# How to conclude that an interaction is attractive from its Fourier transform (momentum space representation)?

Background: In the book by Altland and Simons, Condensed matter field theory, in exercise 4.5.7, one is supposed to use the effective field theory method to integrate out the phonon field in an interacting electron-phonon system and find an attractive electron-electron interaction.

The form found for the effective action is $$S_\text{int} =- \frac{\gamma}{2m}\sum_{\mathbf q, \omega} \frac{q^2}{-\omega^2 + q^2}\rho_{-q}\rho_q$$ where $\gamma > 0$ is a coupling constant, $m$ the electron mass, $q = |\mathbf q|$ is momentum, $\omega$ frequency and $\rho$ is electron density.

Now Altland and Simons write that when $\omega<q$ the interaction is attractive. However, I am not sure how to draw this conclusion, since the action is in momentum space form, that is, we have its Fourier transform. I can't think of an obvious intuitive reason why there should be a relation between a function's sign, and its Fourier transform's sign. Maybe there is one and I can't see it, or Is there a more sophisticated principle at work here?

I found on Google something called Bochners theorem that gives a condition for a function to be the Fourier transform of a positive function. But I can't recall seeing this theorem mentioned in a physics text.

• Take a look at Anthony Zee's book: QFT in a nutshell. He explains why interactions mediated by spin 1 particles are repulsive for like charges and attractive for unlike charges, while they're the other way round for spin 0 or spin 2 mediators. – Siva Mar 30 '14 at 22:20
• I recall seeing this argument, and it might be from Zee's book. I will check it again, but I don't think it's that simple. The phonon field is integrated out so in this action there is no mediating field. Also, Zee might assume Lorentz invariance, but I do not think this problem is Lorentz invariant. – Robin Ekman Mar 31 '14 at 16:06

Indeed there is no relation between the sign of a potential and the sign of its Fourier transformation. But why should we care about this?

In the field theory, the criterion is very simple, an interaction is attractive if its coefficient (in the Hamiltonian) is negative, and is repulsive if its coefficient is positive. According to this criterion, Altland and Simons's conclusion is straight forward.

I think your confusion may arise from the picture of classical mechanics, in which an attractive interaction is a potential that grows with the inter-particle distance, so that the forces between the particles are pointing to each other attractively. But here, in the frequency-momentum space, we see no notion of "inter-particle distance", and no sign of the "growth" of the potential energy. We don't even know what is the "force" between particles, so how can we judge if the interaction is attractive or not? Because the concept of "force" has been discarded in the field theory, the attractiveness is defined in a different way in the field theory context. The energy instead the force becomes our criterion. If the interaction energy decreases with the square of the particle density, then we know the interaction must be attractive such that particles can gain energy by getting together. Then from the expression $\rho_{-q}\rho_q$ is the density squared, so just by looking at the sign of the coefficient in front of it, we can tell if the interaction is attractive or not.

• Yes, precisely since there is no notion of physical distance in momentum space is why I hesitate to draw the conclusion. However, I think what you're saying is if $\rho^\dagger A \rho$ appears where $A$ is pos/neg definite in momentum space, then $A$ is also pos/neg definite in real space, and from this we can draw conclusions. I'm still a little hesitant about this because you could come up with a positive definite interaction energy in real space that isn't minimal at 0 distance. For example, something like $\phi(x)((x-y)^4 - (x-y)^2 + 3)\phi(y)$. – Robin Ekman Apr 3 '14 at 10:04
• However for the case in the question one could expand, when $\omega \ll q$, to first order $\frac{q^2}{q^2-\omega^2} \approx 1 + \frac{\omega^2}{q^2}$ which is in real space is a $\phi^4$ and something similar to a Coulomb interaction and the conclusion follows. Your answer got me thinking about this. I had previously tried to Fourier transform the kernel and apply the convolution theorem, but I'm not good enough with transforms. – Robin Ekman Apr 3 '14 at 10:11
• @RobinEkman No, please forget about anything in the real space, it is totally irrelevant. The sign of A at a particular momentum tell us nothing about the sign of A in the real space. Even if A is positive in the real space, and even A has a local maximum at 0 distance, we can still say that the interaction is attractive at some particular momentum and frequency! And this is the point indeed: the electron-electron interaction is repulsive in the real space, but in the momentum space, the same interaction can be attractive at some particular momentum! – Everett You Apr 4 '14 at 0:26

Firstly: are we sure the numerator is q^2? I believe the BCS Hamiltonian gives q/(q^2-omega^2). Otherwise, it won't be zero at q->infinity. I will assume that.

I have been thinking about the precise same problem for a while. My ongoing conclusion is this: this expression in q-space (relative momentum space) is monotonically decreasing in absolute value, so that it's Fourier transform (real space interaction) is also "localized" (this only works out if you "smear out" that infinity at omega=q with a small imaginary part).

Then, if it is localized and negative, it means it grows monotonically as x increases, where x is the relative coordinate. On the other hand, if it is positive, than it decreases.

Makes sense?

PS: writing in frequency-momentum space shouldn't be a problem for interpreting the concept of attractive and repulsive. It is a mere choice of mathematical representation and does not impact how the forces turn out. The correct way to interpret this is not by moving to classical physics, but by analyzing the relative movement of wave packets -- I learned to tell these two apart thinking about excitons in a crystal, and the difference between Frenkel and Wannier excitons.

I do not think this is a real space statement. The precise statement is: when you have a particle-hole pair with momentum $\vec q$ and another particle-hole pair with the opposite momentum (look back at the definition of $\rho$, you will see it represents a particle-hole pair), when the exchange phonon has a frequency smaller than that momentum, the effective interaction between these two particle-hole pairs is attractive, because Hamiltonian appears in the (imaginary time) action just as itself and now it is negative.

Intuitively, an electron travels by an ion and polarizes the lattice locally, but the excitation of lattice, which is phonon, vibrates so slowly that it sees the second electron passing by after the first electron leaves, and because the interaction between electron and ion is attractive, the second electron effectively feels the attraction from the first one.