# Srednicki QFT chapter 8 question

1. Feb 11, 2010

### 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 cant see how this is?

Thanks

2. Feb 11, 2010

### Avodyne

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

3. Feb 12, 2010

### 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.

4. Feb 12, 2010

### 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.