# Quantum operator catastrophe

Assume we look at an interaction between 2 fermions

$V \sum_{k_i,k_j,k_m,k_n} c_{k_i}^\dagger c_{k_j}^\dagger c_{k_m} c_{k_n} \delta_k$

where $\delta_k$ conserves momentum. We can directly write down a few terms from the sum

$... + \underbrace{c_{k_1}^\dagger c_{k_2}^\dagger c_{k_3} c_{k_4}}_{i=1,j=2,m=3,n=4} + \underbrace{c_{k_2}^\dagger c_{k_1}^\dagger c_{k_3} c_{k_4}}_{i=2,j=1,m=3,n=4} + ... \ .$

and then use the anticommutator relations

$[c_i,c_j]_+ = [c_i^\dagger,c_j^\dagger]_+ = 0$

$[c_i,c_j^\dagger]_+ = c_i c_j^\dagger + c_j^\dagger c_i = \delta_{ij}$

to exchange the first two operators in the second summand ($c_{k_2}^\dagger c_{k_1}^\dagger = -c_{k_1}^\dagger c_{k_2}^\dagger$) such that

$... + c_{k_1}^\dagger c_{k_2}^\dagger c_{k_3} c_{k_4} - c_{k_1}^\dagger c_{k_2}^\dagger c_{k_3} c_{k_4} + ...$

the summation vanishes. For terms that do not vanish as pairs such as $c_{k_1}^\dagger c_{k_1}^\dagger c_{k_3} c_{k_4}$ we can see that due to the anticommutator, these terms vanish individually.

Now where is the mistake?

The mistake is that interaction terms do not correspond to what you wrote, which is indeed $0=0$. You forgot that in general, $V$ depends on the $k_i$'s, with the ad hoc sign changes when its arguments are exchanged. When $V$ is assumed to be a constant, one does not sum over all possible $k_i$'s, but only a subset (see for instance the BCS interaction).
• I agree, $V$ is usually a function of $k$ or the difference between $k$s. Now for BCS interaction, we only sum over half the set of possible $k$ values? Because the BCS interaction for $s$-wave is one example where the interaction potential $V$ I think can be assumed to be symmetric towards the exchange of momenta i.e. $V(k_1-k_2)=V(k_2-k_1)$. Means anytime we have a symmetric interaction potential we have to limit the sumation to a subset? – DrCommando Oct 30 '13 at 2:49
• Moreover I think that physically the processes $c_{k_1}^\dagger c_{k_2}^\dagger c_{k_3} c_{k_4}$ and $c_{k_2}^\dagger c_{k_1}^\dagger c_{k_3} c_{k_4}$ are identical and that even a $k$ dependent interaction potential $V$ won't distinguish between these two. – DrCommando Oct 30 '13 at 3:09
• @DrComando: you forget that the BCS interaction the fermions have opposite spins, so the the exchange of $k_i$ does not mean that the sum is zero. I changed my answer correspondingly. – Adam Oct 30 '13 at 3:30
• If you add another quantum number like spin and you also sum over all spin configurations I think you will end up with the same problem. The total number of terms increases but they still always come in pairs. Actually I am willing to accept your first point of a reduced subset. This is (1) also my personal idea and (2) may be the equivalent to double counting for non-fermionic states. For fermions double counting of states does not give twice the state but gives $0$ because of the anticommutator. To avoid this we cannot just divide by a number but we literally have to reduce the summation. – DrCommando Oct 30 '13 at 5:48
• I think you just need to see by yourself with an example that there is no problem when you take into account the momentum/spin dependence of the interaction potential. For instance, the Coulomb interaction is $\sum_{s,s'}\int_{x,y} V(|x-y|)\psi_s^\dagger(x)\psi_{s'}^\dagger(y)\psi_{s'}(y)\psi_s(x)$, and you can verify that there is no problem, in real space or in momentum space. – Adam Oct 30 '13 at 13:31
As an addition an example to Adam's answer, consider the action for the superfluid normal transition in liquid helium-4: $$F = \int dx \left( \frac{\hbar^2}{2m}|\nabla \phi(x)|^2 - \mu |\phi(x)|^2 + \frac{V_0}{2}|\phi(x)|^4 \right)$$ Fourier transforming the latter interaction term (which by the way comes from assuming a contact potential $V(x,x') = V_0\delta(x-x')$) yields: $$\frac{V_0}{2V}\sum_{K,k,k'}\phi^*_{K-k}\phi^*_{k}\phi_{K-k'}\phi_{k'},$$ which obviously does not vanish.