# Traveling and Standing Waves in Quantum Mechanics

1. Oct 9, 2009

### ourio

1. The problem statement, all variables and given/known data
A wave in quantum mechanics is represented by Aei(kx-$$\omega$$t). Show that a standing wave looks like 2iAe-i$$\omega$$tsin(kx) by subtracting two waves moving in opposite directions. (Hint: make the k negative in one of the waves)

2. Relevant equations
As far as I know, the 2 equations in the problem

3. The attempt at a solution
Honestly, I have no idea where to start. Any help would be greatly appreciated.

2. Oct 9, 2009

### Staff: Mentor

Just follow the hint. How would you represent the wave moving in the opposite direction?

3. Oct 9, 2009

### ourio

That's the thing... I'm not sure. If I had to guess, I would say it would look something like:

Aei(kx-$$\omega$$t) - Bei(-kx-$$\omega$$t)

But how do I relate it to sine?

4. Oct 9, 2009

### Staff: Mentor

Good, but use A for the second wave instead of B.
Hint: Look up Euler's formula.

5. Oct 9, 2009

### ourio

Thanks for the hint about Euler's formula... it really helped. So from the original equation, I now have:

A[cos(kx-wt)+i sin(kx-wt)]-A[cos(-kx-wt)+i sin(-kx-wt)]

I distributed the A term throughout, and found that the cosine terms canceled. I was left with:

Aisin(kx-wt)-Aisin(-kx-wt)

which could be written as

Aisin(kx-wt)+Aisin(kx+wt)

But now I have to somehow re-introduce an exp function and take the 'wt' term from the sine. Hmmmm....

6. Oct 9, 2009

### Staff: Mentor

Another hint: ea+b = eaeb

7. Oct 9, 2009

### ourio

OK... so I found that the original equation can be written as follows:
Aeikxe-iwt
Using Euler's formula on the eikx gives:
Ae-iwt (cos(ikx)+i sin(ikx))

I know that the cosine term will eventually cancel when I subtract. But the complex number in the sine bothers me. Won't that end up being a hyperbolic function? I carried the math through and came up with this:

2iAe-iwt (sin(ikx))

I'm a lot closer, but I don't know how to deal with the complex number in the trig function

Last edited: Oct 9, 2009
8. Oct 10, 2009

### Staff: Mentor

Not exactly. Euler's formula says:
eix = cosx + isinx

9. Oct 10, 2009

### ourio

Oh, I'm an idiot! Of course the complex number isn't in the sine!

Thanks Doc Al for all of your help! I'll never forget about Euler's formula again! :)