I've spent an embarrassingly long time on a seemingly simple homework problem and am starting to believe that either the solution is wrong or that I am a lot stupider than I thought.

The problem is to show that the expression


contains a term which describes completely the time dependence

$$g(x)\cos((\epsilon_2-\epsilon_1)t + \phi)$$

where $g(x)$ is real and $\phi$ does not depend on $x$

It seems to me that generally, the solution relies on the existence of a real function function $g(x)$ and an x-independent $\phi$ such that


(Note: $f(x)$ is real)

I cannot think of any $\phi$ satisfying this equation that would be independent of $x$

Assuming $f(x) = (k_2-k_1)x$, what $g(x)$ and $\phi$ would satisfy the equation $$e^{-if(x)}+e^{if(x)}=g(x)(e^{-i\phi}+e^{i\phi})$$ and the constraints that $g(x)$ is real and $\phi$ is independent of $x$ and non-zero?


migrated from physics.stackexchange.com Sep 22 '18 at 6:01

This question came from our site for active researchers, academics and students of physics.

  • $\begingroup$ You're aware that $e^{ix} + e^{-ix} = 2\cos(x)$? $\endgroup$ – Alfred Centauri Sep 21 '18 at 18:49
  • $\begingroup$ I am, yes. Its important to the larger problem at hand that the x dependence be factored out in the way described. $\endgroup$ – Malcolm Regan Sep 21 '18 at 18:51
  • $\begingroup$ I suppose I'm misreading something then because, by inspection, it looks like $g(x) = \cos(\Delta k\cdot x),\quad \phi = 0$ is a solution $\endgroup$ – Alfred Centauri Sep 21 '18 at 18:56
  • $\begingroup$ It is a solution to the question I asked but not to the larger problem. That's my fault, I will update to include more context. $\endgroup$ – Malcolm Regan Sep 21 '18 at 18:59
  • 2
    $\begingroup$ The problem, as stated, is very much ill-defined (or trivial, depending on how you want to think about it). Consider the simpler problem "prove that the expression $A$ contains a term of the form $B$". The solution is $A=B+(A-B)$, which obviously contains a term of the form $B$, but in an absolutely trivial way. In any case, I voted to migrate this to Math.SE, as this is a pure-math problem, with no physics involved (even if it comes from solving a physics problem). $\endgroup$ – AccidentalFourierTransform Sep 21 '18 at 19:31

Assuming $f(x) = (k_2-k_1)x$, what $g(x)$ and $\phi$ would satisfy the equation $$e^{-if(x)}+e^{if(x)}=g(x)(e^{-i\phi}+e^{i\phi})$$ and the constraints that $g(x)$ is real and $\phi$ is independent of $x$ and non-zero?

Any time you see $e^{-if(x)}+e^{if(x)}$, you can replace it with $2\cos\left(f(x)\right)$. So to answer this question as it stood before the unmotivated addition of the "non-zero" condition, you could take $g(x) = \cos\left(f(x)\right)$ and $\phi=0$. But if you really want it to be nonzero, note that since $e^{-i\phi}+e^{i\phi} = 2\cos\phi$, you can quite generally pick any value for $\phi$ as long as $\cos\phi \neq 0$, and then set \begin{equation} g(x) = \frac{\cos\left(f(x)\right)} {\cos\phi}. \end{equation} This is why AccidentalFourierTransform is saying that the problem is ill defined: there's no single answer to it.

Now, looking at your original expression, and remembering that $e^A e^B = e^{A+B}$, we have \begin{align} e^{-i(k_2-k_1)x}e^{-i(\epsilon_2-\epsilon_1)t} + e^{i(k_2-k_1)x} e^{i(\epsilon_2-\epsilon_1)t} &=e^{-i(k_2-k_1)x-i(\epsilon_2-\epsilon_1)t} + e^{i(k_2-k_1)x+i(\epsilon_2-\epsilon_1)t} \\ &=2\cos \left[(k_2-k_1)x+(\epsilon_2-\epsilon_1)t\right] \\ &=2\cos \left[(k_2-k_1)x\right] \cos\left[(\epsilon_2-\epsilon_1)t\right] -2\sin \left[(k_2-k_1)x\right] \sin\left[(\epsilon_2-\epsilon_1)t\right] \end{align} If you're literally asking to show how that expression "contains a term" like $g(x)\cos\left[(\epsilon_2-\epsilon_1)t + \phi\right]$... well just set $\phi=0$ and $g(x) = 2\cos \left[(k_2-k_1)x\right]$. (Remember that a "term" is one part of a sum, whereas a "factor" is one part of a product, so you can see that "term" in the result above.)

EDIT: Given the OP's comments, I think I can rephrase the original question as:

Find $g(x)$ and $\phi$ such that \begin{equation} g(x) \cos[(\epsilon_2-\epsilon_1)t + \phi] + h(x) = e^{-i(k_2-k_1)x}e^{-i(\epsilon_2-\epsilon_1)t} + e^{i(k_2-k_1)x} e^{i(\epsilon_2-\epsilon_1)t}, \end{equation} where $h(x)$ is an arbitrary function independent of time.

It's not hard to get around the addition of $h(x)$ by simply differentiating both sides of this equation with respect to time. Using the result I showed above, this gives us \begin{equation} -g(x) \sin[(\epsilon_2-\epsilon_1)t + \phi] (\epsilon_2-\epsilon_1) = -2\sin \left[(k_2-k_1)x+(\epsilon_2-\epsilon_1)t\right](\epsilon_2-\epsilon_1), \end{equation} which simplifies to \begin{equation} g(x) \sin[(\epsilon_2-\epsilon_1)t + \phi] = 2\sin \left[(k_2-k_1)x+(\epsilon_2-\epsilon_1)t\right]. \end{equation} Now, this must be true for all values of $t$, so let's just pick two values and see where it leads. First, take $t=0$ and we get \begin{equation} g(x) \sin[\phi] = 2 \sin \left[(k_2-k_1)x\right]. \end{equation} Then, take $t = \pi/[2(\epsilon_2-\epsilon_1)]$ and simplify to find \begin{equation} g(x) \cos[\phi] = 2 \cos \left[(k_2-k_1)x\right]. \end{equation} We can divide the first of these by the second to eliminate $g(x)$ and solve for $\phi$: \begin{equation} \tan\phi = \tan \left[(k_2-k_1)x\right]. \end{equation} So, as you can see, $\phi$ cannot be independent of $x$ unless $k_2=k_1$.

  • $\begingroup$ My fault again. All the time dependence needs to be contained in the term in question. I will update the question. Sorry and thank you. $\endgroup$ – Malcolm Regan Sep 21 '18 at 19:53
  • $\begingroup$ If this question goes dead I'll accept this as the answer because it technically is an answer to the question as I wrote it $\endgroup$ – Malcolm Regan Sep 21 '18 at 19:56
  • $\begingroup$ Maybe I'm bad at writing questions. But this discussion left me convinced that $\phi$ necessarily depends on x for any non-trivial solution. I'll just ask the teacher on monday. Thanks to all involved. I'll work on my question writing skills. $\endgroup$ – Malcolm Regan Sep 21 '18 at 20:03
  • $\begingroup$ With your new clarification, I've edited my answer to show that indeed $\phi$ must depend on $x$ unless $k_1 = k_2$. $\endgroup$ – Mike Sep 21 '18 at 20:41
  • $\begingroup$ Beautiful. Thank you. $\endgroup$ – Malcolm Regan Sep 22 '18 at 2:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.