# Normalization condition and phase constant(/)

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!)

• Are you sure about the last exponential? Maybe it is $e^{-2\alpha x^2}$... – MattG88 Nov 23 '16 at 23:23
• Sorry, e^2αx^2 should read e^-αx^2 – Mike A Nov 23 '16 at 23:26
• Ok then I calculated that the normalization constant should be $\frac{\sqrt{2}}{2}$..is it possible? – MattG88 Nov 23 '16 at 23:28
• Simon's advice below steered me right. I finished the integral and it was equal to 1, which verified that the wave function was normalized. This also means the normalization function was (2α/π)^1/4. Thanks for looking into it for me! – Mike A Nov 23 '16 at 23:39

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}
• I think that the integral is equal to $\frac{\sqrt 2}{2}$, so the wave function must be normalized...please check it – MattG88 Nov 23 '16 at 23:52
• corrected it. The exponent-part should be $e^{-\alpha x^2}$, and after squaring $e^{-2\alpha x^2}$. – Simon Nov 24 '16 at 0:01