Mercurial > dylan
diff org/bk_quandary.org @ 0:f743fd0f4d8b
initial commit of dylan's stuff
author | Robert McIntyre <rlm@mit.edu> |
---|---|
date | Mon, 17 Oct 2011 23:17:55 -0700 |
parents | |
children |
line wrap: on
line diff
1.1 --- /dev/null Thu Jan 01 00:00:00 1970 +0000 1.2 +++ b/org/bk_quandary.org Mon Oct 17 23:17:55 2011 -0700 1.3 @@ -0,0 +1,566 @@ 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.