quantum mechanics - What is the physical significance of the imaginary part when plane waves are represented as $e^{i(kx-omega t)}$?

I've read that plane wave equations can be represented in various forms, like sine or cosine curves, etc. What is the part of the imaginary unit $i$ when plane waves are represented in the form $$f(x) = Ae^{i (kx - \omega t)},$$ using complex exponentials?

Answer

It doesn't really play a role (in a way), or at least not as far as physical results go. Whenever someone says

we consider a plane wave of the form $f(x) = Ae^{i(kx-\omega t)}$,

what they are really saying is something like

we consider an oscillatory function of the form $f_\mathrm{re}(x) = |A|\cos(kx-\omega t +\varphi)$, but:

• we can represent that in the form $f_\mathrm{re}(x) = \mathrm{Re}(A e^{i(kx-\omega t)})=\frac12(A e^{i(kx-\omega t)}+A^* e^{-i(kx-\omega t)})$, because of Euler's formula;


• everything that follows in our analysis works equally well for the two components $A e^{i(kx-\omega t)}$ and $A^* e^{-i(kx-\omega t)}$;


• everything in our analysis is linear, so it will automatically work for sums like the sum of $A e^{i(kx-\omega t)}$ and its conjugate in $f_\mathrm{re}(x)$;


• plus, everything is just really, really damn convenient if we use complex exponentials, compared to the trigonometric hoop-jumping we'd need to do if we kept the explicit cosines;



• so, in fact, we're just going to pretend that the real quantity of interest is $f(x) = Ae^{i(kx-\omega t)}$, in the understanding that you obtain the physical results by taking the real part (i.e. adding the conjugate and dividing by two) once everything is done;


• and, actually, we might even forget to take the real part at the end, because it's boring, but we'll trust you to keep it in the back of your mind that it's only the real part that physically matters.

This looks a bit like the authors are trying to cheat you, or at least like they are abusing the notation, but in practice it works really well, and using exponentials really does save you a lot of pain.

That said, if you are careful with your writing it's plenty possible to avoid implying that $f(x) = Ae^{i(kx-\omega t)}$ is a physical quantity, but many authors are pretty lazy and they are not as careful with those distinctions as they might.

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(As an important caveat, though: this answer applies to quantities which must be real to make physical sense. It does not apply to quantum-mechanical wavefunctions, which must be complex-valued, and where saying $\Psi(x,t) = e^{i(kx-\omega t)}$ really does specify a complex-valued wavefuntion.)