT Re-examining physical constraints
   1.263 +
   1.264 +We have now discovered a flaw: when applied to the state
   1.265 +$|\psi\rangle$, the second method violates the rule that physical
   1.266 +operators must only take in physical states and must only produce
   1.267 +physical states. Let's examine the problem more closely.
   1.268 +
   1.269 +We have said already that physicists require wavefunctions to satisfy
   1.270 +certain properties in order to be deemed \ldquo{}physical\rdquo{}. To
   1.271 +be specific, wavefunctions in the infinitely deep well
   1.272 +- Must be *normalizable*, because they correspond to
   1.273 +  probability amplitudes.
   1.274 +- Must have *smoothly-varying probability*, because if a particle is very
   1.275 +  likely to be at a location, it ought to be likely to be /near/
   1.276 +  it as well.
   1.277 +- Must *not exist outside the well*, because it
   1.278 +  would take an infinite amount of energy to do so.
   1.279 +
   1.280 +We now have discovered an important flaw in the second method: when
   1.281 +applied to the state $|\bar\psi\rangle$, the second method violates
   1.282 +the rule that physical operators must only take in
   1.283 +physical states and must only produce physical states. The problem is
   1.284 +even more serious, however
   1.285 +
   1.286 +
   1.287 +
   1.288 +[fn:1] I'm defining a new variable just to make certain expressions
   1.289 +  look shorter; this cannot affect the content of the answer we'll
   1.290 +  get. 
   1.291 +
   1.292 +[fn:2] For example, in vaccuum (i.e., when the potential of the
   1.293 +  physical system is $V(x)=0$ throughout all space), the momentum
   1.294 +  eigenstates are not normalizable\mdash{}the relevant integral blows
   1.295 +  up to infinity instead of converging to a number. Physicists modify
   1.296 +  the definition of normalization slightly so that
   1.297 +  \ldquo{}delta-normalizable \rdquo{} functions like these are included
   1.298 +  among the physical wavefunctions.
   1.299 +
   1.300 +
   1.301 +
   1.302 +* COMMENT: What I thought I knew
   1.303 +
   1.304 +The following is a list of things I thought were true of quantum
   1.305 +mechanics; the catch is that the list contradicts itself.
   1.306 +
   1.307 +1. For any hermitian operator: Eigenstates with different eigenvalues are orthogonal.
   1.308 +2. For any hermitian operator: Any physically allowed state can be
   1.309 +   written as a linear sum of eigenstates of the operator.
   1.310 +3. The momentum operator and energy operator are hermitian, because
   1.311 +   momentum and energy are measureable quantities.
   1.312 +4. In the vacuum potential, the momentum and energy operators have these eigenstates:
   1.313 +   - the momentum operator has an eigenstate
   1.314 +     \(p(x)=\exp{(ipx/\hbar)}\) for each value of $p$.
   1.315 +   - the energy operator has an eigenstate \(|E\rangle =
   1.316 +     \alpha|p\rangle + \beta|-p\rangle\) for any \(\alpha,\beta\) and
   1.317 +     the particular choice of momentum $p=\sqrt{2mE}$.
   1.318 +5. In the infinitely deep potential well, the momentum and energy
   1.319 +   operators have these eigenstates:
   1.320 +   - The momentum eigenstates and energy eigenstates have the same form
   1.321 +     as in the vacuum potential: $p(x) =
   1.322 +     \exp{(ipx/\hbar)}$ and $|E\rangle = \alpha|p\rangle + \beta|-p\rangle$.
   1.323 +   - Even so, because of the boundary conditions on the
   1.324 +     well, we must make the following modifications:
   1.325 +     + Physically realistic states must be impossible to find outside the well. (Only a state of infinite
   1.326 +       energy could exist outside the well, and infinite energy is not
   1.327 +       realistic.) This requirement means, for example, that momentum
   1.328 +       eigenstates in the infinitely deep well must be
   1.329 +       \(p(x)
   1.330 +       = \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;
   1.331 +       \\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
   1.332 +     + Physically realistic states must vary smoothly throughout
   1.333 +       space. This means that if a particle in some state is very unlikely to be
   1.334 +       /at/ a particular location, it is also very unlikely be /near/
   1.335 +       that location. Combining this requirement with the above
   1.336 +       requirement, we find that the momentum operator no longer has
   1.337 +       an eigenstate for each value of $p$; instead, only values of
   1.338 +       $p$ that are integer multiples of $\pi \hbar/a$ are physically
   1.339 +       realistic. Similarly, the energy operator no longer has an
   1.340 +       eigenstate for each value of $E$; instead, the only energy
   1.341 +       eigenstates in the infinitely deep well
   1.342 +       are $E_n(x)=\sin(n\pi x/ a)$ for positive integers $n$.
   1.343 +
   1.344 +* COMMENT: 
   1.345 +
   1.346 +** Eigenstates with different eigenvalues are orthogonal
   1.347 +
   1.348 +#+begin_quote
   1.349 +*Theorem:* Eigenstates with different eigenvalues are orthogonal.
