1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

QM: Two coupled spins in a magnetic field

  1. Jun 26, 2015 #1
    1. The problem statement, all variables and given/known data
    Consider two spins, L and R, in a magnetic field along the z-axis, i.e. [itex] B = (0, 0, B) [/itex]. The magnetic moments of the two spins are coupled to each other so that the total Hamiltonian reads
    [tex] H = g\mu_B\mathbf{B}\cdot(\mathbf{S}_L + \mathbf{S}_R) + J \mathbf{S}_L\cdot \mathbf{S}_R [/tex]

    Write this Hamiltonian in the basis [itex] \mathbf{\{} \mid \uparrow \uparrow \rangle, \mid \uparrow \downarrow \rangle, \mid \downarrow \uparrow \rangle, \mid \downarrow \downarrow \rangle \mathbf{\}} [/itex]

    2. Relevant equations

    The equations for the Pauli spin matrices


    3. The attempt at a solution
    I know that generally you can write a matrix:
    [tex]
    \newcommand{\unit}{1\!\!1}
    a\unit+ x \hat{\sigma_x} + y\hat{\sigma_y} + z\hat{\sigma_z} =
    \left( \begin{array}{ccc}
    a + z & x-iy \\
    x+iy & a-z \end{array} \right)
    [/tex]

    But other than that I dont know how to start especially with two particles.
     
  2. jcsd
  3. Jun 26, 2015 #2
    Write the Hamiltonian in cartesian components.
     
  4. Jun 26, 2015 #3

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    You have four states, so your Hamiltonian is going to be a 4x4 matrix. Start by examining what you get when you act on one of your states with your Hamiltonian. Use the result to write down part of the Hamiltonian.
     
  5. Jun 26, 2015 #4
    You mean like
    [tex]
    H_x = J\mathbf{S}_{L,x} \mathbf{S}_{R,x} \\
    H_y = J\mathbf{S}_{L,y} \mathbf{S}_{R,y} \\
    H_z = B(\mathbf{S}_{L,z} + \mathbf{S}_{R,z}) + J\mathbf{S}_{L,z} \mathbf{S}_{R,z}
    [/tex]
    ?
     
  6. Jun 26, 2015 #5

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    No, this is not correct. The Hamiltonian is not a vector. He means that you should expand the scalar products in terms of the components of the vectors contained in them.
     
  7. Jun 26, 2015 #6
    Write the Hamiltonian in cartesian spin components, indeed.
     
  8. Jun 26, 2015 #7
    Ok, but then would it still be the same except that the hamiltonian is not a vector so just a summation of all the terms? Like:
    [tex]
    H = J\mathbf{S}_{L,x} \mathbf{S}_{R,x} + J\mathbf{S}_{L,y} \mathbf{S}_{R,y} + B(\mathbf{S}_{L,z} + \mathbf{S}_{R,z}) + J\mathbf{S}_{L,z} \mathbf{S}_{R,z}
    [/tex]
    Or is this also wrong?
    As far as what you suggested, do you mean to evaluate something like [itex] H \mid \uparrow \uparrow \rangle [/itex] ?
    Using H of what I worte above that would be:
    [tex]
    H \mid \uparrow_L \uparrow_R \rangle =B \left( \begin{array}{ccc} 1 & 0 \\ 0 & -1 \end{array} \right)_L \left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_L
    + B \left( \begin{array}{ccc} 1 & 0 \\ 0 & -1 \end{array} \right)_R \left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_R \\
    + J \left( \begin{array}{ccc} 0 & 1 \\ 1 & 0 \end{array} \right)_L \left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_L \left( \begin{array}{ccc} 0 & 1 \\ 1 & 0 \end{array} \right)_R\left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_R
    + J \left( \begin{array}{ccc} 0 & -i \\ i & 0 \end{array} \right)_L \left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_L \left( \begin{array}{ccc} 0 & -i \\ i & 0 \end{array} \right)_R\left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_R
    + J \left( \begin{array}{ccc} 1 & 0 \\ 0 & -1 \end{array} \right)_L \left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_L \left( \begin{array}{ccc} 1 & 0 \\ 0 & -1 \end{array} \right)_R\left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_R
    [/tex]

    This will be I think:
    [tex]
    H \mid \uparrow_L \uparrow_R \rangle = \left(\begin{array}{c} B\\ 0 \\\end{array}\right)_L + \left(\begin{array}{c} B\\ 0 \\\end{array}\right)_R
    + J\left(\begin{array}{c} 0\\ 1 \\\end{array}\right)_L \left(\begin{array}{c} 0\\ 1 \\\end{array}\right)_R
    + J\left(\begin{array}{c} 0\\ i \\\end{array}\right)_L \left(\begin{array}{c} 0\\ i \\\end{array}\right)_R
    + J\left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_L \left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_R
    [/tex]

    Or am I going completely off-track? And what should I do with it?
     
  9. Jun 26, 2015 #8

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    You are going a bit off track. You cannot have a state containing only one of the spins as your first terms do.
     
