The footnote at &7.6 page 329 writes:(adsbygoogle = window.adsbygoogle || []).push({});

'' 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?

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# A question on Weinberg's QFT.

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