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).