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

In summary, the Hamiltonian for a 2 spin 1/2 particle system is given as H = w1 S1z + w2 S2z. The possible values are 2 \hbar^2 and 0. At some later time, the system will look like \Psi(t) = a e{-iE_1 t/ \hbar} + b e{-iE_2 t/ \hbar} where E_1 and E_2 are the coefficients for the spin states. The probabilities of measuring each eigenvalue are given by the product of e1 and e2.
  • #1
cp51
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Homework Statement


I have a two spin 1/2 particles. The Hamiltonian for the system is given as H = w1 S1z + w2 S2z. 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 | [tex]\uparrow[/tex] [tex]\downarrow[/tex] > + b | [tex]\downarrow[/tex] [tex]\uparrow[/tex]>

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?
 
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  • #2
You need to express [itex]|\psi(t)\rangle[/itex] in terms of the eigenstates of S2. Do you know how to do that?
 
  • #3
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.
 
  • #4
Yes, that's what I mean. Do you know how to express those states as linear combinations of [itex]|\uparrow\,\downarrow\,\rangle[/itex] and [itex]|\downarrow\,\uparrow\,\rangle[/itex]? 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.
 
  • #5
Ok, I think I got it,

so using:

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

and

e2 = [TEX]\frac{1}{\sqrt{2}}[/TEX](|+,-> - |-,+>) 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.
 
  • #6
Yup, good job!
 

1. What is the possible range of values for S^2 - Spin?

The possible range of values for S^2 - Spin is from 0 to (s(s+1)), where s is the spin quantum number.

2. How is the probability of measuring S^2 - Spin calculated?

The probability of measuring S^2 - Spin is calculated by taking the absolute square of the spin state vector, also known as the wave function, at a given point in space. This value represents the likelihood of obtaining a specific measurement of S^2 - Spin.

3. Can the probability of measuring S^2 - Spin be greater than 1?

No, the probability of measuring S^2 - Spin cannot be greater than 1. This is because the absolute square of the spin state vector always yields a value between 0 and 1, representing the likelihood of obtaining a specific measurement.

4. How does the probability of measuring S^2 - Spin change with different spin quantum numbers?

The probability of measuring S^2 - Spin increases with higher spin quantum numbers. This is because higher spin values result in a larger range of possible values for S^2 - Spin, increasing the likelihood of obtaining a specific measurement.

5. Does the probability of measuring S^2 - Spin change in different physical systems?

Yes, the probability of measuring S^2 - Spin can vary in different physical systems. This is because the spin quantum number and the allowed values for S^2 - Spin depend on the specific properties of the system, such as the type of particle and the strength of the external magnetic field.

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