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In one book, I have got the following lines which I found myself unable to understand what is effective operator? The paragraph is given below:

The weak interaction describes nuclear beta decay, and at low energy it is given by an effective four fermion interaction. Since the effective operator has dimension 6, the coupling constant has inverse mass-squared dimension.

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3 Answers 3

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To make things clear for the specific case you are talking about, think about what is really happening in beta decay. It occurs via the following process:

enter image description here

It involves the exchange of a W boson and so to model the true process properly using the standard model requires the three point fermion gauge boson vertices (operators): $$\frac{g}{\sqrt{2}} W^\mu\left(\bar\nu \, \gamma_\mu e + \bar u \gamma_\mu d \right).$$ These are interactions between two fermions and the W boson allowing us to draw the above diagram. Note that the coupling constant $g$ is dimensionless. The contribution of the exchange of the W boson is given by $$\frac{g^2}{E^2-M_W^2}$$ where E is the energy transferred by the W boson.

Now at low energies which correspond to small distances the W boson only propagates for a very small distance and it can be approximated by a contact interaction between the fermions:

enter image description here

(Note here that we have replaced the individual quarks by $p$ and $n$ but that is not important to the argument). You see that the contribution of the $W$ is reduced to a contact interaction between four fermions! So the effective operator for this interaction now contains a term with four fermions such as $$G_F \, \bar\nu e \bar p n.$$ At low energies $(E^2 \ll M_W^2)$ we find $$\frac{g^2}{E^2-M_W^2} \to \frac{g^2}{M_W^2}$$ and so the new coupling constant (Fermi constant) is given by $$\frac{G_F}{\sqrt{2}}=\frac{g^2}{8M_W^2}$$ which has dimension of inverse mass squared! (The numerical factors are not important for the argument and come out from doing the full calculation properly).

I am not sure if it was the use of the word operator in the quote that confused you. Remember that in QFT the fields are operators and so a term in the Lagrangian containing a bunch of fields is often referred to as an operator. If four fermion fields are involved (as above for the contact interaction) then it is called a four fermion operator. If the operator appears from integrating out some other fields (the $W$ in this case) then it is known as an effective operator meaning at low energies it effectively captures the correct physics.

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To complement the other answer, you can think of it the other way around. Suppose that you're doing a low-energy experiment, but you are interested in some theory which lives at much higher energies. However, you cannot directly 'see' the physics living at these higher scales - it consists of degrees of freedom you don't have access to. But still, it influences your low-energy experiment, through quantum corrections - so you should 'integrate out' these high-energy DoFs.

In your case, the high-energy DoF is a gauge boson with a mass of order $\mathcal{O}(100 \text{ GeV})$, so you cannot probe it by observing beta decay, although it is responsible for it. The four-fermion operator in your low-energy Lagrangian is thus effective: it describes the DoF you see, but not the fundamental DoFs living at higher scales (in a healthier QFT).

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$\mathcal{O}(100\ \mathrm{GeV})$, not MeV. :) –  Michael Brown Feb 18 '13 at 14:11
A painful mistake ;). I'll correct the error, in order not to confuse other readers. –  Vibert Feb 19 '13 at 10:17
What's three orders of magnitude between friends, eh? –  Michael Brown Feb 19 '13 at 10:25

An effective operator is an operator which appears in the Lagrangian of an effective field theory (EFT). An EFT is a theory which captures low energies features of some other theory and is limited to energies below a certain scale. The operators in your concrete case appear in an approach to Beyond-the-Standard-Model (BSM) physics: it involves the construction of a Lagrangian which captures the low energy features of a BSM theory, and such includes the Standard Model. The effective Lagrangian contains products of fermion bilinears, operators of mass dimension six.

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The operator in question doesn't come from BSM physics. It comes from integrating out the W boson which is part of the SM! –  Mistake Ink Feb 19 '13 at 10:44
I did not claim that it appears exclusively in the effective Lagrangian. –  Frederic Brünner Feb 19 '13 at 12:18

protected by Qmechanic May 15 at 5:26

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