   1.350 +#+end_quote
   1.351 +
   1.352 +** COMMENT :
   1.353 +I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$
   1.354 +and $|b\rangle$ are eigenstates of $\Lambda$. This means that
   1.355 +
   1.356 +
   1.357 +\(
   1.358 +\begin{eqnarray}
   1.359 +\Lambda |a\rangle&=& a|a\rangle,\\
   1.360 +\Lambda|b\rangle&=& b|b\rangle.\\
   1.361 +\end{eqnarray}
   1.362 +\)
   1.363 +
   1.364 +If we take the difference of these eigenstates, we find that
   1.365 +
   1.366 +\(
   1.367 +\begin{eqnarray}
   1.368 +\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle
   1.369 +\qquad \text{(because $\Lambda$ is linear.)}\\
   1.370 +&=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and
   1.371 +$|b\rangle$ are eigenstates of $\Lambda$)}
   1.372 +\end{eqnarray}\)
   1.373 +
   1.374 +
   1.375 +which means that $a\neq b$.
   1.376 +
   1.377 +** Eigenvectors of hermitian operators span the space of solutions
   1.378 +
   1.379 +#+begin_quote
   1.380 +*Theorem:* If $\Omega$ is a hermitian operator, then every physically
   1.381 + allowed state can be written as a linear sum of eigenstates of
   1.382 + $\Omega$.
   1.383 +#+end_quote
   1.384 +
   1.385 +
   1.386 +
   1.387 +** Momentum and energy are hermitian operators
   1.388 +This ought to be true because hermitian operators correspond to
   1.389 +observable quantities. Since we expect momentum and energy to be
   1.390 +measureable quantities, we expect that there are hermitian operators
   1.391 +to represent them.
   1.392 +
   1.393 +
   1.394 +** Momentum and energy eigenstates in vacuum
   1.395 +An eigenstate of the momentum operator $P$ would be a state
   1.396 +\(|p\rangle\) such that \(P|p\rangle=p|p\rangle\).
   1.397 +
   1.398 +** Momentum and energy eigenstates in the infinitely deep well
   1.399 +
   1.400 +
   1.401 +
   1.402 +* COMMENT Can you measure momentum in the infinitely deep well?
   1.403 +In summary, I thought I knew:
   1.404 +1. For any hermitian operator: eigenstates with different eigenvalues
   1.405 +   are orthogonal.
   1.406 +2. For any hermitian operator: any physically realistic state can be
   1.407 +   written as a linear sum of eigenstates of the operator.
   1.408 +3. The momentum operator and energy operator are hermitian, because
   1.409 +   momentum and energy are observable quantities. 
   1.410 +4. (The form of the momentum and energy eigenstates in the vacuum potential)
   1.411 +5. (The form of the momentum and energy eigenstates in the infinitely deep well potential)
   1.412 +
   1.413 +Additionally, I understood that because the infinitely deep potential
   1.414 +well is not realistic, states of such a system  are not necessarily
   1.415 +physically realistic. Instead, I understood
   1.416 +\ldquo{}realistic states\rdquo{} to be those that satisfy the physically
   1.417 +unrealistic Schr\ouml{}dinger equation and its boundary conditions.
   1.418 +
   1.419 +With that final caveat, here is the problem:
   1.420 +
   1.421 +According to (5), the momentum eigenstates in the well are 
   1.422 +
   1.423 +\(p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
   1.424 +
   1.425 +(Additionally, we require that $p$ be an integer multiple of $\pi\hbar/a$.) 
   1.426 +
   1.427 +However, /these/ states are not orthogonal, which contradicts the
   1.428 +assumption that (3) the momentum operator is hermitian and (2)
   1.429 +eigenstates of a hermitian are orthogonal if they have different eigenvalues.
   1.430 +
   1.431 +#+begin_quote 
   1.432 +*Problem 1. The momentum eigenstates of the well are not orthogonal*
   1.433 +
   1.434 +/Proof./ If $p_1\neq p_2$, then 
   1.435 +
   1.436 +\(\begin{eqnarray}
   1.437 +\langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)\,dx\\
   1.438 +&=& \int_0^a p_1^*(x)p_2(x)\,dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{
   1.439 +outside the well.}\\
   1.440 +&=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)\,dx}}\\
   1.441 +&=& \int_0^a \exp{\left[i(p_1-p_2)x/\hbar\right]}\,dx\\
   1.442 +\end{eqnarray}\)
   1.443 +$\square$
   1.444 +
   1.445 +#+end_quote
   1.446 +
   1.447 +
   1.448 +
   1.449 +** COMMENT  Momentum eigenstates
   1.450 +
   1.451 +In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the
   1.452 +momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\).
   1.453 +
   1.454 +In the infinitely deep potential well, the Hamiltonian is the same but
   1.455 +there is a new condition in order for states to qualify as physically
   1.456 +allowed: the states must not exist anywhere outside of well, as it
   1.457 +takes an infinite amount of energy to do so. 
