QM: Two coupled spins in a magnetic field

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SUMMARY

The discussion focuses on the Hamiltonian for two coupled spins, L and R, in a magnetic field along the z-axis, represented as H = gμB(B)(SL + SR) + JSL·SR. Participants detail the process of expressing this Hamiltonian in the basis { |↑↑⟩, |↑↓⟩, |↓↑⟩, |↓↓⟩ }, leading to the construction of a 4x4 matrix representation. Key equations involving Pauli spin matrices and ladder operators are utilized to derive the Hamiltonian's matrix elements, ultimately arriving at a correct matrix form that incorporates the necessary factors for each term.

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  • Understanding of quantum mechanics, specifically spin systems.
  • Familiarity with Hamiltonians and their matrix representations.
  • Knowledge of Pauli spin matrices and ladder operators.
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barefeet
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Homework Statement


Consider two spins, L and R, in a magnetic field along the z-axis, i.e. B = (0, 0, B). The magnetic moments of the two spins are coupled to each other so that the total Hamiltonian reads
H = g\mu_B\mathbf{B}\cdot(\mathbf{S}_L + \mathbf{S}_R) + J \mathbf{S}_L\cdot \mathbf{S}_R

Write this Hamiltonian in the basis \mathbf{\{} \mid \uparrow \uparrow \rangle, \mid \uparrow \downarrow \rangle, \mid \downarrow \uparrow \rangle, \mid \downarrow \downarrow \rangle \mathbf{\}}

Homework Equations



The equations for the Pauli spin matrices

The Attempt at a Solution


I know that generally you can write a matrix:
<br /> \newcommand{\unit}{1\!\!1}<br /> a\unit+ x \hat{\sigma_x} + y\hat{\sigma_y} + z\hat{\sigma_z} = <br /> \left( \begin{array}{ccc}<br /> a + z &amp; x-iy \\<br /> x+iy &amp; a-z \end{array} \right)<br />

But other than that I don't know how to start especially with two particles.
 
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Write the Hamiltonian in cartesian components.
 
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.
 
You mean like
<br /> H_x = J\mathbf{S}_{L,x} \mathbf{S}_{R,x} \\<br /> H_y = J\mathbf{S}_{L,y} \mathbf{S}_{R,y} \\<br /> H_z = B(\mathbf{S}_{L,z} + \mathbf{S}_{R,z}) + J\mathbf{S}_{L,z} \mathbf{S}_{R,z} <br />
?
 
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.
 
Write the Hamiltonian in cartesian spin components, indeed.
 
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:
<br /> 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}<br />
Or is this also wrong?
As far as what you suggested, do you mean to evaluate something like H \mid \uparrow \uparrow \rangle ?
Using H of what I worte above that would be:
<br /> H \mid \uparrow_L \uparrow_R \rangle =B \left( \begin{array}{ccc} 1 &amp; 0 \\ 0 &amp; -1 \end{array} \right)_L \left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_L <br /> + B \left( \begin{array}{ccc} 1 &amp; 0 \\ 0 &amp; -1 \end{array} \right)_R \left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_R \\<br /> + J \left( \begin{array}{ccc} 0 &amp; 1 \\ 1 &amp; 0 \end{array} \right)_L \left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_L \left( \begin{array}{ccc} 0 &amp; 1 \\ 1 &amp; 0 \end{array} \right)_R\left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_R<br /> + J \left( \begin{array}{ccc} 0 &amp; -i \\ i &amp; 0 \end{array} \right)_L \left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_L \left( \begin{array}{ccc} 0 &amp; -i \\ i &amp; 0 \end{array} \right)_R\left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_R<br /> + J \left( \begin{array}{ccc} 1 &amp; 0 \\ 0 &amp; -1 \end{array} \right)_L \left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_L \left( \begin{array}{ccc} 1 &amp; 0 \\ 0 &amp; -1 \end{array} \right)_R\left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_R<br />

This will be I think:
<br /> 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<br /> + J\left(\begin{array}{c} 0\\ 1 \\\end{array}\right)_L \left(\begin{array}{c} 0\\ 1 \\\end{array}\right)_R<br /> + J\left(\begin{array}{c} 0\\ i \\\end{array}\right)_L \left(\begin{array}{c} 0\\ i \\\end{array}\right)_R<br /> + J\left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_L \left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_R<br />

Or am I going completely off-track? And what should I do with it?
 
You are going a bit off track. You cannot have a state containing only one of the spins as your first terms do.
 
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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 B S_{L,z} actually B S_{L,z} 1\!\!1_R. But that wouldn't change much.
Then:
<br /> 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<br /> + J\left(\begin{array}{c} 0\\ 1 \\\end{array}\right)_L \left(\begin{array}{c} 0\\ 1 \\\end{array}\right)_R<br /> + J\left(\begin{array}{c} 0\\ i \\\end{array}\right)_L \left(\begin{array}{c} 0\\ i \\\end{array}\right)_R<br /> + J\left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_L \left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_R<br />
 
  • #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)|++> ?
 
  • #11
barefeet said:
You mean like
barefeet said:
<br /> H_x = J\mathbf{S}_{L,x} \mathbf{S}_{R,x} \\<br /> H_y = J\mathbf{S}_{L,y} \mathbf{S}_{R,y} \\<br /> H_z = B(\mathbf{S}_{L,z} + \mathbf{S}_{R,z}) + J\mathbf{S}_{L,z} \mathbf{S}_{R,z}<br />
?
If H_x etc are just names and not components of a vector as the notation suggests, this is ok.
Use ladder operators.
 
