I Derivation of Eigenfunctions/Eigenvalues of the Momentum Operator

[TRB8985]

TL;DR

Just trying to understand the derivation of the eigenvalues & eigenfunctions of the momentum operator.

Good afternoon all,

In David Griffiths' "Intro to Quantum Mechanics", I'm looking through Example 3.2 on page 115 that shows how to get the eigenfunctions and eigenvalues of the momentum operator.

I completely understand everything up until this part:

$\int_{-\infty}^{\infty} f_p'^*(x) f_p(x) dx = |A|^2 \int_{-\infty}^{\infty} e^{(i(p-p')x/\hbar)}dx = |A|^2 2\pi \hbar \delta(p-p')$

Where $f_p(x) = Ae^{ipx/\hbar}.$

I'm not really understanding how the first parts follow into the last. When I try to do the integration explicitly of the middle part, I end up with the following:

$|A|^2 \frac {\hbar} {i(p-p')} \int_{-\infty}^{\infty}e^{(i(p-p')x/\hbar)}dx$

I can see how $\hbar$ ends up in the numerator on the end, but the factor of $2 \pi$ seemingly comes out of nowhere, and the delta function just appears without any factor of $i$ attached somehow.

Could anyone explain the missing link between the middle part of the equation and the last? Thank you so very much.

 

[TRB8985]

I think I've figured this out, so I thought I'd post what I found so if anyone else gets stuck they can reference back here. Please feel free to point out if this is incorrect!

Referring back to the definition of the Dirac delta function on page 89:

$$ \delta(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ikx} dk $$

If I pull up the expression that was giving me trouble in the middle, we basically have the same kind of form, but with integration to x instead of k:

$$ |A|^2 \int_{-\infty}^{\infty} e^{i(p-p')x/\hbar} dx$$

Thus, I'm not going have have a delta in the form of $\delta(x)$ like when the integration variable was with respect to k, but now, since the variable has switched to x, my delta function essentially "switches feet" in the exponent to $\delta(\frac{p-p'}{\hbar})$.

Additionally, we have the following quality for the delta function:

$$ \delta(cx) = \frac{1}{|c|}\delta(x) $$

Which means,

$$ \delta(\frac{p-p'}{\hbar}) = \frac{1}{|1/\hbar|}\delta(p-p') = \hbar \delta(p-p')$$

Recall the hidden $1/2\pi$ buried in the definition at the top of this post. To keep the equation balanced (not off by a factor of $1/2\pi$), we'll need to multiply by $2\pi$. In doing so, we match the right-hand side of the equation at the top of my original post:

$$ \implies |A|^2 \int_{-\infty}^{\infty} e^{i(p-p')x/\hbar} dx = |A|^2 2\pi \hbar \delta(p-p')$$

Thank you for reading!