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Spin inversion under dipole-dipole interaction of fermions

  1. Oct 9, 2014 #1
    1. The problem statement, all variables and given/known data

    Consider two spin 1/2 particles interacting through a dipole-dipole potential

    [tex]\hat{V} = A\frac{(\hat{\sigma_1} \cdot \hat{\sigma_2})r^2 - (\sigma_1 \cdot \vec{r})(\sigma_2 \cdot \vec{r})}{r^5} [/tex]

    If both spins are fixed at a distance d between each other, and at t = 0 one of them is parallel to the radius vector r, whereas the other one is anti-parallel to r, find the first moment when the orientations of the spins have inverted.

    (Please note that the wording of the problem doesn't mention what does A nor sigma mean, but I assume A must be a normalisation constant and sigma/2 the corresponding spin operators)

    2. Relevant equations

    Time-dependent Schrödinger equation:

    [tex]-\frac{\hbar^2}{2m}\nabla^2\Psi + V(\mathbf{r})\Psi
    = i\hbar \frac{\partial\Psi}{\partial t}[/tex]

    3. The attempt at a solution

    I don't really know where to start from. This exercise is included in the Identical Particles homework series of my Quantum Mechanics II course. I'm not sure what does "find the first moment" really mean. How can there be "a first moment" when in general all we tan talk about is probabilities of measuring certain observables?

    As I have to deal with the time variable, I was guessing it might have something to do with the time-dependent Schödinger equation. I guess the wave function needs to evolve with time, and the orientation of the spins would have to change either continuously or spontaneously. Plus the wave function must be anti-symmetric because we are dealing with fermions... Uhm... I am kind of lost. Any kind of help or reference would be greatly appreciated.
     
  2. jcsd
  3. Oct 10, 2014 #2

    DrClaude

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    Staff: Mentor

    ##A## is not a normalization constant, but rather parametrizes the strength of the interaction.

    The particles being fixed, I don't know why you have a momentum operator there. The first part of such a problem is determining what the Hamiltonian is, and then you can write the proper Schrödinger equation.

    But what if the probabilities are time dependent?

    Exactly. You have to start by writing the initial state of the system, then find its time evolution (which will be continuous, I don't know why it would be "spontaneous"). As for the parity of teh wave function, just make sure that the initial wave function is anti-symmetric, and the time evolution will preserve the parity.
     
  4. Oct 11, 2014 #3
    Thank you very much for your reply. I am getting some problems with the algebra:

    As the particles are fixed, the hamiltonian is going to be equal to the potential energy:

    [itex]\hat{H} = \hat{V}[/itex]

    Generally speaking,

    [itex]\hat{\sigma_1} = \sigma_{1x} \hat{i} + \sigma_{1y} \hat{j} + \sigma_{1z} \hat{k}[/itex]
    [itex]\hat{\sigma_2} = \sigma_{2x} \hat{i} + \sigma_{2y} \hat{j} + \sigma_{2z} \hat{k}[/itex]

    Let [itex]\vec{r}[/itex] point in the [itex]\hat{k}[/itex] direction, i.e. [itex]\vec{r}=z\hat{k}[/itex]

    At t = 0, we have:

    [itex]\hat{\sigma_1} = \sigma_{1z} \hat{k}[/itex]
    [itex]\hat{\sigma_2} =-\sigma_{2z} \hat{k}[/itex]

    [itex] (\hat{\sigma_1} \cdot z\hat{k})(\hat{\sigma_2} \cdot z\hat{k}) = -z^2 \sigma_{1z} \sigma_{2z}[/itex]

    whereas

    [itex] \hat{\sigma_1} \cdot \hat{\sigma_2} = - \sigma_{1z} \sigma_{2z}[/itex]

    thus the potential vanishes at t = 0. I might be messing up at some very basic mistake, right?

    Thanks in advance!
     
  5. Oct 11, 2014 #4
    Sorry, I think my reasoning was erroneous (it still might be)

    I think it is better to say that the initial state of the system can be described by the singlet state (because they point in opposite directions at t = 0, and the wave function needs to be antisymmetric):

    [itex] \Psi(0) = 1/ \sqrt{2} ( |+ \rangle |- \rangle - |-\rangle |+ \rangle)[/itex]

    Now, to construct the time evolution operator, the hamiltonian of the system is given by the dipole interaction. As I said previously, in general we have

    [itex] \hat{\sigma_1} \cdot \hat{\sigma_2} = \sigma_{1x}\sigma_{2x} + \sigma_{1y}\sigma_{2y} + \sigma_{1z}\sigma_{2z}[/itex]

    and (assuming that [itex] \vec{r} [/itex] points in the k direction

    [itex] \hat{\sigma_i} \cdot \vec{r} = d \sigma_{iz}[/itex]

    so the second term in the potential would be [itex]d^2 \sigma_{1z}\sigma_{2z}[/itex], while the first term corresponds to [itex] d^2(\sigma_{1x}\sigma_{2x} + \sigma_{1y}\sigma_{2y} + \sigma_{1z}\sigma_{2z})[/itex]

    This way, the hamiltonian would read
    [itex] \hat{H} = \frac{A}{d^3}(\sigma_{1x}\sigma_{2x} + \sigma_{1y}\sigma_{2y} + \sigma_{1z}\sigma_{2z} - \sigma_{1z}\sigma_{2z})[/itex]
    [itex] \hat{H} = \frac{A}{d^3}(\sigma_{1x}\sigma_{2x} + \sigma_{1y}\sigma_{2y})[/itex]

    I am not sure how to proceed from here. I was hoping to get a single-matrix expression for the hamiltonian so I could diagonalise it and construct the evolution operator. Does this procedure make some sense at all?
     
  6. Oct 11, 2014 #5

    vela

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    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    The standard trick is to use ##\vec{S}^2 = (\vec{S}_1+\vec{S}_2)^2 = \vec{S}_1^2+\vec{S}_2^2+2\vec{S}_1\cdot\vec{S}_2##.
     
  7. Oct 11, 2014 #6
    do you suggest using that relation on the last part, or did I make a wrong choice from the beginning?
     
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