# Many body problem -> non-interacting particles

1. Apr 19, 2008

### JustinLevy

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.

2. Apr 21, 2008

### JustinLevy

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?

3. Apr 22, 2008

### lbrits

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)

4. Apr 22, 2008

### olgranpappy

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.

5. Apr 22, 2008

### olgranpappy

I.e.,

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