# Fermionic Condensates

1. Jan 3, 2009

Doing some theory research on Superconductivity, I stumbled on a few of the monsters hidden in the closet of QFT. Among which is the fact that BEC (Bose Einstein condensation) has not been proven to happen (not even to bosons), and that there is a no go theorem forbidding spontaneous breaking of the phase symmetry for non degenerate ground states... Everything seems very messy.

Superconductivity is often discussed as the BEC of cooper pairs or spontaneous breaking of the U(1) symmetry. Now I have a few questions:

1) We know BEC of fermions has been done in the lab in that case the fermions pair to form bosons. Are there any suggestions what a creation operator of a boson consisting of fermions could look like. Whenever I try to stuff more then one electron in one state the fermionic operators cancel out.

2) Since such a creation operator doesn't seem to be a simple product of a number of raising operators, is the underlying Fock space even a good ground to build these bosonic operators on.

3) I read somewhere in relation to Bloch waves, and symmetry groups, that a symmetry of a Hamiltonian forces the same symmetry upon its Eigenstates as a consequence of Schur's lemma (which I couldn't see). When I look at the cheap explanation of spontaneous symmetry breaking with the Mexican hat potential, the ground state breaks the Symmetry of the Hamiltonian. How does it cheat its way around group theory?

Sorry that the questions are probably full of bad implicit assumptions. I'd be glad if someone could elucidate this even just a bit, or give me some pointers to helpful literature.

2. Jan 5, 2009

### strangerep

Wiki says otherwise. (See the BEC entry.)
http://en.wikipedia.org/wiki/Bose-Einstein_condensate

I'm not sure what you're doing. Maybe post some of your math?

3. Jan 5, 2009

About the wikipedia artikle. The remark, that BEC is not proven comes directly from our theory department. I think it relates to interacting QFTs, I am aware of the quantum statistics argument.

2) I did nothing fancy. At first I looked at bcs theory and thought that the resulting raising operators after the Bogolubov transform would be my cooper pair creation operators. Sorry if I don't have my stuff with me but it roughly looks like this:

$$b^+ = u c + vc^+$$

where the c are fermionic creation and anihilation operators for the electrons. Now I was trying to prove to myself that these create bosons, so I wouldn't get 0 when applying them twice. But alas

$$b^+ b^+ \left| 0 \right> = u^2 cc \left| 0 \right> + uv c c^+ \left| 0 \right> + uv c^+ c \left| 0 \right> + v^2 c^+ c^+ \left| 0 \right>$$
$$= 0 + \kappa \left| 0 \right> + 0 + 0$$

The operator creates nothing. Now that I think of it, I think the creation and anihilation operators even had incompatible indices, so they completely cancel out (no vacuum just 0). So I was told this operator was not the cooper pair creation operator, but something more complicated was.

I think whenever an operator decomposes into a series of products of fermionic creation and anihilation operators, it will fail, because you can simply never put two electrons into the same state. But I have never seen anything else applied to a Fock space, which makes me doubt if the Fock space is even ok.

4. Jan 5, 2009

### strangerep

Oh, you meant "exists" in the mathematical sense (whereas I thought you meant physically).
Yes, no interacting QFT in 4D has yet been proven rigorously to exist mathematically.

The B-transform just takes you to a different (unitarily inequivalent) Fock space (ie with
a different vacuum). But it can't turn fermions into bosons.

That sounds right.

The operator you want consists of even products of fermionic ones. Cooper pairs involve
non-local correlations, mediated by interactions with ions in the crystal lattice, so you
need to create fermions at different places in the lattice. Among other things, a B-transform
helps to diagonalize the full original Hamiltonian.

5. Jan 6, 2009

What makes you so confident, that these bosonic operators can be written as a product of fermionic ones? If I understand you correctly you say that my creation operator $$a^+$$ can be written as an even product of fermionic creation and annihilation ones
$$a^+ = \sum_j \prod _{i \in \left{ (\mathbf{k},\mathrm{spin}) \right}}^{2N_j} c_i^{\pm}$$
But to me this still looks as if a second application of such a raising operator $$a^+a^+\left| 0 \right>$$ cannot increase the number of involved electrons, since we just use the same operators again. If a creation operator has been used before it will just cancel the second time because it is fermionic. I have to think about it, but I still doubt that an operator consisting of a fixed set of fermionic creation and anihilation operators can be build such that upon double application it will produce more occupied states than the first time.

Your statement about different places though makes me wonder. I have never gotten the hang of creation operators in real space, only in boxes when the k-space becomes discrete I have the feeling that I know what I am doing. So if I am creating particles in one place in real space or destroy them in another I don't really understand what is happening, or if that fourier transform of operators does really make mathematical sense. If you can point me to a good text I would be grateful.

6. Jan 9, 2009

### genneth

Note that the pair $$P_k = c_k c_{-k}$$ isn't a boson; i.e. $$[P_k, P_{k'}] \neq 1$$.

7. Jan 11, 2009

That this operator doesn't act as a boson creation operator is something I had already found. I thought that I had conveyed that. That is why I am asking, because I can see no way how a boson creation operator, could be build from fermionic operators. I'd be happy for any example as synthetic or stupid as it may be.

8. Jan 12, 2009

I found a way. Its easy and uses even powers of operators. Now I feel stupid :)