A question about effective action in QFT?

Dec 2010
25
0
They say that for constant fields the effective potential in the tree approximately is given simply by minus the non-derivative terms in the Lagrangian density.I do not understand why the derivative terms are ''kinematic energy'' so that the potential equalling minimum expectation energy must abandons the kinematic energy(so abandons derivative terms).Because for example in Klein-Gordon Lagrangian the second order derivative(of four-vector) is proportional with square of mass(?) not being kinematic energy.
 
Dec 2010
25
0
Because of constant field,in effective action it has not derivative term,but I still do not understand why the effective action has only minus non-derivative terms of Lagrangian.B/c quantum effective action also contains the current J corresponding with the value of field.
 
Dec 2010
25
0
Now I can understand with tree approximation the quantum effective action is coincidence with Lagrangian density.
 
Dec 2010
25
0
At the moment,I think that this is true for Phi-4 theory
 

topsquark

Forum Staff
Apr 2008
3,106
659
On the dance floor, baby!
I think I'll be polite and let you know that the lack of response isn't that we are ignoring you. I can't answer you simply because I've never really understood the point of derivative couplings. As far as I can tell a derivative coupling is simply is a new field which we can handle in the usual way. I can't see any reason why that would necessarily become a contribution to the kinetic energy terms, unless you need to actually add a symmetry restoring kinetic energy term that is.

-Dan