What are the eigenstates of the anti-ferromagnetic dimer Hamiltonian?

In summary, the conversation discusses the calculation of eigenvalues and eigenvectors for the Hamiltonian of a simple anti-ferromagnetic dimer. The hint given is to rewrite the spin operator term in terms of the total spin squared and individual spin squared. This leads to the calculation of four eigenvalues, which correspond to the singlet and triplet state energies with an added magnetic field term.
  • #1
opaka
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Homework Statement


The hamiltonian of a simple anti-ferromagnetic dimer is given by

H=JS(1)[itex]\bullet[/itex]S(2)-μB(Sz(1)+Sz(2))

find the eigenvalues and eigenvectors of H.

Homework Equations





The Attempt at a Solution


The professor gave the hint that the eigenstates are of S2=(S(1)+S(2))2, S(1)2, S(2)2, and Sz. So I know I should have four eigenvalues. but I still have no Idea how to get this into a form that I recognize as being able to get eigenvalues from. (a matrix, a DiffEQ, etc.)

Please help
 
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  • #2
The hint is to help you to deal with the ##\vec{S}_1\cdot\vec{S}_2## term. Rewrite that term in terms of ##\vec{S}^2##, ##\vec{S}_1^2##, and ##\vec{S}_2^2##.
 
  • #3
When I do that, and apply the spin operators, S2 ket (S,Sz)=s(s+1) ket (s,sz) and Sz ket (S,Sz) = szket (s,sz)(sorry, couldn't find the ket symbol in latex)
I get
H = J/2 (s(s+1) - s1(s1+1)-s2(s2+1))-μB(s1z+s2z)

Is this correct?
 
  • #4
Yes. Now you can calculate what H does to simultaneous eigenstates of ##\vec{S}^2##, ##S_z##, ##\vec{S}_1^2##, and ##\vec{S}_2^2##. Recall that these are exactly the states that you got from adding angular momenta.
 
  • #5
I get four answers : J/4 + μB, J/4 -μB, J/4 and - 3J/4. Is this right? These look like the singlet and triplet state energies, but with an added B term.
 
  • #6
Yeah, that looks right.
 
  • #7
Thanks so much Vela! you've been a wonderful help.:smile:
 

FAQ: What are the eigenstates of the anti-ferromagnetic dimer Hamiltonian?

What are eigenstates of Hamiltonian?

Eigenstates of Hamiltonian refer to the set of states in a quantum system that are the solutions to the time-independent Schrödinger equation, Hψ = Eψ, where H is the Hamiltonian operator, ψ is the wavefunction, and E is the corresponding energy. These states are stable and do not change over time, making them important for understanding the behavior of a quantum system.

How are eigenstates of Hamiltonian related to energy levels?

Each eigenstate of Hamiltonian corresponds to a unique energy level in a quantum system. The energy of a state is given by the eigenvalue E in the Schrödinger equation, and the state itself is represented by the eigenvector ψ. The higher the energy level, the greater the energy of the state.

Can a system have multiple eigenstates of Hamiltonian?

Yes, a system can have multiple eigenstates of Hamiltonian. In fact, most physical systems have infinitely many eigenstates, each with a unique energy level. The number of eigenstates depends on the size and complexity of the system.

How are eigenstates of Hamiltonian used in quantum mechanics?

Eigenstates of Hamiltonian are used to describe the behavior and properties of quantum systems. They provide information about the energy levels and stability of a system, and are essential in calculating the probabilities of different outcomes in quantum experiments. They also play a crucial role in understanding the principles of quantum mechanics, such as superposition and entanglement.

Can eigenstates of Hamiltonian change over time?

No, eigenstates of Hamiltonian do not change over time. They are stationary states that remain constant and do not evolve. However, the coefficients of the eigenstates in a wavefunction can change over time, representing the probability amplitudes of different energy levels. This allows for the observed changes and fluctuations in a quantum system.

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