# Fermion Self-Interaction

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I'm trying to think of a theory with a Fermion self-interaction, similar to the $\phi^4$ theory.

The first difficulty is of course that such a theory would have a non-renormalizable mass dimension: for example, $[(\bar{\psi}\psi)^2]=mass^6$, so to get to $mass^4$ you need a coupling constant with $[g]=mass^{-2}$ which is not-renormalizable using the formula $$SDoD=4-\sum_{f}E_{f}\left(1+s_{f}\right)-\sum_{i}N_{i}\left[g_{i}\right]$$ where $E_f$ is the number of external legs of field $f$, $s_f=0$ for Bosons and $\frac{1}{2}$ for Fermions, $N_i$ is the number of vertices of operator of type $i$ and $g_i$ is the coupling constant of operator of type $i$.

Then I started wondering, why does the Fermion field need to have a mass dimension of $\frac{3}{2}$? Because of the term $\bar{\psi}\partial\psi$. So what I'm wondering about is: is there any way to fix this and get a renormalizable theory? Is it worthwhile to study Fermion self-interactions even for non-renormalizable theories? (non-renormalizable theories could still be predictive, as effective theories?)

This post imported from StackExchange Physics at 2014-08-07 15:35 (UCT), posted by SE-user PPR
retagged Aug 7, 2014
Does your theory need to be relativistic? The theory of non-relativistic fermions with a self-interaction is quite well known. It is called condensed matter physics ;)

This post imported from StackExchange Physics at 2014-08-07 15:35 (UCT), posted by SE-user Mark Mitchison
"So what I'm wondering about is: is there any way to fix this and get a renormalizable theory?". You invent a (electro-) weak force quantum field theory with its $Z, W^\pm$ (massive) gauge bosons, and you get a renormalizable theory with $3$-particles vertex and coupling constants having no mass dimensions.

This post imported from StackExchange Physics at 2014-08-07 15:35 (UCT), posted by SE-user Trimok

What does push you to advance "a theory with a Fermion self-interaction"? An analogy with other studies, fruitless ones. And silly too. Because you first advance a self-interaction and then, if you are lucky enough, you remove it by renormalization. In terms of Lagrangian, you must add counterterms to compensate the wrongness, and in your case a perfect counterterm is just the same self-interacting term, but with the opposite sign.

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You can get such a theory starting from the Yukawa Lagrangian

$L=\frac{1}{2}(\partial\phi)^2-\frac{1}{2}m^2\phi^2+\bar\psi(i\gamma\cdot\partial-g\phi-m_0)\psi.$
Your equation of motions for the scalar part are given by

$\Box\phi=-g\bar\psi\psi$

that can be immediately integrated away to

$\phi=\phi'-g\int d^4yG(x-y)\bar\psi(y)\psi(y)$

being $G(x-y)$ the Green function satisfying $(\partial^2+m^2)G(x-y)=\delta^4(x-y)$ and $\phi'$ a solution of the homogeneous equation that we assume to be 0 for the sake of simplicity. So, after integration by part

$L=-\frac{1}{2}\phi(\partial^2\phi+m^2\phi)+\bar\psi(i\gamma\cdot\partial-g\phi-m_0)\psi$

and after substitution one has

$L=-\frac{g^2}{2}\int d^4y\bar\psi(y)\psi(y)G(x-y)\bar\psi(x)\psi(x)+\bar\psi\left(i\gamma\cdot\partial+g^2\int d^4yG(x-y)\bar\psi(y)\psi(y)-m_0\right)\psi$.

After a proper reshuffling of the Dirac fields this becomes

$L=\bar\psi(i\gamma\partial-m_0)\psi+\frac{g^2}{2}\int d^4y\bar\psi(x)\psi(x)G(x-y)\bar\psi(y)\psi(y)$

and we are just a step away from the answer to your question. Just note that your propagator is

$G(p)=-\frac{1}{p^2-m^2+i\epsilon}$

and when you are in the deep infrared, $p^2\ll m^2$ you are reduced to the simple result

$G(x-y)=\frac{1}{m^2}\delta^4(x-y)$

a contact interaction. Your model now

$L=\bar\psi(i\gamma\partial-m_0)\psi+\frac{G}{2}\bar\psi(x)\psi(x)\bar\psi(x)\psi(x)$

is the celebrated Nambu-Jona-Lasinio model, with $G=\frac{g^2}{m^2}$, that is the equivalent of a quartic scalar field for fermions. Indeed, also in this case you will get symmetry breaking and a mass gap equation.

answered Aug 7, 2014 by (345 points)
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By starting from the Fermi theory and requirement of the tree-unitarity of this theory (it is similar to renormalizability, but only on a tree level) you may build theory of electroweak interactions (even with Higgs boson). I'm only show you how does it work.

