Thread: expected value of x in Gaussian wave packet

1. expected value of x in Gaussian wave packet

This problem is killing me.

$\Psi(x) = Ae^{-\frac{(x-x_0)^2}{2K^2}}$

I normalized the wave packet

$\int^\infty_{-\infty} \Psi\Psi*\,dx = 1$ where * is the complex conjugate

using a table integral:

$\int^{\infty}_{-\infty}{e^{-\frac{(x-b)^2}{c^2}}}\,dx=c\sqrt{\pi}$

$A^2 = \frac{1}{K\sqrt{\pi}}$

Now I need the expected value of x

$=\int^\infty_{-\infty} x\Psi\Psi*\,dx$
$=A^2\int^\infty_{-\infty} xe^{-\frac{(x-x_0)^2}{K^2}}\,dx$

which I expect to be $x_0$

but I can't get it to work out. I tried substitution with u = the exponent but looking back over it I don't think it will work right because I can't seem to get rid of all the references to x when I substitute.

Any pointers?

2. Originally Posted by stevedave
I need the expected value of x

$=\int^\infty_{-\infty} x\Psi\Psi*\,dx$
$=A^2\int^\infty_{-\infty} xe^{-\frac{(x-x_0)^2}{K^2}}\,dx$

which I expect to be $x_0$

but I can't get it to work out.
Substitute $y = x-x_0$. Then $\int^\infty_{-\infty}\!\!\! xe^{-\frac{(x-x_0)^2}{K^2}}\,dx = \int_{-\infty}^\infty\!\!\!(y+x_0)e^{-y^2/K^2}dy = \int_{-\infty}^\infty\!\!\! ye^{-y^2/K^2}dy\ +\ x_0\int_{-\infty}^\infty\!\!\! e^{-y^2/K^2}dy.$ In the first of those two integrals the integrand is an odd function, so the integral will be 0. The second integral gives the answer that you want.

3. Thanks, that works out nice. I think my problem was not changing the limits of integration when I tried to substitute for the entire exponent. The square relationship would have square the limits making them both positive and giving the same answer.