Expectation values of spin operators

[InsertName]

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

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

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

hbar/2 (([tex]\psi[/tex]up)^*[tex]\psi[/tex]down + ([tex]\psi[/tex]down)^*[tex]\psi[/tex]up)

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

Similarly, for S_y it gives the expectation value of hbar Im{([tex]\psi[/tex]up)^*[tex]\psi[/tex]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 [itex]z=x+iy[/itex]. Its conjugate is [itex]\bar{z}=x-iy[/itex]. What are [itex]z+\bar{z}[/itex] and [itex]z-\bar{z}[/itex] equal to?

 

[InsertName]

Thanks for the reply.

So, is it just a case of:

<S_x> = [tex]\frac{\hbar}{2}[/tex](([tex]\psi_up[/tex])^*[tex]\psi[/tex]_down + ([tex]\psi[/tex]_down)^*[tex]\psi[/tex]_up))

Let z = ([tex]\psi_up[/tex])^*[tex]\psi[/tex]_down
Let z^* = ([tex]\psi[/tex]_down)^*[tex]\psi[/tex]_up)

z+z^* = 2Re(z)

Therefore, [tex]\left\langle[/tex]S_x[tex]\right\rangle[/tex] = 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.

 

[paranormal]

Hello,

Thank you for sharing your findings and questions about expectation values of spin operators. It seems like you have a good understanding of the formula for calculating expectation values, <Si> = <\psi|Si\psi>, where i can be x, y, or z. However, I can see where you might be confused about the solution for Sx in your lecture notes.

The key to understanding the difference between your solution and the one in the notes lies in the use of complex numbers. In quantum mechanics, we often use complex numbers to represent the spin states of particles. The spinors you mention, \psiup and \psidown, are actually complex numbers representing the spin states "up" and "down". This means that when we take the inner product <\psiup|\psidown>, we are actually taking the complex conjugate of \psiup and multiplying it by \psidown. This is why your solution for Sx includes terms like (\psiup)*\psidown and (\psidown)*\psiup.

To simplify the expression for Sx, we can use the fact that the complex conjugate of a complex number is equal to its conjugate multiplied by -1. This means that we can rewrite <Sx> as:

hbar/2 ((\psiup)*\psidown + (\psidown)*\psiup) = hbar/2 ((\psiup)*\psidown - (\psiup)*\psidown) = 0

This is why the final answer for <Sx> in your lecture notes is only hbar Re{(\psiup)*\psidown}. Since the imaginary terms cancel out, we are left with the real part of the expression.

I hope this explanation helps to clarify the concept of complex numbers in quantum mechanics and how they relate to the expectation values of spin operators. If you have any further questions, please don't hesitate to ask. Keep up the good work in your studies!