Expectation values of spin operators

[InsertName]

Hi, I've found the expectation value of S_z, which is hbar/2 (|$$\psi$$up|² - |$$\psi$$down|²) by using the formula:

<S_i> = <$$\psi$$|S_i$$\psi$$> where i can bex, y or z and $$\psi$$ is the 'spinor' vector.

I tried to find S_x using the same formula, however, I could only get as far as:

hbar/2 (($$\psi$$up)^*$$\psi$$down + ($$\psi$$down)^*$$\psi$$up)

In my lecture notes, it has the (final) answer (only) as hbar Re{($$\psi$$up)^*$$\psi$$down}.

Similarly, for S_y it gives the expectation value of hbar Im{($$\psi$$up)^*$$\psi$$down}.

I'm not sure how to get from my answer for <S_x> to the one in the notes. I'm assuming it's just a lack of knowledge of some identity with complex numbers.

Any help is appreciated, thanks in advance.

 

[vela]

Hint: Let $z=x+iy$. Its conjugate is $\bar{z}=x-iy$. What are $z+\bar{z}$ and $z-\bar{z}$ equal to?

 

[InsertName]

Thanks for the reply.

So, is it just a case of:

<S_x> = $$\frac{\hbar}{2}$$(($$\psi_up$$)^*$$\psi$$_down + ($$\psi$$_down)^*$$\psi$$_up))

Let z = ($$\psi_up$$)^*$$\psi$$_down
Let z^* = ($$\psi$$_down)^*$$\psi$$_up)

z+z^* = 2Re(z)

Therefore, $$\left\langle$$S_x$$\right\rangle$$ = 2Re(z).

This is why I hate LaTeX!

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<S_x> = (hbar/2)((psi-up)*(psi-down) + (psi-down)*(psi-up))

Let z = (psi-up)*(psi-down)
Let z* = (psi-down)*(psi-up)

z+z* = 2Re(z)

Therefore <S_x> = 2Re((psi-up)*(psi-down))

Thank you.

 

[vela]

Yup.