- #1

LCSphysicist

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- 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.

= \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.