Srednicki QFT chapter 8 question

[LAHLH]

Hi,

In chapter 8 Srednicki employs the 1-i \epsilon trick. He multiplies the Hamiltonian desity,

H=\frac{1}{2} \Pi^2+\frac{1}{2}(\nabla\phi)^2+\frac{1}{2}m^2\phi^2

by this 1-i \epsilon, and says it's equivalent to if we replaced m^2 with m^2-i \epsilon. I can't see how this is?

Thanks

 

[Avodyne]

He explains this in detail in the earlier chapter on the harmonic oscillator.

 

[LAHLH]

In chptr 7, it's completely different,

H(P,Q)=\frac{P^2 m^{-1}}{2}+\frac{1}{2}m\omega^2 Q^2

So multiplying by 1-i\epsilon, gives :

H(P,Q)=\frac{P^2 (1-i\epsilon)m^{-1}}{2}+\frac{1}{2}(1-i\epsilon) m\omega^2 Q^2

So directly you can see this is equivalent to if we used m^{-1} \rightarrow (1-i\epsilon)m^{-1} and m\omega^2 \rightarrow (1-i\epsilon)m\omega^2

I fail to see how to perform in a similar process to get Srednicki's mass substitution for the H in my first post. Thanks.

 

[Physics Monkey]

Hi LAHLH,

I would recommend not trying to take Srednicki's comment too literally. A useful point of view is the following. The purpose of the i \epsilon is to get the pole structure right in the propagator. To make sense of what Srednicki is saying, try to determine if modifying only the mass by i \epsilon is sufficient to modify the pole structure so that integrating along the real line is the "right" contour.