# How to find this spin component

1. Sep 16, 2007

### indigojoker

We know that $$S_x = \frac{\hbar}{2} \left( |+ \rangle \langle - | + | - \rangle \langle+| \right)$$

But what is $$|S_x ; + \rangle$$?

I think my text says $$|S_x ; + \rangle = \frac{1}{\sqrt{2}} \left( |+ \rangle + | - \rangle \right)$$ but i dont know how they got this.

I feel like this is a trivial question but I'm not sure how one finds $$|S_x ; + \rangle$$

2. Sep 17, 2007

### dextercioby

But WHAT is $|S_{x},+\rangle$ ?? I've never seen this notation before...And it's not that i've looked into one book... I haven't looked in your book, apparently, you might share with us the title and the author...

3. Sep 17, 2007

### genneth

I imagine that it's the positive spin direction for S_x.

OP: it's just an eigenvector -- so you find it in the same way that you find any eigenvectors. If it helps, write S_x as a matrix, in the |+>, |-> basis that you've got things in.

4. Sep 17, 2007

### indigojoker

I just figured it out.

I am using * as the dot product

$$S * \hat n | S * \hat n ; + \rangle = \frac{ \hbar}{2} | S * \hat n ; + \rangle$$
$$| S * \hat n ; + \rangle = \cos \frac{\beta}{2} |+ \rangle + \sin \frac{\beta}{2} e^{i \alpha} | - \rangle$$

where beta is the polar angle and alpha is the azimuthal angle.

therefore, an S_x measurement would be where beta = pi/2 and alpha =0

since the S_x measurement would yield +hbar/2, we get:

$$| S_x; + \rangle = \cos \frac{\pi/2}{2} |+ \rangle + \sin \frac{\pi/2}{2} e^{0} | - \rangle$$

therefore:
$$|S_x ; + \rangle = \frac{1}{\sqrt{2}} \left( |+ \rangle + | - \rangle \right)$$

Last edited: Sep 17, 2007
5. Sep 17, 2007

### nrqed

They want the state which is an eigenstate of Sx with the eigenvalue +hbar/2.

So you could write $$|S_x ; + \rangle = \alpha |+ \rangle + \beta |- \rangle$$

and apply S_x, imposing $S_x |S_x; + > = \frac{\hbar}{2} |S_x;+>$ and then solve for alpha and beta (and normalize at the end)