Quantum mechanics Dirac delta representation with integral

Question

So I’m doing QM and found bunch of problems for beginners and I’m struggling with this one: $$\lim_{a\rightarrow 0}\int^{\infty}_{-\infty}e^{\frac{ip x}{\hbar}-a x^2}dx=2\pi\hbar\delta(p).$$ If I swap the limit and the integral I get the correct answer, but can I do that?

Answer 1

Using $\int_{-\infty}^{\infty}e^{-\alpha x^2+\beta x}=\sqrt{\frac{\pi}{\alpha}}\exp{\frac{\beta^2}{4\alpha}}$
put $n=1/a$ we get

$\lim_{n \to \infty}\sqrt{\pi n \cdot\frac {\pi 4 \hbar^2}{\pi 4 \hbar^2}}\exp{\frac{-p^2n}{4\hbar^2}}$ Now define $n=>n/4\hbar^2$

Now,$e^{-nx^2}\sqrt{\frac{n}{\pi}}$ defines a generalised function $\delta(x)$ such that

$\int_{-\infty}^{\infty}\delta(x)F(x)=F(o)$ in limit $n\rightarrow\infty$ can be easily proved (ref.M.J.Lighthill Fourier and Generalised Functions for more)

We get your answer $2\pi\hbar\delta(p)$

Answer 2

You can just rewrite the integral first(make substitutions $x = -x$, $\frac{p}{\hbar}=k$), say $$I=\int_{-\infty}^{\infty}dx \ e^{-i k.x} e^{-ax^2}$$ Now, this is a like a Fourier transform of a Gaussian function.

Where $$I = \sqrt{\frac{\pi}{a}}e^{-\frac{k^2}{4a}}$$

Therefore, you can look at the definitions of dirac delta as limit, and write $$ 2\pi \ \lim_{a\rightarrow 0} \frac{1}{2\sqrt{{\pi a}}}e^{\frac{-k^2}{4a}} = 2\pi \delta(k) $$ Now, using the property of dirac delta $\delta(ax) = \frac{1}{|a|}\delta(x)$, you can get to your final form.

PS: But, it would be interesting to think of the physical motivation of this manipulation for the context in which OP might be looking at this question.