# Meaning of Zeros of Partition function

1. Apr 15, 2008

### mhill

given a partition function of the form

$$Z= \prod Z_{i}$$

$$Z_{i} = \sum_{n=-\infty}^{\infty}e^{iuE_{n}^{i}}$$

what is the meaning of zeros ? i mean the values that make Z=0 and how could we calculate these zeros ??

2. Apr 15, 2008

### genneth

At the zero of the partition function, the free energy (or whatever the associated 0'th lagrangian multiplier is) would have a logarithmic divergence. Not too sure what physical effect this has. Do you have a system in mind?

As far as calculating the zeros, it would be just the same as solving any equation...

3. Apr 15, 2008

### mhill

thanks genneth .. i was thinking about ' Lee-*Yang circle theorem' that was related to this topic and said that all the zeros (complex ??) of Partition function lie on a certain circle of the complex plane.

4. Apr 15, 2008

### genneth

Looking up the theorem and associated literature, the key paper appears to be http://prola.aps.org/abstract/PR/v87/i3/p404_1 where Lee and Yang showed what physical properties are connected with the zeros of the grand partition function. It's pretty much what you would have guessed --- phase transitions. The surprise I would have said was that the information is pretty much *entirely* in the location of the zeros. The 2nd paper in that series gives a proof of the so-called circle theorem.

5. Apr 15, 2008

### JustinLevy

I have wondered about this as well, for the zeros seem to have a very special meaning. I just can't figure out how to make physical sense of it yet.
Here is a specific example:

Consider a potential with two sites for interacting fermions.
$$H = \epsilon c_1^\dagger c_1 + \epsilon c_2^\dagger c_2 + t (c_2^\dagger c_1 + c_1^\dagger c_2)$$
Where t is a tunnelling term coupling the two sites.

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 $$H|2\rangle = (2 \epsilon) |2\rangle$$

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

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

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!

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.

6. Apr 15, 2008

### JustinLevy

Oops... I took too long to post.
The answers are probably in that paper, I'll check it out.
Thanks genneth!

7. Apr 16, 2008

### JustinLevy

That paper was very interesting, but it didn't answer the follow up questions I wrote.

Can anyone help?

8. Apr 17, 2008

### genneth

I don't think your problem here is actually to do with the stated thread. Expressing interacting systems as non-interacting ones is the basis of the entire field of condensed matter. However, it is not true that it's *always* possible. For example, multiple spinless, massive particles with mutual Coulomb repulsion. For the specific case that you have stated, it is possible, and does indeed give a nice answer; but you could have seen that without doing the statistical mechanics --- just changes basis for the hamiltonian itself (notice that it is a quadratic form, so diagonalise it).

9. Apr 17, 2008

### JustinLevy

I'm quite willing to accept that it may not be always true. But I can't think of a convincing argument either way.

In particular, the partition function can't have zeros at the origin, and thus it seems you could always factor it the way I suggested. So what are these seemingly non-interacting quasi-particle states? Even if it doesn't work for some reason I am missing, it is still not clear to me what it means when it does work (and it isn't just linear combinations of single particle states).

Can you give me a specific Hamiltonian to play with? I think I need to convince myself of this. Thank you for pointing this out.

Would it be something like this? (written in the position basis)
$$H = \int \epsilon \ a^\dagger(r_1) a(r_1) \ d^3 r_1 + \frac{1}{2} \int \frac{A}{|r_1 - r_2|} (a^\dagger(r_1) a^\dagger(r_2) a(r_1) a(r_2) + a^\dagger(r_2) a^\dagger(r_1) a(r_2) a(r_1)) \ d^3 r_1 \ d^3 r_2$$

Does the problem still arise when considering just a finite number of locations (like on a couple lattice positions)? That would make it a bit easier for me to conceptualize and play with.

Yes, I guess I made the example too simple.
Add a repulsion term, and now we have:
$$H = \epsilon c_1^\dagger c_1 + \epsilon c_2^\dagger c_2 + t (c_2^\dagger c_1 + c_1^\dagger c_2) + A (c_1^\dagger c_2^\dagger c_1 c_2 + c_2^\dagger c_1^\dagger c_2 c_1)$$

Maybe again this is simple, but I don't see how to put this in the form:
$$H = \sum_i \epsilon_i b_i^\dagger b_i$$
with just a simple basis change. Yet the factorization above still holds. What are these non-interacting states: what do they mean physically? and how are they related to the original creation operators?

Last edited: Apr 17, 2008
10. Apr 17, 2008

### genneth

Basically, consider a bunch of electrons moving in some overall potential and their mutual repulsion. The problem is not exactly easy --- various phase transitions, etc. If you through in more complicated mutual interactions things quickly get out of hand.

This general method of turning interacting systems into non-interacting ones is very powerful, and where we get quasi-particles from. In fact, it's almost a mantra of condensed matter that particle are just things which don't interact very much. Now there's not really any particular requirement that these minimal excitations are free in the sense that they have a quadratic action in momentum, but just that you can account for their quantum properties completely.