# Diagonalizing the all-to-all quantum spin model (quantum Curie-Weiss) with uniform couplings

I am interested in diagonalizing the all-to-all quantum spin model

\begin{align} \hat{H} = \frac{1}{2}\sum_{i,j \neq i} \hat{S}_i \cdot \hat{S}_j \end{align}

or, if possible, a more general form involving an arbitrary coupling $$\tilde{H}$$ between the spin states on sites $$i$$ and $$j$$:

\begin{align} \hat{H} = \frac{1}{2} \sum_{i, j\neq i} \tilde{H}_{ij} \end{align}

There are numerous treatments for how to approach this sort of problem using mean field techniques, but for the applications I'm interested in I would like to obtain the spectrum exactly. Is this a tractable problem? Is the answer known? Is it trivial?

I am imagining something along the lines of an adaptation of the Bethe Ansatz to this problem-- for instance, in the case of the first Hamiltonian when the couplings are instead restricted to be nearest-neighbor instead of all to all, the solution is well known.

• Why spin “glass”? Your original Hamiltonian is antiferromagnetic… Don’t you rather mean a quantum analogue of the Sherrington-Kirpatrick model?
– LPZ
Jan 29 at 22:19
• @LPZ you're right, I've removed "glass" from the title. I had in mind the SK model when I wrote it but for my problem the fact that the couplings are fixed nonrandom is actually a core part so I would not expect glassy behavior. Jan 29 at 22:29

Assuming that each spin has total spin $$s$$, then you can rewrite the Hamiltonian: $$H = -\frac J2\hat S{}^2 +\frac {JN}2s(s+1)\\ \hat S = \sum_i \hat s_i$$ You therefore just need to diagonalize $$\hat S{}^2$$. For this, you'll need to decompose the product space of $$N$$ $$s$$ spin spaces into irreducible representations. The spectrum of $$\hat S{}^2$$ will therefore be of the form $$S(S+1)$$ with $$S$$ ranging the possible total spin number, namely $$0, 1 ... Ns$$. To diagonalise the Hamiltonian, you'll need to use Clebsch-Gordon coefficients.