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Spin System

  1. Oct 29, 2004 #1
    Hello.

    Consider a two object spin system. For t>0 the Hamiltonian is given as:
    H= 4*LAPLACE/h_^2 * S_1*S_2

    The initial state is given as |+->. I need to calculate the probabilities of finding the system in one of the four possible states (|++>,|+->,|-->,|-+>)

    a) by solving the problem exactly
    b) by using the time dependent perturbation theory (H being the perturbation)

    Perhaps somebody could give me a hint? I could write S_1*S_2 as S_1z*S_2z + 0.5 (S_1+*S_2- + S_1-*S_2+) to use the known eigen-value equations for S_z.

    Thanks in advance.
     
  2. jcsd
  3. Oct 30, 2004 #2
    Nobody? :)

    PS: Wow, just discovered the tex code function. Here's the Hamiltonian in a more readable format:

    [tex]H=(\frac{4\Delta}{\hbar^2})\vec{S_1}\cdot\vec{S_2}[/tex]
     
    Last edited: Oct 31, 2004
  4. Oct 30, 2004 #3
    Hello

    The hamiltonian you show is not very readable for me.
    What is the meaning of "LAPLACE"? I assume it is a 2 spin system with 4 possible basis states.
    LAPLACE, for me, indicates a spatial operator and this is strange.

    To answer further, I still assume your system is a two spin system.
    I also assume you have to calculate these probabilities as a function of time.
    Then, the general formula is

    [tex]\psi(t) = e^\frac{Ht}{i\hbar}\;\psi(0) = U(t)\;\psi(0)[/tex]​

    The exponential is easy to calculate in a basis where H is diagonal.
    Assume

    [tex]H = M^{-1}DM[/tex]​
    where D is diagonal

    then

    [tex]U(t) = M^{-1}e^\frac{Dt}{i\hbar}M[/tex]​

    In the last expression, the matrix exponential of Dt/ihb is easy to calculate.
    As usually, in principle, this is all a matter of eigenvalues and eigenvectors.

    So, a good hint would be to calculate to full 4x4 hamiltonian matrix.
    May be you can post the matrix, this could help.
    It is difficult to go further without clarification for H.
    A shorter way cannot be excluded. But it will amount to calculating the above faster.
     
  5. Oct 31, 2004 #4
    Hello.

    I had the same feeling when I looked at the Hamiltonian in this problem. For clarification I've emailed my tutor but I'm still waiting for a reply. For the moment, let's just think of the same Hamiltonian without the [tex]\Delta[/tex] operator:

    [tex]H = (\frac{4}{\hbar^2})\;\vec{S_1}\cdot\vec{S_2} = (\frac{4}{\hbar^2})\; (S_{1x}\cdot S_{2x} + S_{1y}\cdot S_{2y} + S_{1z}\cdot S_{2z})[/tex]

    OK. I calculated the full 4x4 hamiltonian matrix using tensor products, e.g.:

    [tex]S_{1z} = S^{(1)}_z \otimes E
    = \left(\begin{array}{cc}+\frac{\hbar}{2}\cdot\left(\begin{array}{cc}1&0\\ 0&1\end{array}\right) & \quad\!0\cdot\left(\begin{array}{cc}1&0\\ 0&1\end{array}\right)\\
    \quad\!0\cdot\left(\begin{array}{cc}1&0\\ 0&1\end{array}\right)&-\frac{\hbar}{2}\cdot\left(\begin{array}{cc}1&0\\ 0&1\end{array}\right)\end{array}\right)
    &=& \frac{\hbar}{2}\left(\begin{array}{cccc}
    +1&0&0&0\\
    0&+1&0&0\\
    0&0&-1&0\\
    0&0&0&-1\\
    \end{array}\right)\\
    [/tex]

    If I do that for all spin matrices I come up with the following Hamiltonian:

    [tex]H =
    \left(\begin{array}{cccc}
    +1&0&0&0\\
    0&-1&+2&0\\
    0&+2&-1&0\\
    0&0&0&+1\\
    \end{array}\right)
    [/tex]

    The eigen-values of this matrix are 1 (three times) and -3 (once) which leads to four eigen-states which are known as the triplets and the singlet:

