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Consider the Hamiltonian $H = -J_\text{F}S^{(1)}_zS^{(2)}_z + J_{AF}S^{(1)}_zS^{(2)}_z$, describing the graph

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Here, F means ferromagnetic and AF means antiferromagnetic interactions. I am having problem with the value of $S^{(1)}_zS^{(2)}_z$.Someone suggested to me that $$S^{(1)}_z=\frac{1}{2}\begin{pmatrix} -1 & 0 &0 &0 \\ 0&-1 &0 &0 \\ 0 &0 &1 &0 \\ 0 &0 &0 &1 \end{pmatrix},\quad S^{(2)}_z=\frac{1}{2}\begin{pmatrix} -1 & 0 &0 &0 \\ 0&1 &0 &0 \\ 0 &0 &-1 &0 \\ 0 &0 &0 &1 \end{pmatrix},$$ and therefore $$S^{(1)}_z\cdot S^{(1)}_z=\frac{1}{4}\begin{pmatrix} 1 & 0 &0 &0 \\ 0&-1 &0 &0 \\ 0 &0 &-1 &0 \\ 0 &0 &0 &1 \end{pmatrix}.$$

On the other hand, from page 7 of these notes on Pauli spin matrices, I know that for two spin systems $$\Sigma_z = \begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -2 \\ \end{pmatrix}.$$ I asked the person but never got a reply. I don't see what is the difference between $S^{(1)}_z\cdot S^{(1)}_z$ and $\Sigma_z$; when I should use what?

Emilio Pisanty
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Omar Shehab
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2 Answers2

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The operator $\Sigma_z$ is the $z$ component of the total spin, and therefore is the sum of the quantities $S^{(1)}_z$ and $S^{(2)}_z$, whereas the operator $S^{(1)}_z\cdot S^{(2)}_z$ is their product. You can easily check that for the matrices in your question the identity $$\Sigma_z=-2(S^{(1)}_z+ S^{(2)}_z)$$ holds; the factor of $-2$ is due to differing conventions in sign and normalization between the ones in your post and the ones in the link.

For your particular system you want the product, as that will have the states $|\uparrow\rangle|\uparrow\rangle$, $|\downarrow\rangle|\downarrow\rangle$, $|\uparrow\rangle|\downarrow\rangle$ and $|\downarrow\rangle|\uparrow\rangle$ as eigenstates, with eigenvalues $+1,+1,-1$ and $-1$ resp., which is how you want to model (anti)ferromagnetic interactions in the first place.

The sum is useful in situations where you want the total number of excitations. This is the case, for example, in the Jaynes-Cummings model, where you use spin matrices to model two-level atoms even though the states are not necessarily spin states (and more probably electronic eigenstates). Thus taking $|\uparrow\rangle=|\text{excited}\rangle$ and $|\downarrow\rangle=|\text{ground}\rangle$, the operator $S^{(1)}_z+ S^{(2)}_z$ counts the total number of excited atoms, which is of course the atomic part of the Jaynes-Cummings hamiltonian.

You should also notice that your hamiltonian simplifies to $H=(J_\text{AF}-J_\text{F})S^{(1)}_z\cdot S^{(2)}_z$: only a ferromagnetic or an antiferromagnetic interaction is possible between any two spins. They either like being parallel or not. The two edges in your graph combine to a single one, of either F or AF character. The graph formalism is useful when you have multiple spins with varying interactions.

Emilio Pisanty
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  • Yes. I have chosen random values $J_F = -0.5$ and $J_{AF} = 0.5$. So, $H = \frac{1}{4}\begin{pmatrix} 1 & 0 &0 &0 \ 0&-1 &0 &0 \ 0 &0 &-1 &0 \ 0 &0 &0 &1 \end{pmatrix}$. – Omar Shehab Mar 13 '13 at 19:55
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    The point of the signs in your hamiltonian is that you keep both $J$s positive, or you change the character of the interaction. $J_\text{F}$ and $J_\text{AF}$ are meaningless here; it's only $J=J_\text{F}-J_\text{AF}$ and its sign that matters. – Emilio Pisanty Mar 13 '13 at 20:44
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Essentialy $S^{(1)}_{z}.S^{(2)}_{z}=\sigma_{z}^{1}\otimes\sigma_{z}^{2}$ and you can easily check that $\sigma_{z}^{1}\otimes\sigma_{z}^{2}=\sigma_{z}^{1}\otimes I_{2\times 2}. I_{2\times 2}\otimes\sigma_{z}^{2}$. And $\Sigma_{z}=\sigma_{z}\otimes I_{2\times2}+I_{2\times2}\otimes\sigma_{z}$.

Abrin
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