# Ground states of the Hamiltonian of a two spin system

For the spin system shown in this graph (http://i.stack.imgur.com/3lg1R.png), the Hamiltonian is $$S^{(1)}_z\cdot S^{(1)}_z=\frac{1}{4}\begin{pmatrix} 1 & 0 &0 &0 \\ 0&-1 &0 &0 \\ 0 &0 &-1 &0 \\ 0 &0 &0 &1 \end{pmatrix}.$$ The eigenvalues of this Hamiltonian are $-\frac{1}{4}$ when the eigenstate is $$\begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}$$ or $$\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}$$ and $\frac{1}{4}$ when the eigenstate is $$\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$ or $$\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}.$$ Moreover, the expectation value of the Hamiltonian for $$\begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}$$ and $$\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}$$ is $-\frac{1}{4}$ and for $$\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$ and $$\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}$$ is $\frac{1}{4}$.

Based on these information can we say that $$\begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}$$ and $$\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}$$ are the ground states of the Hamiltonian while $$\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$ and $$\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}$$ are the first excited states?

-
Note that whenever a state is an eigenstate of an observable, the expectation value of that observable in that state is always the eigenvalue. You don't need to specify that seperately. – Michael Brown Mar 14 '13 at 23:47

## 1 Answer

Ok, but be aware that since the eigenvalues are degenerate, you have freedom of choice in defining a basis. You might choose

$\begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}$

but someone else might like

$\begin{pmatrix} 0 \\ {1 / \sqrt{2} }\\ {1 / \sqrt{2} } \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ {1 / \sqrt{2} } \\ -{1 / \sqrt{2} } \\ 0 \end{pmatrix}$

-
I calculated expectation values of the basis states you mentioned for the same Hamiltonian, $\frac{1}{4}\begin{pmatrix} 1 & 0 &0 &0 \\ 0&-1 &0 &0 \\ 0 &0 &-1 &0 \\ 0 &0 &0 &1 \end{pmatrix}$. Can we say that $\begin{pmatrix} 0 \\ {1 / \sqrt{2} }\\ {1 / \sqrt{2} } \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ {1 / \sqrt{2} } \\ -{1 / \sqrt{2} } \\ 0 \end{pmatrix}$ are ground states as the expectation value is -0.25? – Omar Shehab Mar 14 '13 at 23:53
Moreover, $\begin{pmatrix} {1 / \sqrt{2} } \\ 0\\ 0 \\ {1 / \sqrt{2} } \end{pmatrix}$ and $\begin{pmatrix} {1 / \sqrt{2} } \\ 0 \\ 0 \\ -{1 / \sqrt{2} } \end{pmatrix}$ are first excited states as the expectation value is 0.25? – Omar Shehab Mar 14 '13 at 23:55