Spin-1/2 Particle Transformation: Spin-Z to Spin-X

[RedX]

I got a quick question about the transformation matrix from the spin-z basis to the spin-x basis for spin-1/2 particles.

Would the matrix be:

$$<br /> \left(\begin{array}{ccc}<br /> \frac{e^{i\theta}}{\sqrt{2}} &\frac{e^{i\delta}}{\sqrt{2}} \\<br /> \frac{e^{i\theta}}{\sqrt{2}} & -\frac{e^{i\delta}}{\sqrt{2}}<br /> \end{array}\right) <br />$$

I put that down as an answer and got it marked wrong, and what the grader wrote down as the answer is what you get when you set all angles in the matrix to zero.

Do you get different physical results if you choose different phase factors?

 

[Avodyne]

Did you try your matrix out and see what you get?

If you do, you'll find that it only works for $\delta=\theta$, which then just becomes an overall phase for the entire matrix. An overall phase has no physical relevance.

 

[RedX]

Did you try your matrix out and see what you get?

If you do, you'll find that it only works for $\delta=\theta$, which then just becomes an overall phase for the entire matrix. An overall phase has no physical relevance.

Yeah I guess it doesn't work. The matrix I wrote down does diagonalize the spin operator in the x-direction (since the eigenvectors of that operator are in the columns of the transformation matrix, but with arbitrary phase factor which shouldn't matter), but the phase factor matters in calculations of other quantities.

Something that's interesting is that the Pauli matrices P are usually written in the z-basis. If you want to convert them to the x-basis via $$U^{\dagger}PU$$, then if you use the matrix I wrote down but with the angle in the first column equal to zero, and the second angle equal to $$\frac{-\pi}{2}$$, then the new set of matrices have the same entries as the old set of matrices. If instead you set all angles equal to 0, then you get 3 new matrices which don't look like the original Pauli matrices at all.