# Find Eigenstates of a Hamiltonian that lets two spin 1/2 interact but also acts on one of them

I have the following Hamiltonian describing two spin 1/2 systems, represented by the pauli matrices $\sigma_1$ and $\sigma_2$:

$H = D \sigma_{1z} + J (\sigma_1 \cdot \sigma_2)$.

The two spins are coupled by $J$ but at the same time the first spin $\sigma_1$ is also under the influence of $D$ that influences it's z-component. My questions now are:

1. Is this Hamiltonian even allowed in a physical way? I.e. can I just "divide" the spin $\sigma_1$ like this and claim that on the one hand it is interacting with another spin but on the other hand it is also influenced by something in which I do not take the second spin into account anymore and just add the Hamiltonians up. Somehow I feel that a perturbation is more appropiate.

2. Now of course I want to find the eigenstates and eigenvalues of this Hamiltonian, for this I couple the spins and try the following 4x4 Hamiltonian with the new 4x4 spin operators: $$H = D (\sigma_{1z} \otimes \mathbb{1}^{(2x2)}) + J (\sigma_{1x} \otimes \sigma_{2x} + \sigma_{1y} \otimes \sigma_{2y} + \sigma_{1z} \otimes \sigma_{2z})$$ I then naively diagonalize this Hamiltonian with Matlab and get some eigenvalues. The eigenvalues seem to agree with the expected energies. But my problem now is that I'm not sure if I this is correct based on my first question and I have no idea what the eigenstates are supposed to mean, I feel like I need to do a unitary transformation or a base change to make sense out of the eigenstates that I get. I also can't find operators that now commute with this Hamiltonian, i.e. $\sigma^2$ does not commute anymore. Where I have used new operator for the total spin in the following form (here only for the z-component): $$S_z = \sigma_{1z} \otimes \mathbb{1}^{(2x2)} + \mathbb{1}^{(2x2)} \otimes \sigma_{2z}.$$ Is there a way to find good quantum numbers again that are conserved since I'm really confused now how I am supposed to label the eigenstates, i.e. which m_z value and total spin they have.

Thanks!

• For spin-$1/2$, $\sigma_z^2=1$. So the first term is a constant :) – Meng Cheng May 18 '15 at 23:37
• Haha, true... Let's make it a higher spin then, e.g. spin-1 or just leave it without the square, thanks for pointing this out. – magforce May 18 '15 at 23:39
• Doesn't $\sigma_{1z} \otimes \sigma_{2z}$ commute with the Hamiltonian? – Peter Shor May 19 '15 at 13:39
• The first term that @Meng Cheng is talking about has been edited out. I have no idea whether your calculations are correct or not; if you did them right (where the first term is not squared), they should be. – Peter Shor May 19 '15 at 13:41
• Why do you expect the eigenstates to have definite $m_z$ or total spin values? It seems fairly evident from the Hamiltonian that they won't (unless you have some kind of accidental degeneracy.) – Michael Seifert May 19 '15 at 14:04

To do this, it is clever to first analyze the easier Hamiltonian $H_0 = 2g (\vec L \cdot \vec S)$, where the $L_i$ and $S_j$ fulfill independent $SU(2)$-algebrae $$[L_i, L_j] = i \epsilon_{ijk} L_k\\ [S_i, S_j] = i \epsilon_{ijk} S_k.$$ This Hamiltonian can be written as $$H_0 = g(J^2 - L^2 - S^2),$$ where we have defined the following operators: $$L^2 = \sum_{i=1}^3 L_i^2 \otimes \mathbb{1},\\ S^2 = \sum_{i=1}^3 \mathbb{1} \otimes S_i^2, \\ J_i = L_i\otimes \mathbb{1} + \mathbb{1}\otimes S_i,\\J^2 = \sum_{i=1}^3 J_i^2.$$ Now as $L^2$ and $S^2$ are equal to $\frac{1}{2} (1+\frac{1}{2})$ on the subspace we are interested in (namely the one of a spin-1/2-particle), we can write the Hamiltonian as $$H_0 = g(J^2 - 3/2)$$ and by simple addition of angular momenta, we find the eigenstates $|j, m\rangle$: $$|1, 1\rangle = \left|\uparrow\uparrow\right\rangle\\ |1, 0\rangle = \frac{1}{\sqrt{2}}(\left|\uparrow\downarrow\right\rangle+\left|\downarrow\uparrow\right\rangle) \\ |1, -1\rangle = \left|\downarrow\downarrow\right\rangle\\ |0, 0\rangle = \frac{1}{\sqrt{2}}(\left|\uparrow\downarrow\right\rangle-\left|\downarrow\uparrow\right\rangle)$$ with energies $E_0 = -3g/2$ and $E_1 = g/2$ respectively (and obvious notation for the product base of the two particle hilbert space).
Now let's proceed to the real problem and add the second term. As you said, $$[J^2, L_z] = 2i (\vec L\times \vec S)_z$$ and so the eigenvalue of $J^2$ is no good quantum number anymore. But $$[H, J_z] = 0,$$ so we can label the states of the system by their energy and their $J_z$ component. Indeed, if we calculate the action of $H$ on the former eigenbase, we see, that it only mixes $|1, 0\rangle$ and $|0, 0\rangle$. By diagonalizing the full Hamiltonian we find a new base of eigenstates:
$$|m=\pm 1, E_{\pm1} \rangle = |1, \pm 1\rangle\\ |m=0, E_{0,\pm}\rangle = \frac{1}{C_\pm}\left(\left(g\pm\sqrt{g^2+d^2}\right) |1, 0\rangle + d|0, 0\rangle\right),$$ where $C_\pm^2 = 2\left(g^2 \pm g \sqrt{g^2+d^2} + d^2\right)$ and corresponding energies $$E_{\pm1} = \frac{g}{2}\pm d\\ E_{0, \pm} = -\frac{g}{2} \pm \sqrt{g^2 + d^2},$$ which of course reduce to the eigenvalues of $H_0$ if you turn $d$ to zero.