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I have a system of $4$ spin $\frac{1}{2}$ particles, whose Hamiltonian looks like this:

$$ H = \alpha (s_1 \cdot s_2 + s_1 \cdot s_4 +s_2 \cdot s_3 +s_3 \cdot s_4 )$$

In order to find its eigenvalues and respective degeneracies I tried the typical method:

$$ \hat{\overrightarrow{S}} = s_1+ s_2+s_3+s_4 $$

$$ \hat{S}^2 = (s_1+ s_2+s_3+s_4) \cdot (s_1+ s_2+s_3+s_4)$$

which leads me to:

$$ \frac{1}{2} \left( \hat{S}^2 - \sum_{i=1}^4 |s_i|^2\right) = s_1 \cdot s_2 + s_1 \cdot s_4 +s_2 \cdot s_3 +s_3 \cdot s_4 + s_1 \cdot s_3 +s_2 \cdot s_4 $$

There are two extra terms $s_1 \cdot s_3$ and $s_2 \cdot s_4$ that will still appear in the Hamiltonian. How do I handle these in order to get the possible eigenvalues and degeneracies of this particular system?

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You are composing four doublets, hence you are inspecting a $2^4\times 2^4= 16\times 16$ matrix. This reduces to the familiar spin 2, spin 1, and spin 0 blocks, specifically a quintet, three triplets, and two singlets, 16 states in all, $$ 1/2\otimes 1/2\otimes 1/2\otimes 1/2= 2\oplus (3) ~~1\oplus (2) ~~0. $$

But your hamiltonian $\alpha(𝑠_1+𝑠_3)⋅(𝑠_2+𝑠_4)$ is special: it does not care how spins 1 & 3 compose together symmetrically or anti symmetrically, in a triplet or a singlet, and likewise for 2 & 4. These two options are degenerate, respectively, in both cases, so you are really composing just $$ \vec S_a = {\vec s_1+\vec s_3} ~~~\hbox{with} \\ \vec S_b = {\vec s_2+\vec s_4}, $$ where the spins of each $S_a$ and $S_b$ are 0 and 1; so, four combinations in all for $\vec S= \vec S_a+\vec S_b$, $$ H= {\alpha\over 2} (S^2-S_a^2-S_b^2). $$

Recalling that the Casimirs for spin 0; 1; and 2 are 0; 2; and 6, respectively: you get the singlet-singlet combination (singlet) to have 0 energy eigenvalue; the two singlet-triplet combinations (triplets) to have 0 energy eigenvalue, as well.

The triplet-triplet combination reduces to the quintet, with eigenvalue α; a triplet with eigenvalue ; and a singlet with eigenvalue -2α.

Not all products of the reduction with a common spin are treated identically by the (a)symmetric hamiltonian.

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    $\begingroup$ Shouldn´t the possible values for $S_a$ and $S_b$ be $0$ or $1$? Using the theory of addition of angular momentum, that's what it should be, right? $\endgroup$ Dec 7, 2021 at 15:13
  • $\begingroup$ Also, for the eigenvalues I used: $H = \frac{\alpha}{2} (S^2 - S_a^2- S_b^2) = \frac{\alpha}{2} (S(S+1) - S_a(S_a +1)- S_b(S_b+1))$ and tried the different cases: $S=2, S_a=1 , S_b=1$; $S= 1 , S_a=1 , S_b=0$, etc... and got $3\alpha$, $2 \alpha$, $\alpha$ and $0$ for the eigenvalues. $\endgroup$ Dec 7, 2021 at 15:17
  • $\begingroup$ Pardon my typo; indeed, the intermediate components spins should be 0 or 1, but my typing was thinking "triplet" for the latter. The last paragraph is detailing $S_a=1=S_b$ for $S=2,1,0$, hence eigenvalues α, -α, -2α, as it states. Recall, quintet is spin 2, triplet is spin 1, and singlet is spin 0. $\endgroup$ Dec 7, 2021 at 15:25
  • $\begingroup$ Sorry for being annoying but, in the triplet-singlet combination (which I presume is for $S_a=1$, $S_b=0$ and $S_b=1$, $S_a=0$ ), when plugging in the values $S=0,1$ I found that the eigenvalue associated for $S=0$ is $H=- \alpha$ and for $S=1$ is $H=0$. Am I right? $\endgroup$ Dec 7, 2021 at 17:20
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    $\begingroup$ Nevermind, I got it now. Thank you! $\endgroup$ Dec 7, 2021 at 19:24

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