  10. Jun 26, 2015 #9
    Where do I go wrong? Is my expression for the Hamiltonian(first line) at least correct?
    And am I missing an identity operator in the expressions and is [itex] B S_{L,z} [/itex] actually [itex] B S_{L,z} 1\!\!1_R [/itex]. But that wouldnt change much.
    Then:
    [tex]
    H \mid \uparrow_L \uparrow_R \rangle = \left(\begin{array}{c} B\\ 0 \\\end{array}\right)_L\left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_R + \left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_L\left(\begin{array}{c} B\\ 0 \\\end{array}\right)_R
    + J\left(\begin{array}{c} 0\\ 1 \\\end{array}\right)_L \left(\begin{array}{c} 0\\ 1 \\\end{array}\right)_R
    + J\left(\begin{array}{c} 0\\ i \\\end{array}\right)_L \left(\begin{array}{c} 0\\ i \\\end{array}\right)_R
    + J\left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_L \left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_R
    [/tex]
     
  11. Jun 27, 2015 #10
    You must use two spin basis functions. Use the basis you mention in the question.
    Treat the hamiltonian term by term.
    What is B(Sz1+Sz2)|++> ?
     
  12. Jun 27, 2015 #11
    If H_x etc are just names and not components of a vector as the notation suggests, this is ok.
    Use ladder operators.
     
  13. Jun 27, 2015 #12
    Ah ok so I have to construct H as:
    [tex]
    H = \left( \begin{array}{ccc}
    \langle \uparrow \uparrow \mid H \mid \uparrow \uparrow \rangle & \langle \uparrow \uparrow \mid H \mid \uparrow \downarrow \rangle & \langle \uparrow \uparrow \mid H \mid \downarrow \uparrow \rangle & \langle \uparrow \uparrow \mid H \mid \downarrow \downarrow \rangle\\
    \langle \uparrow \downarrow \mid H \mid \uparrow \uparrow \rangle & \langle \uparrow \downarrow \mid H \mid \uparrow \downarrow \rangle & \langle \uparrow \downarrow \mid H \mid \downarrow \uparrow \rangle & \langle \uparrow \downarrow \mid H \mid \downarrow \downarrow \rangle\\
    \langle \downarrow \uparrow \mid H \mid \uparrow \uparrow \rangle & \langle \downarrow \uparrow \mid H \mid \uparrow \downarrow \rangle & \langle \downarrow \uparrow \mid H \mid \downarrow \uparrow \rangle & \langle \downarrow \uparrow \mid H \mid \downarrow \downarrow \rangle\\
    \langle \downarrow \downarrow \mid H \mid \uparrow \uparrow \rangle & \langle \downarrow \downarrow \mid H \mid \uparrow \downarrow \rangle & \langle \downarrow \downarrow \mid H \mid \downarrow \uparrow \rangle & \langle \downarrow \downarrow \mid H \mid \downarrow \downarrow \rangle\\
    \end{array} \right)
    [/tex]
    Using the ladder operators [itex] S_x = \frac{1}{2}(S_+ + S_-) \quad S_y = \frac{1}{2i}(S_+ - S_-) [/itex] and:
    [tex]
    S_z \mid \uparrow \rangle = \frac{h}{2} \mid \uparrow \rangle \quad \quad S_z \mid \downarrow \rangle = \frac{-h}{2} \mid \downarrow \rangle \\
    S_+ \mid \uparrow \rangle = 0 \quad \quad S_+ \mid \downarrow \rangle = h \mid \uparrow \rangle \\
    S_- \mid \uparrow \rangle = h \mid \downarrow \rangle \quad \quad S_- \mid \downarrow \rangle = 0
    [/tex]

    I get:
    [tex]
    H \mid \uparrow \uparrow \rangle = g\mu_BBh \mid \uparrow \uparrow \rangle + \frac{Jh^2}{4} \mid \uparrow \uparrow \rangle + \frac{Jh^2}{2} \mid \downarrow \downarrow \rangle + \frac{Jh^2}{2i} \mid \downarrow \downarrow \rangle \\
    H \mid \uparrow \downarrow \rangle = 0 + \frac{-Jh^2}{4} \mid \uparrow \downarrow \rangle + \frac{Jh^2}{2} \mid \downarrow \uparrow \rangle + \frac{-Jh^2}{2i} \mid \downarrow \uparrow \rangle \\
    H \mid \downarrow \uparrow \rangle = 0 + \frac{-Jh^2}{4} \mid \downarrow \uparrow \rangle + \frac{Jh^2}{2} \mid \uparrow \downarrow \rangle + \frac{-Jh^2}{2i} \mid \uparrow \downarrow \rangle \\
    H \mid \downarrow \downarrow \rangle = g\mu_BBh \mid \downarrow \downarrow \rangle + \frac{Jh^2}{4} \mid \downarrow \downarrow \rangle + \frac{Jh^2}{2} \mid \uparrow \uparrow \rangle + \frac{Jh^2}{2i} \mid \uparrow \uparrow \rangle
    [/tex]

