Eigenstates, Eigenvalues & Multicplity of Hamiltonian w/ Spin 1/2

[LCSphysicist]

Homework Statement

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Relevant Equations

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> Consider two particle with spin 1/2 interacting via the hamiltonian $H
= \frac{A}{\hbar^2}S_{1}.S_{2}$, Where A is a constant. What aare the eigenstates, eigenvalues and its multicplity?

$H = \frac{A}{\hbar^2}S_{1}.S_{2} = A\frac{(SS-S_{1}S_{1}-S_{2}S_{2})}{2\hbar^2 } = A\frac{(S^2-S_{1}^2-S_{2}^2)}{2\hbar^2 }$

Now, $S_{1}²$, for example, has the same eigenvectors as S1z, that is, $11,10,1-1,00$
And all these states are eigenvectors of S², so we have:

$$H|11\rangle= A\frac{(S^2-S_{1}^2-S_{2}^2)}{2\hbar^2 }|11\rangle = A\frac{(2 \hbar^2- 3 \hbar^2/4 -3 \hbar^2/4 )}{2\hbar^2 }|11\rangle = \frac{A}{4}|11\rangle$$

$$H|10\rangle= A\frac{(S^2-S_{1}^2-S_{2}^2)}{2\hbar^2 }|10\rangle = A\frac{(2 \hbar^2- 3 \hbar^2/4 -3 \hbar^2/4 )}{2\hbar^2 }|10\rangle = \frac{A}{4}|10\rangle$$

$$H|1-1\rangle= A\frac{(S^2-S_{1}^2-S_{2}^2)}{2\hbar^2 }|1-1\rangle = A\frac{(2 \hbar^2- 3 \hbar^2/4 -3 \hbar^2/4 )}{2\hbar^2 }|1-1\rangle = \frac{A}{4}|1-1\rangle$$

$$H|00\rangle= A\frac{(S^2-S_{1}^2-S_{2}^2)}{2\hbar^2 }|00\rangle = A\frac{(- 3 \hbar^2/4 -3 \hbar^2/4 )}{2\hbar^2 }|00\rangle = \frac{-3A}{4}|00\rangle$$

I want to know i this is right. Is it? To be honest, i think it is, but what worries me is that i am not sure i these are all the eigenvalues/eigenvectors. I believe H would be something like a "4x4" matrix, so i think it is. But want to hear your answer too.

 

[TSny]

Your work looks correct to me.

For inline Latex, use double hashtag rather than single dollar sign.

 

[EigenState137]

Just making it easier to read> Consider two particle with spin 1/2 interacting via the hamiltonian $$H
= \frac{A}{\hbar^2}S_{1}.S_{2}$$, Where A is a constant. What aare the eigenstates, eigenvalues and its multicplity?

$$H = \frac{A}{\hbar^2}S_{1}.S_{2} = A\frac{(SS-S_{1}S_{1}-S_{2}S_{2})}{2\hbar^2 } = A\frac{(S^2-S_{1}^2-S_{2}^2)}{2\hbar^2 }$$

Now, $$S_{1}²$$, for example, has the same eigenvectors as S1z, that is, $$11,10,1-1,00$$
And all these states are eigenvectors of S², so we have:

 

[kuruman]

I want to know i this is right. Is it? To be honest, i think it is, but what worries me is that i am not sure i these are all the eigenvalues/eigenvectors. I believe H would be something like a "4x4" matrix, so i think it is. But want to hear your answer too.

The Hamiltonian is indeed a 4×4 matrix. You have found 4 eigenvalues, so what's your concern? A quick test on your eigenvalues to check whether if the Hamiltonian matrix is traceless (the sum of diagonal elements is zero), the sum of the eigenvalues must be zero. That is true in this case which does not guarantee the correctness of your eigenvalues but at least they pass this test so there is no error in the arithmetic.