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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!)

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  • $\begingroup$ Are you sure about the last exponential? Maybe it is $e^{-2\alpha x^2}$... $\endgroup$ – MattG88 Nov 23 '16 at 23:23
  • $\begingroup$ Sorry, e^2αx^2 should read e^-αx^2 $\endgroup$ – Mike A Nov 23 '16 at 23:26
  • $\begingroup$ Ok then I calculated that the normalization constant should be $\frac{\sqrt{2}}{2}$..is it possible? $\endgroup$ – MattG88 Nov 23 '16 at 23:28
  • $\begingroup$ 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! $\endgroup$ – Mike A Nov 23 '16 at 23:39
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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:

$\begin{align} \int_{-\infty}^\infty dx\ \left|\left(\frac{2\alpha}{\pi}\right)^{1/4} e^{-ikx} e^{-\alpha x^2}\right|^2 &= \left(\frac{2\alpha}{\pi}\right)^{1/2}\int_{-\infty}^\infty dx\ e^{-2\alpha x^2} = 1 \end{align}$

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  • $\begingroup$ Yeah, I've clearly been starting off wrong. That's perfect, thank you Simon, I will give it another go! $\endgroup$ – Mike A Nov 23 '16 at 23:30
  • $\begingroup$ The new integral gave me an answer of 1 as required. Thanks again! $\endgroup$ – Mike A Nov 23 '16 at 23:39
  • $\begingroup$ glad to help. quantum mechanics is fun :) $\endgroup$ – Simon Nov 23 '16 at 23:40
  • $\begingroup$ I think that the integral is equal to $\frac{\sqrt 2}{2}$, so the wave function must be normalized...please check it $\endgroup$ – MattG88 Nov 23 '16 at 23:52
  • $\begingroup$ corrected it. The exponent-part should be $e^{-\alpha x^2}$, and after squaring $e^{-2\alpha x^2}$. $\endgroup$ – Simon Nov 24 '16 at 0:01

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