# Quantum Mechanics: Basis

1. Mar 3, 2015

### Robben

1. The problem statement, all variables and given/known data

How can I find the matrix representation of $\mathbb{S}_+$ and $\mathbb{S}_-$ in the $|\pm y\rangle$ or $|\pm x\rangle$ basis?

2. Relevant equations

$\mathbb{\hat{S}}_+|s,m\rangle = \sqrt{s(s+1)-m(m+1)}\hbar|s,m+1\rangle$

3. The attempt at a solution

The book almost always ask to find something in the $z-$basis but rarely asks to find something in $x$ or $y$ basis. I know how this matrix representation will look with the $z-$basis, so I am wondering how the matrix representation will look in the $x$ or $y$ basis instead of $z$?

2. Mar 3, 2015

### Simon Bridge

You find the projections along the unit vectors like normal.
If you already have the projection along the z axis, then what will the projections along the x and y axes look like?

3. Mar 3, 2015

### Robben

Can you elaborate please? Will it just then be a reflection about the axis?

4. Mar 4, 2015

### Simon Bridge

"the z basis" does not really mean anything.

You can find, for eg, the z component of angular momentum right?
You know how the x and y components are given the z component.
If you don't, look it up.

5. Mar 4, 2015

### Robben

So I will just use $|+x\rangle = \frac{1}{\sqrt{2}}(|+z\rangle +|-z\rangle)$ as my basis and solve it the way I solved for the $z$ basis?

6. Mar 4, 2015

### Simon Bridge

Well, again, there is no such thing as a z basis that I know of.
I think you need to revisit the definitions - what is it that S+ and S- do?

7. Mar 4, 2015

### Staff: Mentor

It's a common expression when talking about spin. The z basis for a spin-1/2 particle is the set $\left\{ |+z\rangle, |-z\rangle\right\}$, with "spin up" corresponding to a spin along +z, and "spin down" to a spin along -z.

You can express an operator $\hat{O}$ in, e.g., the x basis, as a matrix using
$$\mathbf{O}_x = \begin{pmatrix} \langle +x | \hat{O} | +x \rangle & \langle +x | \hat{O} | -x \rangle \\ \langle -x | \hat{O} | +x \rangle & \langle -x | \hat{O} | -x \rangle \end{pmatrix}$$
For that, you need of course to find how to apply the operator in the x basis, i.e., $\hat{O} | +x \rangle$. This you can find by expressing $| +x \rangle$ in terms of $| +z \rangle$ and $| -z \rangle$.

Alternatively, you can calculate the rotation matrix $\mathbf{R}$ that brings you from the z basis to the x basis, and find the operator matrix using
$$\mathbf{O}_x = \mathbf{R} \mathbf{O}_z \mathbf{R}^T$$

8. Mar 4, 2015

### Robben

Very helpful! Thank you very much!