# Quantum mechanics - normalisation of wavefunction

1. Sep 22, 2009

### tigger88

1. The problem statement, all variables and given/known data
Normalise the wavefunction:
$$\Psi$$(x) = C exp(-mwx$$^{2}$$/(2h))
for the 1-D harmonic oscillator.

2. Relevant equations

$$\int$$$$\Psi$$*$$\Psi$$dx = 1

3. The attempt at a solution
I used the following integral from -inf to inf:
¦C¦^2$$\int$$exp(-ax^2)dx = srqt(pi/a) where a = const. >0
where a = mw/h

I solved for C and got C = ((mw/h pi)^(1/4) exp(i$$\theta$$)

I just want to check if it's the right answer as there's a follow-on question and it looks quite awkward.

Thanks for any help!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 22, 2009

### Feldoh

C = (mw/h pi)^(1/4), it's not clear how you're getting an additional exp(i * theta)

3. Sep 22, 2009

### tigger88

I got the i*theta bit from some old notes... the lecturer seemed to wap it on the end so that both real and imaginary components (if there are any) of the normalisation factor are included. I don't entirely understand it...

Thanks very much!

4. Sep 22, 2009

### gabbagabbahey

Surely you mean

$$|C|^2\int_{-\infty}^{\infty} e^{-ax^2}dx=|C|^2\sqrt{\frac{\pi}{a}}$$

right?

And shouldn't you have $a=\frac{m\omega}{2\hbar}$, leaving you with $C=\left(\frac{m\omega}{2\hbar \pi}\right)^{1/4}e^{i\theta}$?

In any case, you should recall that a wavefunction's overall phase factor has no physical effect, and so you are free to choose any particular value of $\theta$ you want. (I'd choose $\theta=0$ for obvious reasons!)

5. Sep 22, 2009

### tigger88

I thought that when I squared the wavefunction (or multiplied by the complex conjugate, which in this case would be synonymous with squaring) the exponential would do this:
exp(-2mwx^2/(2h)) which would simplify to exp(-mwx^2/h).
Is this not the case?

6. Sep 22, 2009

### Feldoh

You're right tigger88, it does simplify to $$e^{-mwx^2/h}$$ in the integrand which gives you a value of $$C=\left(\frac{m\omega}{\hbar \pi}\right)^{1/4}$$

I'm not even sure you have to put a phase since the wave function is a stationary state (ground state of harmonic oscillator wave function).

Last edited: Sep 22, 2009
7. Sep 22, 2009

### gabbagabbahey

Yes, my mistake...oops!

8. Sep 22, 2009

### gabbagabbahey

I'm not sure what you mean by this. An overall phase factor is an unobservable quantity. Choosing any value of $\theta$ will produce exactly the same physics.

9. Sep 22, 2009

### Feldoh

You said it yourself, it doesn't matter, we don't really need to tack it on in this case since it doesn't change anything.