# Normalization of a wavefunction

• dapias09

#### dapias09

Hello, I'm trying to find out the normalization constant in a given wavefunction but I cannot. I think that this is a math problem because I cannot solve the integral of the probability density but your experience could help; I was trying several steps and I tried in the software "derive" but the answer didn't like to me.

The wavefunction is ψ(x,t)= [b^2-(x-vt)^2]^(-1) , b is the normalization constant, and I must integrate it with respect to x in the interval -∞<x<∞.

Thanks in advance for any help!

Recall that the integral is of the square wavefunction, we are assuming that b and t are real constants.

$$\psi(x,t)= \frac{1}{b^2-(x-vt)^2}$$

something is wrong b/c this wave function is singular; what about

$$\psi(x,t)= \frac{1}{b^2+(x-vt)^2}$$

?

Hi. Thanks for your help. You are right, with the first one we get a singularity. My teacher changed it for the second one. Anyway I cannot solve it, I tried to solve it with a software and I know the answer but the steps are very difficult. Thanks again. Bye .

If you think a simple integration by a change of variable is very difficult, then you should, perhaps, leave that QM book aside for good... You have to integrate

$$|\psi(x,t)|^2= \left|\frac{1}{b^2+(x-vt)^2}\right|^2 = \left|\frac{1}{b^2+y^2}\right|^2 = \frac{1}{(\bar{b}^2+y^2)(b^2+y^2)}$$

where I used y = x-vt

You can do this using partial fractions plus contour integrals or using partial fractions plus elementary integrals continued to complex parameters ib.

The denominators for the partial fractions will be something like

$$\frac{1}{y \pm ib}$$

As dextercioby said: this is standard knowledge for doing quantum mechanics.