Traveling and Standing Waves in Quantum Mechanics

[ourio]

Homework Statement

A wave in quantum mechanics is represented by Ae^{i(kx-\omegat)}. Show that a standing wave looks like 2iAe^-i\omegatsin(kx) by subtracting two waves moving in opposite directions. (Hint: make the k negative in one of the waves)

Homework Equations

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

The Attempt at a Solution

Honestly, I have no idea where to start. Any help would be greatly appreciated.

 

[Doc Al]

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

 

[ourio]

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

Ae^{i(kx-\omegat)} - Be^{i(-kx-\omegat)}

But how do I relate it to sine?

 

[Doc Al]

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

Ae^{i(kx-\omegat)} - Be^{i(-kx-\omegat)}

Good, but use A for the second wave instead of B.

But how do I relate it to sine?

Hint: Look up Euler's formula.

 

[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...

 

[Doc Al]

Another hint: e^a+b = e^ae^b

 

[ourio]

OK... so I found that the original equation can be written as follows:
Ae^ikxe^-iwt
Using Euler's formula on the e^ikx 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

 

[Doc Al]

Using Euler's formula on the e^ikx gives:
Ae^-iwt (cos(ikx)+i sin(ikx))

Not exactly. Euler's formula says:
e^ix = cosx + isinx

 

[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! :)