I'm studying Quantum Mechanics right now and working through an example in the book of an eigenfunction with a continuous spectrum - the momentum operator, $\hat{p} = -i\hbar\frac{d}{dx}$.
In the example we have our eigenfunctions $f_{p}(x) = Ae^{ipx/\hbar}$ with eigenvalues $p$ and have shown thus far that if $p \in \mathbb{C}$ then the eigenfunctions are not square integrable (i.e., $\int_{-\infty}^{\infty} |f_{p}(x)|^{2}dx$ diverges)
- My first question is why necessarily is it that when $p$ is complex that $f_{p}(x)$ is not square integrable? I imagine it has something to do with having $p^{*} \neq p$ but the book just glosses over this result as though it was something that should be obvious to us.
We then proceed to show that we can recover an ersatz orthonormality if we restrict our eigenvalues to be real.
$\int_{-\infty}^{\infty} [f_{p'}(x)]^{*}f_{p}(x)dx = |A|^{2}\int_{-\infty}^{\infty} e^{-ip'x/\hbar}e^{ipx/\hbar}dx = |A|^{2}\int_{-\infty}^{\infty} e^{i(p-p')x/\hbar}dx = |A|^{2} \cdot 2\pi\hbar\delta (p-p')$
- Now I'm having trouble seeing where we're getting the result from the final integral. I'm aware of the following properties of the diract delta function: $\delta(cx) = \frac{1}{|c|} \delta(x)$ and $\delta (x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ikx}dk$. Given these properties I would imagine that the $c$ in property $1$ would just be $\frac{1}{\hbar}$ which would give us an $\hbar$ out front as a coefficient and then for property $2$ we would take $x$ in $\delta (x)$ to be $p-p'$ and then we associate the $k$ and $dk$ with $\frac{x}{\hbar}$ and $dx$? And then since our resultant integral doesn't have a $2\pi$ in it we can surmise that it's equivalent to the definition of $\delta (x)$ only multiplied by $2\pi$? A walkthrough of this would be very much appreciated.
Finally, given the definition of the dirac delta I was able to find online (assumed knowledge in the book, I guess) where:
$\delta (x-x_{0}) = \begin{cases} \infty&\text{if}\, x = x_{0}\\ 0&\text{otherwise}\\ \end{cases} $
- How would this be a sort of orthonomality? This would be analogous to $\langle \psi_m | \psi_n \rangle = \infty$ when $m = n$ in the usual discrete case which wouldn't make sense because that's the normalization condition and we would expect that the inner product of $2$ normalized functions to be 1. The other case in the piecewise function makes sense as the orthogonality condition.
I appreciate your time and effort in helping me to understand some of these things.
Practically speaking, you cannot know the exact momentum of a particle. If you could, then the system would be in an eigenstate of momentum. However, eigenstates must be in $L^2(\mathbb{R})$, and you don't have that. You can contrive pure states by using delta functions, but that introduces added complexity into the mathematical model, and it can appear to be inconsistent if not handled extremely carefully. With that in mind, it is reasonable to look at a distribution density $\rho(x)$ and look at the corresponding momentum state $$ C\int_{-\infty}^{\infty}\rho(x)e^{ipx/\hbar}dx,\;\;\int_{-\infty}^{\infty}\rho(x)dx=1. $$ (Here $C$ is a normalization constant associated with the Fourier transform.) This can be as sharply concetrated near some $p_0$ as you like, but the physical reality does not allow you to have it perfectly concentrated at a single point because $e^{ipx}\notin L^2$. If you want to take a logical approach to understanding this physical model, then you must work within this framework. Once you understand it, then you can conceive of allowing $\rho$ to evolve into a $\delta$-function. And, as you do that, you can explore the effect on the position distribution. Then you'll understand the Physicist's handling of the limiting cases where either position or momentum distributions are approaching delta-function distributions
Physicists are very good at reasoning through the limiting cases by using constructs such as the Dirac delta function, but I don't think you can really get good at that type of reasoning until you can understand them as limiting cases. For example, consider $$ \left\langle \frac{1}{\sqrt{2\pi}}\frac{1}{2h}\int_{p_0-h}^{p_0+h}e^{-ipx}dp ,\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\rho(p)e^{-ipx}dp\right\rangle \\ = \frac{1}{2h}\int_{p_0-h}^{p_0+h}\rho(p)dp\rightarrow\rho(p_0)\mbox{ as } h\rightarrow 0. $$