#+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.