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Phileas.Fogg
Feb28-09, 07:15 AM
Hello,
I read chapter 7.1 of "An Introduction to Quantum Field Theory" by Peskin and Schröder and have two questions.

They derive the two point function for the interacting case.
On page 213 they manipulate the matrix element, after insertion of the complete set of eigenstates.

<\Omega | \Phi (x) | \lambda_p >
= < \Omega | e^{iPx} \Phi (0) e^{- iPx} | \lambda_p >
= < \Omega | \Phi (0) | \lambda_p > e^{- ipx} \end{array}

with E_p = p^0

1. Could anyone explain to me, how they get the third line above (also see P & S: equation (7.4) ) ?

2. Later on page 215 they make a Fourier Transform of the spectral decomposition. I don't know, how they derive equation 7.9

\int d^4 x e^{ipx} < \Omega | T \Phi (x) \Phi(0) | \Omega> = \frac{iZ}{p^2 - m^2 + i \epsilon} + \int_{~4m^2}^{\infty} \frac{d M^2}{2 \pi} \rho(M^2) \frac{i}{p^2 - M^2 + i \epsilon}

Regards,
Mr. Fogg

JosephButler
Feb28-09, 11:23 AM
Hello,
<\Omega | \Phi (x) | \lambda_p >
= < \Omega | e^{iPx} \Phi (0) e^{- iPx} | \lambda_p >
= < \Omega | \Phi (0) | \lambda_p > e^{- ipx} \end{array}

with E_p = p^0

1. Could anyone explain to me, how they get the third line above (also see P & S: equation (7.4) ) ?


The momentum operator acting to the left on the vacuum gives zero since the vacuum is assumed to be Lorentz invariant .

-JB

Phileas.Fogg
Feb28-09, 11:55 AM
And where does the change e^{-iPx} \rightarrow e^{-ipx} come from?

xepma
Feb28-09, 03:34 PM
Don't have the book with me, but my best guess would be that they set the state |\lambda_p\rangle to be a momentum eigenstate. This means the momentum operator P acts on this state and is replaced by the momentum eigenvalue p (which is ofcourse a 4-vector).

You can also check that when the operator e^P acts on the state the P is also replaced by the eigenvalue p turning it into e^p.