# Many body problem -> non-interacting particles

## Main Question or Discussion Point

many body problem --> non-interacting particles

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.
$$H = \epsilon c_1^\dagger c_1 + \epsilon c_2^\dagger c_2 + t (c_2^\dagger c_1 + c_1^\dagger c_2) + A (c_1^\dagger c_2^\dagger c_1 c_2 + c_2^\dagger c_1^\dagger c_2 c_1)$$
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:
$$|0\rangle$$, energy is $$H |0\rangle = 0 |0\rangle$$

one fermion eigen solutions:
let $$|1+\rangle = \frac{1}{\sqrt{2}}(c_1^\dagger + c_2^\dagger)|0\rangle$$, energy is $$H|1+\rangle = (\epsilon+t) |1+\rangle$$
let $$|1-\rangle = \frac{1}{\sqrt{2}}(c_1^\dagger - c_2^\dagger)|0\rangle$$, energy is $$H|1-\rangle = (\epsilon-t) |1-\rangle$$

two fermion eigen solution:
let $$|2\rangle = \frac{1}{\sqrt{2}}(c_1^\dagger c_2^\dagger - c_2^\dagger c_1^\dagger)|0\rangle$$, energy is $$H|2\rangle = (2 \epsilon+A) |2\rangle$$

The grand partition function is therefore:
$$\mathcal{Z} = 1 + e^{-\beta ((\epsilon+t)-\mu)} + e^{-\beta ((\epsilon-t)-\mu)} + e^{-\beta (2\epsilon+A-2\mu)}$$

let $$\lambda = e^{-\beta (\epsilon-\mu)}$$, then we can clearly see the polynomial:
$$\mathcal{Z} = 1 + e^{-\beta t}\lambda + e^{-\beta (-t)} \lambda + e^{-\beta A} \lambda^2}$$
$$\mathcal{Z} = 1 + (e^{-\beta t}+ e^{\beta t})\lambda + e^{-\beta A} \lambda^2$$

And of course we can factor this polynomial to rewrite the partition function as:
$$\mathcal{Z} = \prod_i Q_i$$
Where $$Q_i = (\lambda - \lambda_i)$$ and $$\lambda_i$$ are the zero's of the polynomial. Notice that for this example, $\lambda_i$ are real, less than zero, and $$\prod_i \lambda_i = 1$$.

So I can rewrite this as:
$$\mathcal{Z} = \prod_i Z_i$$
Where $$Z_i = (1 + e^{-\beta (\epsilon_i-\mu)})$$. And now the partition functions are of non-interacting quasiparticle states!

So the Hamiltonian should be able to be rewritten as such:
$$H = \sum_i \epsilon_i b_i^\dagger b_i$$
Where $b_i^\dagger$ is the creation operator for non-interacting state $i$ with energy $\epsilon_i$.

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 $$c_1^\dagger$$ and $$c_2^\dagger$$ ?

Any insight people can provide on this would be appreciated.

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I've thought about this some more, and started wondering: what is the commutation relation for these new creation/annihilation operators? Are they still fermions? I'm not sure how to check. Basically I want to someone prove whether or not the Hamiltonian is still Hermitian. If not, then clearly there is a mistake and using this can help me track down the mistake (hopefully).

Any ideas anyone?

Those operators will probably not diagonalize the Hamiltonian in the interacting case, so calling them "particles" isn't a good idea. In other words, particle number will not be conserved (doesn't commute with H)

olgranpappy
Homework Helper
hmmm... phys forums isn't letting me quote and post... so maybe this will work.

anyway, you have to factor out the e^{\beta A} in order to write your polynomial in the way yuo wrote it. and then I don't see how you rewrite it in the "quasiparticle" form. the algebra doesnt work.

olgranpappy
Homework Helper
I.e.,

you have to factor our the coefficient of the lambda^2 term.
$$Z=e^{-\beta A}(\lambda-\lambda_+)(\lambda-\lambda_-)$$
where
$$\lambda_{\pm}=\frac{-\cosh(\beta t)e^{\beta A}\pm\sqrt{\cosh^2(\beta t)e^{2\beta A}-e^{\beta A}}}{}$$