QM: Two coupled spins in a magnetic field

In summary, we start by considering two spins, L and R, in a magnetic field along the z-axis. The total Hamiltonian is the sum of the coupling between the magnetic moments of the two spins and the external magnetic field, and the interaction between the spins themselves. We can write this Hamiltonian in the basis of four states, namely the spin-up/spin-up, spin-up/spin-down, spin-down/spin-up, and spin-down/spin-down states. Using the ladder operators and the equations for the Pauli spin matrices, we can express the Hamiltonian in terms of the spin components and use it to act on the four basis states to obtain their respective energies.
  • #1
barefeet
59
2

Homework Statement


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]

Homework Equations



The equations for the Pauli spin matrices

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 don't know how to start especially with two particles.
 
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  • #2
Write the Hamiltonian in cartesian components.
 
  • #3
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.
 
  • #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]
?
 
  • #5
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.
 
  • #6
Write the Hamiltonian in cartesian spin components, indeed.
 
  • #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?
 
  • #8
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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  • #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 wouldn't 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]
 
  • #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:
[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]
?
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:
[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.
 
  • #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 [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:
  • #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,
[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?
 
  • #18
Are you sure about the final factor 1/2 ?
 
  • #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?
 
  • #20
barefeet said:
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]
This is correct.
 

Related to QM: Two coupled spins in a magnetic field

1. What is a coupled spin?

A coupled spin refers to the interaction between two particles or systems in which their individual spins become entangled or correlated. This means that the spin of one particle or system is dependent on the spin of the other.

2. How does a magnetic field affect coupled spins?

A magnetic field can cause the spins of two particles or systems to become coupled, meaning they will interact and influence each other's spin orientations. This is known as the Zeeman effect and is a result of the interaction between the magnetic field and the particles' intrinsic magnetic moments.

3. What is the significance of studying coupled spins in a magnetic field?

Understanding the behavior of coupled spins in a magnetic field is crucial in various areas of physics, including quantum computing and magnetic resonance imaging (MRI). It also has practical applications in materials science and chemistry, as the coupling between spins can affect the physical and chemical properties of materials.

4. How is the Hamiltonian used to describe the behavior of coupled spins in a magnetic field?

The Hamiltonian is a mathematical operator that describes the total energy of a system. In the case of coupled spins in a magnetic field, the Hamiltonian includes terms for the individual spin energies and the interaction energy between the spins and the magnetic field. It is used to calculate the expected behavior of the system and the probabilities of different spin states.

5. What is the Bloch sphere and how is it related to coupled spins in a magnetic field?

The Bloch sphere is a geometric representation of the possible spin states of a coupled spin system. It is used to visualize and understand the complex behavior of coupled spins in a magnetic field. The north and south poles of the sphere represent the two possible spin states of a single particle, while the equator represents the entangled states of the two coupled spins.

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