I have been given a wave function and been tasked with verifying it has been normalized.
$$\psi(x, 0) = \left(\frac{2\alpha}{\pi}\right)^{1/4} e^{-ikx} e^{-\alpha x^2}$$.
I've normalized it by taking the integral of its square from -infinity to infinity. I used $u=\sqrt{2α}x$ in the integral and manipulated it until I could use a given integral
$$e^{-ibu} e^{-u^2} = \sqrt{\pi} e^{-\frac{b^2}{4}}$$.
I cancelled out the normalization constant and anything else i could expecting it to equal 1, but instead i end up with:
$$e^{-b^2/4}$$, where $$b=(2k/2α)^2$$.
Have I gone wrong somewhere along the way or is it a phase constant which would equal 1? In my book, they are supposed to have an i and my answer doesn't.
Any help would be appreciated.
(Thanks for the edit Simon!)
normalization of a wave function means $\int |\psi|^2 = 1$. The absolute-value is important, because this means that the $e^{-ikx}$ term in your wave function simply disappears. The first step in your calculation should thereofre look like this:
$\begin{align} \int_{-\infty}^\infty dx\ \left|\left(\frac{2\alpha}{\pi}\right)^{1/4} e^{-ikx} e^{-\alpha x^2}\right|^2 &= \left(\frac{2\alpha}{\pi}\right)^{1/2}\int_{-\infty}^\infty dx\ e^{-2\alpha x^2} = 1 \end{align}$