Superposition of two plane waves

[ballzac]

Hi,
I'm not after much help here. I already have an answer, but I want to make sure that I haven't made any stupid mistakes and that I understand the question. I also have a query about the second part of the question.

Homework Statement

What is the probability density of the superposition of two plane waves of differing angular frequencies, whose wave vectors point in different directions, and what is the frequency at which this probability density oscillates.

Homework Equations

The two plane waves are
$$\Psi_1(\textbf{r},t)=exp[i(\textbf{k}\cdot \textbf{r}-\omega t)]$$
and
$$\Psi_2(\textbf{r},t)=exp[i(\textbf{k}'\cdot \textbf{r}-\omega' t)]$$

The Attempt at a Solution

$$\rho=|\Psi_1+\Psi_2|^2$$
$$=|exp[i(\textbf{k}\cdot \textbf{r}-\omega t)]+exp[i(\textbf{k}'\cdot \textbf{r}-\omega' t)]|^2$$
$$=2+exp[i[(\textbf{k}\cdot \textbf{r}-\omega t)-(\textbf{k}'\cdot \textbf{r}-\omega' t)]]+exp[i[(\textbf{k}'\cdot \textbf{r}-\omega' t)-(\textbf{k}\cdot \textbf{r}-\omega t)]]$$
$$=2+cos[(\textbf{k}\cdot \textbf{r}-\omega t)-(\textbf{k}'\cdot \textbf{r}-\omega' t)]+isin[(\textbf{k}\cdot \textbf{r}-\omega t)-(\textbf{k}'\cdot \textbf{r}-\omega' t)]+cos[(\textbf{k}'\cdot \textbf{r}-\omega' t)-(\textbf{k}\cdot \textbf{r}-\omega t)]+isin[(\textbf{k}'\cdot \textbf{r}-\omega' t)-(\textbf{k}\cdot \textbf{r}-\omega t)]$$
Due to the properties of even and odd functions, this simplifies to
$$2(1+cos[(\textbf{k}\cdot \textbf{r}-\omega t)-(\textbf{k}'\cdot \textbf{r}-\omega' t)])$$
$$=2(1+cos[(\textbf{k}-\textbf{k}')\cdot\textbf{r}-(\omega+\omega')t])$$
This seems to be simplified as much as possible, and answers the first part of the question. As for the second part, it seems that the angular frequency is just $$\omega +\omega'$$
making the frequency
$$\nu=\frac{\omega +\omega'}{2\pi}$$
My question is, if I have the right answer here, then can't this be violated? For instance, if one angular frequency is an integer multiple of the other, won't the resultant frequency just be the lower frequency?

Thanks for your help.

P.S It is also required that I find the time-average of the probability density as a function of position. I would assume that for a cosine function with
$$\omega \neq 0$$
the average would be zero, which would mean in this case the answer is simply
$$\rho_{av} (\textbf{r}) = 2$$
Does this sound right? It sounds plausible to me that the time-average probability is constant over all space, and of course this is not a normalised wave function, so I don't see any problem with having a probability over 1. Do I understand this correctly?

 

[Brian_C]

You made a small error in your derivation. The resultant angular frequency should be the difference of the two frequencies, which is the beat frequency you learned about in lower level physics. The rest of your conclusions sound accurate. The probability density is greater than one because you're only using two plane waves. You would need to superpose an infinite number of plane waves with different wavelengths to obtain a normalizable function. This is done in standard QM texts.

 

[ballzac]

ah, okay. It did seem kind of odd that it was a sum and not difference of angular frequencies. I'll have a look through it to see where I went wrong. Thanks heaps.

 

[ballzac]

I've fixed up my arithmetic and arrived at the result you suggested. Just another quick question. The value of $$\rho$$ is the same whether $$\omega$$ is larger or $$\omega'$$ is larger, so does that mean that the resultant frequency is the absolute value of the difference? i.e.

$$\nu_\rho=\frac{|\omega-\omega'|}{2\pi}$$

I guess it has to be, otherwise I wouldn't know which way around to put the two omegas in the difference. Thanks again for your help.