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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!

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  • $\begingroup$ For spin-$1/2$, $\sigma_z^2=1$. So the first term is a constant :) $\endgroup$ – Meng Cheng May 18 '15 at 23:37
  • $\begingroup$ 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. $\endgroup$ – magforce May 18 '15 at 23:39
  • $\begingroup$ Doesn't $\sigma_{1z} \otimes \sigma_{2z}$ commute with the Hamiltonian? $\endgroup$ – Peter Shor May 19 '15 at 13:39
  • $\begingroup$ 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. $\endgroup$ – Peter Shor May 19 '15 at 13:41
  • $\begingroup$ 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.) $\endgroup$ – Michael Seifert May 19 '15 at 14:04
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I'm not so sure, if this is really, what you're looking for, but you can of course solve this easy problem analytically.

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.

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