Question

Normalization of the Gaussian wavepacket

I haven't studied complex analysis and I can't figure out why I have to include \[e^{i\phi}\]

Answer 1

The wavepacket is defined as \[\Psi(x)=\Psi_0e^{\frac{ -(x-x_0)^2 }{ 4\sigma^2 }}\]

Answer 2

So, what I do is to square the wavepacket to give:
\[|\Psi_0|^2 \int\limits_{-\infty}^{\infty}e^\frac{ -(x-x_0)^2 }{ 2\sigma^2 }dx=1\]
So, I set \[y = \frac{ -(x-x_0) }{ \sqrt2\sigma }\] and \[dx = \sqrt2\sigma dy\]
And the probability becomes \[\sqrt2\sigma |\Psi_0|^2 \int\limits_{-\infty}^{-\infty}e^{-y^2}dy=1\]
And after replacing the Gaussian integral, I get \[|\Psi_0|^2 = \frac {1}{\sqrt{2 \pi} \sigma }\]
And \[|\Psi_0| = \frac {1}{\ (2 \pi \sigma^2 )^{1/4}} \]

Answer 3

But then I don't know why when replacing \[ \Psi_0 \] back in the wavepacket definition, I'm supposed to also put \[ e^{i \phi } \] in the numerator of the normalization constant.

Answer 4

The e^iθ is the phase factor and defines phase for the wave packet. It should be there before normalization as every wave function has a phase factor. It doesn't describe the evolution of phase for the wave packet but rather phase offset in the beginning. Since it's a function of θ, which in this case is a constant, it doesn't change with the integration you've shown.

Answer 5

Right, thank you!

Answer 6

You're welcome.