Eigenvalues of Hamiltonian operator

  • #1
Juli
21
5
Homework Statement
Consider a system with two spin-1 particles, which is described by the Hamiltonian operator

## H = \lambda \vec{S}_1 \cdot \vec{S}_2 ##

with ##\lambda \in \mathbb{R} ##.

1. Express H in terms of the total spin ## \vec{S} = \vec{S}_1 + \vec{S}_2 ##.

2. What eigenvalues does H have and how are these degenerate?
Relevant Equations
##\vec{S}^2 = S(S+1)\hbar^2##
##\vec{S_1}^2 = S_1(S_1+1)\hbar^2##
##\vec{S_2}^2 = S_2(S_2+1)\hbar^2##
Hello, I try to solve this problem, and I think a) wasn't too hard, I have the following solution:

##H = \lambda (\frac{\vec{S^2-(\vec{S_1}^2+\vec{S_2}^2)}{2})##.
I struggle with 2. I find it very abstract. When I have H as a matrix I know how to calculate eigenvalues, but I don't know how to proceed with this general approach.

I tried to go somewhere with the eigenvalues of S, but I didn't get far...

Can someone help me solve this?
 
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  • #2
Expand the right-hand side of the operator ##~S^2=(\vec S_1+\vec S_2)\cdot(\vec S_1+\vec S_2)~## and solve for the operator ##\vec S_1\cdot \vec S_2##. Its eigenvalues are the eigenvalues of the operator on the right-hand side.
 
  • #3
I did the first task and got this ##H = \lambda (\frac{\vec{S^2-(\vec{S_1}^2+\vec{S_2}^2)}{2})##
But I don't know how to get the eigenvalues of the operator on the right-hand side. Are they the eigenvalues of the individual spins?
 

FAQ: Eigenvalues of Hamiltonian operator

What are eigenvalues of a Hamiltonian operator?

The eigenvalues of a Hamiltonian operator are the possible energy levels that a quantum system can have. They are obtained by solving the Schrödinger equation, where the Hamiltonian operator acts on a wavefunction to produce an eigenvalue, which corresponds to the energy associated with that wavefunction.

How are the eigenvalues of a Hamiltonian operator determined?

The eigenvalues of a Hamiltonian operator are determined by solving the time-independent Schrödinger equation, \( H \psi = E \psi \), where \( H \) is the Hamiltonian operator, \( \psi \) is the wavefunction, and \( E \) is the eigenvalue. This typically involves finding solutions to a differential equation or solving a matrix eigenvalue problem.

Why are the eigenvalues of a Hamiltonian operator important in quantum mechanics?

The eigenvalues of a Hamiltonian operator are important because they represent the allowed energy levels of a quantum system. These energy levels determine the possible states the system can occupy and play a crucial role in predicting the behavior and properties of the system, such as spectral lines in atomic physics and energy bands in solid-state physics.

Can a Hamiltonian operator have complex eigenvalues?

In general, the eigenvalues of a Hamiltonian operator are real numbers if the Hamiltonian is Hermitian (self-adjoint), which is a requirement for physical observables in quantum mechanics. Non-Hermitian Hamiltonians can have complex eigenvalues, but such cases typically represent open systems or non-conservative forces, rather than closed, isolated quantum systems.

What is the physical significance of the ground state eigenvalue of a Hamiltonian operator?

The ground state eigenvalue of a Hamiltonian operator represents the lowest possible energy that a quantum system can have. This state is of particular importance because it is the most stable state of the system, and many physical properties and behaviors of the system at low temperatures are determined by the ground state and its corresponding eigenvalue.

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