Fermi theory predicts growth the matrix element of neutrino-lepton scattering as $E^{2}$ (or $\left(\frac{s}{4}\right)$). This tells us that Fermi theory is only the effective theory. But it is very similar to the low-energy limit of the second order of perturbative theory with lagrangian $$L = g \bar{l}_{L}\gamma_{\mu}\nu_{L}W^{\mu} + h.c.$$ Indeed, if we discuss neutrino-lepton scattering, we will get $$M = -g^{2}\bar{l}_{L}(p_{4})\gamma^{\mu}\nu_{L} (p_{2})\bar{\nu}_{L}(p_{3})\gamma^{\nu}l_{L}(p_{1})\frac{\left( g_{\mu \nu} - \frac{q_{\mu}q_{\nu}}{m_{w}^{2}}\right)}{q^{2} - m_{W}^{2}}.$$ At the limit $E^{2} <<m_{W}^{2}$ you will get $$M = \frac{g^{2}}{m_{W}^{2}}\bar{l}_{L}(p_{4})\gamma^{\mu}\nu_{L} (p_{2})\bar{\nu}_{L}(p_{3})\gamma_{\mu}l_{L}(p_{1}).$$ So we have made the first step from effective theory to renormalizing theory.

The second one is to add EM interactions for $W$-boson. But the requirement of the unitarity in the process $W^{-}W^{+}$ tells us that we can't restrict ourselves to the minimal Lagrangian (the minimal lagrangian can be earned by the elongation of the derivative $\partial_{\mu} \to \partial_{\mu} - ieA_{\mu}$ in the free one for $W$-boson); we also must add the term $-iW^{\mu}(W^{\dagger})^{\nu}F_{\mu \nu}$.

The third step is to discuss the unitarity of some processes included leptons and W-bosons. For example, we can see that both of processes $W^{-}W^{-} \to W^{-}W^{-}$ and $l^{+}l^{-} \to W^{+}W^{-}$ violate the unitarity (first is proportional to $E^{4}$ while the second one is proportional to $E^{2}$). We can add the interaction with real spin-one massive particle $Z$, with real spin-zero massive particle $h$ and/or with charged spin-one-half massive particle. As it can be shown, the last case contradicts the experimental data (because it doesn't predict neutral current and the correct value of mass of particle), while the first two delete $E^{4}$- and $E^{2}$-terms respectively.

But how exactly to build this interaction? We need to include all of the terms in the lagrangian which don't have coupling constant with negative dimension. For example, $W,Z$ interaction may be written in a form $$g_{1}W^{2}Z^{2} + g_{2}(W^{+} \cdot Z )(W^{-} \cdot Z) + g_{WWZ}(EM-type),$$ where "EM-type"-terms coincide with corresponding $W,A$-three-vertex terms;

$Z, W, A$-interaction we may written in a form $$g_{3}(W^{-} \cdot A )(W^{+} \cdot Z) + g_{4}W^{2}(A \cdot Z) + g_{5}(W^{-}\cdot Z)(W^{+} \cdot A),$$ $W, h$ (similarly as $Z, h$) - in a form $$g_{6}W^{2}h^{2} + g_{7}W^{2}h.$$ Also we need the self-interaction terms (for scalar boson and for $W$-boson): $$g_{8}(W^{+} \cdot W^{-})^{2} + g_{9}(W^{-})^{2}(W^{+})^{2} + g_{10}h^{4} + g_{11}h^{3}.$$ The interaction of $Z, h$ with leptons may be written in a form $$g_{12}\bar{\nu}_{L}\gamma^{\alpha}\nu_{L}Z_{\alpha} + (g_{L}\bar{l}_{L}\gamma^{\alpha} l_{L} + g_{R}\bar{l}_{R}\gamma^{\alpha}l_{R})Z_{\alpha}$$ I repeat once more: all these terms are invented not accidental. Each of them delete some violating-unitary terms in the amplitude of some process (by fixing the value of constants). For example, requirement of the unitarity of processes $e^{+}e^{-} \to W^{+}W^{-}, \bar{\nu}\nu \to W^{+}W^{-}, e\bar{\nu} \to W^{-}Z$ leads us to the system of equations: $$-\frac{1}{2}g^{2}+ g_{WWZ}g_{12} = 0, \quad -\frac{1}{2}g^{2} + e^{2} - g_{L}g_{WWZ} = 0,$$ $$e^{2} -g_{R}g_{WWZ} = 0 = 0, \quad -g_{L} + g_{12} - g_{WWZ} = 0,$$ $$g_{R} - g_{12}+ g_{WWZ}\left( 1 - \frac{m_{Z}^{2}}{2m_{W}^{2}}\right) = 0.$$ From this set of equations we can determine constants $g_{L},g_{R}, g_{12},g_{WWZ}$ and get relation $m_{W} = m_{Z}\sqrt{1 - \frac{e^{2}}{g^{2}}}$.

This post imported from StackExchange Physics at 2014-08-07 15:35 (UCT), posted by SE-user Andrew McAddams
answered Aug 7, 2014 by (340 points)

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