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     1.4 +#+TITLE: Bugs in quantum mechanics
     1.5 +#+AUTHOR: Dylan Holmes
     1.6 +#+SETUPFILE: ../../aurellem/org/setup.org
     1.7 +#+INCLUDE:   ../../aurellem/org/level-0.org
     1.8 +
     1.9 +#Bugs in Quantum Mechanics
    1.10 +#Bugs in the Quantum-Mechanical Momentum Operator
    1.11 +
    1.12 +
    1.13 +I studied quantum mechanics the same way I study most subjects\mdash{}
    1.14 +by collecting (and squashing) bugs in my understanding. One of these
    1.15 +bugs persisted throughout two semesters of
    1.16 +quantum mechanics coursework until I finally found
    1.17 +the paper 
    1.18 +[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
    1.19 +mechanics/]], which helped me stamp out the bug entirely. I decided to
    1.20 +write an article about the problem and its solution for a number of reasons:
    1.21 +
    1.22 +- Although the paper was not unreasonably dense, it was written for
    1.23 +  teachers. I wanted to write an article for students.
    1.24 +- I wanted to popularize the problem and its solution because other
    1.25 +  explanations are currently too hard to find. (Even Shankar's
    1.26 +  excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it.)
    1.27 +- I wanted to check that the bug was indeed entirely
    1.28 +  eradicated. Attempting an explanation is my way of making
    1.29 +  sure.
    1.30 +
    1.31 +* COMMENT
    1.32 + I recommend the
    1.33 +paper not only for students who are learning
    1.34 +quantum mechanics, but especially for teachers interested in debugging
    1.35 +them. 
    1.36 +
    1.37 +* COMMENT
    1.38 +On my first exam in quantum mechanics, my professor asked us to
    1.39 +describe how certain measurements would affect a particle in a
    1.40 +box. Many of these measurement questions required routine application
    1.41 +of skills we had recently learned\mdash{}first, you recall (or
    1.42 +calculate) the eigenstates of the quantity
    1.43 +to be measured; second, you write the given state as a linear
    1.44 +sum of these eigenstates\mdash{} the coefficients on each term give
    1.45 +the probability amplitude.
    1.46 +
    1.47 +
    1.48 +* What I thought I knew
    1.49 +
    1.50 +The following is a list of things I thought were true of quantum
    1.51 +mechanics; the catch is that the list contradicts itself.
    1.52 +
    1.53 +- For any hermitian operator: Eigenstates with different eigenvalues are orthogonal.
    1.54 +- For any hermitian operator: Any physically allowed state can be
    1.55 +  written as a linear sum of eigenstates of the operator.
    1.56 +- The momentum operator and energy operator are hermitian, because
    1.57 +  momentum and energy are measureable quantities.
    1.58 +- In vacuum,
    1.59 +  - the momentum operator has an eigenstate
    1.60 +    \(p(x)=\exp{(ipx/\hbar)}\) for each value of $p$.
    1.61 +  - the energy operator has an eigenstate \(|E\rangle =
    1.62 +    \alpha|p\rangle + \beta|-p\rangle\) for any \(\alpha,\beta\) and
    1.63 +    the particular choice of momentum $p=\sqrt{2mE}$.
    1.64 +- In the infinitely deep potential well,
    1.65 +  - the momentum operator has eigenstates with the same form $p(x) =
    1.66 +    \exp{(ipx/\hbar)}$, but because of the boundary conditions on the
    1.67 +    well, the following modifications are required.
    1.68 +    - The wavefunction must be zero everywhere outside the well. That
    1.69 +      is, \(p(x) = \begin{cases}\exp{(ipx/\hbar)},& 0\lt{}x\lt{}a;
    1.70 +      \\0, & \text{for }x<0\text{ or }x>a \\ \end{cases}\)
    1.71 +#0,&\text{for }x\lt{}0\text{ or }x\gt{}a\end{cases}\)
    1.72 +    - no longer has an eigenstate for each value
    1.73 +      of $p$. Instead, only values of $p$ that are integer multiples of
    1.74 +      $\pi a/\hbar$ are physically realistic. 
