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Homework Help: Addition of Spin 1/2

  1. Jan 19, 2008 #1
    [SOLVED] Addition of Spin 1/2

    1. The problem statement, all variables and given/known data
    A system is composed of two spin 1/2. The system's hamiltonian is [itex]H=w_1 S_1z + w_2 S_2z[/itex]. The initial state of the system is, at t=0,
    [itex] | \psi (0) > = \frac{1}{\sqrt(2)}[|+ -> + |-+>] [/itex].

    a) At time t, we measure [itex] S^2 = (S1+S2)^2 [/itex] What results can we find and with what probabilities?

    2. Relevant equations
    [itex] H\psi = E\psi [/itex]

    3. The attempt at a solution
    So, I think the first step is to find the Hamiltonian of my system to know how the system will evolve, as we need the eigenvalues of the energy in the exponential.

    So I thought of doing

    [itex] (w_1 S_1z + w_2 S_2z)|\psi (0) > = E|\psi(0)> [/itex]. But, here, I'm not so sure of what to do. I mean, [itex] S_1z [/itex] is a two-D operator while psi(0) is a 4x1 matrice (0 1 1 0) (placed vertically). This matrix product doesn't seem correct to my senses.
  2. jcsd
  3. Jan 19, 2008 #2
    hum... instead, using [itex] S_1z|+ - > = \frac{\hbar}{2} |+-> [/itex] and so on, I have the eigenvalue [itex] E=\frac{\hbar}{2}(w_1 - w_2). [/itex]. The problem is... am I not supposed to have more than one eigenvalue of energies? Because the eigenvalue equation [itex] H\psi = E\psi [/itex] then gives me

    [itex]\frac{\hbar}{2\sqrt 2}(w1-w2)[|+-> + |-+>]=E\psi [/itex].

    I seem to have advanced to the solution. But I don't know where to go from there :S?
    Last edited: Jan 19, 2008
  4. Jan 19, 2008 #3


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    What is the time evolution of the state

    [itex] | \psi (0) > = \frac{1}{\sqrt(2)}[|+ -> + |-+>] [/itex]

    If the hamiltonian is given by :
    [itex]H=w_1 S_{1z} + w_2 S_{2z}[/itex] ?

    (I assume you meant this, so i fixed your latex code)

    So first find [tex] \psi (t) [/tex], by applying the time evolution operator to [tex] \psi (0) [/tex].

    then you apply the operator [itex] S^2 = (S1+S2)^2 [/itex] and find the possible outcomes and probs.


    So you have [tex] \psi (0) [/tex], decompose it, a hint;)
    Last edited: Jan 19, 2008
  5. Jan 19, 2008 #4


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    [itex] S^2 = (S_1+S_2)^2 [/itex] You manipulate a bit.
  6. Jan 19, 2008 #5
    Hi malawi, thanks for your help,
    Yes, that is what I was trying to do (I need energy eigenvalues in order to apply the time evolution operator). The problem is actually finding energy eigenvalues...

    Using your idea of decomposing [itex] \psi(0) [/itex], maybe it would be a nice idea to express [itex] \Psi(0)[/itex] in the hamiltonian basis and then actually find the eigenvalues of [itex] [\psi(0)]_E [/itex].

    Then, I would have energy eigenvalues... right...?
  7. Jan 19, 2008 #6


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    Oki, if I start, then you can fill in the rest.. (we shall now find the evolved state [tex] \psi (t) [/tex])

    Time-ev op: [tex] U(t,0) = \exp (-iHt/\hbar ) [/tex]

    initial state: [itex] | \psi (0) > = \frac{1}{\sqrt(2)}|+ -> + \frac{1}{\sqrt(2)}|-+> [/itex]

    So the first ket on the RHS: evolves according to:

    [tex] U(t,0) |+ -> = \exp (-i(w_1 S_{1z} + w_2 S_{2z})t/\hbar )|+ -> = [/tex]

    [tex] \exp (-i(w_1 \hbar / 2 - w_2 \hbar /2)t/\hbar )|+ -> = [/tex] ...

    I hope you got the idea, now continue to get the final [tex] \psi (t) [/tex].
    Last edited: Jan 19, 2008
  8. Jan 19, 2008 #7
    Got it! I forgot that subtility of "decomposing" the [itex] \psi [/itex] function. I forgot that the function was a linear combination of the two independant states. I feel somehow ashamed.

    \psi(t)>=\frac{1}{\sqrt 2}|+ -> \exp(-\frac{-i}{2}(w_1 - w_2)t) + \frac{1}{\sqrt 2} |- +>\exp(\frac{-i}{2}(-w_1 +w_2)t))[/itex]

    Now, I just have to apply [itex] S^2 [/itex] to my [itex] \psi(t) [/itex] and see what eigenvalues I can have...

