How to find this spin component

[indigojoker]

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

But what is [tex]|S_x ; + \rangle[/tex]?

I think my text says [tex]|S_x ; + \rangle = \frac{1}{\sqrt{2}} \left( |+ \rangle + | - \rangle \right)[/tex] but i don't know how they got this.

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

 

[dextercioby]

But WHAT is [itex] |S_{x},+\rangle [/itex] ?? 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...

 

[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.

 

[indigojoker]

I just figured it out.

I am using * as the dot product

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

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:

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

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

 

[nrqed]

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

But what is [tex]|S_x ; + \rangle[/tex]?

I think my text says [tex]|S_x ; + \rangle = \frac{1}{\sqrt{2}} \left( |+ \rangle + | - \rangle \right)[/tex] but i don't know how they got this.

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

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

So you could write [tex]|S_x ; + \rangle = \alpha
|+ \rangle + \beta |-
\rangle [/tex]

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