# Ground and first excited state of non interacting spin system Hamiltonian

For a non interacting spin system containing two $\frac{1}{2}$ spin particles I am trying to determine its Hamiltonian. If the energy of a up spin is $+\mu {\bf B}$ and a down spin is $-\mu {\bf B}$, the possible energies are $+2\mu {\bf B}$ for $|UU\rangle$, $-2\mu {\bf B}$ for $|DD\rangle$ and 0 for both $|UD\rangle$ and $|DU\rangle$. I am confused what will be the Hamiltonian which will allow me to determine the energy of this system. If $|DD\rangle = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}$, $|DU\rangle = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}$, $|UD\rangle = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}$ and $|UU\rangle = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}$, I am not sure how to determine the matrix representation of the Hamiltonian. Although I suspect that the Hamiltonian matrix will remain same if I use an alternative set of bases i.e. $\begin{pmatrix} 0 \\ {1 / \sqrt{2} }\\ {1 / \sqrt{2} } \\ 0 \end{pmatrix}$, $\begin{pmatrix} 0 \\ {1 / \sqrt{2} }\\ {-1 / \sqrt{2} } \\ 0 \end{pmatrix}$, $\begin{pmatrix} {1 / \sqrt{2} } \\ 0\\ 0 \\ {1 / \sqrt{2} } \end{pmatrix}$ and $\begin{pmatrix} {1 / \sqrt{2} } \\ 0\\ 0 \\ {-1 / \sqrt{2} } \end{pmatrix}$.

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Are you claiming that $|UU\rangle,|UD\rangle,|DU\rangle,|DD\rangle$ are the energy eigenstates of the Hamiltonian? If so that's trivial, since the Hamiltonian is diagonal in the basis of its eigenstates. (Generally it won't be if you use another basis set.) –  Michael Brown Mar 15 '13 at 7:05
@MichaelBrown, I am not sure what is the eigenstates of the Hamiltonian of this system. If it were a spin system of interacting Hamiltonian I understand that the Hamiltonian would be $S^{(1)}_z\cdot S^{(1)}_z=\frac{1}{4}\begin{pmatrix} 1 & 0 &0 &0 \\ 0&-1 &0 &0 \\ 0 &0 &-1 &0 \\ 0 &0 &0 &1 \end{pmatrix}$. With this Hamiltonian, the set of eigenstates are $\left\{ \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \right\}$ or –  Omar Shehab Mar 15 '13 at 16:24
$\left\{ \begin{pmatrix} 0 \\ {1 / \sqrt{2} }\\ {1 / \sqrt{2} } \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ {1 / \sqrt{2} } \\ -{1 / \sqrt{2} } \\ 0 \end{pmatrix}, \begin{pmatrix} {1 / \sqrt{2} } \\ 0\\ 0 \\ {1 / \sqrt{2} } \end{pmatrix}, \begin{pmatrix} {1 / \sqrt{2} } \\ 0 \\ 0 \\ -{1 / \sqrt{2} } \end{pmatrix} \right\}$. I think for a non interacting spin system the Hamiltonian and the eigenstates should be different as the energy of the system is different. –  Omar Shehab Mar 15 '13 at 16:27
@MichaelBrown, can we consider the formula of the Hamiltonian as $H = -h \sum_{i=1}^2 S^{(i)}_z$? In that case, it will be $$H = -h S^{(1)}_z -h S^{(2)}_z$$ $$H = -h \frac{1}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} -h \frac{1}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$ $$H = -h \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$. Do you think that this Hamiltonian can determine the energy of this two non interacting spin system? –  Omar Shehab Mar 16 '13 at 5:13
@MichaelBrown, if $H = -h \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ is the appropriate Hamiltonian, the eigenvalue is -h for the eigenvector \begin{pmatrix} 1 \\ 0 \end{pmatrix} and the eigenvalue is h for the eigenvector \begin{pmatrix} 0 \\ 1 \end{pmatrix}. So, in that case, \begin{pmatrix} 1 \\ 0 \end{pmatrix} should be the ground energy state and \begin{pmatrix} 0 \\ 1 \end{pmatrix} should be the first excited state for the Hamiltonian, $-h \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ of this two non interacting spin system. I expected a 4 X 4 matrix and 4 X 1 vectors. –  Omar Shehab Mar 16 '13 at 5:22
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