# If an observer changes their velocity, is it equivalent to a momentum shift? [duplicate]

Suppose we have a world with one massive free particle and one observer. If the observer changes their velocity with respect to the particle, will this have the same effect on $$\lvert\Psi(r,t)\rvert^2$$ as if $$\langle p\rangle$$ was changed locally (for example, using compton scattering if we added another particle)?

Part of me says, yes: the "wavelength of $$\Psi$$" is related to momentum by the de Broglie relation, and momentum is relative (assuming we have just a Galilean transform here).

Part of me says, no: due to the uncertainty principle, the probability of finding a free particle anywhere would seem to be uniform and negligible over all space. From this argument, the wave function shouldn't change, because for a free particle, it wouldn't have a well-defined wavelength to begin with.

I know I'm missing some conceptual connection, because these can't both be true.

I am currently studying an intro QM course, so please take my undergraduate level of understanding into account!

• "Part of me says, no: due to the uncertainty principle, the probability of finding a free particle anywhere would seem to be uniform and negligible over all space." How does the uncertainty principle say this? – BioPhysicist Apr 4 at 2:28
• To my understanding, the momentum of a free particle is well-defined because its velocity is well-defined in terms of the group velocity (its 'classical velocity'). Hence, we can know the momentum of a free particle precisely from the wave equation, but - according to the uncertainty principle - not it's position. I initially thought that this meant that the uncertainty in position is infinite, such that the probability density has to be infinitesimally small everywhere for the wave-function to even be normalizable. That's my rationale. – sven Apr 4 at 2:56

As other answers have mentioned, the SE is invariant under Galilean transformation, and there are other PSE posts that cover this. However, I wanted to address some specific things in your question.

Part of me says, no: due to the uncertainty principle, the probability of finding a free particle anywhere would seem to be uniform and negligible over all space. From this argument, the wave function shouldn't change, because for a free particle, it wouldn't have a well-defined wavelength to begin with.

First, a change in relative velocity will just change $$\langle p\rangle$$, not $$\Delta p$$, so we don't need to worry about any changes due to the uncertainty principle.

But also something to keep in mind is that systems with a certain Hamilton do not have to be eigenstates of that Hamiltonian. Yes, eigenstates of the free particle Hamiltonian have $$\Delta p=0$$ and are not physically valid states, but you can still have a free particle system that is in a superposition of such states so that $$\Delta p\neq 0$$.

To use another example, think of the particle in a box. Eigenfunctions of the Hamiltonian take the form of standing sine waves, but that doesn't mean all particle in a box systems look like this. Your system can be a superposition of these states, and hence $$\psi(x)$$ will not be a sine wave nor an eigenfunction of the Hamiltonian.

• Answers on the galilean transform make sense, but this addresses my root confusion about Δp and the hamiltonian. – sven Apr 4 at 14:31
• @sven Glad I could help clear something up for you – BioPhysicist Apr 4 at 14:33

No $$|\psi(x,t)|^2$$ will not change.

Reason Schrodinger equation is invariant under Galilean transformation.

The relation between the two cases is given by a unitary transformation. This means that $$|\psi(x,t)|^2$$ is necessarily preserved.