joneall said:
Sorry, I just do not get it.
What basic QM textbooks have you read? What I am telling you is basic QM.
joneall said:
It seems to me if wave function ##\ket{z+}## with z-spin up, with which you agree, then the rotation to get wave function ##- \ket{z+}## very definitely represents spin down.
It may seem that way to you, but it's wrong. If we have a state ##\ket{\psi}##, then ##c \ket{\psi}##, where ##c## is any complex number, represents the
same state. Normally we require states to have unit norm, so we would restrict ##c## to be a complex exponential ##e^{i \alpha}##, where ##\alpha## is a real number--i.e., a phase.
Again, this is basic QM, so if you are not familiar with it, I strongly suggest that you take the time to work through a QM textbook.
So we have the state ##\ket{z+}## that represents z spin up, and we have the state ##\ket{z-}## that represents z spin down. From the above it should be evident that we cannot multiply either one of these states by
any complex number and get the other. Whatever operation might exist that transforms one of these states into the other, it can't be as simple as just multiplying by some complex number.
joneall said:
we've turned something pointing up thru 180°.
No, we haven't. We've turned it thru 360 degrees. We rotated by ##2 \pi##, not ##\pi##. Once again you are confusing the rotation angle with the effect on the wave function. They're not the same.
Also see further comments below.
joneall said:
Please just explain what is going on.
I am trying to, but you do not appear to have the necessary background knowledge of basic QM. You need to fix that if a discussion like this is going to be fruitful. For a QM textbook I would recommend Ballentine, although there are many and different people have different preferences, so others might recommend a different one.
That said, I'll expand briefly on what I said above for the case of spinors, i.e., spin-1/2 particles. (More details about what I am going to say can be found in Chapter 7 of Ballentine.) Remember that I said the Hilbert space for the spin degree of freedom of these particles is complex 2-vectors, i.e., column vectors like this:
$$
\begin{bmatrix}
a \\
b
\end{bmatrix}
$$
where ##a## and ##b## are complex numbers, and if we want to restrict to vectors of unit norm, as is common in QM, we require ##|a|^2 + |b|^2 = 1##. Note that in writing down this column vector we have to choose a basis, which for spin degrees of freedom means choosing a rotation axis. We make the common choice of the ##z## axis, and this means that the two basis column vectors represent z spin up and z spin down, i.e., we have
$$
\ket{z+} = \begin{bmatrix}
1 \\
0
\end{bmatrix}
$$
$$
\ket{z-} = \begin{bmatrix}
0 \\
1
\end{bmatrix}
$$
This makes it obvious that you can't multiply one of these by any complex number and get the other.
The matrices that represent rotations of spinors are elements of SU(2) in its 2 x 2 matrix representation. You seem to be reasonably familiar with these, and their normal representation relies on the normal representation of the Pauli matrices, which is in the ##z## axis basis in which I wrote down the above column vectors. So you should be able to verify that rotating either of the above column vectors by ##2 \pi## about any axis just multiplies the vector by ##-1##. You should also be able to verify that rotating either of the above column vectors by ##\pi## about either the ##x## or the ##y## axis turns it into minus the other.
So it
is possible to turn ##\ket{z+}## into ##\ket{z-}## (but with a minus sign) by rotating through 180 degrees--but you have to pick a rotation axis that is orthogonal to the ##z## axis to do it. Rotating by 180 degrees about the ##z## axis itself doesn't change ##\ket{z+}## (or ##\ket{z-}##) physically--although it does multiply it by ##i##.