    [itex] S^2| \psi(t)> \frac{1}{\sqrt 2} ( \exp(\frac{-i}{2}(w_1 - w_2)t) \hbar ^2 [|+-> +[-+> ] + \exp(\frac{-i}{2}(w_2 - w_1)t) \hbar ^2 [|+-> +[-+> ])...... [/itex]

    Which eventually leads to

    [itex] S^2| \psi (t) > = 2\hbar ^2 | \psi (t)> [/itex]

    ... Which makes me wonder why there's only one eigenvalue...? :S I must be forgetting something somewhere...

    THanks for your help.
  9. Jan 19, 2008 #8


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    It is not THAT straightforward to operate with [tex] S^2 [/tex]

    Did you write [tex] S^2 = (S_1 + S_2)^2 = S_1^2+S_2^2 + 2S_{1z}\cdot S_{2z} + ...[/tex] ?
  10. Jan 19, 2008 #9
    Well, in my course, the teacher developped the expression you wrote and concluded that[itex]
    S^2|++>=\frac{3}{4}\hbar^2|++> + \frac{3}{4}\hbar^2|++> + \frac{1}{2}\hbar^2 |++> +0 = 2\hbar^2 |++>
    and by similar development :

    ... Did I apply the operator too naively?

    I used these identities to calculate the effet of S^2 on my function... Was it wrong :S ?
  11. Jan 19, 2008 #10


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    if you write [tex] \psi (t) = A(t)|+-> + B(t)|-+> [/tex]

    where [tex] A(t) = \exp (-i(w_1 \hbar / 2 - w_2 \hbar /2)t/\hbar )[/tex]

    and B(t) is the other exp..

    Then see what you get.
  12. Jan 19, 2008 #11


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    is there a part b) and c) on this question also?

    The total spin of a two spin ½ system is either 1 or 0, so S^2 should give you the value 2 or 0, but the problem here is to find out their probabilties as a function of time.

    So i think you should take these functions:
    [tex] \frac{1}{\sqrt(2)}[|+ -> + |-+>] [/tex] (the spin S = 1, M_s = 0 state)

    and [tex] \frac{1}{\sqrt(2)}[|+ -> - |-+>][/tex] (the spin S = 0, M_s = 0 state)

    and projectet onto [tex] \psi (t) [/tex]

    But I must admit that this was a kinda tricky problem :S
  13. Jan 19, 2008 #12
    So, knowing that, we can write that

    [itex] S^2|\psi(t)>=A(t) \hbar^2 [|+->+|-+>] + B(t) \hbar^2[|+->+|-+>][/itex].

    That means that it is equal to (if I regroup the terms)

    [itex](A(t)\hbar^2|+->+B(t)\hbar^2|-+>) + (A(t)\hbar^2|-+> + B(t)\hbar^2|+->) [/itex]

    [itex]=(\hbar^2 \psi(t) ) + (\hbar^2 \phi(t)) [/itex]

    Where [itex] \phi(t) [/itex] represend the last term.
  14. Jan 19, 2008 #13
    Yes, there's a part b) which is to find out the relation between w1 and w2 so that the probabilities are independant of time. Which, I expect, should be fairly easy once part a) is done ;).

    So you are saying that I should try, instead of doing what I wrote in my last message, to write psi(t) in function of the two functions you wrote. OK, but I do not have any minus coming from the relations I used... :S.
  15. Jan 19, 2008 #14
    Ok, I got it.

    So, to do a small résumé, I

    1) Had to find the possible energies to use the time evolution operator. For this, I applied H on |+-> and |-+>.
    2) Then, I was able to write my function psi(t)
    3) As this function evolve in time, it's not a surprise it would be a linear combination of the two possibles values of spin (1 0) (0 0). So I can write my psi(t) function in a linear combination of (1 0) and (0 0).
    4) Happily enough, this is quite simple. I wrote |+-> and |-+> in function of |0 0> and |1 0> and put it in psi(t).
    5) Then, I can find quite easily the possible values of S^2, especially if I use some trig. identities involving cos and sin in function of complex exponential.
    6) The probabilities are then proportional to cos^2 and sin^2 and the values are 2hbar^2 and 0.

    Thank you very much, malawi_glenn. Your help was very much appreciated! :)

    P.S. I don't know how to put [Solved] beside my thread's title. But it is pretty much solved.
    Last edited: Jan 19, 2008
  16. Jan 20, 2008 #15


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    That looks ok and reasonable :) Good work!
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