Momentum: eigenvalues and normalization

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Section 3.3.2; Problem 3.9.

The example of a periodic function which we studied earlier had discrete eigenvalues for both the first and second derivative of the periodic variable. In particular, for the operator ${id/d\phi}$ we found that the eigenvalues are all integers, with eigenfunctions ${e^{in\phi}}$ since

$\displaystyle i\frac{d}{d\phi}e^{in\phi}=-ne^{in\phi} \ \ \ \ \ (1)$

This operator bears a strong resemblance to the momentum operator in one dimension, which is ${\hat{p}=-i\hbar d/dx}$. However, if we try to find the eigenvalues and eigenfunctions of ${\hat{p}}$, we run into a bit of a problem. We try to solve, for some eigenvalue ${p}$:

 $\displaystyle \hat{p}f$ $\displaystyle =$ $\displaystyle pf\ \ \ \ \ (2)$ $\displaystyle -i\hbar\frac{d}{dx}f$ $\displaystyle =$ $\displaystyle pf \ \ \ \ \ (3)$

This has the solution

$\displaystyle f_{p}(x)=Ae^{ipx/\hbar} \ \ \ \ \ (4)$

for some constant ${A}$. Ordinarily, at this stage, we would impose some boundary condition on the solution to obtain acceptable values of ${p}$. The problem is that we’d like to define this function over all ${x}$ and, if we try to do this, the function is not normalizable for any value of ${p}$. At first glance, we might think that if we chose ${p}$ to be purely imaginary as in ${p=\alpha i}$, it might work since we get

$\displaystyle f(x)=Ae^{-\alpha x/\hbar} \ \ \ \ \ (5)$

but of course this tends to infinity at large negative ${x}$ so that doesn’t work. In fact if ${p}$ has a non-zero imaginary part, ${f(x)}$ goes to infinity at one end of its domain. So we’re restricted to looking at real values of ${p}$.

In that case, ${f(x)}$ is periodic and thus is still not normalizable. Thus there are no eigenfunctions of the momentum operator that lie in Hilbert space (which, remember, is the vector space of square-integrable functions).

What happens if do the normalization integral anyway? That is, we try

$\displaystyle \int_{-\infty}^{\infty}f_{p_{1}}^*\left(x\right)f_{p_{2}}\left(x\right)dx=\left|A\right|^{2}\int_{-\infty}^{\infty}e^{i\left(p_{2}-p_{1}\right)x/\hbar}dx \ \ \ \ \ (6)$

By using the variable transformation ${\xi\equiv x/\hbar}$, we get

$\displaystyle \int_{-\infty}^{\infty}f_{p_{1}}^*\left(x\right)f_{p_{2}}\left(x\right)dx=\left|A\right|^{2}\hbar\int_{-\infty}^{\infty}e^{i\left(p_{2}-p_{1}\right)\xi}d\xi \ \ \ \ \ (7)$

It’s at this point that we invoke the dodgy formula involving the Dirac delta function that we obtained a while back. Using this, we can write the integral as a delta function, and we get

$\displaystyle \int_{-\infty}^{\infty}f_{p_{1}}^*\left(x\right)f_{p_{2}}\left(x\right)dx=2\pi\left|A\right|^{2}\hbar\delta\left(p_{2}-p_{1}\right) \ \ \ \ \ (8)$

This is sort of like a normalization condition, in that the integral is zero when ${p_{1}\ne p_{2}}$ (that is, if you believe that the integral really does evaluate to a delta function), and non-zero (infinite, in fact) if ${p_{1}=p_{2}}$. In fact, if we take the constant ${A}$ to be

$\displaystyle A=\frac{1}{\sqrt{2\pi\hbar}} \ \ \ \ \ (9)$

and use the bra-ket notation for the integral, we can write

$\displaystyle \left\langle \left.f_{p_{1}}\right|f_{p_{2}}\right\rangle =\delta\left(p_{2}-p_{1}\right) \ \ \ \ \ (10)$

We can also express an arbitrary function ${g(x)}$ as a Fourier transform over ${p}$ by writing

 $\displaystyle g(x)$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}c\left(p\right)f_{p}\left(x\right)dp\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}c\left(p\right)e^{ipx/\hbar}dp\ \ \ \ \ (12)$ $\displaystyle g\left(\hbar\xi\right)$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}c\left(p\right)e^{ip\xi}dp \ \ \ \ \ (13)$

From Plancherel’s theorem, we can invert this relation to get ${c\left(p\right)}$:

 $\displaystyle c\left(p\right)$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar}{2\pi}}\int_{-\infty}^{\infty}g\left(\hbar\xi\right)e^{-ip\xi}d\xi\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}g\left(x\right)e^{-ipx/\hbar}dx\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \left.f_{p}\right|g\right\rangle \ \ \ \ \ (16)$

In general, hermitian operators with continuous eigenvalues don’t have normalizable eigenfunctions and have to be analyzed in this way. In particular, the hamiltonian (energy) of a system can have an entirely discrete spectrum (infinite square well or harmonic oscillator), a totally continuous spectrum (free particle, delta function barrier or finite square barrier) or a mixture of the two (delta function well or finite square well).