# Why is it possible to choose an arbitrary zero energy level when dealing with frequencies of a wave function?

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)?

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.