    This gives me the following matrix:
    [tex]
    H =
    \left( \begin{array}{ccc}
    g\mu_BB + \frac{Jh^2}{4} & 0 & 0 & \frac{Jh^2}{2} + \frac{Jh^2}{2i} \\
    0 & \frac{-Jh^2}{4} & \frac{Jh^2}{2} + \frac{-Jh^2}{2i} & 0\\
    0 & \frac{Jh^2}{2} + \frac{-Jh^2}{2i} & \frac{-Jh^2}{4} & 0\\
    \frac{Jh^2}{2} + \frac{Jh^2}{2i} & 0 & 0 & g\mu_BB + \frac{Jh^2}{4}
    \end{array} \right)
    [/tex]

    This gets me in the right direction but is still wrong in the off diagonal terms.
     
  14. Jun 27, 2015 #13
    You are definitely getting closer.
    Write S1xS2x+S1y+S2y in trems of S+ and S-.
    What is (S1z+S2z)|--> ?
     
  15. Jun 27, 2015 #14
    The trace of all operators in your hamiltonian is zero,
    so the trace of H should be zero. Check if it is.
     
  16. Jun 27, 2015 #15
    Ah I see, I am missing a factor [itex] \frac{1}{2}[/itex] in the [itex] S_x [/itex] term and a [itex] \frac{1}{2i}[/itex] in the [itex] S_y [/itex] term
    So it will be:
    [tex]
    H \mid \uparrow \uparrow \rangle = g\mu_BhBh \mid \uparrow \uparrow \rangle + \frac{Jh^2}{4} \mid \uparrow \uparrow \rangle + \frac{Jh^2}{4} \mid \downarrow \downarrow \rangle - \frac{Jh^2}{4} \mid \downarrow \downarrow \rangle \\
    H \mid \uparrow \downarrow \rangle = 0 + \frac{-Jh^2}{4} \mid \uparrow \downarrow \rangle + \frac{Jh^2}{4} \mid \downarrow \uparrow \rangle + \frac{Jh^2}{4} \mid \downarrow \uparrow \rangle \\
    H \mid \downarrow \uparrow \rangle = 0 + \frac{-Jh^2}{4} \mid \downarrow \uparrow \rangle + \frac{Jh^2}{4} \mid \uparrow \downarrow \rangle + \frac{Jh^2}{4} \mid \uparrow \downarrow \rangle \\
    H \mid \downarrow \downarrow \rangle = - g\mu_BhBh \mid \downarrow \downarrow \rangle + \frac{Jh^2}{4} \mid \downarrow \downarrow \rangle +\frac{Jh^2}{4} \mid \uparrow \uparrow \rangle - \frac{Jh^2}{4} \mid \uparrow \uparrow \rangle
    [/tex]

    Giving the matrix:
    [tex]
    H =
    \left( \begin{array}{ccc}
    g\mu_BhB + \frac{Jh^2}{4} & 0 & 0 & 0 \\
    0 & \frac{-Jh^2}{4} & \frac{Jh^2}{2} & 0\\
    0 & \frac{Jh^2}{2} & \frac{-Jh^2}{4} & 0\\
    0 & 0 & 0 & - g\mu_BhB + \frac{Jh^2}{4}
    \end{array} \right)
    [/tex]
     
    Last edited: Jun 27, 2015
  17. Jun 27, 2015 #16
    The matrix is close to the correct answer. The equations differ from it.
    You should write S1xS2x+S1yS2y in terms of S1+ etc.
     
  18. Jun 27, 2015 #17
    Ok,
    [tex]
    S_{x,L}S_{x,R} + S_{y,L}S_{y,R} = \frac{1}{2}(S_+ + S_-)_L\frac{1}{2}(S_+ + S_-)_R + \frac{1}{2i}(S_+ - S_-)_L\frac{1}{2i}(S_+ - S_-)_R \\
    =\frac{1}{2}(S_{+,L}S_{-,R} + S_{-,L}S_{+,R} )
    [/tex]
    But I still get the same answer. Whiech term exactly should I recheck?
     
  19. Jun 27, 2015 #18
    Are you sure about the final factor 1/2 ?
     
  20. Jun 27, 2015 #19
    I guess:
    [tex]
    S_{x,L}S_{x,R} + S_{y,L}S_{y,R} = \frac{1}{2}(S_+ + S_-)_L\frac{1}{2}(S_+ + S_-)_R + \frac{1}{2i}(S_+ - S_-)_L\frac{1}{2i}(S_+ - S_-)_R \\
    = \frac{1}{4} ( S_+S_+ + S_+S_- + S_-S_+ + S_-S_- - S_+S_+ + S_+S_- + S_+S_- - S_-S_- )
    =\frac{1}{2}(S_{+,L}S_{-,R} + S_{-,L}S_{+,R})
    [/tex]
    Or do you mean another 1/2 in the matrix?
     
  21. Jun 28, 2015 #20
    This is correct.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: QM: Two coupled spins in a magnetic field
Loading...