This is a followup of my previous Why don't the De Broglie dispersion relation contain a constant term? question.

Answerers pointed out that only differences in energy matter I can understand that in most of the cases. But this would mean that you can freely bias the dispersion relation of the particles, so:

$$ \omega = \omega_0(k) + C $$

That's interesting. Let's see we have a particle that have 100Hz frequency at ground state so when it stand still. It's in a superposition state which makes possible to have a moving version of it with 2 additional speeds with a corresponding frequency of 120Hz and 140Hz.

I can accept frequency stretching. Since it depends on how one measures time. So it seems reasonable to convert it to 50,60 and 70Hz or 200, 240 and 280Hz.

But I don't see why is it correct to shift the energy levels so I can turn the frequencies to 0, 20 and 40Hz respectively.

I think the waves generated by mixing a 100, 120 and 140Hz source is totally different than waves generated by mixing a 20 and 40Hz source. Isn't it?

Or with symbols. Let's say you have the following plane wave (stripped down):

$$e^{i(kx-\omega t)}$$

Shifting the frequency you'll have:

$$e^{i(kx-(\omega + C) t)} = e^{iCt}e^{i(kx-\omega t)}$$

Although the original function remains there, it's still multiplied with another time dependent function, not a constant term.

Why is this frequency shifting possible? Why don't this frequency shift affect how does the system behave (even if the wave function maybe totally different)?


1 Answer 1


When we are talking about wave functions, the physically relevant quantities are expectation values, which are calculated as $$ \int \psi(x) \mathcal{O} \psi^*(x) dx $$ and do not change when adding a phase factor $e^{i(kx-wt)}$ $$ \Rightarrow \int \psi(x) e^{i(kx-wt)} \mathcal{O} \psi^*(x) e^{-i(kx-wt)} dx= \int \psi(x) \mathcal{O} \psi^*(x) dx. $$ This is again the expectation value of the observable $\mathcal{O}$. Adding a constant as you did, is called gauge invariance and a very important feature of any expectation value.

  • $\begingroup$ Thanks for the remark, I did not really think about what to calculate, but of course physically relevant things are expectations values of some observables. Edited the post accordingly. $\endgroup$
    – Remo
    Jan 7, 2015 at 13:59
  • $\begingroup$ OK, +1. I am deleting my comments now as they are no longer relevant. $\endgroup$ Jan 7, 2015 at 14:47

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