Quantum mechanics - normalisation of wavefunction

[tigger88]

Homework Statement

Normalise the wavefunction:
$$\Psi$$(x) = C exp(-mwx$$^{2}$$/(2h))
for the 1-D harmonic oscillator.

Homework Equations

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

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!

 

[Feldoh]

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

 

[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!

 

[gabbagabbahey]

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$$\theta$$)

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!)

 

[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?

 

[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).

 

[gabbagabbahey]

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?

Yes, my mistake...oops!

 

[gabbagabbahey]

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

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.

 

[Feldoh]

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.

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.