    1.75 +
    1.76 +
    1.77 +
    1.78 +* COMMENT: 
    1.79 +
    1.80 +** Eigenstates with different eigenvalues are orthogonal
    1.81 +
    1.82 +#+begin_quote
    1.83 +*Theorem:* Eigenstates with different eigenvalues are orthogonal.
    1.84 +#+end_quote
    1.85 +
    1.86 +** COMMENT :
    1.87 +I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$
    1.88 +and $|b\rangle$ are eigenstates of $\Lambda$. This means that
    1.89 +
    1.90 +
    1.91 +\(
    1.92 +\begin{eqnarray}
    1.93 +\Lambda |a\rangle&=& a|a\rangle,\\
    1.94 +\Lambda|b\rangle&=& b|b\rangle.\\
    1.95 +\end{eqnarray}
    1.96 +\)
    1.97 +
    1.98 +If we take the difference of these eigenstates, we find that
    1.99 +
   1.100 +\(
   1.101 +\begin{eqnarray}
   1.102 +\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle
   1.103 +\qquad \text{(because $\Lambda$ is linear.)}\\
   1.104 +&=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and
   1.105 +$|b\rangle$ are eigenstates of $\Lambda$)}
   1.106 +\end{eqnarray}\)
   1.107 +
   1.108 +
   1.109 +which means that $a\neq b$.
   1.110 +
   1.111 +** Eigenvectors of hermitian operators span the space of solutions
   1.112 +
   1.113 +#+begin_quote
   1.114 +*Theorem:* If $\Omega$ is a hermitian operator, then every physically
   1.115 + allowed state can be written as a linear sum of eigenstates of
   1.116 + $\Omega$.
   1.117 +#+end_quote
   1.118 +
   1.119 +
   1.120 +
   1.121 +** Momentum and energy are hermitian operators
   1.122 +This ought to be true because hermitian operators correspond to
   1.123 +observable quantities. Since we expect momentum and energy to be
   1.124 +measureable quantities, we expect that there are hermitian operators
   1.125 +to represent them.
   1.126 +
   1.127 +
   1.128 +** Momentum and energy eigenstates in vacuum
   1.129 +An eigenstate of the momentum operator $P$ would be a state
   1.130 +\(|p\rangle\) such that \(P|p\rangle=p|p\rangle\).
   1.131 +
   1.132 +** Momentum and energy eigenstates in the infinitely deep well
   1.133 +
   1.134 +
   1.135 +
   1.136 +* Can you measure momentum in the infinite square well?
   1.137 +
   1.138 +
   1.139 +
   1.140 +** COMMENT  Momentum eigenstates
   1.141 +
   1.142 +In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the
   1.143 +momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\).
   1.144 +
   1.145 +In the infinitely deep potential well, the Hamiltonian is the same but
   1.146 +there is a new condition in order for states to qualify as physically
   1.147 +allowed: the states must not exist anywhere outside of well, as it
   1.148 +takes an infinite amount of energy to do so. 
   1.149 +
   1.150 +Notice that the momentum eigenstates defined above do /not/ satisfy
   1.151 +this condition.
   1.152 +
   1.153 +
   1.154 +
   1.155 +* COMMENT
   1.156 +For each physical system, there is a Schr\ouml{}dinger equation that
   1.157 +describes how a particle's state $|\psi\rangle$  will change over
   1.158 +time.
   1.159 +
   1.160 +\(\begin{eqnarray}
   1.161 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
   1.162 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
   1.163 +
   1.164 +This is a differential equation; each solution to the
   1.165 +Schr\ouml{}dinger equation is a state that is physically allowed for
   1.166 +our particle. Here, physically allowed states are
   1.167 +those that change in physically allowed ways. However, like any differential
   1.168 +equation, the Schr\ouml{}dinger equation can be accompanied by
   1.169 +/boundary conditions/\mdash{}conditions that further restrict which
   1.170 +states qualify as physically allowed.
