# 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?

## 2 Answers

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).

Usually, the interaction also depends on spins, so even if the interaction is symmetric in momenta, it is not necessarily zero.

• 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.