  • #12
Ah ok so I have to construct H as:
<br /> H = \left( \begin{array}{ccc}<br /> \langle \uparrow \uparrow \mid H \mid \uparrow \uparrow \rangle &amp; \langle \uparrow \uparrow \mid H \mid \uparrow \downarrow \rangle &amp; \langle \uparrow \uparrow \mid H \mid \downarrow \uparrow \rangle &amp; \langle \uparrow \uparrow \mid H \mid \downarrow \downarrow \rangle\\<br /> \langle \uparrow \downarrow \mid H \mid \uparrow \uparrow \rangle &amp; \langle \uparrow \downarrow \mid H \mid \uparrow \downarrow \rangle &amp; \langle \uparrow \downarrow \mid H \mid \downarrow \uparrow \rangle &amp; \langle \uparrow \downarrow \mid H \mid \downarrow \downarrow \rangle\\ <br /> \langle \downarrow \uparrow \mid H \mid \uparrow \uparrow \rangle &amp; \langle \downarrow \uparrow \mid H \mid \uparrow \downarrow \rangle &amp; \langle \downarrow \uparrow \mid H \mid \downarrow \uparrow \rangle &amp; \langle \downarrow \uparrow \mid H \mid \downarrow \downarrow \rangle\\ <br /> \langle \downarrow \downarrow \mid H \mid \uparrow \uparrow \rangle &amp; \langle \downarrow \downarrow \mid H \mid \uparrow \downarrow \rangle &amp; \langle \downarrow \downarrow \mid H \mid \downarrow \uparrow \rangle &amp; \langle \downarrow \downarrow \mid H \mid \downarrow \downarrow \rangle\\ <br /> \end{array} \right)<br />
Using the ladder operators S_x = \frac{1}{2}(S_+ + S_-) \quad S_y = \frac{1}{2i}(S_+ - S_-) and:
<br /> S_z \mid \uparrow \rangle = \frac{h}{2} \mid \uparrow \rangle \quad \quad S_z \mid \downarrow \rangle = \frac{-h}{2} \mid \downarrow \rangle \\<br /> S_+ \mid \uparrow \rangle = 0 \quad \quad S_+ \mid \downarrow \rangle = h \mid \uparrow \rangle \\<br /> S_- \mid \uparrow \rangle = h \mid \downarrow \rangle \quad \quad S_- \mid \downarrow \rangle = 0<br />

I get:
<br /> 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 \\<br /> 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 \\<br /> 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 \\<br /> 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<br />

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

This gets me in the right direction but is still wrong in the off diagonal terms.
 
  • #13
You are definitely getting closer.
Write S1xS2x+S1y+S2y in trems of S+ and S-.
What is (S1z+S2z)|--> ?
 
  • #14
The trace of all operators in your hamiltonian is zero,
so the trace of H should be zero. Check if it is.
 
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  • #15
Ah I see, I am missing a factor \frac{1}{2} in the S_x term and a \frac{1}{2i} in the S_y term
So it will be:
<br /> 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 \\<br /> 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 \\<br /> 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 \\<br /> 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<br />

Giving the matrix:
<br /> H =<br /> \left( \begin{array}{ccc}<br /> g\mu_BhB + \frac{Jh^2}{4} &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; \frac{-Jh^2}{4} &amp; \frac{Jh^2}{2} &amp; 0\\<br /> 0 &amp; \frac{Jh^2}{2} &amp; \frac{-Jh^2}{4} &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp; - g\mu_BhB + \frac{Jh^2}{4}<br /> \end{array} \right)<br />
 
Last edited:
  • #16
The matrix is close to the correct answer. The equations differ from it.
You should write S1xS2x+S1yS2y in terms of S1+ etc.
 
  • #17
Ok,
<br /> 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 \\<br /> =\frac{1}{2}(S_{+,L}S_{-,R} + S_{-,L}S_{+,R} ) <br />
But I still get the same answer. Whiech term exactly should I recheck?
 
  • #18
Are you sure about the final factor 1/2 ?
 
  • #19
I guess:
<br /> 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 \\<br /> = \frac{1}{4} ( S_+S_+ + S_+S_- + S_-S_+ + S_-S_- - S_+S_+ + S_+S_- + S_+S_- - S_-S_- )<br /> =\frac{1}{2}(S_{+,L}S_{-,R} + S_{-,L}S_{+,R})<br />
Or do you mean another 1/2 in the matrix?
 
  • #20
barefeet said:
Ah I see, I am missing a factor \frac{1}{2} in the S_x term and a \frac{1}{2i} in the S_y term
So it will be:
<br /> 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 \\<br /> 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 \\<br /> 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 \\<br /> 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<br />

Giving the matrix:
<br /> H =<br /> \left( \begin{array}{ccc}<br /> g\mu_BhB + \frac{Jh^2}{4} &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; \frac{-Jh^2}{4} &amp; \frac{Jh^2}{2} &amp; 0\\<br /> 0 &amp; \frac{Jh^2}{2} &amp; \frac{-Jh^2}{4} &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp; - g\mu_BhB + \frac{Jh^2}{4}<br /> \end{array} \right)<br />
This is correct.
 

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