dylan: org/bk4.org comparison

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initial commit of dylan's stuff

author	Robert McIntyre <rlm@mit.edu>
date	Mon, 17 Oct 2011 23:17:55 -0700
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+#+TITLE: Bugs in quantum mechanics
+#+AUTHOR: Dylan Holmes
+#+SETUPFILE: ../../aurellem/org/setup.org
+#+INCLUDE:   ../../aurellem/org/level-0.org
+#Bugs in Quantum Mechanics
+#Bugs in the Quantum-Mechanical Momentum Operator
+I studied quantum mechanics the same way I study most subjects\mdash{}
+by collecting (and squashing) bugs in my understanding. One of these
+bugs persisted throughout two semesters of
+quantum mechanics coursework until I finally found
+the paper
+[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
+mechanics/]], which helped me stamp out the bug entirely. I decided to
+write an article about the problem and its solution for a number of reasons:
+- Although the paper was not unreasonably dense, it was written for
+teachers. I wanted to write an article for students.
+- I wanted to popularize the problem and its solution because other
+explanations are currently too hard to find. (Even Shankar's
+excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it.)
+- I wanted to check that the bug was indeed entirely
+eradicated. Attempting an explanation is my way of making
+sure.
+* COMMENT
+I recommend the
+paper not only for students who are learning
+quantum mechanics, but especially for teachers interested in debugging
+them.
+* COMMENT
+On my first exam in quantum mechanics, my professor asked us to
+describe how certain measurements would affect a particle in a
+box. Many of these measurement questions required routine application
+of skills we had recently learned\mdash{}first, you recall (or
+calculate) the eigenstates of the quantity
+to be measured; second, you write the given state as a linear
+sum of these eigenstates\mdash{} the coefficients on each term give
+the probability amplitude.
+* What I thought I knew
+The following is a list of things I thought were true of quantum
+mechanics; the catch is that the list contradicts itself.
+1. For any hermitian operator: Eigenstates with different eigenvalues are orthogonal.
+2. For any hermitian operator: Any physically allowed state can be
+written as a linear sum of eigenstates of the operator.
+3. The momentum operator and energy operator are hermitian, because
+momentum and energy are measureable quantities.
+4. In the vacuum potential, the momentum and energy operators have these eigenstates:
+- the momentum operator has an eigenstate
+\(p(x)=\exp{(ipx/\hbar)}\) for each value of $p$.
+- the energy operator has an eigenstate \(|E\rangle =
+\alpha|p\rangle + \beta|-p\rangle\) for any \(\alpha,\beta\) and
+the particular choice of momentum $p=\sqrt{2mE}$.
+5. In the infinitely deep potential well, the momentum and energy
+operators have these eigenstates:
+- The momentum eigenstates and energy eigenstates have the same form
+as in the vacuum potential: $p(x) =
+\exp{(ipx/\hbar)}$ and $|E\rangle = \alpha|p\rangle + \beta|-p\rangle$.
+- Even so, because of the boundary conditions on the
+well, we must make the following modifications:
++ Physically realistic states must be impossible to find outside the well. (Only a state of infinite
+energy could exist outside the well, and infinite energy is not
+realistic.) This requirement means, for example, that momentum
+eigenstates in the infinitely deep well must be
+\(p(x)
+= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;
+\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
++ Physically realistic states must vary smoothly throughout
+space. This means that if a particle in some state is very unlikely to be
+/at/ a particular location, it is also very unlikely be /near/
+that location. Combining this requirement with the above
+requirement, we find that the momentum operator no longer has
+an eigenstate for each value of $p$; instead, only values of
+$p$ that are integer multiples of $\pi a/\hbar$ are physically
+realistic. Similarly, the energy operator no longer has an
+eigenstate for each value of $E$; instead, the only energy
+eigenstates in the infinitely deep well
+are $E_n(x)=\sin(n\pi x/ a)$ for positive integers $n$.
+* COMMENT:
+** Eigenstates with different eigenvalues are orthogonal
+#+begin_quote
+*Theorem:* Eigenstates with different eigenvalues are orthogonal.
+#+end_quote
+** COMMENT :
+I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$
+and $|b\rangle$ are eigenstates of $\Lambda$. This means that
+\(
+\begin{eqnarray}
+\Lambda |a\rangle&=& a|a\rangle,\\
+\Lambda|b\rangle&=& b|b\rangle.\\
+\end{eqnarray}
+\)
+If we take the difference of these eigenstates, we find that
+\(
+\begin{eqnarray}
+\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle
+\qquad \text{(because $\Lambda$ is linear.)}\\
+&=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and
+$|b\rangle$ are eigenstates of $\Lambda$)}
+\end{eqnarray}\)
+which means that $a\neq b$.
+** Eigenvectors of hermitian operators span the space of solutions
+#+begin_quote
+*Theorem:* If $\Omega$ is a hermitian operator, then every physically
+allowed state can be written as a linear sum of eigenstates of
+$\Omega$.
+#+end_quote
+** Momentum and energy are hermitian operators
+This ought to be true because hermitian operators correspond to
+observable quantities. Since we expect momentum and energy to be
+measureable quantities, we expect that there are hermitian operators
+to represent them.
+** Momentum and energy eigenstates in vacuum
+An eigenstate of the momentum operator $P$ would be a state
+\(|p\rangle\) such that \(P|p\rangle=p|p\rangle\).
+** Momentum and energy eigenstates in the infinitely deep well
+* Can you measure momentum in the infinitely deep well?
+In summary, I thought I knew:
+1. For any hermitian operator: eigenstates with different eigenvalues
+are orthogonal.
