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Hamiltonian of two identical spin-1/2 particles

  1. Dec 21, 2014 #1
    1. The problem statement, all variables and given/known data

    Two identical spin-1/2 particles of mass m moving in one dimension have the Hamiltonian $$H=\frac{p_1^2}{2m} + \frac{p_2^2}{2m} + \frac{\lambda}{m}\delta(\mathbf r_1-\mathbf r_2)\mathbf s_1\cdot\mathbf s_2,$$ where (pi, ri, si) are the momentum, position, and spin operators for the ith particle.

    (a) What operators, besides the Hamiltonian, are constants of motion and provide good quantum numbers for the stationary states?

    (b) What are the symmetry requirements for the spin and spatial wavefunctions?

    (c) If λ > 0, find the energy and quantum numbers for the bound state.

    2. Relevant equations

    An operator A is a constant of motion if [H, A] = 0.

    ##(\mathbf s_1 + \mathbf s_2)^2 = s_1^2 + s_2^2 + 2\mathbf s_1\cdot\mathbf s_2##.

    Total mass ##M = m_1+m_2##. Reduced mass ##\mu = \dfrac{m_1m_2}{M}##.

    Center of mass ##R = \dfrac{m_1r_1 + m_2r_2}{M}##. Separation ##r=r_1-r_2##.

    3. The attempt at a solution

    First off, if it's a 1-D problem, I have no idea why the position operators are vectors. But I guess I can get over that.

    (a) Operators which commute with H. I think p1 should be safe, since p1 commutes with itself, with p2, and with that whole interaction term. Likewise p2. But I don't know about the positions and spins. Do s1 and s1 · s2 commute? What about r1 and δ(r1r2)?

    My inclination is to rewrite the Hamiltonian in terms of better variables. I can introduce better spatial variables for the problem by considering the center of mass R and the separation r, as defined above. And I can introduce the total spin S, also defined above, to decompose the dot-product.
    $$H=\frac{p_R^2}{2M} + \frac{p_r^2}{2\mu} + \frac{\lambda}{m}\delta(r)\frac12(S^2-s_1^2-s_2^2).$$
    Looking at it this way, it seems like pR , pr , s1, s2, S, and R all commute with H. Still not sure about r1, r2, and r.

    (b) I think I'm good on this one. Spin-1/2 particles means fermions. Thus the states must be overall anti-symmetric. So if the spatial wavefunction is symmetric, the spin wavefunction must be antisymmetric (the "para" state); and vice versa (the "ortho" state).

    (c) ???
     
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  3. Dec 21, 2014 #2

    Orodruin

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    Does it though?
     
  4. Dec 21, 2014 #3
    Well, yes, I suppose you have a point there. So is pR the only safe one, then?
     
  5. Dec 21, 2014 #4
    Or no, I suppose S12 and S22 commute with each other, don't they.
     
  6. Dec 21, 2014 #5

    Orodruin

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    ##s_1^2## and ##s_2^2## act on different spin spaces, so yes, they commute with each other. ##S^2## should also be a good quantum number, the reason ##p_r## is not is the ##\delta(r)## in the Hamiltonian.
     
  7. Dec 21, 2014 #6
    Okay, so that makes pR, s12, s22, and S2. How do I know if that's all of them, though? I think, for instance, I could safely add Sz to that list, ya?

    Or remove S2 and Sz but add s1z and s2z?

    In classical mechanics there are as many constants of the motion as there are degrees of freedom. In quantum mechanics, how do I ever know when I'm done finding them?

    PS: Any small hints on part (c)?
     
  8. Dec 21, 2014 #7

    Orodruin

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    ##s_1^2## and ##s_2^2## are just the individual spins of the particles and nothing is going to change these, naturally.

    Of course, you can make new constants of motion by taking linear combinations of others - you can do this also in classical mechanics.

    On (c), the bound state exists only if the pre factor of the ##\delta## is negative. This tells you something about the spin state of the bound state, and thus about the symmetry of the wave function.
     
  9. Dec 21, 2014 #8
    Thank you for all the assistance. Here's what I have written as my answer. Any final feedback is welcome.

    (a) Constants of motion must commute with ##H##. It is simple to see right away that ##r_1## is no good; it won't commute with the ##p_1## term. Likewise, ##r_2##, ##p_1##, and ##p_2## are all out.

    It is illuminating to rewrite the given Hamiltonian in terms of more convenient variables. Given that both masses are equal, the total mass is ##M=2m## and the reduced mass is ##\mu=m/2##. Define ##R## to be the position of the center of mass and ##r## to be the distance between the two particles:
    $$
    R = \frac{mr_1+mr_2}{M} = \frac{r_1+r_2}{2}, \qquad r = r_1 - r_2.
    $$
    Additionally, define the total spin ##\mathbf S=\mathbf s_1+\mathbf s_2##. Then we can write
    $$
    S^2 = (\mathbf s_1+\mathbf s_2)^2 = s_1^2 + s_2^2 + 2\mathbf s_1\cdot\mathbf s_2.
    $$
    With all of these definitions, we can rewrite the given Hamiltonian as
    $$
    H=\frac{p_R^2}{2M} + \frac{p_r^2}{2\mu} + \frac{\lambda}{m}\delta(r)\frac12(S^2-s_1^2-s_2^2).
    $$

    From this new formulation, ##R##, ##r##, and ##p_r## are also all ruled out as constants of motion. But ##p_R## (the momentum of the system's COM) is a good one, since no ##R## appears anywhere in the Hamiltonian. Furthermore, ##s_1^2##, ##s_2^2##, and ##S^2## each commute with every term in ##H##, so they are all good constants of motion as well.

    (b) Spin-1/2 particles means fermions. Thus the states must be overall antisymmetric. So if the spatial wavefunction is symmetric, the spin wavefunction must be antisymmetric (the “para” state with ##S=0##); and if the spatial wavefunction is antisymmetric, the spin wavefunction must be symmetric (the “ortho” state with ##S=1##).

    (c) There will only exist a bound state if the energy of such a state is lower than the total energies of the two particles individually. So we require the contribution of the interaction term to be negative. Since ##\lambda## is stated to be positive, the parenthesized difference of spins must be negative. For spin-1/2 particles, ##s_1=s_2=1/2##. So the interaction energy is positive in the “ortho” state for which ##S=1## — no good! — but negative in the “para” state where ##S=0##. So at the final tally:

    ##s_1=1/2##

    ##s_2=1/2##

    ##S=0##

    ##p_R={}##anything

    ##E = \dfrac{p_R^2}{4m} + \dfrac{\langle p_r\rangle^2}{m} - \dfrac{\lambda}{4m}##
     
  10. Dec 21, 2014 #9

    Orodruin

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    Something like that. But. The energy you have given is not really an answer to the question. First of all you need to compute the energy contribution of the ##p_r^2## term. Second, the interaction term is not simply a constant. You must find the wave function of the bound eigenstate and the action of the Hamiltonian on this state. (This is simply finding the energy of the bound state of a negative delta potential. It is a standard exercise in QM and can be done by solving the SE on both sides of the delta and then using a matching condition. Note that this state has a symmetric wave function. If the coupling λ changes sign, there is no bound state as the S=1 state is symmetric in spin space and thus requires an anti symmetric wave function.)
     
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