    [tex]
    |1,1\rangle := |++\rangle\\
    [/tex]
    [tex]
    |1,0\rangle := \frac{1}{\sqrt{2}}(|+-\rangle + |-+\rangle)\\
    [/tex]
    [tex]
    |1,-1\rangle := |--\rangle\\
    [/tex]
    [tex]
    |0,0\rangle := \frac{1}{\sqrt{2}}(|+-\rangle - |-+\rangle)
    [/tex]

    [tex]|+-\rangle[/tex] and [tex]|-+\rangle[/tex] are no eigen-states so the probability of finding these states is zero. Am I wrong? And what about the other states?
     
    Last edited: Oct 31, 2004
  6. Oct 31, 2004 #5
    My tutor emailed me: The [tex]\Delta[/tex] is just a variable for the intensity of the spin coupling so the above assumption of leaving out the [tex]\Delta[/tex] doesn't matter at all. Feel free to post your thoughts about the above calculations. :smile:
     
  7. Nov 1, 2004 #6
    Hello

    I have the feeling you are on the track now.
    I have some difficulties to understand what you wrote:

    • you did not indicate the label for the rows and columns of H (is it ++,+-,-+,-- for example?)
    • what is the meaning of E, is it the identity matrix?

    Concerning your last remark: if the initial state is not an eigenstate of H, then it is a superposition of eigenstates of H anyway. Then, this superposition will change with time since each part of the superposition involves a different frequency(energy). If the initial state is an eigenstate, then the system remains in this state for all times. All this is contained in the formal solution I wrote in my firt post.

    Note: I have no idea about the track followed in your quantum physics course regarding time evolution of quantum states. I guess that you have learned the Schrodinger equation and the Dirac notation. It could be good that you explain, in short, what else you are supposed to know to solve your exercise. Have you seen something about the exponential of operators? Are you maybe supposed to solve the Schrödinger equation as a sytem of first order differential equations? There are several equivalent ways to develop the solution.
     
  8. Nov 1, 2004 #7
    Hi

    Concerning your questions:
    - What do you mean by labeling the matrix? For me, |++> = (1,0,0,0), |+-> = (0,1,0,0) and so on. But I guess that was not your question.
    - E is the identity matrix

    OK, my quantum physics course somehow follows Sakurai's books "Modern Quantum Mechanics" and "Advanced Quantum Mechanics". We're just about to finish the first book so I should know about the Schrödinger equation, momentum, spin, Dirac notation, Schrödinger/Heisenberg/Interaction picture, perturbation theory (both stationary and time dependent),...

    I'm pretty sure we can use the interaction picture to solve problem a). The main equation in the interaction picture is:

    [tex]i\hbar \dot{c_k} = \sum\limits_{n=1}^{k}V_{kn}e^{i\omega_{kn}t}c_n
    [/tex]

    With
    [tex]\omega_{kn} = \frac{E_k-E_n}{\hbar}[/tex]

    [tex]|c_k|^2[/tex] gives you the probability of finding the state k. The formulation in the problem indicates that [tex]H_0 = 0[/tex] which means that [tex]H = H_0 + V = V[/tex] and [tex]\omega_{kn} = 0[/tex]. I'm not so sure about the last conclusion but anything else doesn't make sense when I try to solve the differential equations.

    If you would like to have a look at my calculations in detail, click here (Doh! German...). Anyway, my results are:

    [tex]
    c_1(t) &= a_1e^{-i\frac{\Delta}{\hbar}t} & \Rightarrow |c_1(t)|^2 = a_1^2[/tex]

    [tex]
    c_2(t) &=\frac{1}{\sqrt{2}}(a_2e^{-i\frac{\Delta}{\hbar}t}+a_4e^{3i\frac{\Delta}{\hbar}t}) & \Rightarrow |c_2(t)|^2 =
    \frac{1}{2}(a_2^2+a_2a_42\cos(\frac{4\Delta}{\hbar}t)+a_4^2)
    [/tex]

    [tex]
    c_3(t) &= \frac{1}{\sqrt{2}}(a_2e^{-i\frac{\Delta}{\hbar}t}-a_4e^{3i\frac{\Delta}{\hbar}t}) & \Rightarrow |c_3(t)|^2 =
    \frac{1}{2}(a_2^2-a_2a_42\cos(\frac{4\Delta}{\hbar}t)+a_4^2)
    [/tex]

    [tex]
    c_4(t) &= a_3e^{-i\frac{\Delta}{\hbar}t} & \Rightarrow |c_4(t)|^2 = a_3^2
    [/tex]

    Now, my problem is: How do I assign these results to the |++>,|+->,|-+>.|--> states?
     
