# College Math Teaching

## August 11, 2011

### Quantum Mechanics and Undergraduate Mathematics XIII: simplifications and wave-particle duality

In an effort to make the subject a bit more accessible to undergraduate mathematics students who haven’t had much physics training, we’ve made some simplifications. We’ve dealt with the “one dimensional, non-relativistic situation” which is fine. But we’ve also limited ourselves to the case where:
1. state vectors are actual functions (like those we learn about in calculus)
2. eigenvalues are discretely distributed (e. g., the set of eigenvalues have no limit points in the usual topology of the real line)
3. each eigenvalue corresponds to a unique eigenvector.

In this post we will see what trouble simplifications 1 and 2 cause and why they cannot be lived with. Hey, quantum mechanics is hard!

Finding Eigenvectors for the Position Operator
Let $X$ denote the “position” operator and let us seek out the eigenvectors for this operator.
So $X\delta = x_0 \delta$ where $\delta$ is the eigenvector and $x_0$ is the associated eigenvalue.
This means $x\delta = x_0\delta$ which implies $(x-x_0)\delta = 0$.
This means that for $x \neq x_0, \delta = 0$ and $\delta$ can be anything for $x = x_0$. This would appear to allow the eigenvector to be the “everywhere zero except for $x_0$” function. So let $\delta$ be such a function. But then if $\psi$ is any state vector, $\int_{-\infty}^{\infty} \overline{\delta}\psi dx = 0$ and $\int_{-\infty}^{\infty} \overline{\delta}\delta dx = 0$. Clearly this is unacceptable; we need (at least up to a constant multiple) for $\int_{-\infty}^{\infty} \overline{\delta}\delta dx = 1$

The problem is that restricting our eigenvectors to the class of functions is just too restrictive to give us results; we have to broaden the class of eigenvectors. One way to do that is to allow for distributions to be eigenvectors; the distribution we need here is the dirac delta. In the reference I linked to, one can see how the dirac delta can be thought of as a sort of limit of valid probability density functions. Note: $\overline{\delta} = \delta$.

So if we let $\delta_0$ denote the dirac that is zero except for $x = x_0$, we recall that $\int_{\infty}^{\infty} \delta_0 \psi dx = \psi(x_0)$. This means that the probability density function associated with the position operator is $P(X = x_0) = |\psi(x_0)|^2$

This has an interesting consequence: if we measure the particle’s position at $x = x_0$ then the state vector becomes $\delta_0$. So the new density function based on an immediate measurement of position would be $P( X = x_0) = |\langle \delta_0, \delta_0 \rangle|^2 = 1$ and $P(X = x) = 0$ elsewhere. The particle behaves like a particle with a definite “point” position.

Momentum: a different sort of problem

At first the momentum operator $P\psi = -i \hbar \frac{d\psi}{dx}$ seems less problematic. Finding the eigenvectors and eigenfunctions is a breeze: if $\theta_0$ is the eigenvector with eigenvalue $p_0$ then:
$\frac{d}{dx} \theta_0 = \frac{i}{\hbar}p_0\theta_0$ has solution $\theta_0 = exp(i p_0 \frac{x}{\hbar})$.
Do you see the problem?

There are a couple of them: first, this provides no restriction on the eigenvalues; in fact the eigenvalues can be any real number. This violates simplification number 2. Secondly, $|\theta_0|^2 = 1$ therefore $|\langle \theta_0, \theta_0 \rangle |^2 = \infty$. Our function is far from square integrable and therefore not a valid “state vector” in its present form. This is where the famous “normalization” comes into play.

Mathematically, one way to do this is to restrict the domain (say, limit the non-zero part to $x_0 < x < x_1$ ) and multiply by an appropriate constant.

Getting back to our state vector: $exp(ip_0 \frac{x}{\hbar}) = cos(\frac{p_0 x}{\hbar}) + i sin(\frac{p_0 x}{\hbar})$. So if we measure momentum, we have basically given a particle a wave characteristic with wavelength $\frac{\hbar}{p_0}$.

Now what about the duality? Suppose we start by measuring a particle’s position thereby putting the state vector in to $\psi = \delta_0$. Now what would be the expectation of momentum? We know that the formula is $E(P) = -i\hbar \int-{-\infty}^{infty} \delta_0 \frac{\partial \delta_0}{\partial x} dx$. But this quantity is undefined because $\frac{\partial \delta_0}{\partial x}$ is undefined.

If we start in a momentum eigenvector and then wish to calculate the position density function (the expectation will be undefined), we see that $|\theta_0|^2 = 1$ which can be interpreted to mean that any position measurement is equally likely.

Clearly, momentum and position are not compatible operators. So let’s calculate $XP - PX$
$XP \phi = x(-i\hbar \frac{d}{dx} \phi) = -xi\hbar \frac{d}{dx} \phi$ and $PX\phi = -i \hbar\frac{d}{dx} (x \phi) = -i \hbar (\phi + x \frac{d}{dx}\phi)$ hence $(XP - PX)\phi = i\hbar \phi$. Therefore $XP-PX = i\hbar$. Therefore our generalized uncertainty relation tells us $\Delta X \Delta P \geq \frac{1}{2}h$
(yes, one might object that $\Delta X$ really shouldn’t be defined….) but this uncertainty relation does hold up. So if one uncertainty is zero, then the other must be infinite; exact position means no defined momentum and vice versa.

So: exact, pointlike position means no defined momentum is possible (hence no wave like behavior) but an exact momentum (pure wave) means no exact pointlike position is possible. Also, remember that measurement of position endows a point like state vector of $\delta_0$ which destroys the wave like property; measurement of momentum endows a wave like state vector $\theta_0$ and therefore destroys any point like behavior (any location is equally likely to be observed).