# Homework Help: Problem from Intro to Quantum Mechanics by Griffiths (I know near nothing)

1. Dec 14, 2008

### WraithM

Okay, here's a quantum mechanics problem. I am just starting with quantum, so I have no idea about this stuff. I am requesting guidence or a push in the right direction, not nessicarily a complete answer, please. Also, I don't know if this fits into the Advanced Physics help forum. I've never posted on the homework help forum before. If this is too simple, please let me know.

1. The problem statement, all variables and given/known data

A particle of mass m is in the state

$$\Psi (x, t) = A e^{-a[(m x^2/\hbar) + i t]}$$,

where A and a are positive real constants.

(a) Find A.

2. Relevant equations

Shroedinger's equation?

$$\int_{-\infty}^{+\infty} \mid \psi \mid ^2 dx = 1$$

Maybe?

3. The attempt at a solution

I guess my plan was to take the complex conjugate of the wave function there, multiply them, and then intregrate with $$\int_{-\infty}^{+\infty} \mid \psi \mid ^2 dx = 1$$ to get the answer for A. I ran into troubles along the way, and I have some theories as to where I messed up.

Firstly, I think I took the complex conjugate incorrectly. My knowledge of complex numbers is extremely rusty :( I haven't really studied that since perhaps my sophomore year in high school...

Here's what I said it was:

$$\Psi^* = A e^{-a[(m x^2/\hbar) - i t]}$$

So, assuming I did that correctly (which is not a very good assumption ), I moved foreward.

$$\mid \psi \mid ^2 = A^2 e^{-2 a m x^2/\hbar}$$

Okay, and that's a gaussian, so the integration is tricky. I assume that this is not the proper way of moving foreward because when I tried to think of how to do the integration I got confused.

Does anybody have a good idea of how to move foreward with this problem?

-WraithM

Last edited: Dec 14, 2008
2. Dec 14, 2008

### dextercioby

Can you integrate a gaussian ? If not, then use a table of integrals. I'm sure you can find one in every QM book.

3. Dec 14, 2008

### WraithM

The integral is in the answer of an "error function" which I frankly have no idea how to deal with, but I like your idea of a table of integrals. I have a really solid table of integrals with 728 of them :D I guess I'll try that.

If anybody knows how to do the integral

$$\int_{-\infty}^{+\infty} A^2 e^{-2 a m x^2/\hbar} dx$$

it would help me very much.

4. Dec 14, 2008

### dextercioby

There's no 'error' function answer to this one, because the limits are + and - infinity. The solving of the integral

$$\int_{-\infty} ^{+\infty} e^{-x^2} {} dx$$

is one of the oldest tricks in any analysis book. However, I'm sure you can find its answer in your QM book.

5. Dec 14, 2008

### GDogg

use substitution and then:

http://en.wikipedia.org/wiki/Gaussian_integral#Brief_proof

6. Dec 14, 2008

### WraithM

Win!

$$\int_{0}^{\infty} e^{-b^2 x^2} dx = \frac{\sqrt{\pi}}{2b}$$

My 728 best friends never fail.

$$\int_{-\infty}^{+\infty} A^2 e^{-2 a m x^2/\hbar} dx = 2 \int_{0}^{\infty} A^2 e^{-2 a m x^2/\hbar} dx$$

So,

$$b^2 = 2 a m/\hbar$$

And

$$2 \int_{0}^{\infty} A^2 e^{-2 a m x^2/\hbar} dx = A^2 \sqrt{\frac{\pi \hbar}{2 a m}} = 1$$

Therefore,

$$A = (\frac{2 a m}{\pi \hbar})^{1/4}$$

My math is rusty :( I'm not going to do well figuring out this quantum mechanics, lol.

Thank you very much for the help!