Introducing cut-off in a renormalisation procedure for quantum mechanics I've been reading a paper on renormalisation theory as applied to a simple one-particle Coulombic system with a short-range potential.
In the process of renormalisation, the authors introduce an ultraviolet cutoff into the Coulomb potential through its Fourier transform:
$$
\frac{1}{r} \xrightarrow{\text{F.T.}} \frac{4\pi}{q^{2}} \xrightarrow{\text{cutoff}} \frac{4\pi}{q^{2}} e^{-q^{2}a^{2}/2} \xrightarrow{\text{F.T.}} \frac{erf(r/\sqrt{2}a)}{r} $$
It would really help me out if you could explain in plain language what is going on here.
I am a fourth year undergraduate student with only a basic knowledge of quantum mechanics, so you might have to dumb down a bit.
 A: What you describe is usually called regularization, as distinct from renormalization, although the terms are related. It could help to cite the paper that you are reading, but in any case, it often happens that long wavelength physics do not depend exactly on the details of short distance physics. For example, if you are scattering a particle off of a coulomb potential, then with finite momentum (finite wavelength) there is not enough resolution to see what is happening deep inside the potential well. For example, in scattering one proton off of another (Rutherford Scattering), the amplitude for the two protons to get close enough for us to see sub-nuclear details is vanishingly small at low momentum. On the other hand, if we just try to plug the Coulomb potential into our formulas, the calculations blow up.
The idea then is to pick a length scale and agree that all momenta will be such that wavelengths are greater than this length scale. We then deform the Coulomb potential so that our calculations don't misbehave at $r=0$. Because our wavelengths are much longer than our cut-off, the details of how we deform the potential shouldn't affect the answer (this is not always guaranteed). If this is indeed the case, taking the cut-off length scale to $0$ is equivalent to our answers converging to some limit.
The reason we perform the cut-off in momentum space is because this is the natural space where short distance physics separates from long distance physics (the Fourier transform separates wave functions according to their wavelength). It is not always the way things are done. In quantum field theory, for example, one way is to take the dimension of spacetime as a free parameter. The exact choice of regularization procedure depends on what is convenient for the calculation. The divergence in your calculation is ultimately connected to the fact that the photon is massless. The regulator that is used is to give the photon a finite but small mass.
To summarize: it's a mathematical trick that depends on a decoupling between short distance and long distance physics. The latter is called renormalizability.
