Correlation function poles in Peskin's derivation of LSZ formula

[Fysicus]

Hi folks,

Been trying to fill some of the more formal gaps in my knowledge by tackling the more technical stuff in P&S Chapter 7. Their derivation of the LSZ formula is quite different to those of books like, say, Srednicki, as they basically Fourier transform the whole argument as I understand it to concentrate more interpreting the pole structure of a general correlation function.

My question is this - having Fourier transformed the n-point function with respect to one field co-ordinate:

$\int d^4x e^{ip \cdot x} \left\langle \Omega \left| T{ \phi (x) \phi (z_{1}) \phi (z_{2}) ... } \right| \Omega \right\rangle$
​

And splitting the $x^{0}$ integral into three zones (I, II, III):

$\int dx^{0} = \int^{\infty}_{T_{+}} dx^{0} + \int^{T_{+}}_{T_{-}} dx^{0} + \int^{T_{-}}_{\infty} dx^{0}$
​

They insert a complete set of states from the interacting theory, and choose to evaluate region I and do the momentum integrals that come with that, along with the position integrals from our Fourier Transform to arrive at the very ugly Eq. 7.36:

$\int_{I} d^4x e^{ip \cdot x} \left\langle \Omega \left| T{ \phi (x) \phi (z_{1}) \phi (z_{2}) ... } \right| \Omega \right\rangle = \sum \frac{1}{2E_\textbf{p}(\lambda)} \frac{i e^{i(p^{0}-E_{\textbf{p}}+i\epsilon)T_{+}}}{p^{0}-E_{\textbf{p}}+i \epsilon} \left\langle \Omega \left| \phi (0) \right| \lambda_{0} \right\rangle \left\langle \lambda_{\textbf{p}} \left| T \phi (z_{1}) \phi (z_{2}) \ldots \right| \Omega \right\rangle$
​

Where the sum runs over the different mass states $\lambda$ and the 'I' on the integral just means $dx^{0}$ integrated over the region: $\int^{\infty}_{T_{+}} dx^{0}$.

At this point they state " the denominator is just that [as]$p^2 - m^2_{\lambda}$ " . I can't see why this is the case - yes, the pole in $p^{0}$ is in the same location as one of the poles in $p^2 - m^2_{\lambda}$ , but since all the integrals are done I can't see any way to massage this into this form exactly (by that i mean playing with delta functions and the E).

In other words, how can this statement true? Are we to literally claim that:

$\frac{1}{2E_\textbf{p}} \frac{1}{p^{0}-E_{\textbf{p}}+i\epsilon} = \frac{1}{p^{2}-m_{\lambda}^{2}}$ ?
​

Since, close to the pole, the exponential factor goes to 1, and the leftmost matrix element is the square root of the field strength renormalization factor $\sqrt{Z}$, do they just mean that this quantity has a pole with residue $\sqrt{Z}$ at $p^{0} = E_{\textbf{p}}$ much like the piece of the full propagator containing the mass singularities? Is this possibly what they mean by the next statement (7.37):

$\int d^4x e^{ip \cdot x} \left\langle \Omega \left| T{ \phi (x) \phi (z_{1}) \phi (z_{2}) ... } \right| \Omega \right\rangle \sim \frac{i}{p^{2}-m^{2}+i\epsilon} \sqrt{Z} \left\langle \textbf{p} \left| T \phi (z_{1}) \ldots \right| \Omega \right\rangle$

as

$p^{0} \rightarrow +E_{\textbf{p}}$
​

Had a look at the P&S questions reference thread: https://www.physicsforums.com/showthread.php?t=400073 , and it seems no-one has asked about this derivation before, so either I'm missing something, overthinking it, or any explanations might be useful to others for future reference. Cheers!

 

[p-brane]

$${p^0} = {E_{\bf{p}}}\left( \lambda \right) = \sqrt {{{\left| {\bf{p}} \right|}^2} + m_\lambda ^2} \,\, \Rightarrow \,\,\,{E_{\bf{p}}}\left( \lambda \right){p^0} - E_{\bf{p}}^2\left( \lambda \right) = {\left( {{p^0}} \right)^2} - {\left| {\bf{p}} \right|^2} - m_\lambda ^2 = {p^2} - m_\lambda ^2$$

 

[Fysicus]

Hey p-brane, thanks for the reply.

So implicit in the first steps you make there is that at the pole region only:

$p^{0} \rightarrow E_{\textbf{p}}$
​

Basically you can argue 'backwards' and reinterpret $E_{\textbf{p}}p^{0}$ where instead:

$E_{\textbf{p}} \rightarrow p_{0}$?
​

Or to write it another way,

$2E_{\textbf{p}}(p^{0}-E_{\textbf{p}}) \rightarrow (p^{0}+E_{\textbf{p}})(p^{0}-E_{\textbf{p}})$
​

Cheers!

 

[p-brane]

they state " the denominator is just that [as]$p^2 - m^2_{\lambda}$ " . I can't see why this is the case - yes, the pole in $p^{0}$ is in the same location as one of the poles in $p^2 - m^2_{\lambda}$

7.36 is a "sum over singularities" indexed by λ. If the masses are those of one-particle states, the singularities are poles. If the masses are those of multiparticle states, the singularities are branch cuts. That's why I displayed the energy as a function of the index.

 

[Fysicus]

That's true, so just so I'm understanding it correctly a quick recap, thinking of the whole thing as a function in $p^{0}$:

We've got an expression which is a sum over poles in $p^{0}$ - with positions determined by the energies $E_{\textbf{p}}$ corresponding to given states $\lambda_{\textbf{p}}$.

Within this there's one isolated pole defining the one-particle state, then a branch cut beyond the 2 particle threshold which is basically like a continuum of 'poles' - and it's generally true that at the region of the isolated pole this relationship that the denominator equals $p^{2}-m^{2}_{\lambda}$ holds as we may manipulate it as above, and furthermore that on the branch cut, at the point corresponding any given state $\lambda_{\textbf{p}}$ it also holds as each point there will have the value of $m_{\lambda}$ to keep it generally true.

However in the void for ('unphysical') values of $p^{0}$ between the isolated pole and branch-cut, this statement about the denominator of 7.36 being $p^{2}-m^{2}_{\lambda}$ is not true.

Thanks again!

 

[p-brane]

Within this there's one isolated pole defining the one-particle state…

By “isolated” is meant that there’s a gap between the mass m of any one-particle state and the mass of the lightest multiparticle state, which is the two-particle state with mass 2m. In other words, the mass of anyone particle state is isolated from the continuum of masses of multiparticle states.