QM: Two coupled spins in a magnetic field

[barefeet]

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:
$$\newcommand{\unit}{1\!\!1} a\unit+ x \hat{\sigma_x} + y\hat{\sigma_y} + z\hat{\sigma_z} = \left( \begin{array}{ccc} a + z & x-iy \\ x+iy & a-z \end{array} \right)$$

But other than that I don't know how to start especially with two particles.

 

[my2cts]

Write the Hamiltonian in cartesian components.

 

[Orodruin]

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.

 

[barefeet]

You mean like
$$H_x = J\mathbf{S}_{L,x} \mathbf{S}_{R,x} \\ H_y = J\mathbf{S}_{L,y} \mathbf{S}_{R,y} \\ H_z = B(\mathbf{S}_{L,z} + \mathbf{S}_{R,z}) + J\mathbf{S}_{L,z} \mathbf{S}_{R,z}$$
?

 

[Orodruin]

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.

 

[my2cts]

Write the Hamiltonian in cartesian spin components, indeed.

 

[barefeet]

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:
$$H = J\mathbf{S}_{L,x} \mathbf{S}_{R,x} + J\mathbf{S}_{L,y} \mathbf{S}_{R,y} + B(\mathbf{S}_{L,z} + \mathbf{S}_{R,z}) + J\mathbf{S}_{L,z} \mathbf{S}_{R,z}$$
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:
$$H \mid \uparrow_L \uparrow_R \rangle =B \left( \begin{array}{ccc} 1 & 0 \\ 0 & -1 \end{array} \right)_L \left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_L + B \left( \begin{array}{ccc} 1 & 0 \\ 0 & -1 \end{array} \right)_R \left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_R \\ + J \left( \begin{array}{ccc} 0 & 1 \\ 1 & 0 \end{array} \right)_L \left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_L \left( \begin{array}{ccc} 0 & 1 \\ 1 & 0 \end{array} \right)_R\left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_R + J \left( \begin{array}{ccc} 0 & -i \\ i & 0 \end{array} \right)_L \left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_L \left( \begin{array}{ccc} 0 & -i \\ i & 0 \end{array} \right)_R\left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_R + J \left( \begin{array}{ccc} 1 & 0 \\ 0 & -1 \end{array} \right)_L \left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_L \left( \begin{array}{ccc} 1 & 0 \\ 0 & -1 \end{array} \right)_R\left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_R$$

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

Or am I going completely off-track? And what should I do with it?

 

[Orodruin]

You are going a bit off track. You cannot have a state containing only one of the spins as your first terms do.

 

[barefeet]

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:
$$H \mid \uparrow_L \uparrow_R \rangle = \left(\begin{array}{c} B\\ 0 \\\end{array}\right)_L\left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_R + \left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_L\left(\begin{array}{c} B\\ 0 \\\end{array}\right)_R + J\left(\begin{array}{c} 0\\ 1 \\\end{array}\right)_L \left(\begin{array}{c} 0\\ 1 \\\end{array}\right)_R + J\left(\begin{array}{c} 0\\ i \\\end{array}\right)_L \left(\begin{array}{c} 0\\ i \\\end{array}\right)_R + J\left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_L \left(\begin{array}{c} 1\\ 0 \\\end{array}\right)_R$$

 

[my2cts]

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)|++> ?

 

[my2cts]

You mean like

$$H_x = J\mathbf{S}_{L,x} \mathbf{S}_{R,x} \\ H_y = J\mathbf{S}_{L,y} \mathbf{S}_{R,y} \\ H_z = B(\mathbf{S}_{L,z} + \mathbf{S}_{R,z}) + J\mathbf{S}_{L,z} \mathbf{S}_{R,z}$$
?

If H_x etc are just names and not components of a vector as the notation suggests, this is ok.
Use ladder operators.

 

[barefeet]

Ah ok so I have to construct H as:
$$H = \left( \begin{array}{ccc} \langle \uparrow \uparrow \mid H \mid \uparrow \uparrow \rangle & \langle \uparrow \uparrow \mid H \mid \uparrow \downarrow \rangle & \langle \uparrow \uparrow \mid H \mid \downarrow \uparrow \rangle & \langle \uparrow \uparrow \mid H \mid \downarrow \downarrow \rangle\\ \langle \uparrow \downarrow \mid H \mid \uparrow \uparrow \rangle & \langle \uparrow \downarrow \mid H \mid \uparrow \downarrow \rangle & \langle \uparrow \downarrow \mid H \mid \downarrow \uparrow \rangle & \langle \uparrow \downarrow \mid H \mid \downarrow \downarrow \rangle\\ \langle \downarrow \uparrow \mid H \mid \uparrow \uparrow \rangle & \langle \downarrow \uparrow \mid H \mid \uparrow \downarrow \rangle & \langle \downarrow \uparrow \mid H \mid \downarrow \uparrow \rangle & \langle \downarrow \uparrow \mid H \mid \downarrow \downarrow \rangle\\ \langle \downarrow \downarrow \mid H \mid \uparrow \uparrow \rangle & \langle \downarrow \downarrow \mid H \mid \uparrow \downarrow \rangle & \langle \downarrow \downarrow \mid H \mid \downarrow \uparrow \rangle & \langle \downarrow \downarrow \mid H \mid \downarrow \downarrow \rangle\\ \end{array} \right)$$
Using the ladder operators $S_x = \frac{1}{2}(S_+ + S_-) \quad S_y = \frac{1}{2i}(S_+ - S_-)$ and:
$$S_z \mid \uparrow \rangle = \frac{h}{2} \mid \uparrow \rangle \quad \quad S_z \mid \downarrow \rangle = \frac{-h}{2} \mid \downarrow \rangle \\ S_+ \mid \uparrow \rangle = 0 \quad \quad S_+ \mid \downarrow \rangle = h \mid \uparrow \rangle \\ S_- \mid \uparrow \rangle = h \mid \downarrow \rangle \quad \quad S_- \mid \downarrow \rangle = 0$$

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

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

This gets me in the right direction but is still wrong in the off diagonal terms.