   1.171 +
   1.172 +
   1.173 +
   1.174 +
   1.175 +** Eigenstates of momentum
   1.176 +
   1.177 +
   1.178 +
   1.179 +
   1.180 +#In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger
   1.181 +
   1.182 +#\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\)
   1.183 +
   1.184 +
   1.185 +
   1.186 +
   1.187 +
   1.188 +
   1.189 +
   1.190 +* COMMENT
   1.191 +
   1.192 +#* The infinite square well potential
   1.193 +
   1.194 +A particle exists in a potential that is
   1.195 +infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the
   1.196 +particle exists in a potential[fn:coords][fn:infinity]
   1.197 +
   1.198 +
   1.199 +\(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for
   1.200 +}\;x<0\text{ or }x>a.\end{cases}\)
   1.201 +
   1.202 +The Schr\ouml{}dinger equation describes how the particle's state 
   1.203 +\(|\psi\rangle\) will change over time in this system.
   1.204 +
   1.205 +\(\begin{eqnarray}
   1.206 +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
   1.207 +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
   1.208 +
   1.209 +This is a differential equation; each solution to the
   1.210 +Schr\ouml{}dinger equation is a state that is physically allowed for
   1.211 +our particle. Here, physically allowed states are
   1.212 +those that change in physically allowed ways. However, like any differential
   1.213 +equation, the Schr\ouml{}dinger equation can be accompanied by
   1.214 +/boundary conditions/\mdash{}conditions that further restrict which
   1.215 +states qualify as physically allowed.
   1.216 +
   1.217 +
   1.218 +Whenever possible, physicists impose these boundary conditions:
   1.219 +- A physically allowed state ought to be a /smoothly-varying function of position./ This means
   1.220 +  that if a particle in the state  is likely to be /at/ a particular location,
   1.221 +  it is also likely to be /near/ that location.
   1.222 +
   1.223 +These boundary conditions imply that for the square well potential in
   1.224 +this problem,
   1.225 +
   1.226 +- Physically allowed states must be totally confined to the well,
   1.227 +  because it takes an infinite amount of energy to exist anywhere
   1.228 +  outside of the well (and physically allowed states ought to have
   1.229 +  only finite energy).
   1.230 +- Physically allowed states must be increasingly unlikely to find very
   1.231 +  close to the walls of the well. This is because of two conditions: the above
   1.232 +  condition says that the particle is /impossible/ to find
   1.233 +  outside of the well, and the smoothly-varying condition says
   1.234 +  that if a particle is impossible to find at a particular location,
   1.235 +  it must be unlikely to be found nearby that location.
   1.236 +
   1.237 +#; physically allowed states are those that change in physically
   1.238 +#allowed ways.
   1.239 +
   1.240 +
   1.241 +#** Boundary conditions
   1.242 +Because the potential is infinite everywhere except within the well,
   1.243 +a realistic particle must be confined to exist only within the
   1.244 +well\mdash{}its wavefunction must be zero everywhere beyond the walls
   1.245 +of the well.
   1.246 +
   1.247 +
   1.248 +[fn:coords] I chose my coordinate system so that the well extends from
   1.249 +\(0<x<a\). Others choose a coordinate system so that the well extends from
   1.250 +\(-\frac{a}{2}<x<\frac{a}{2}\). Although both coordinate systems describe the same physical
   1.251 +situation, they give different-looking answers.
   1.252 +
   1.253 +[fn:infinity] Of course, infinite potentials are not
   1.254 +realistic. Instead, they are useful approximations to finite
   1.255 +potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
   1.256 +of the well\rdquo{} are close enough for your own practical
   1.257 +purposes. Having introduced a physical impossibility into the problem
   1.258 +already, we don't expect to get physically realistic solutions; we
   1.259 +just expect to get mathematically consistent ones. The forthcoming
   1.260 +trouble is that we don't.