Normalisation of a free particle with Gaussian wave packet

Question

The Gaussian wave packet where the $x$-dependence is given by the wave function $$\Phi(x) = N\exp\bigg(ikx - \frac{x^2}{2\Delta^2}\bigg)$$ $N$ is a normalisation constant. $k$ is the wave number. I need to find the case where: $$\int^{\infty}_{-\infty} |\Phi(x)|^2\,\mathrm{d}x = \int^{\infty}_{-\infty} \Phi^*(x)\Phi(x)\,\mathrm{d}x = 1$$ $\Phi^*(x)$ is the complex conjugate. I want to find the normalisation constant $N$.

I have some integrals to use but I'm not sure how to do it: $$\int^{\infty}_{-\infty} e^{-\alpha x^2}\,\mathrm{d}x = \sqrt{\frac{\pi}{\alpha}},\quad \int^{\infty}_{-\infty} xe^{-\alpha x^2}\,\mathrm{d}x = 0,\quad \int^{\infty}_{-\infty} x^2e^{-\alpha x^2}\,\mathrm{d}x = \frac{1}{2}\sqrt{\frac{\pi}{\alpha^3}}$$ I have thought about rearranging the given wave function to get: $$\Phi(x) = N\exp(ikx)\exp\Big(-\frac{x^2}{2\Delta^2}\Big)$$ But I am also unsure what to make of the $\Delta$ part.

Answer 1

With the assumption that $\Delta$ is just a variable, I have the following solution:

\begin{align} \int^{\infty}_{\infty} |\Phi(x)|^2\,\mathrm{d}x &= \int^{\infty}_{\infty} N^2\exp\bigg(-\frac{2x^2}{2\Delta^2}\bigg)\cdot\frac{e^{ikx}}{e^{ikx}}\,\mathrm{d}x = N^2\int^{\infty}_{\infty} \exp\bigg(-\frac{x^2}{\Delta^2}\bigg)\,\mathrm{d}x\\ &= N^2\cdot \sqrt{\frac{\pi}{\frac{1}{\Delta^2}}} = N^2\sqrt{\pi}\Delta = 1 \Rightarrow N = \frac{1}{\sqrt[4]\pi \sqrt\Delta} \end{align} I don't know if Delta means something else in quantum mechanics so if it does then please share. If not then could someone check this please?

EDIT: Yey! I have a confirmation email from my tutor that $\Delta$ is a constant in this case.

Answer 2

Assume that $\Delta^2$ is real and positive and that $k$ is real. Then $$\Phi^*(x)\Phi(x)=N^2\mathrm{e}^{-x^2/\Delta^2}$$ can be integrated using your table.

If $\Delta^2$ is complex and satisfies $\Re \Delta^2 >0$, then $$\Phi^*(x)\Phi(x)=N^2\mathrm{e}^{-x^2\Re \Delta^{-2}}$$ can be integrated using your table.