Correlation function poles in Peskin's derivation of LSZ formula

1. May 26, 2014

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!

2. May 26, 2014

p-brane

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

3. May 27, 2014

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!

4. May 27, 2014

p-brane

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.

Last edited: May 27, 2014
5. May 27, 2014

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!

6. May 27, 2014

p-brane

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 any one particle state is isolated from the continuum of masses of multiparticle states.

Last edited: May 27, 2014