A question on Weinberg's QFT.

  • #1
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The footnote at &7.6 page 329 writes:
'' Recall that by canonical transformation, we mean a transformation from a set of phase space coordinates [itex]\Psi[/itex][itex]^{a}[/itex],[itex]\Pi[/itex][itex]_{a}[/itex] to some other phase space [itex]\tilde{\Psi}[/itex][itex]^{a}[/itex],[itex]\tilde{\Pi}[/itex][itex]_{a}[/itex] such that
[[itex]\tilde{\Psi}[/itex][itex]^{a}[/itex],[itex]\tilde{\Pi}[/itex][itex]_{b}[/itex]][itex]_{P}[/itex]=[itex]\delta[/itex][itex]^{a}_{b}[/itex] and [[itex]\tilde{\Psi}[/itex][itex]^{a}[/itex],[itex]\tilde{\Psi}[/itex][itex]^{b}[/itex]][itex]_{P}[/itex]=[[itex]\tilde{\Pi}[/itex][itex]_{a}[/itex],[itex]\tilde{\Pi}[/itex][itex]_{b}[/itex]][itex]_{P}[/itex]=0.It follows that the
Poisson brackets for any functions A,B are the same whether calculated interms of ψ,∏ or in terms
of [itex]\tilde{\Psi}[/itex] and [itex]\tilde{\Pi}[/itex].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?
 

Answers and Replies

  • #2
1,024
32
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.
 

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