Suppose one considers a multi-component free fermions field theory with field $\psi_{q_i}$ with a give global symmetry (such as U(1)). We can say that every component of fermions carry some U(1) charge $q_i$. Now, let us introduce Yukawa-Higgs term to mediate fermion-fermion interactions. Explicit Higgs charge are assigned to maintain the U(1) symmetry in Yukawa-Higgs term. We also consider the Higgs field strongly-disordered so do NOT break global symmetry (here U(1)) via the normal Higgs mechanism. This can be done by adding kinetic terms to Higgs field, and in principle $\langle \Phi \rangle=0$ but $ (\Phi-\langle \Phi \rangle)^2\neq0$, i.e. the Higgs field does not condense, but Higgs field strongly fluctuates.

Here is the Question:

whether including all symmetry-allowed Yukawa-Higgs term of a single Higgs field ($\Phi$) can induce any symmetry-allowed multi-fermion interactions?

I can imagine this may be true if we have:

including all symmetry-allowed Yukawa-Higgs term of (multi-component and multi-species) Higgs fields ($\Phi_1$,$\Phi_2$,$\Phi_3$,$\Phi_4$,$\dots$)

$\Leftrightarrow$ (iff: correct?)

induces all symmetry-allowed multi-fermion interactions.

But let us say if we have only ONE single component scalar Higgs field, say all symmetry-allowed Yukawa-Higgs term of $\psi_{q_i}^{n_i} \Phi^{m} \psi_{q_j}^{n_j}$ (with $\psi$ are fermions field and $\Phi$ is the only Higgs, with some power ${n_i},{n_j},m$ perhaps with complex conjugate $(\psi_{q_i}^{\dagger})^{|n_i|}$ if ${n_i}<0$.) can one prove or disprove this is true or no that it can generate all multi-fermion interactions, such as $$V_{interaction}=\big( ({\psi}_{q_1} \nabla^1_x {\psi}_{q_1} \nabla^2_x {\psi}_{q_1} \dots \nabla^{N_{q_1}}_x {\psi}_{q_1}) \dots ({\psi}_{q_n} \nabla^1_x {\psi}_{q_n} \nabla^2_x {\psi}_{q_n} \dots \nabla^{N_{q_n}}_x {\psi}_{q_n}) \dots ) \big) + \text{hermitian conjugate}$$ with potential higher derivative terms?


1 Answer 1


All terms that do not break explicitly the U(1) symmetry will be generated under the RG (in a Wilsonian approach). To see that, integrate out the scalar field. Then you have a theory with a four-fermions interaction. This interaction by itself will generate all possible interaction terms (with arbitrary number of fermionic fields and derivative terms). The same thing arises in the case of a $\phi^4$ theory.

  • $\begingroup$ Thanks Adam! are you sure just one single Higgs scalar field is enough? Plus, your argument contains a loop-hole: it is not necessarily 4-fermion interactions. To maintain the U(1) symmetry, depending on each species fermion charge, it can have general n-fermion interactions (to maintain the U(1) unbroken). $\endgroup$
    – Angie38750
    Commented Feb 3, 2014 at 3:47
  • $\begingroup$ Usually, a Yukawa coupling is a the form $\phi \bar\psi\psi$, so that will generate a 4 fermions coupling (between all the species) when integrated out. $\endgroup$
    – Adam
    Commented Feb 3, 2014 at 4:53

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