- #1

Jillds

- 22

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## \psi (x,t) = e^{ik_{0}x-iw(k_{0})t} \Big( \frac{\pi}{\alpha+i\beta t} \Big)^{\frac{1}{2}} e^{-\big[\frac{(x-v_g t)^2}{4(\alpha +i\beta t)}\big]}##

Next we take the square of the absolute of the wave function. I know the first exponent would equal to 1 as a square, as it is phasefactor. And if I calculate the rest as a square I have.

## |\psi (x,t)|^2 = \frac{\pi}{\alpha+i\beta t} e^{-\big[\frac{(x-v_g t)^2}{2(\alpha +i\beta t)}\big]}##

My course, however, has a different result, and I haven't got a clue what my professor did to get that result:

## |\psi (x,t)|^2 = \Big( \frac{\pi^2}{\alpha^2+\beta^2 t^2} \Big)^{\frac{1}{2}} e^{-\alpha \big[\frac{(x-v_g t)^2}{2(\alpha^2 +\beta^2 t^2)}\big]}##

How does one get ##\alpha^2 +\beta^2 t^2## ?