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normalization of a gaussian wavefunction

skatenerd

Active member
Oct 3, 2012
114
I'm given a wavefunction (I think it's implied this is some sort of solution to the Schrodinger equation) in my quantum mechanics class, and I need to normalize it to find its constant coefficient.
So I have
$$\psi(x)=Ne^{-\frac{|x-x_o|}{2a}}$$
And the formula for normalizing this to find \(N\) would be
$$\int_{-\infty}^{\infty}\bar{\psi(x)}\psi(x){dx}=1$$
Plugging in \(\psi(x)\) gives
$$1=\int_{-\infty}^{\infty}N^{2}e^{-\frac{|x-x_o|}{a}}dx$$

At first I was thinking I could just take the derivative of the exponent and divide by that to solve the integral but I realized that wouldn't work out right, and this integral behaves somewhat like a gaussian integral like when you need to integrate \(e^{-x^2}\).

I know the process of how to integrate \(e^{-x^2}\) from negative infinity to infinity (defining the integral as I and then squaring it, changing to polar coordinates, u-subbing and then taking the root of that solution to get \(\sqrt{\pi}\)) but when I tried to do that with \(\frac{|x-x_o|}{a}\) instead of \(x^2\) I end up with a weird expression in the exponent that I don't know what to do with. I was hoping changing to polar coordinates would work but I don't see how to do that with this.
Any guidance would be really appreciated! Thanks
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
Separate the integral to integrate $(-\infty, x_0]$ and $[x_0, \infty)$ separately. Then the absolute value disappears, you can factor out the constant $x_0$ and are left with a standard exponential. Or is $x$ a complex number or something like that? It's been a while since my introductory quantum mechanics class.
 

skatenerd

Active member
Oct 3, 2012
114
I don't think we are expected to know how to work with complex variables in this class so \(x\) is probably real.
But yeah I see what you're saying, and breaking up the bounds to make two integrals will be helpful if I can figure out how to integrate this crazy integrand...
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
I don't think we are expected to know how to work with complex variables in this class so \(x\) is probably real.
But yeah I see what you're saying, and breaking up the bounds to make two integrals will be helpful if I can figure out how to integrate this crazy integrand...
With complex numbers it could probably be worked about the same since the integral would be spherically symmetric, but I wouldn't know. Anyway once you've broken it up it becomes simple because the annoying absolute values disappear, as:

$$|x - x_0| = \begin{cases}x - x_0 ~ ~ ~ \mathrm{if} ~ x > x_0 \\ x_0 - x ~ ~ ~ \mathrm{if} ~ x < x_0\end{cases}$$

Furthermore since $|x - x_0|$ is symmetric you only need to compute one side of the integral, the whole integral is just twice that.
 

skatenerd

Active member
Oct 3, 2012
114
Ahhh I see what you're saying now! Thanks that helps a lot, neat little trick...