  9. Nov 1, 2004 #8
    Peperone,

    I have checked your Hamiltonian, your eigenvalues and your eigenvectors.
    I get the same results as you, assuming the labels for the rows in H are as you mentioned:

    |++> = (1,0,0,0)
    |+-> = (0,1,0,0)
    |-+> = (0,0,1,0)
    |--> = (0,0,0,1) ​

    Now, I am a little bit perplexed by your start for the solution.
    You assume that H=Ho+V.
    Do you mean that you have to solve a perturbation problem?
    I guessed that you have to calculate the evolution of the system assuming the Hamiltonian is the one you gave first:
    [tex]H = (\frac{4}{\hbar^2})\;\vec{S_1}\cdot\vec{S_2}[/tex]​

    The easiest way is to calculate the time evolution in the diagonal basis.
    Then, the given initial condition has to be translated. The initial state can be expressed as:

    [tex]|\phi(t=0)\rangle = |+-\rangle = \frac{1}{\sqrt{2}}(|1,0\rangle + |0,0\rangle)[/tex]​

    and the time evolution is then simply:

    [tex]|\phi(t)\rangle = \frac{1}{\sqrt{2}}(e^{\frac{E_{10}t}{i\hbar}}|1,0\rangle + e^{\frac{E_{00}t}{i\hbar}}|0,0\rangle)[/tex]​

    (where you have to replace E00 and E10 by the ad-hoc eigenvalues)

    The probabilities of finding the system in the different states above is easily obtained by the usual projections. Note that the state is a mixture of |+-> and |-+> for all times. Therefore, the probabilities to find the system in the states |++> or |--> is always 0.

    What remains unclear is that question of "perturbation". If you have to solve a perturbed system things will be somewhat similar but you need some data on Ho.
     
  10. Nov 1, 2004 #9
    Hm... The following would make sense (mathematically).

    Let's say:

    [tex]|c_1(t)|^2 \Leftrightarrow p(|++\rangle)[/tex]
    [tex]|c_2(t)|^2 \Leftrightarrow p(|+-\rangle)[/tex]
    [tex]|c_3(t)|^2 \Leftrightarrow p(|-+\rangle)[/tex]
    [tex]|c_4(t)|^2 \Leftrightarrow p(|--\rangle)[/tex]

    Then, the initial conditions would be:
    [tex]|c_1(0)|^2 = |c_3(0)|^2 = |c_4(0)|^2 = 0 \Rightarrow a_1 = a_3 = 0[/tex]
    [tex]|c_2(0)|^2 = 1 \Rightarrow a_2 = a_4 = \frac{1}{\sqrt{2}}[/tex]

    Which leads to the final result:
    [tex]|c_1(t)|^2 = |c_4(t)|^2 = 0[/tex] (no |++> and no |-->)
    [tex]|c_2(t)|^2 = 0,5 (1+\cos(\frac{4\Delta}{\hbar }t))[/tex]
    [tex]|c_3(t)|^2 = 0,5 (1-\cos(\frac{4\Delta}{\hbar }t))[/tex]

    As I said, this fits nicely (mathematically) but I'd like to be able to give reasons for this... Any ideas?
     
  11. Nov 1, 2004 #10
    Heh, that nearly was a simultaneous post. :wink:

    The problem formulation says:
    H = 0 for t<0 and H = the Hamiltonian posted in the beginning for t>0. This made me think of H as a perturbation which is "switched on" at t=0.

    Your solution looks nice, too.
    Edit: Infact, it leads to exactly the same solution which is a good thing. :biggrin:
     
    Last edited: Nov 1, 2004
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