# Renormalization in non-relativistic quantum mechanics

I read many articles about renormalization in the Internet, but as I currently don't know much of QFT (currently just studying classical field theory and QM), and as all this looks quite interesting, I'd like to still get some bit of understanding and feeling of it — in the context of non-relativistic quantum mechanics.

So, my question is: are there any (hopefully simple) examples of quantum mechanical problems, applying perturbation theory to which will give divergent series, which can then be regularized by renormalization procedure? What are they, what does this process of renormalizing them look like?

• If I recall correctly, the infinities only starts to pop out when one considers loops in the Feynman diagrams. – Your Majesty Nov 1 '14 at 11:24
• It's not quantum mechanics, but it will give you a feel for renormalization and associated methods: arxiv.org/abs/0812.3578. – JamalS Nov 1 '14 at 11:52
• It should be possible to construct a good QM example. E.g. you could consider some Coulomb scaterring problem where you need to regulate the Coulomb potential by making it a Yukawa potential, do the perturbative computation and then take the limit back to Coulomb potential. – Count Iblis Nov 3 '14 at 17:44
• Some advice: The appearance of infinities is a red herring in renormalization anyways. The central physical idea is that you're you can ignore degrees of freedom that aren't excited at low energy, if you adjust the constants in equations of motion that describes the low energy degrees of freedom. This is analogous to treating the hydrogen atom as an electron in a Coulomb potential. It works well, but you need to use the reduced mass $\mu = \frac{m_p}{m_e+m_p} m_e$. – user1504 Oct 13 '16 at 14:20

To understand the essence of perturbative renormalization you don't need any quantum field theory nor any quantum mechanics. A simple toy model suffices.

Suppose your theory makes a prediction for two distinct observables $F$ and $G$ in terms of a perturbative parameter $g$:

$$F = g + g^2 (S+1) + g^3 (S+1)^2 + g^4 (S+1)^3 + ...$$ $$G = g + g^2 (S-1) + g^3 (S-1)^2 + g^4 (S-1)^3 + ...$$

Here, $S$ represents some divergent mathematical sum. For instance, $S=1+2+3+4+...$. As in both expressions each term beyond first order in $g$ diverges, your perturbative theory seems pretty useless. What to do?

Under perturbative renormalization, one attempts to eliminate the non-observable parameter $g$ from the theory, and rewrite the perturbative series in terms of the observable $G$. So, we write:

$$g= \alpha_1 G + \alpha_2 G^2 + \alpha_3 G^3 + ...$$

and substitute this expression in the perturbation expansion for $G$. This gives us (up to order $G^3$):

$$G = \alpha_1 G + \alpha_2 G^2 + ... + (\alpha_1 G + ...)^2 (S-1)+ ...$$

It follows that $\alpha_1 =1$ and $\alpha_2 = -(S-1)$:

$$g= G - G^2 (S-1) + ...$$

Substituting this series in the expression for the observable $F$ yields:

$$F= G - G^2 (S-1) + ... + (G-...)^2 (S+1)+ ...$$

Which gives us the end result:

$$F = G + 2G^2 + ...$$

Note that this perturbative expression for $F$ is written solely in terms of the observable $G$, and no longer contains the divergent quantity $S$. You conclude that your perturbative theory for the observables $F$ and $G$ is renormalizable.

Note: when performing a perturbative renormalization and all infinities exactly cancel each other, it feels a bit like magic. In this toy example the magic is a direct consequence of the hidden non-perturbative relationship $F-G = 2FG$.

• It always looks very arbitrary to me. Indeed, for me, you just did some magic, some ARBITRARY MANIPULATION, without any mathematical rigour. If someone other do it, he/she might gets something different. This is inevitable, as far as i can see. – Jiang-min Zhang Nov 2 '14 at 22:57
• The point is, what is the mathematics for the manipulation called 'renormalization'? For quantum mechanics, the mathematics is functional analysis. I would like to ask for one for 'renormalization'. – Jiang-min Zhang Nov 2 '14 at 23:00
• @Jiang-minZhang - "If someone other do it, he/she might gets something different. This is inevitable, as far as i can see." - I challenge you to demonstrate your claim. Use the above example and derive an alternative convergent expansion of $F$. – Johannes Nov 3 '14 at 13:34
• Another example of what renormalization means at a very basic level, without QFTs: arxiv.org/abs/hep-th/0212049 – Adam Nov 3 '14 at 17:48
• Similarly useful: Regularization, renormalization, and dimensional analysis: Dimensional regularization meets freshman E&M Fredrick Olness and Randall Scalise Citation: American Journal of Physics 79, 306 (2011); doi: 10.1119/1.3535586 View online: dx.doi.org/10.1119/1.3535586 – Peter Morgan Dec 7 '14 at 0:51

Normally renormalization is necessary when the phenomenological constants acquire unnecessary perturbative corrections, not obligatorily divergent. But there may be other cases. Read, for example, http://www.physics.umd.edu/courses/Phys851/Luty/notes/renorm.pdf for QM.

http://www.roma1.infn.it/~amelino/appunti1.pdf

Here, page 16: "Aside on perturbative renormalizability". You can find a quite simple but enlightening example of renormalization applied to a non-relativistic theory where you also have the exact energy spectrum and eigenfunctions to be compared to.