# A question on Weinberg's QFT.

1. Sep 24, 2013

### ndung200790

The footnote at &7.6 page 329 writes:
'' Recall that by canonical transformation, we mean a transformation from a set of phase space coordinates $\Psi$$^{a}$,$\Pi$$_{a}$ to some other phase space $\tilde{\Psi}$$^{a}$,$\tilde{\Pi}$$_{a}$ such that
[$\tilde{\Psi}$$^{a}$,$\tilde{\Pi}$$_{b}$]$_{P}$=$\delta$$^{a}_{b}$ and [$\tilde{\Psi}$$^{a}$,$\tilde{\Psi}$$^{b}$]$_{P}$=[$\tilde{\Pi}$$_{a}$,$\tilde{\Pi}$$_{b}$]$_{P}$=0.It follows that the
Poisson brackets for any functions A,B are the same whether calculated interms of ψ,∏ or in terms
of $\tilde{\Psi}$ and $\tilde{\Pi}$.It also follows that the Hamintonian equation of motions are the same for ψ,∏ before and after being transformed.
The Lagrangian is changed by canonical transformation,but only by a time-derivative,which does not affect the action.''
To calculate the action we must to time integrate the Lagrangian,but by the transformation the Lagrangian changed by time derivative of a function of field(?),so the action changed by the function being calculated at infinite past and future.Then why we know field vanish at infinite time?

2. Sep 25, 2013

### andrien

So far it is an assumption that in infinite past or future,the fields will not give any contribution.As far as why the lagrangian changes by a time derivative,you can show in a simple way that for the hamiltonian eqn. of motion to be same after canonical transformation the lagrangian can change only by a time derivative.So far however it is a restricted version.