# Hamiltonian of a simple graph

I have a spin system:

As shown in the picture, there are two spins S1 and S2, and a pair of interactions between them. One is a ferromagnetic interaction and the other is anti ferromagnetic interaction. I am trying to calculate the Hamiltonian of this system.

The Hamiltonian of the system is:

$$H = -J_F S1_z S2_z +J_{AF} S1_z S2_z$$

$S1_z$ is the spin matrix for Z direction for spin 1 and $S2_z$ is the spin matrix for Z direction for spin 2. If we allow two random values for $J_F$ and $J_{AF}$, -0.5 and 0.5 respectively the Hamiltonian of the system is as follows.

$$H = 0.5 S1_z S2_z + 0.5 S1_z S2_z$$ $$= S1_z S2_z$$ $$= \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix} \times \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}$$ $$= \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}$$

Am I able to calculate the Hamiltonian correctly?

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The Hamiltonian of this system lives in a 4-dimensional Hilbert space since you have two spin $1/2$. Therefore, you should represent the spin matrix in this four dimensional space like this:

$S_1^z=\begin{pmatrix} -0.5 & 0 &0 &0 \\ 0&-0.5 &0 &0 \\ 0 &0 &0.5 &0 \\ 0 &0 &0 &0.5 \end{pmatrix}$ , $S_2^z=\begin{pmatrix} -0.5 & 0 &0 &0 \\ 0&0.5 &0 &0 \\ 0 &0 &-0.5 &0 \\ 0 &0 &0 &0.5 \end{pmatrix}$

The order of the four states along the rows and columns is $|DD\rangle,|DU\rangle, |UD\rangle, |UU\rangle$ where $U$ stands for spin up and $D$ stands for spin down.

In this case $S_1^z.S_2^z=\begin{pmatrix} 0.25 & 0 &0 &0 \\ 0&-0.25 &0 &0 \\ 0 &0 &-0.25 &0 \\ 0 &0 &0 &0.25 \end{pmatrix}$

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I am little bit confused here. Let me define a single bit $\frac{1}{2}$ spin as $|Z\rangle = \begin{pmatrix} 1 \\ 0 \\ \end{pmatrix}$ or logic zero and a single bit $-\frac{1}{2}$ spin as $|O\rangle = \begin{pmatrix} 0 \\ 1 \\ \end{pmatrix}$ or logic one. (contd...) – Omar Shehab Mar 12 '13 at 2:13
So, for a two spin spin system, the state vectors are $|ZZ\rangle = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ \end{pmatrix}$, $|ZO\rangle = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \\ \end{pmatrix}$, $|OZ\rangle = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \\ \end{pmatrix}$ and $|OO\rangle = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \\ \end{pmatrix}$. To know the spins I operate your operators on the state vectors. – Omar Shehab Mar 12 '13 at 2:14
When I operate $S_1^z$ on $|ZZ\rangle$ I get the following result. $$S_1^z . |ZZ\rangle = \begin{pmatrix} -0.5 & 0 & 0 & 0 \\ 0 & -0.5 & 0 & 0 \\ 0 & 0 & 0.5 & 0 \\ 0 & 0 & 0 & 0.5 \\ \end{pmatrix} . \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ \end{pmatrix} = \begin{pmatrix} -0.5 \\ 0 \\ 0 \\ 0 \\ \end{pmatrix}$$ I don't understand what to make of it. But when I calculate the expected value, i.e., $\langle ZZ | S_1^z | ZZ \rangle$ (contd...) – Omar Shehab Mar 12 '13 at 2:15
I get $\begin{pmatrix} -0.5 \\ \end{pmatrix}$. I am not sure if the expectation value I am getting is the $-\frac{1}{1}$-spin. If so, I think I almost understand you. But in page 7 of web.uconn.edu/~ch351vc/pdfs/spin1.pdf, the operator is different. It is $\Sigma_z = \begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -2 \\ \end{pmatrix}.$ Any clue? – Omar Shehab Mar 12 '13 at 2:15
$\Sigma_z=S_1^z+S_2^z$ The first matrix you wrote for $\Sigma_z$ is consistent with the definitions for$S_1^z$ and $S_2^z$. What is the problem in $S_1^z |ZZ\rangle=-0.5|ZZ\rangle$ ? Isn't this expected? – Tarek Mar 13 '13 at 18:13