# Help with QM harmonic Oscillator Question

1. Nov 6, 2004

### Rachael_Victoria

Ok so i have been instructed to normalize N*x*exp(-ax^2), so i squared the function and trying to take the integral. I am a) assuming that the integration should run from the negative value of the amplitude or -A to the positive value of the Amplitude of A, i have a formula for A. My assumption about integrating over the amplitude is based upon my understanding that the largest values of x are going to be at the maximum and minimum amplitude respectively.
My main question is how the heck do you integrate (x^2)(exp(-a*x^2)) this is making me crazy, in tried integration by parts but that doesn't work as you just end up with the exponential back which i can't integrate in the first place. U sub would also make me insane. If someone can explain how to integrate this evil thing i would really appreciate it. I am willing to use any method of integration people come up with.
Also am I right about the integrating over the respective amplitudes? All space does not make sense to me, as we are talking about the stretching of an atom and that does not stretch infinitely.
If anyone can help I would REALLY appreciate it,
thanks
rachael

2. Nov 6, 2004

### AKG

Integrate by parts:

$$\int _{-\infty} ^{\infty} udv = (uv)|_{-\infty} ^{\infty} - \int _{-\infty} ^{\infty}vdu$$

$$u = N^2x,\ dv = xe^{-2ax^2}dx$$

3. Nov 7, 2004

### Rachael_Victoria

Hey thanks for your quick response, I tried integration by parts, the only problem that i have with it is that whether i assign exp(-ax^2) to dv or u i end up having to do the integral of exp(-ax^2). Cause you take the derivative for du and you get (-4ax)exp(-2ax^2). use it as dv and you still have to integrate it, either way you end up integrating this incredibly annoying function. I am growing to hate the once beloved exponetial function.

4. Nov 7, 2004

### AKG

Re-read my post. Notice the last line.

5. Nov 7, 2004

### cepheid

Staff Emeritus
That's driving me nuts too now...at first I thought you could do this:

$$\int{x^{2}e^{-ax^2}dx}$$

$$u = x^2$$
$$\frac{du}{2} = xdx$$

So you have:

$$\int{xe^{-ax^2}xdx} = \int{u^{\frac{1}{2}}e^{-au}du}$$

"Aha!" I thought..."Now I can integrate by parts."

But it does not turn out any nicer than before

6. Nov 7, 2004

### Rachael_Victoria

Ok so I found a formula

Which gives you the normalization constant for the QM harmonic Oscillator. New question. How do you calculate the uncertainty in position for a QM harmonic oscillator?

7. Nov 7, 2004

### Rachael_Victoria

The formula they gave us to find the normalization constant for a QM harmonic oscillator is Nv=((1/pi)^1/2(1/((2^v)v!)) where v is the wavenumber if that is any consolation. I am not interested in math thoroughly enough to lose any sleep over how you integrate exp(-ax^2) but i know some people are. So if you are doing QM for a harmonic oscillator use the formula above and don't even attempt to integrate it cause it will make you crazy.

8. Nov 7, 2004

### Rachael_Victoria

right but to get v i would have to integrate dv, which brings me back to integrating exp(-2ax^2).

9. Nov 7, 2004

### AKG

Yeah, I've been looking over it and that doesn't seem to be much help either. I was hoping that once you got an integral that was just "e" to the "something" then it would be easy, but it wasn't, and I've encountered this before. The answer has something to do with a function called "erf" and that's certainly not what they want you to do. In my QM book, they have some integrals like that, where you have something like $e^{x^2}$ as the integrand, and they just pull the solution out of nowhere. Look through your QM book, see if they have any examples or anything and just mimic their solution.

10. Nov 7, 2004

### AKG

Using this, I got:

$$\int _{-\infty} ^{\infty}N^2x^2e^{-2ax^2}dx$$

$$= -\frac{N^2}{4a}\left ( xe^{-2ax^2}|^{\infty} _{-\infty} - \int _{-\infty} ^{\infty}e^{-2ax^2}dx \right )$$

$$= \frac{N^2}{4a}\left ( \int _{-\infty} ^{\infty}e^{-2ax^2}dx \right )$$

Let $t = \sqrt{2a}x$, so $dx = 1/\sqrt{2a} dt$. Assuming a is positive:

$$= \frac{N^2}{2^{2.5}a^{1.5}}\left ( \int _{-\infty} ^{\infty}e^{-t^2}dx \right )$$

$$= \frac{N^2 \pi ^{0.5}}{2^{3.5}a^{1.5}}\lim _{z \to -\infty}\mathop{\rm erfc}\nolimits (z)$$

I figured that last step using the link at the top of this post. Looking at the graph on that page, we get:

$$= \frac{N^2 \pi ^{0.5}}{2^{2.5}a^{1.5}}$$

You can figure out how to normalize the function from here.

EDIT: looking at my book, they give something similar to the normalization formula you have. My book integrates over $\mathbb{R}$, not just over [-A, A].

Last edited: Nov 7, 2004
11. Nov 7, 2004

### Rachael_Victoria

WOW, thank you, i seriously do not have the math skills to pull that off. thanks for taking the time.