Possible values and their Probability of Measuring S^2 - Spin

[cp51]

Homework Statement

I have a two spin 1/2 particles. The Hamiltonian for the system is given as H = w₁ S_1z + w₂ S_2z. I need to find the possible values and their probabilities when I measure S^2 at some later time T. Also the Initial state \Psi (0) = a | \uparrow \downarrow > + b | \downarrow \uparrow>

Homework Equations

The Attempt at a Solution

Now I know for a 2 spin 1/2 particle system, s = 1 and 0.

The eigenvalue equation for S² is S²|sm> = hbar² ( s ( s+1) )|sm>

So the possible values are 2 \hbar^2 and 0.

I know at some later time, the state will look like \Psi(t) = a e^{{-iE_1 t/ \hbar}} + b e^{{-iE_2 t/ \hbar}}

and I can find E_1 and E_2

However, how do i find the probabilities?

If I was just looking for S_z probabilities, I know it would be a² for spin up and b² for spin down. I also know that if I was looking for S_x I would need to evolve the coefficients in time. However, how do I measure the probabilities of S²?

 

[vela]

You need to express |\psi(t)\rangle in terms of the eigenstates of S². Do you know how to do that?

 

[cp51]

Hmm, I'm not positive... is that writing it in |10> and |00> states? I am not exactly sure how to do this. Can you help get me moving in the right direction?

thanks for the help.

 

[vela]

Yes, that's what I mean. Do you know how to express those states as linear combinations of |\uparrow\,\downarrow\,\rangle and |\downarrow\,\uparrow\,\rangle? If not, you should figure out how to do that. It's probably covered in your textbook as it's a pretty common example of the addition of angular momentum.

 

[cp51]

Ok, I think I got it,

so using:

e1 = \frac{1}{\sqrt{2}}(|+,-> + |-,+>) with eigenvalue 2hbar^2

and

e2 = \frac{1}{\sqrt{2}}(|+,-> - |-,+>) with eigenvalue 0

I rewrite: |+,-> as (e1 + e2)*(sqrt(2)/2) and |-,+> as (e1 - e2)*(sqrt(2)/2)

Combine e1 and e2 terms. And then the coefficients squared give me the probabilities of measuring each eigenvalue as a function of time.

Sound good?

Thanks for your help.

 

[vela]

Yup, good job!