   1.458 +
   1.459 +Notice that the momentum eigenstates defined above do /not/ satisfy
   1.460 +this condition.
   1.461 +
   1.462 +
   1.463 +
   1.464 +* COMMENT
   1.465 +For each physical system, there is a Schr\ouml{}dinger equation that
   1.466 +describes how a particle's state $|\psi\rangle$  will change over
   1.467 +time.
   1.468 +
   1.469 +\(\begin{eqnarray}
   1.470 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
   1.471 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
   1.472 +
   1.473 +This is a differential equation; each solution to the
   1.474 +Schr\ouml{}dinger equation is a state that is physically allowed for
   1.475 +our particle. Here, physically allowed states are
   1.476 +those that change in physically allowed ways. However, like any differential
   1.477 +equation, the Schr\ouml{}dinger equation can be accompanied by
   1.478 +/boundary conditions/\mdash{}conditions that further restrict which
   1.479 +states qualify as physically allowed.
   1.480 +
   1.481 +
   1.482 +
   1.483 +
   1.484 +** Eigenstates of momentum
   1.485 +
   1.486 +
   1.487 +
   1.488 +
   1.489 +#In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger
   1.490 +
   1.491 +#\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\)
   1.492 +
   1.493 +
   1.494 +
   1.495 +
   1.496 +
   1.497 +
   1.498 +
   1.499 +* COMMENT
   1.500 +
   1.501 +#* The infinite square well potential
   1.502 +
   1.503 +A particle exists in a potential that is
   1.504 +infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the
   1.505 +particle exists in a potential[fn:coords][fn:infinity]
   1.506 +
   1.507 +
   1.508 +\(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for
   1.509 +}\;x<0\text{ or }x>a.\end{cases}\)
   1.510 +
   1.511 +The Schr\ouml{}dinger equation describes how the particle's state 
   1.512 +\(|\psi\rangle\) will change over time in this system.
   1.513 +
   1.514 +\(\begin{eqnarray}
   1.515 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
   1.516 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
   1.517 +
   1.518 +This is a differential equation; each solution to the
   1.519 +Schr\ouml{}dinger equation is a state that is physically allowed for
   1.520 +our particle. Here, physically allowed states are
   1.521 +those that change in physically allowed ways. However, like any differential
   1.522 +equation, the Schr\ouml{}dinger equation can be accompanied by
   1.523 +/boundary conditions/\mdash{}conditions that further restrict which
   1.524 +states qualify as physically allowed.
   1.525 +
   1.526 +
   1.527 +Whenever possible, physicists impose these boundary conditions:
   1.528 +- A physically allowed state ought to be a /smoothly-varying function of position./ This means
   1.529 +  that if a particle in the state  is likely to be /at/ a particular location,
   1.530 +  it is also likely to be /near/ that location.
   1.531 +
   1.532 +These boundary conditions imply that for the square well potential in
   1.533 +this problem,
   1.534 +
   1.535 +- Physically allowed states must be totally confined to the well,
   1.536 +  because it takes an infinite amount of energy to exist anywhere
   1.537 +  outside of the well (and physically allowed states ought to have
   1.538 +  only finite energy).
   1.539 +- Physically allowed states must be increasingly unlikely to find very
   1.540 +  close to the walls of the well. This is because of two conditions: the above
   1.541 +  condition says that the particle is /impossible/ to find
   1.542 +  outside of the well, and the smoothly-varying condition says
   1.543 +  that if a particle is impossible to find at a particular location,
   1.544 +  it must be unlikely to be found nearby that location.
   1.545 +
   1.546 +#; physically allowed states are those that change in physically
   1.547 +#allowed ways.
   1.548 +
   1.549 +
   1.550 +#** Boundary conditions
   1.551 +Because the potential is infinite everywhere except within the well,
   1.552 +a realistic particle must be confined to exist only within the
   1.553 +well\mdash{}its wavefunction must be zero everywhere beyond the walls
   1.554 +of the well.
   1.555 +
   1.556 +
   1.557 +[fn:coords] I chose my coordinate system so that the well extends from
   1.558 +\(0<x<a\). Others choose a coordinate system so that the well extends from
   1.559 +\(-\frac{a}{2}<x<\frac{a}{2}\). Although both coordinate systems describe the same physical
   1.560 +situation, they give different-looking answers.
   1.561 +
   1.562 +[fn:infinity] Of course, infinite potentials are not
   1.563 +realistic. Instead, they are useful approximations to finite
   1.564 +potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
   1.565 +of the well\rdquo{} are close enough for your own practical
   1.566 +purposes. Having introduced a physical impossibility into the problem
   1.567 +already, we don't expect to get physically realistic solutions; we
   1.568 +just expect to get mathematically consistent ones. The forthcoming
   1.569 +trouble is that we don't.