I'm having trouble rationalizing a particular statement that is, surely, present in many quantum mechanics textbooks. The following statement comes from the orthnormalization condition for eigenfunctions of the wavefunction, $\Psi (x, t) $, subject to the momentum operator, $\hbar/i (d/dx)$, with REAL eigenvalues, $\lambda \in R$. The eigenfunctions are of the form:
$\psi_\lambda = e^{i \lambda x / \hbar} $
and then, taking their inner product
$ < \psi_\lambda' | \psi_\lambda > = |A|^2 \int_{-\infty}^{\infty} e^{i (\lambda – \lambda')x/\hbar} dx = |A|^2 2 \pi \hbar \ \delta(\lambda – \lambda') $
So I'm quite unclear about that last part. Of course, if $\lambda' \neq \lambda$ , you are integrating a sinusoid, and the answer is zero, however, isn't:
$ \left. \int_{-\infty}^{\infty} e^{i (\lambda – \lambda')x/\hbar} dx \right|_{\lambda = \lambda'} = \int_{-\infty}^{\infty} 1 dx = \infty $
I mean, $ e^{i (\lambda – \lambda')z/\hbar}$ is analytic everywhere, so you could just complex integrate, treating i as a constant and take limits, (same answer, no?). I also thought about Cauchy's Integral Theorem, and doing a contour integral, and using the maximum modulus principle:
$ \lim_{p \rightarrow \infty} \left. \int_{-p}^{p} e^{i (\lambda – \lambda')x/\hbar} dx \right|_{\lambda = \lambda'} = 2 \pi \sum_i f(z_i) – lim_{p \rightarrow \infty} p \left| e^{i(\lambda – \lambda') p / \hbar} \right| $
where $z_i$ is the location of singularity i. However, as far as I can tell…there are no singularities, and the term on the right is unbounded, ( $\rightarrow \infty$). So why does that evaluate to the delta function?
Thanks
Best Answer
When you see a delta function, you should always understand any equation, identity or expression in a distributional sense, which means that
$$\int_{-\infty}^{\infty} f(\lambda')\delta(\lambda-\lambda')d\lambda'=f(\lambda)$$
for any sufficiently smooth $f(\lambda)$.
Now
$$\int_{-\infty}^{\infty} f(\lambda')\int_{-\infty}^{\infty} e^{i (\lambda - \lambda')x/\hbar} dx d\lambda'$$
is, provided the integrations can be exchanged,
$$\int_{-\infty}^{\infty} e^{i\lambda x/\hbar}\int_{-\infty}^{\infty} f(\lambda')e^{- i \lambda'x/\hbar} d\lambda' dx = \int_{-\infty}^{\infty} e^{i\lambda x/\hbar}\tilde{f}(x/(2\pi\hbar)) dx = 2\pi\hbar f(\lambda) $$
where I successively applied the Fourier transform and the inverse Fourier transform.