MichPod said:
choosing the particular and vectors in the Hilbert space to represent spin-up and spin-down (along the Z-axis) physical states does define X and Y axises.
No, it doesn't. You said in response to my earlier post...
MichPod said:
...but you apparently failed to grasp the fact that it makes your claim quoted at the start of this post false.
I'll illustrate why in response to this from
@PeroK:
PeroK said:
What are my z-spin-up and z-spin-down vectors?
I say they are ##(1, 0)## and ##(0, 1)## like they always are!
You do have one degree of phase freedom here. In other words, you could choose ##e^{i \phi} (1, 0)## and ##e^{i \phi} (0, 1)## (note the
same ##\phi## in both cases) as your z-spin up and z-spin down vectors and everything would be the same as far as physical predictions. In particular, the spin-x up superposition, once we have chosen which direction is the physical x axis, would then be written as ##(1 / \sqrt{2}) e^{i \phi} \left[ (1, 0) + (0, 1) \right]##, which represents the same physical state as plain ##(1 / \sqrt{2}) \left[ (1, 0) + (0, 1) \right]## without the phase factor.
Now let's look at what happens if, as
@MichPod keeps saying, we try to pick the phases of the vectors we use to represent z-spin up and z-spin down, ##(1, 0)## and ##(0, 1)##, independently. We then have the vectors ##e^{i \phi_1} (1, 0)## and ##e^{i \phi_2} (0, 1)##. Superposing these two then gives ##(1 / \sqrt{2}) \left[ e^{i \phi_1} (1, 0) + e^{i \phi_2} (0, 1) \right] = (1 / \sqrt{2}) e^{i \phi_1} \left[ (1, 0) + e^{i (\phi_2 - \phi_1)} (0, 1) \right]##, which is physically equivalent to ##(1 / \sqrt{2}) \left[ (1, 0) + e^{i (\phi_2 - \phi_1)} (0, 1) \right]##.
@MichPod is saying that what we have done is to
define the vector ##(1 / \sqrt{2}) \left[ (1, 0) + e^{i (\phi_2 - \phi_1)} (0, 1) \right]## as "x-spin up", and therefore to have chosen the direction of the x axis. But that's not what we have done. Actually, what has happened is that we
misdescribed the earlier operation of picking the two phases ##\phi_1## and ##\phi_2## independently. We described that operation as choosing the vectors that would represent z-spin up and z-spin down. But it was actually a
combination of choosing the phase ##\phi_1## as the phase factor for z-spin representatives,
and choosing the
relative phase ##\phi_2 - \phi_1## of the superposition we constructed. The first phase choice has no physical meaning, but the second does. Which in turn means that the superposition we constructed is
not x-spin up; it does
not describe that physical state. It describes a
superposition of x-spin up and x-spin down, or of y-spin up and y-spin down if we choose to look at it in terms of y-spin state. It does not represent an eigenstate of
any of the three x, y, or z spin operators.
So what
does represent "choosing the direction of the x axis" in the math? The answer is that the question is backwards. By
definition the superposition ##(1 / \sqrt{2}) \left[ (1, 0) + (0, 1) \right]## (with whatever overall phase factor we want in front of it, since such an overall phase factor has no physical meaning) represents the physical state "x-spin up", and the superposition ##(1 / \sqrt{2}) \left[ (1, 0) - (0, 1) \right]## (with
the same overall phase factor we chose for x-spin up) represents the physical state "x-spin down". In other words, we have to choose the physical direction of the x-axis
first, and
that choice then determines how the states in the math relate to actual physical measurements. There is
no way to choose the direction of the x-axis by making phase choices in the math; you have to do it the other way around.
