# Change of Basis For Pauli Matrix From Z Diagonal to X Diagonal Basis

1. Oct 14, 2014

### bohrpiphi

I want to find a matrix such that it takes a spin z ket in the z basis,

$| \; S_z + >_z$

and operates on it, giving me a spin z ket in the x basis,

$U \; | \; S_z + >_z = | \; S_z + >_x$

I would have thought that I could find this transformation operator matrix simply by using the following argument:

$U \; | \; S_z >_z = | \; S_z >_x$

$_z< S_z \; | U \; | \; S_z >_z = _z< S_z \; | \; S_z >_x$

Therefore, elements of U are given by the inner product $_z< S_z \; | \; S_z >_x$

However, to compute the inner product $_z< S_z \; | \; S_z >_x$ , I need to know $| \; S_z >_x$, which is exactly what I am trying to find.

Where is my misunderstanding?

I have shown that the matrix given by the inner products $_z< S_z \; | \; S_x >_z$ gives the matrix:

$\begin{matrix} 1/\sqrt{2} & 1/\sqrt{2} \\ 1/\sqrt{2} & -1/\sqrt{2} \end{matrix}$,

which cannot be correct, since $_z< S_z \; | \; S_x >_z \neq _z< S_z \; | \; S_z >_x$.

However, starting in the x basis and calculating $| \; S_z >_x$ shows that the above matrix works. I imagine this is not a coincidence, but it seems to be implying that $| \; S_z >_x = | \; S_x + >_z$.

This is not a homework question.

2. Oct 14, 2014

### bohrpiphi

I also just realized that this should have been posted in the Advanced Homework section.