# Changing the basis of Pauli spin matrices

• Phruizler

## Homework Statement

Find the matrix representation of $S_z$ in the $S_x$ basis for spin $1/2$.

## Homework Equations

I have the Pauli matrices, and I also have the respective kets derived in each basis. There aren't really any relevant equations, other than the eigenvalue equations for the operators.

## The Attempt at a Solution

In the $S_x$ basis, the $S_x$ operator is just

$$\frac{\hbar}{2} \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}$$

So isn't the $S_z$ operator in the $S_x$ basis just equal to the $S_x$ operator in the $S_z$ basis? That is,

$$S_z= \frac{\hbar}{2} \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$$

in the $S_x$ basis? If so, I don't really understand how to show this in any meaningful way... I've also tried solving the eigenvalue equations for $S_z$ and converting the kets to the $S_x$ basis but that just lead me in circles. I can see that the question is a very basic one but something is just not letting me get a satisfactory answer. Any help would be much appreciated, thanks!

After some thought, I've considered that maybe I can just use the vector representation of $S_z$ kets in the $S_x$ basis. That is,

$$|+>_x \doteq \begin{pmatrix} 1\\ 0 \end{pmatrix}$$

in the $S_x$ basis, and the same for the spin down ket, so I can just plug these vector representations into the eigenvalue equation and solve for the $S_z$ matrix. This will indeed give me the matrix which I asked about above (namely, the same as the $S_x$ the $S_z$ basis). I'm going to assume this is correct unless anyone tells me otherwise!

Phruizler - to find the required matrix, you just have to find the four matrix elements involving the eigenvectors of $S_x$.

That is, you need to find $\langle x,\pm \lvert S_z \lvert x,\pm \rangle$.

Thanks! This yields the answer I got doing it the above way but is much more satisfying. I didn't even think of solving the matrix elements individually like that for some reason. Quantum mechanical problem solving simply doesn't adhere to the same intuition as other problems in physics. I can tell it will be a while before I will have useful insight into even some of the easier problems!

Thanks again!