Probability of non-normalizable states

Question

In the book Quantum Field Theory by Jakob Schwichtenberg, he is discussing about non-normalizable states in chapter 8. If you compute the normalization in $(8.67)$ you just get infinity. He said one way out of this is to just assume a finite volume $V$ so that the normalization is now a finite $V$ in $(8.68)$.

The part I'm not so sure how to think about is $(8.69)$, where you can compute probabilities of non-normalizable states only if you consider probability ratios. Can anyone give me an example where a non-normalizable state $\lvert i \rangle$ time evolves to a non-normalizable state $\lvert f \rangle$ through $U$ and that in the end there is a volume $V$ factor that comes out to cancel $\langle i \lvert i \rangle$ and $\langle f \lvert f \rangle$ which are just equal to $V$?

Update 1.

First, let $\vert f \rangle = \vert \vec{k} \rangle$ and $\vert i \rangle = \vert \vec{q} \rangle$

$$\langle \vec{k} \vert \vec{k} \rangle = \int_{-L/2}^{L/2}\int_{-L/2}^{L/2}\int_{-L/2}^{L/2} d^3x \langle \vec{k} \vert \vec{x} \rangle \langle \vec{x} \vert \vec{k} \rangle = \int_{-L/2}^{L/2}\int_{-L/2}^{L/2}\int_{-L/2}^{L/2} d^3x e^{-i\vec{k} \cdot \vec{x}} e^{i\vec{k} \cdot \vec{x}} = L^3 = V.$$

Similar computations are done for $\langle \vec{q} \vert \vec{q} \rangle$.

For $|\langle f \vert \mathrm{U} \vert i \rangle|^2 = |\langle \vec{k} \vert \mathrm{U} \vert \vec{q} \rangle|^2 = \langle \vec{k} \vert \mathrm{U} \vert \vec{q} \rangle \langle \vec{q} \vert \mathrm{U}^* \vert \vec{k} \rangle$,

$\langle \vec{k} \vert \mathrm{U} \vert \vec{q} \rangle = \int_{-L/2}^{L/2}\int_{-L/2}^{L/2}\int_{-L/2}^{L/2} d^3x \langle \vec{k} \vert \vec{x} \rangle \langle \vec{x} \vert \mathrm{U} \vert \vec{q} \rangle = \int_{-L/2}^{L/2}\int_{-L/2}^{L/2}\int_{-L/2}^{L/2} d^3x \langle \vec{k} \vert \vec{x} \rangle \langle \vec{x} \vert e^{-i H t} \vert \vec{q} \rangle =\\ = \int_{-L/2}^{L/2}\int_{-L/2}^{L/2}\int_{-L/2}^{L/2} d^3x e^{-i \vec{k} \cdot \vec{x}} e^{-i \omega t} e^{i \vec{q} \cdot \vec{x}} = e^{-i \omega t} \int_{-L/2}^{L/2}\int_{-L/2}^{L/2}\int_{-L/2}^{L/2} d^3x e^{i (\vec{q}-\vec{k}) \cdot \vec{x}}.$

$\langle \vec{q} \vert \mathrm{U}^* \vert \vec{k} \rangle = e^{i \omega t} \int_{-L/2}^{L/2}\int_{-L/2}^{L/2}\int_{-L/2}^{L/2} d^3x e^{i (\vec{k}-\vec{q}) \cdot \vec{x}}.$

$\lim_{L \rightarrow \infty} \left[ \lim_{\vec{q} \rightarrow \vec{k}} \frac{\langle \vec{k} \vert \mathrm{U} \vert \vec{q} \rangle \langle \vec{q} \vert \mathrm{U}^* \vert \vec{k} \rangle}{ \langle \vec{k} \vert \vec{k} \rangle \langle \vec{q} \vert \vec{q} \rangle}\right] = \frac{(2 \pi)^3 \delta(0) (2 \pi)^3 \delta(0)}{V^2} = \frac{V^2}{V^2}$

where I have applied eq. (8.71).

