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Okay, this is a question I brought up elsewhere, but I guess I was off topic from the original thread so I decided to post here. Any help would be appreciated.

It appears to me that the grand partition function indicates any problem can be rewritten in terms of non-interacting quasi-particles. Below is an example to "demonstrate" the method, and also to give specifics in which to discuss. I haven't been able to find a flaw or make physical sense of it yet, which is where my questions lie. Any help understanding this would be much appreciated.

Consider a potential with two sites for interacting fermions.

[tex]H = \epsilon c_1^\dagger c_1 + \epsilon c_2^\dagger c_2 + t (c_2^\dagger c_1 + c_1^\dagger c_2) + A (c_1^\dagger c_2^\dagger c_1 c_2 + c_2^\dagger c_1^\dagger c_2 c_1) [/tex]

Where t is a tunnelling term coupling the two sites, and A is a repulsion term. t is real. A is real and greater than zero.

zero fermion eigen solution:

[tex]|0\rangle[/tex], energy is [tex]H |0\rangle = 0 |0\rangle[/tex]

one fermion eigen solutions:

let [tex] |1+\rangle = \frac{1}{\sqrt{2}}(c_1^\dagger + c_2^\dagger)|0\rangle[/tex], energy is [tex] H|1+\rangle = (\epsilon+t) |1+\rangle [/tex]

let [tex] |1-\rangle = \frac{1}{\sqrt{2}}(c_1^\dagger - c_2^\dagger)|0\rangle[/tex], energy is [tex] H|1-\rangle = (\epsilon-t) |1-\rangle [/tex]

two fermion eigen solution:

let [tex] |2\rangle = \frac{1}{\sqrt{2}}(c_1^\dagger c_2^\dagger - c_2^\dagger c_1^\dagger)|0\rangle[/tex], energy is [tex] H|2\rangle = (2 \epsilon+A) |2\rangle [/tex]

The grand partition function is therefore:

[tex]\mathcal{Z} = 1 + e^{-\beta ((\epsilon+t)-\mu)} + e^{-\beta ((\epsilon-t)-\mu)} + e^{-\beta (2\epsilon+A-2\mu)}[/tex]

let [tex]\lambda = e^{-\beta (\epsilon-\mu)}[/tex], then we can clearly see the polynomial:

[tex]\mathcal{Z} = 1 + e^{-\beta t}\lambda + e^{-\beta (-t)} \lambda + e^{-\beta A} \lambda^2}[/tex]

[tex]\mathcal{Z} = 1 + (e^{-\beta t}+ e^{\beta t})\lambda + e^{-\beta A} \lambda^2[/tex]

And of course we can factor this polynomial to rewrite the partition function as:

[tex]\mathcal{Z} = \prod_i Q_i [/tex]

Where [tex]Q_i = (\lambda - \lambda_i)[/tex] and [tex]\lambda_i[/tex] are the zero's of the polynomial. Notice that for this example, [itex]\lambda_i[/itex] are real, less than zero, and [tex]\prod_i \lambda_i = 1[/tex].

So I can rewrite this as:

[tex]\mathcal{Z} = \prod_i Z_i [/tex]

Where [tex]Z_i = (1 + e^{-\beta (\epsilon_i-\mu)})[/tex]. And now the partition functions are ofnon-interactingquasiparticle states!

So the Hamiltonian should be able to be rewritten as such:

[tex]H = \sum_i \epsilon_i b_i^\dagger b_i [/tex]

Where [itex]b_i^\dagger[/itex] is the creation operator for non-interacting state [itex]i[/itex] with energy [itex]\epsilon_i[/itex].

This should probably work for any interacting system of particles. You can always rewrite the polynomial as a product of the zeros. So you can always rewrite the paritition function as a system of non-interacting quasi-particles. But what are these states physically? And how do they relate to the original creation operators [tex]c_1^\dagger[/tex] and [tex]c_2^\dagger[/tex] ?

Any insight people can provide on this would be appreciated.

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# Many body problem -> non-interacting particles

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