Math involved in Quantum Field Theory I recently asked a question on physics stackexchange on the math in Quantum Field Theory, but have yet to receive an answer:
https://physics.stackexchange.com/questions/714183/inconsistency-in-mathematics-of-qft
I feel like my question boils down to a question of mathematical rigor, so I post a related question here at math stackexchange, but a much more concrete one:
Consider the wave function:
\begin{align}
\psi (x, t) = [\frac{1}{\sigma[1+i(t/\tau)]\sqrt{2\pi}}]^{\frac{1}{2}}\exp(-\frac{1}{4}\frac{x^2}{\sigma^{2}[1+i(t/\tau)]}), 
\end{align}
where $\sigma, \tau$ are positive finite constants. If you consider its norm squared, namely:
\begin{align}
|\psi|^2 = \frac{1}{[2\pi\sigma^2(1+t^2/
\tau^2)]^{\frac{1}{2}}} \exp(-\frac{x^2}{2\sigma^2[1+t^2/\tau^2]})
\end{align}
and consider its integration over the entire real line over x and then taking this integral to the limit of $t \to \infty$:
\begin{align}
\lim_{t \to \infty}(\int_{-\infty}^{\infty}|\psi|^2dx) = \lim_{t \to \infty} 1 = 1.
\end{align}
The second eaquality is because the Gaussian distribution is correctly normalized.
However, if one exchange the order of taking the limit in time, t and integration over space x, the result is different:
\begin{align}
\int_{-\infty}^{\infty}(\lim_{t \to \infty}|\psi|^2dx) = \int_{-\infty}^{\infty} (0) \,dx = 0.
\end{align}
There is definitely some math involving the definition of the infinity that I misused here, but can someone with math background point out exactly how? Please keep in mind that I have a physics background, and only have a very little bit of knowledge in real analysis, etc, so if you can, please explain in less technical terms, or source some background materials if technical terms are inevitable.
 A: Note that a priori there is no reason why two limits of a certain expression (the integral is also a limit) should be interchangable. For instance, consider the double sequence
$$
\mathbb{N}\times\mathbb{N}\to\mathbb{R},~a_{n,m}=\frac{n}{m}.
$$
Whenever you first take the $n$-limit it diverges and whenever you first take the $m$-limit it converges to 0. You can even recover any arbitrary rational number $\frac{p}{q}$ as a limit of simultaneously $n\to\infty,m\to\infty$ if you consider it along the "ray" $k\mapsto(p\cdot k,q\cdot k)$ (or many other possibilities), yielding $a_{pk,qk}\to\frac{p}{q}$ as $k\to\infty$.
However, there are some theorems which allow to interchange the integral with the point-wise limit of a family of functions (as your function family $x\mapsto|\psi(x,t)|^2$ indexed by, say, $t>0$). For example the theorem of Lebesgues dominated convergence. The latter states that if there exists an integrable $g$ on $\mathbb{R}$ such that $|\psi(x,t)|^2\le g(x)$ for all $t>0$ (for the limit $t\to \infty$ it would suffice that the estimate holds above some lower bound, i.e. for all $t>t_0$ for some $t_0$), then you can integrate the pointwise limit (which of course must exist in the first place).
If you want to prove that limit and integral are not interchangeable in this particular example, you would just do your computation and show that the results do not coincide.
But you can even see why the hypothesis of Lebesgues dominated convergence is false: If you want to dominate the function family $\{x\mapsto|\psi(x,t)|^2\}_{t>0}$, say, by the neatest bound
$$
x\mapsto\sup_{t>0}|\psi(x,t)|^2
$$
you find that for $|x|\ge1$ this supremum is taken at $t=\sqrt{x^2-1}$ with
$$
\sup_{t>t_0}|\psi(x,t)|^2=|\psi(x,\sqrt{x^2-1})|^2=\text{constant}\cdot\frac{1}{x},
$$
a non-integrable function. In fact your example of a decaying Gaussian curve is a nice example in which you see why a common bound must exist and be integrable. Here, where it doesn't exist, you can distribute the contribution of $|\psi(\cdot,t)|^2$ to its integral, which is constantly 1, arbitrarily wide to still maintain a pointwise limit of 0.
In the end, for many examples in physics you can interchange integral and limit, but actually you need to show why. However, for sure physics professors know that but only talk about that if in a particular case of interest it goes wrong.
