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The scalar field lagrangian

by qft1
Tags: field, lagrangian, scalar
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qft1
#1
Apr14-05, 02:26 AM
P: 1
Hi,

I have a question about a statement I've seen in many a Quantum Field Theory book (e.g. Zee). They say that the general form of the Lagrangian density for a scalar field, once two conditions are imposed:
(1) Lorentz invariance, and
(2) At most two time derivatives,
is:

L = 1/2(d\phi)^2 - V(\phi)

where V(\phi) is a polynomial in \phi.

Why is this? I can understand how the conditions restrict the kinetic energy term to being what it is, but I don't understand why V has to be _polynomial_ in \phi.
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Haelfix
#2
Apr14-05, 03:11 AM
Sci Advisor
P: 1,677
Good question.

The reason is dimensional analysis. The action must be dimensionless to be a lorentz invariant, so the lagrangian has to have dimension mass^4 .

So you can simply power count your fields to tabulate all renormalizable interactions. You've probably seen this before.. You know the spinor field has dimension 3/2, scalar fields dimension 1 etc

So for scalar fields you can only have a (phi)^3, or b phi^4 where a is dimension 1 and b is dimensionless.. Anything higher than that would lead to negative mass dimension coupling constants and a badly nonrenormalizable theory.
dextercioby
#3
Apr14-05, 10:53 AM
Sci Advisor
HW Helper
P: 11,896
It's not a requirement for max.2 time-derivatives.Think about Weyl gravity.The Hamiltonian formalism can be externded to an arbitrary # of time derivatives in the "kinetic" term...


Daniel.


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