# Represent Action of Stern Gerlach Operator as a Matrix

1. Jan 31, 2015

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

Given the series of three Stern-Gerlach devices:

Represent the action of the last two SG devices as matrices $\hat{A}$ and $\hat{B}$ in the $|+z\rangle, |-z\rangle$ basis.

2. Relevant equations

$|+n\rangle = cos(\frac{\theta}{2})|+z\rangle + e^{i\phi}sin(\frac{\theta}{2})|-z\rangle$

3. The attempt at a solution

A Stern-Gerlach device with one beam blocked off can be represented as a projection of the incoming state onto the outgoing state. A projection operator can be written as the outer product of the selected basis state with itself, e.g. $|+z\rangle\langle\,+\,z\,|$ would be the projection operator for an SGz device with the minus size blocked.

The projection matrix of the first SGn is then:

$\hat{A}\rightarrow{|+n\rangle\langle\,+\,n\,|} = \begin{pmatrix} cos(\frac{\theta}{2}) \\ e^{i\phi}sin(\frac{\theta}{2}) \end{pmatrix} \begin{pmatrix} cos(\frac{\theta}{2}) & e^{-i\phi}sin(\frac{\theta}{2}) \end{pmatrix} = \begin{pmatrix} cos^{2}(\frac{\theta}{2}) & e^{-i\phi}sin(\frac{\theta}{2})cos(\frac{\theta}{2}) \\ e^{i\phi}sin(\frac{\theta}{2})cos(\frac{\theta}{2}) & sin^{2}(\frac{\theta}{2}) \end{pmatrix}$

Therefore, the beam coming out of the positive side of the SGn should be:

$|+n\rangle = \hat{A}|+z\rangle \rightarrow \begin{pmatrix} cos^{2}(\frac{\theta}{2}) & e^{-i\phi}sin(\frac{\theta}{2})cos(\frac{\theta}{2}) \\ e^{i\phi}sin(\frac{\theta}{2})cos(\frac{\theta}{2}) & sin^{2}(\frac{\theta}{2}) \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} cos^{2}(\frac{\theta}{2}) \\ e^{i\phi}sin(\frac{\theta}{2})cos(\frac{\theta}{2}) \end{pmatrix}$

Should this be equal to the given $|+n\rangle$?

I feel as though I'm missing something completely obvious or misunderstanding the question.

2. Jan 31, 2015

### TSny

Hello and welcome to PF!
Yes, the output of SGn should be |+n>, up to a normalization factor.

3. Jan 31, 2015

So would $\frac{1}{cos(\frac{\theta}{2})}$ make for a suitable normalization factor? Is there a more formal way of determining this normalization factor?

4. Jan 31, 2015

### TSny

Yes
Not that I can think of at the moment. The usual way to find the normalization factor for a state |ψ>is to just look at <ψ|ψ>. But, in your example, you can spot the normalization factor by inspection.

5. Jan 31, 2015

Thanks for you help. I still don't see how $\langle\,+\,n\,|+n\rangle$ would give me the normalization factor though. I'm getting $cos^{2}(\frac{\theta}{2}) + sin^{2}(\frac{\theta}{2}) = 1$ which is what is expected.

6. Jan 31, 2015

### TSny

Let |ψ> be the output of GSn.

7. Jan 31, 2015

Ahh I see and I set this equal to 1. Thank you for your help!

8. Jan 31, 2015

This didn't work. I got a factor of $\frac{1}{cos^{2}(\frac{\theta}{2})}$ when evaluating $\langle\psi|\psi\rangle$ with $|\psi\rangle$ being the output of SGn.

$\begin{pmatrix} cos^{2}(\frac{\theta}{2}) & e^{-i\phi}sin(\frac{\theta}{2})cos(\frac{\theta}{2}) \end{pmatrix} \begin{pmatrix} cos^{2}(\frac{\theta}{2}) \\ e^{i\phi}sin(\frac{\theta}{2})cos(\frac{\theta}{2}) \end{pmatrix} = cos^{2}(\frac{\theta}{2})[cos^{2}(\frac{\theta}{2}) + sin^{2}(\frac{\theta}{2})] = cos^{2}(\frac{\theta}{2})$

Last edited: Jan 31, 2015
9. Jan 31, 2015

### Staff: Mentor

The normalization factor gets applied to both occurences of ψ, so you need the square root of $\langle\psi|\psi\rangle$ as normalization factor.

10. Feb 1, 2015

Thank you. How does this then relate back to the matrix representation? Would it just be $\frac{1}{cos(\frac{\theta}{2})}\begin{pmatrix}cos^{2}(\frac{\theta}{2}) & e^{-i\phi}sin(\frac{\theta}{2})cos(\frac{\theta}{2}) \\ e^{i\phi}sin(\frac{\theta}{2})cos(\frac{\theta}{2}) & sin^{2}(\frac{\theta}{2})\end{pmatrix}$

11. Feb 1, 2015

### Staff: Mentor

I don't think it makes sense to normalize the matrix. The norm smaller than 1 indicates that some particles get lost, which has a physical meaning.