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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

$$e^{-i(k_2-k_1)x}e^{-i(\epsilon_2-\epsilon_1)t}+e^{i(k_2-k_1)x}e^{i(\epsilon_2-\epsilon_1)t}$$

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

$$e^{-if(x)}+e^{if(x)}=g(x)(e^{-i\phi}+e^{i\phi})$$

(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?

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

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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 $k_2=k_1$.

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  • $\begingroup$ 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

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