Integration Involving a Gaussian Function

[ríomhaire]

This is a University quantum physics problem but I'm posting it in this section as what I'm having trouble with is the integration.

Homework Statement

The first excited state of the harmonic oscillator has a wavefunction of the form
$$\psi(x) = Axe^{- \frac{x^{2}}{2a^{2}}}$$
...
Find the constant A from the normalisation condition.

Homework Equations

Normalisation condition: $$\int ^{\infty}_{-\infty} \left| \psi(x)^2 \right| dx = 1$$

The Attempt at a Solution

So the integral that I'm having trouble with is
$$\int ^{\infty}_{-\infty} \left| x^{2}e^{- \frac{x^{2}}{a^{2}}} \right| dx$$


I tried integration by parts but I really don't know how to handle the $$e^{- \frac{x^{2}}{2a^{2}}}$$ part. After some initial confusion I realize this is a Gaussian and the integral of it from minus infinity to infinity is $$a \sqrt{\pi}$$ which would be fine if it wasn't being multiplied by $$x^{2}$$.


So integration by parts, setting $$u = x^{2}$$ and $$dv = e^{- \frac{x^{2}}{2a^{2}}} dx$$ doesn't work as then I have to integrate dv without limits, which gives the http://mathworld.wolfram.com/Erf.html" which I only just heard of by putting this integral into Wolfram Alpha and don't know what to do with.


Putting $$u = e^{- \frac{x^{2}}{2a^{2}}}$$ is even worse as when I differentiate $$du = \frac{x}{a^{2}} e^{- \frac{x^{2}}{2a^{2}}} dx$$ which means I would have to do integration by parts on this new integral and I think you can see why that's not going to work.


Of course I can just get the answer but I want to know how to solve this problem myself. Do I have to work with the error function or is there a way of getting around it by an application of the limits (or some other trick) that I'm not seeing?

 

[awkward]

Try integration by parts with

$$u = x$$

and

$$dv = x \exp(- \frac{x^2}{2a^2}}) \, dx$$

 

[Gunthi]

How to calculate integrals in the form $$I(n)=\int_{0}^{\infty}x^ne^{-ax^2}dx$$

This first part is to demonstrate to reach the method, but you won't have to reproduce it, just the result.

Starting with $$I=\int_{-\infty}^{\infty}e^{\alpha^2}d\alpha$$ you can see that
$$I^2=\int_{-\infty}^{\infty}e^{x^2}dx\int_{-\infty}^{\infty}e^{y^2}dy$$, because x is a mute variable.

So you get $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dxdy$$

Changing to polar coordinates one gets $$\int_{0}^{2\pi}\int_{0}^{\infty}e^-r^2}drd\theta = \sqrt{\pi}$$.

If that is true, then $$\int_{0}^{\infty}e^-u^2}du = \frac{\sqrt{\pi}}{2}$$ also holds.

Now, if we have something like $$\int_{0}^{\infty}e^{-ax^2}dx$$ we can change the variables to $$z^2=-ax^2, dz=\sqrt{a}dx, dx=\frac{dz}{\sqrt{a}}$$ and then calculate the integral in the form
$$\int_{0}^{\infty}e^{-z^2}\frac{dz}{\sqrt{a}}=\sqrt{\frac{\pi}{a}}\frac{1}{2}$$ which in the form $I(n)$ is $I(0)$.

We need to calculate one more integral in order to have the complete set.

$$I(1)=\int_{0}^{\infty}xe^{-ax^2}, z^2=ax^2$$

$$\int_{0}^{\infty}\frac{z}{\sqrt{a}}e^{z^2}\frac{dz}{\sqrt{a}}=\frac{1}{a}\int_{0}^{\infty}ze^{-z^2} = \frac{1}{2a}$$

So, in order to calculate any integral $I(n)$ all you have to is calculate
$$-\frac{d}{da}I(n-2)$$ because $$\int_{0}^{\infty}-\frac{d}{da}x^{n-2}e^{-ax^2}=\int_{0}^{\infty}-(-x^2)x^{n-2}e^{-ax^2}dx$$.

It's obvious now that both integrals are the same because $$x^2x^{n-2}=x^n$$.

So to recap all that was done here, $$I(n)=-\frac{d}{da}I(n-2), I(0)=\sqrt{\frac{\pi}{a}}\frac{1}{2}, I(1)=\frac{1}{2a}$$ and this is all you need to solve any integral in $I(n)$ form.

If I wasn't clear in any of the steps just say so, I hope this helps anyone with this sort of problem.

 

[vela]

One way to do it is to note that when the gaussian has a mean of zero, E(x²) is equal to the variance:

$$\sigma^2 = \frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty} x^2 e^{-\frac{x^2}{2\sigma^2}}dx$$

Comparing this integral to yours, you can identify $a^2$ with $2\sigma^2$ so that $\sigma=a/\sqrt{2}$; therefore,

$$\int_{-\infty}^{\infty} x^2 e^{-\frac{x^2}{a^2}}dx = \sqrt{2\pi}(a/\sqrt{2})^3 = \frac{\sqrt{\pi}a^3}{2}$$

Another way to do it is to use this trick:

$$\int_{-\infty}^{\infty} x^2 e^{-\alpha{x^2}} dx = \int_{-\infty}^{\infty} \left( -\frac{d}{d\alpha}\right) e^{-\alpha{x^2}} dx=-\frac{d}{d\alpha} \int_{-\infty}^{\infty} e^{-\alpha{x^2}} dx=-\frac{d}{d\alpha}\left(\sqrt{\frac{\pi}{\alpha}\right)= \frac{1}{2}\sqrt{\frac{\pi}{\alpha^3}$$

Setting $\alpha=1/a^2$ gives

$$\int_{-\infty}^{\infty} x^2 e^{-\frac{x^2}{a^2}} dx = \frac{\sqrt{\pi}a^3}{2}$$