When people say that the phase doesn't matter, they mean the *overall*, "global" phase. In other words, the state $|0 \rangle$ is equivalent to $e^{i \theta} |0 \rangle$, the state $|1\rangle$ is equivalent to $e^{i \theta'} |1 \rangle$, and the state $|0\rangle + |1 \rangle$ is equivalent to $e^{i \theta''} (|0 \rangle + |1 \rangle)$.

Note that "equivalence" is not preserved under addition, since $e^{i \theta} |0 \rangle + e^{i \theta'} |1 \rangle$ is not equivalent to $|0 \rangle + |1 \rangle$, because there can be a relative phase $e^{i (\theta - \theta')}$. If we wanted to describe this very simple fact with unnecessarily big words, we could say something like "the complex projective Hilbert space of rays, the set of equivalence classes of nonzero vectors in the Hilbert space under multiplication by complex phase, cannot be endowed with the structure of a vector space".

Because the equivalence doesn't play nicely with addition, it's best to just ignore the global phase ambiguity whenever you're doing real calculations. Finally, when you're done with the entire calculation, and arrive at a state, you are free to multiply that final result by an overall phase.

## Best Answer

If you take the two well-known formulae 1. E = hν (Einstein) 2. pλ = h (de Broglie) and write the usual formula for phase velocity (v'), and do a bit of algebra, you end up with v' = E/p Now p is the momentum, which is well known to be = mv, v the particle velocity, m the mass, but what is E?

If you woodenly put in E = mc^2, you end up with v' = c^2/v, which is in your question. I object to that because while the Einstein formula is technically correct, the energy is unavailable while the particle stays in existence. Heisenberg put E = mv^2/2, i.e. he put E equal to the kinetic energy, but now we have the rather odd result that the wave travels at half the velocity of the particle. So, what you make of this is up to you. Most people ignore this relationship.

If you are only learning, you probably don't want the following. I have my own interpretation, which I call the guidance wave. It is like the pilot wave, but in the above I argue that if there is a wave that causes diffraction in the two-slit experiment, it has to travel at the same velocity as the particle. If so, the energy has to be mv^2, or twice the kinetic energy. That, in turn, means there is energy in the wave, or put it another way, the wave transmits energy (which is what every other wave does.) That has the ugly consequence of requiring effectively at least one more dimension in which this energy resides.

So, you see, there are three options for the phase velocity, and each has their problems. Take your pick. If you are studying for exams, I would strongly recommend you put this problem to one side. If you are trying to understand nature, then in the long term you don't really have that option, but in the short term, I would still put it to one side.