Meaning of `bare Lagrangian'

[metroplex021]

Hi there - I've been confused for a long time about the following. When we learn how to mop up divergences in QFT, we learn two methods: the Feynman method, and the method of counterterms. In the latter, we add to a Lagrangian containing physical values for the parameters a Lagrangian containing counterterms to kill the divergences we get in the Green's functions computed from the physical Lagrangian alone. This total Lagrangian obtained in this way is then called the 'bare Lagrangian'. In the former, we start with a Lagrangian containing bare (unphysical) quantities, interpreted as the mass and charge particles would have 'at infinity' where no interactions are present, and compute vertex functions from that and use some redefinitions to get back to the physical values. My question is: is the 'bare Lagrangian' that is referred to in the method of counterterms mathematically / physically equivalent to the Lagrangian you start with in the Feynman procedure that contains the 'bare quantities'? I know we use the same word in each case, but I can't convince myself one way or the other that 'bare mass', 'bare Lagrangian', etc, really mean the same thing in each context. Any help (greatly) appreciated!

 

[azntoon]

The answer is no; the bare Lagrangian referred to in the method of counterterms is not equivalent to the Lagrangian you start with in the Feynman procedure. The bare Lagrangian contains physical values for the parameters, whereas the Feynman Lagrangian contains unphysical quantities. The difference between the two is that the Feynman Lagrangian is being used to compute vertex functions which are then used to redefine the physical values. This means that the Feynman Lagrangian does not contain the physical values directly, but rather the unphysical values which must be redefined in order to obtain the physical values.