# Mean field Hamiltonians

1. Nov 26, 2013

### aaaa202

My book has a chapter on the general mean field approach. I have attached a page of it I don't understand very well. I do understand the general idea of meanfield theory:
Expand operators around their average and assume small fluctuations and neglect the second order term.
But in the attached page they use a form of the Hamiltonian I am only just getting used to. And for mean field theory to work I need to be able to separate the product of the a-operators and b-operators (in the given hamiltonian they appear as abba with daggers on some of them of course). Can I assume that a and b operators commute? They describe different particles but on the other hand it is interaction between a and b particles that is relevant. So what do I do?

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2. Nov 26, 2013

### Hypersphere

If the operators represent different particles, then yes, they will commute. For creation and annihilation operators, this is apparent through the commutation relation
$$\left[ a_i, a_j^\dagger \right] = \delta_{ij}$$
The reason this works is that each particle has its own "Hilbert space" (a Fock space in this case), and the operators a act in one of them, and the operators b in the other. The interaction term means that the spaces aren't independent (they're coupled in some way), but does not change what the operators actually do.

There is an analogous example for just one particle. We know that its orbital angular momentum $\hat{L}$ and spin $\hat{S}$ commute, since these live in separate spaces (see this thread for more on this) and represent distinct physical concepts. Yet, we know that atoms will have spin-orbit coupling. This interaction is of the form $\hat{L} \cdot \hat{S}$, and couples the spin and the orbital angular momenta. If you write down the total angular momentum $\hat{J}=\hat{L}+\hat{S}$ in matrix form, the interaction is represented by the off-diagonal elements. It'll look the same way for the $a$:s and $a^\dagger$:s.