Why is $\frac{1}{\pi ^{1/4} \sqrt{a}}e^{-x^2/(2a^2)}$ not a gaussian function?

Question

In an quantum mechanics exercise, we were asked to find the ground-state wavefunction of a perturbed harmonic system. The resulting wave-function is $$\psi_0(x) = \frac{1}{\pi ^{1/4} \sqrt{a}}e^{-x^2/(2a^2)}$$

After that, there is a true or false statement:

The ground-state wave function of the perturbed system is a gaussian. -> False.

May I ask you why this is not a gaussian ? When looking on Wikipedia https://en.wikipedia.org/wiki/Gaussian_function , this seems to be a gaussian for me.

Thanks for your help !

Answer 1

The PDF is $|\psi_0|^2=\frac{1}{a\sqrt{\pi}}\exp\frac{-x^2}{a^2}$, that of an $N(0,\,a^2/2)$ distribution, but $\psi_0$ is not itself a Gaussian PDF. In particular, $\int_{\Bbb R}\psi_0(x)=\pi^{1/4}\sqrt{2a}$.