 

[my2cts]

You are definitely getting closer.
Write S1xS2x+S1y+S2y in trems of S+ and S-.
What is (S1z+S2z)|--> ?

 

[my2cts]

The trace of all operators in your hamiltonian is zero,
so the trace of H should be zero. Check if it is.

 

[barefeet]

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:
$$H \mid \uparrow \uparrow \rangle = g\mu_BhBh \mid \uparrow \uparrow \rangle + \frac{Jh^2}{4} \mid \uparrow \uparrow \rangle + \frac{Jh^2}{4} \mid \downarrow \downarrow \rangle - \frac{Jh^2}{4} \mid \downarrow \downarrow \rangle \\ H \mid \uparrow \downarrow \rangle = 0 + \frac{-Jh^2}{4} \mid \uparrow \downarrow \rangle + \frac{Jh^2}{4} \mid \downarrow \uparrow \rangle + \frac{Jh^2}{4} \mid \downarrow \uparrow \rangle \\ H \mid \downarrow \uparrow \rangle = 0 + \frac{-Jh^2}{4} \mid \downarrow \uparrow \rangle + \frac{Jh^2}{4} \mid \uparrow \downarrow \rangle + \frac{Jh^2}{4} \mid \uparrow \downarrow \rangle \\ H \mid \downarrow \downarrow \rangle = - g\mu_BhBh \mid \downarrow \downarrow \rangle + \frac{Jh^2}{4} \mid \downarrow \downarrow \rangle +\frac{Jh^2}{4} \mid \uparrow \uparrow \rangle - \frac{Jh^2}{4} \mid \uparrow \uparrow \rangle$$

Giving the matrix:
$$H = \left( \begin{array}{ccc} g\mu_BhB + \frac{Jh^2}{4} & 0 & 0 & 0 \\ 0 & \frac{-Jh^2}{4} & \frac{Jh^2}{2} & 0\\ 0 & \frac{Jh^2}{2} & \frac{-Jh^2}{4} & 0\\ 0 & 0 & 0 & - g\mu_BhB + \frac{Jh^2}{4} \end{array} \right)$$

 

[my2cts]

The matrix is close to the correct answer. The equations differ from it.
You should write S1xS2x+S1yS2y in terms of S1+ etc.

 

[barefeet]

Ok,
$$S_{x,L}S_{x,R} + S_{y,L}S_{y,R} = \frac{1}{2}(S_+ + S_-)_L\frac{1}{2}(S_+ + S_-)_R + \frac{1}{2i}(S_+ - S_-)_L\frac{1}{2i}(S_+ - S_-)_R \\ =\frac{1}{2}(S_{+,L}S_{-,R} + S_{-,L}S_{+,R} )$$
But I still get the same answer. Whiech term exactly should I recheck?

 

[my2cts]

Are you sure about the final factor 1/2 ?

 

[barefeet]

I guess:
$$S_{x,L}S_{x,R} + S_{y,L}S_{y,R} = \frac{1}{2}(S_+ + S_-)_L\frac{1}{2}(S_+ + S_-)_R + \frac{1}{2i}(S_+ - S_-)_L\frac{1}{2i}(S_+ - S_-)_R \\ = \frac{1}{4} ( S_+S_+ + S_+S_- + S_-S_+ + S_-S_- - S_+S_+ + S_+S_- + S_+S_- - S_-S_- ) =\frac{1}{2}(S_{+,L}S_{-,R} + S_{-,L}S_{+,R})$$
Or do you mean another 1/2 in the matrix?

 

[my2cts]

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:
$$H \mid \uparrow \uparrow \rangle = g\mu_BhBh \mid \uparrow \uparrow \rangle + \frac{Jh^2}{4} \mid \uparrow \uparrow \rangle + \frac{Jh^2}{4} \mid \downarrow \downarrow \rangle - \frac{Jh^2}{4} \mid \downarrow \downarrow \rangle \\ H \mid \uparrow \downarrow \rangle = 0 + \frac{-Jh^2}{4} \mid \uparrow \downarrow \rangle + \frac{Jh^2}{4} \mid \downarrow \uparrow \rangle + \frac{Jh^2}{4} \mid \downarrow \uparrow \rangle \\ H \mid \downarrow \uparrow \rangle = 0 + \frac{-Jh^2}{4} \mid \downarrow \uparrow \rangle + \frac{Jh^2}{4} \mid \uparrow \downarrow \rangle + \frac{Jh^2}{4} \mid \uparrow \downarrow \rangle \\ H \mid \downarrow \downarrow \rangle = - g\mu_BhBh \mid \downarrow \downarrow \rangle + \frac{Jh^2}{4} \mid \downarrow \downarrow \rangle +\frac{Jh^2}{4} \mid \uparrow \uparrow \rangle - \frac{Jh^2}{4} \mid \uparrow \uparrow \rangle$$

Giving the matrix:
$$H = \left( \begin{array}{ccc} g\mu_BhB + \frac{Jh^2}{4} & 0 & 0 & 0 \\ 0 & \frac{-Jh^2}{4} & \frac{Jh^2}{2} & 0\\ 0 & \frac{Jh^2}{2} & \frac{-Jh^2}{4} & 0\\ 0 & 0 & 0 & - g\mu_BhB + \frac{Jh^2}{4} \end{array} \right)$$

This is correct.