Thus, $P(i \rightarrow f) = 1$

I’m not sure if my calculation is correct because it assumes $q \rightarrow k$ so that the final state is the same as the initial state even after some time evolution $U$. How about for plane waves that have different initial and final states?

Update 2.

Following Ruslan's answer,

$|\langle f \vert \mathrm{U} \vert i \rangle|^2 = \langle f \vert \mathrm{U} \vert i \rangle \langle i \vert \mathrm{U}^* \vert f \rangle$

$\langle f \vert \mathrm{U} \vert i \rangle = \int_{-L/2}^{L/2}\int_{-L/2}^{L/2}\int_{-L/2}^{L/2} d^3x \langle f \vert \vec{x} \rangle \langle \vec{x} \vert \mathrm{U} \vert i \rangle = \int_{-L/2}^{L/2}\int_{-L/2}^{L/2}\int_{-L/2}^{L/2} d^3x \langle f \vert \vec{x} \rangle \langle \vec{x} \vert e^{-i H t} \vert i \rangle \\ = \int_{-L/2}^{L/2}\int_{-L/2}^{L/2}\int_{-L/2}^{L/2} d^3x e^{-i \vec{k} \cdot \vec{x}} e^{-i \omega t} e^{i \vec{k} \cdot \vec{x}} = e^{-i \omega t} \int_{-L/2}^{L/2}\int_{-L/2}^{L/2}\int_{-L/2}^{L/2} d^3x = e^{-i \omega t} V.$

$\langle i \vert \mathrm{U}^* \vert f \rangle = e^{i \omega t} V$

where when we impose the boundary conditions, we get $k\frac{L}{2} = N\pi$. Also, note that $V=L^3$.

Thus, $P(i \rightarrow f) = \frac{|\langle f \vert \mathrm{U} \vert i \rangle|^2}{\langle f \vert f \rangle \langle i \vert i \rangle} = \frac{V^2}{V^2} = 1$

Now, we can take $\lim_{V \rightarrow \infty}$ and still maintain a finite probability, which is 1.

Answer 1

Can anyone give me an example where a non-normalizable state $\lvert i \rangle$ time evolves to a non-normalizable state $\lvert f \rangle$ through $U$ and that in the end there is a volume $V$ factor that comes out to cancel $\langle i \lvert i \rangle$ and $\langle f \lvert f \rangle$ which are just equal to $V$?

What about the simplest example: a plane wave, with free-particle evolution? Take e.g.

\begin{align} \langle x|i\rangle&=\exp(ikx),\\ \langle x|f\rangle&=\exp\left(ikx-it\frac{k^2}{2m}\right). \end{align}

These are normalizable if we take a finite interval $L=\frac{2\pi N}k$ with $N\in\mathbb Z$ (assuming Born-von Karman boundary conditions). The norms of both states are $\sqrt L.$ As we take $N\to\infty$, the states become non-normalizable.

With these states we have (taking $\mathcal U=I$ for simplicity):

$$\langle i|\mathcal{U}|f\rangle=\int_0^L \langle i|x\rangle\langle x|f\rangle\,\mathrm{d}x=L\exp\left(-it\frac{k^2}{2m}\right).$$

Then the expression $(8.69)$ in your textbook indeed cancels the $L$, yielding a finite $P(i\to f).$

Answer 2

In an actual physical measurement one will always have a finite precision, i.e., final volume: e.g., a particle detected in a bubble chamber leaves a trace of finite width. The non-normalizability is important to deal with, in order to be able, e.g., to use Fourier transform and other related mathematical apparatus, but it is just a mathematical artefact.

In a similar way, all particles emitted in experiments are created in a form of wave packets, as they are generated with instruments having finite precision and created over finite time interval (certainly shorter than the lifetime of the Universe, and typically shorter than duration of the experiment).