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