+2. For any hermitian operator: any physically realistic state can be
+written as a linear sum of eigenstates of the operator.
+3. The momentum operator and energy operator are hermitian, because
+momentum and energy are observable quantities.
+4. (The form of the momentum and energy eigenstates in the vacuum potential)
+5. (The form of the momentum and energy eigenstates in the infinitely deep well potential)
+Additionally, I understood that because the infinitely deep potential
+well is not realistic, states of such a system  are not necessarily
+physically realistic. Instead, I understood
+\ldquo{}realistic states\rdquo{} to be those that satisfy the physically
+unrealistic Schr\ouml{}dinger equation and its boundary conditions.
+With that final caveat, here is the problem:
+According to (5), the momentum eigenstates in the well are
+\(p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
+However, /these/ states are not orthogonal, which contradicts the
+assumption that (3) the momentum operator is hermitian and (2)
+eigenstates of a hermitian are orthogonal if they have different eigenvalues.
+#+begin_quote
+*Problem 1. The momentum eigenstates of the well are not orthogonal*
+/Proof./ If $p_1\neq p_2$, then
+\(\begin{eqnarray}
+\langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)dx\\
+&=& \int_0^a p_1^*(x)p_2(x)dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{
+outside the well.}\\
+&=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)dx}}
+\end{eqnarray}\)
+$\square$
+#+end_quote
+** COMMENT  Momentum eigenstates
+In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the
+momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\).
+In the infinitely deep potential well, the Hamiltonian is the same but
+there is a new condition in order for states to qualify as physically
+allowed: the states must not exist anywhere outside of well, as it
+takes an infinite amount of energy to do so.
+Notice that the momentum eigenstates defined above do /not/ satisfy
+this condition.
+* COMMENT
+For each physical system, there is a Schr\ouml{}dinger equation that
+describes how a particle's state $|\psi\rangle$  will change over
+time.
+\(\begin{eqnarray}
+i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
+H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
+This is a differential equation; each solution to the
+Schr\ouml{}dinger equation is a state that is physically allowed for
+our particle. Here, physically allowed states are
+those that change in physically allowed ways. However, like any differential
+equation, the Schr\ouml{}dinger equation can be accompanied by
+/boundary conditions/\mdash{}conditions that further restrict which
+states qualify as physically allowed.
+** Eigenstates of momentum
+#In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger
+#\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\)
+* COMMENT
+#* The infinite square well potential
+A particle exists in a potential that is
+infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the
+particle exists in a potential[fn:coords][fn:infinity]
+\(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for
+}\;x<0\text{ or }x>a.\end{cases}\)
+The Schr\ouml{}dinger equation describes how the particle's state
+\(|\psi\rangle\) will change over time in this system.
+\(\begin{eqnarray}
+i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
+H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
+This is a differential equation; each solution to the
+Schr\ouml{}dinger equation is a state that is physically allowed for
+our particle. Here, physically allowed states are
+those that change in physically allowed ways. However, like any differential
+equation, the Schr\ouml{}dinger equation can be accompanied by
+/boundary conditions/\mdash{}conditions that further restrict which
+states qualify as physically allowed.
+Whenever possible, physicists impose these boundary conditions:
+- A physically allowed state ought to be a /smoothly-varying function of position./ This means
+that if a particle in the state  is likely to be /at/ a particular location,
+it is also likely to be /near/ that location.
+These boundary conditions imply that for the square well potential in
+this problem,
+- Physically allowed states must be totally confined to the well,
+because it takes an infinite amount of energy to exist anywhere
+outside of the well (and physically allowed states ought to have
+only finite energy).
+- Physically allowed states must be increasingly unlikely to find very
+close to the walls of the well. This is because of two conditions: the above
+condition says that the particle is /impossible/ to find
+outside of the well, and the smoothly-varying condition says
+that if a particle is impossible to find at a particular location,
+it must be unlikely to be found nearby that location.
+#; physically allowed states are those that change in physically
+#allowed ways.
+#** Boundary conditions
+Because the potential is infinite everywhere except within the well,
+a realistic particle must be confined to exist only within the
+well\mdash{}its wavefunction must be zero everywhere beyond the walls
+of the well.
+[fn:coords] I chose my coordinate system so that the well extends from
+\(0<x<a\). Others choose a coordinate system so that the well extends from
+\(-\frac{a}{2}<x<\frac{a}{2}\). Although both coordinate systems describe the same physical
+situation, they give different-looking answers.
+[fn:infinity] Of course, infinite potentials are not
+realistic. Instead, they are useful approximations to finite
+potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
+of the well\rdquo{} are close enough for your own practical
+purposes. Having introduced a physical impossibility into the problem
+already, we don't expect to get physically realistic solutions; we
+just expect to get mathematically consistent ones. The forthcoming
+trouble is that we don't.

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