# Quantum Mechanics: Matrix Representation

1. Feb 1, 2015

### Robben

1. The problem statement, all variables and given/known data
What is the matrix representation of $\mathbb{\hat J}_z$ using the states $|+y\rangle$ and $|-y\rangle$ as a basis?

2. Relevant equations

$|\pm y\rangle =\frac{1}{\sqrt{2}}|+z\rangle \pm \frac{i}{\sqrt{2}}|-z\rangle$

3. The attempt at a solution

A solution was given:

$\mathbb{\hat J}_z =\left[{\begin{array}{cc} \langle +y|+z\rangle & \langle +y|-z\rangle \\ \langle -y|+z\rangle & \langle -y|-z\rangle \\\end{array}}\right]\frac{\hbar}{2}\left[{\begin{array}{cc} 1 & 0 \\ 0 & -1 \\\end{array}}\right] \left[{\begin{array}{cc} \langle +z|+y\rangle & \langle +z|-y\rangle \\ \langle -z|+y\rangle & \langle -z|-y\rangle \\\end{array}}\right] = \frac{\hbar}{2}\left[{\begin{array}{cc} 0 & 1 \\ 1 & 0 \\\end{array}}\right],$

but I am confused on what is going on? Can anyone explain exactly what is going on, please?

Last edited: Feb 1, 2015
2. Feb 1, 2015

### barefeet

You have the operator $\mathbb{\hat J}_z$. Normally you represent that in the z-basis, so in matrix notation the 4 matrix elements that you already know are $\langle \pm z \mid \mathbb{\hat J}_z \mid \pm z \rangle$. If you want to represent $\mathbb{\hat J}_z$ in another basis like y you have to know the matrix elements $\langle \pm y \mid \mathbb{\hat J}_z \mid \pm y \rangle$ . One way to do this is to insert a unit operator/identity matrix on both sides as this has no effect:
$$\langle \pm z \mid \mathbb{I} \mathbb{\hat J}_z \mathbb{I} \mid \pm z \rangle$$

Now if you know that you can write the identity matrix as $\sum \mid z \rangle \langle z \mid$ using the completeness relation you are practically done:

$$\langle \pm y \mid \mathbb{I} \mathbb{\hat J}_z \mathbb{I} \mid \pm y \rangle = \langle \pm y \mid z \rangle \langle z \mid \mathbb{\hat J}_z \mid z \rangle \langle z \mid \pm y \rangle$$

So the middle part $\langle z \mid \mathbb{\hat J}_z \mid z \rangle$ is the matrix in the already known basis
$\frac{\hbar}{2}\left[{\begin{array}{cc} 1 & 0 \\ 0 & -1 \\\end{array}}\right]$ And the outer parts are the matrices to transform it into an y basis.

(Note: I mix up the notation for matrix elements and the matrix itself. I probably miss a summation here and there, but I hope you understand the difference)

3. Feb 1, 2015

### Robben

I think I understand now. Thank you very much!