How to replace $T$-product with retarded commutator in LSZ formula?

Question

I am reading Itzykson and Zuber's Quantum Field Theory book, and am unable to understand a step that is made on page 246:

Here, they consider the elastic scattering of particle $A$ off particle $B$:

$$A(q_1) + B(p_1) ~\rightarrow~ A(q_2) + B(p_2)$$

and proceed to write down the $S$-matrix element using the LSZ formula, with the $A$ particles reduced:

$$S_{fi}=-\int d^4x\, d^4y e^{i(q_2.y-q_1.x)}(\square_y+m_a^2)(\square_x+m_a^2)\langle p_2|T \varphi^\dagger(y) \varphi(x)|p_1 \rangle \tag{5-169}$$

Then they say that because $q_1$ and $q_2$ are in the forward light cone, the time-ordered product can be replaced by a retarded commutator:

$$T \varphi^\dagger(y) \varphi(x) ~\rightarrow~ \theta(y^0-x^0)[\varphi^\dagger(y),\,\varphi(x)]\,.$$

This justification for this replacement completely eludes me. What is the mathematical reason for this?

Answer 1

I) Before we start let us briefly recall certain aspects of the formalism from Ref. 1. The Minkowski sign convention is $(+,-,-,-)$. The momentum space measure for a particle $A$ is

$$ \widetilde{dk}~:=~ \frac{d^3k}{(2\pi)^3 2\omega_{k,A}} ~=~ \frac{d^4k}{(2\pi)^3} \delta(k^2-m^2_A)\theta(k^0), $$ $$ \omega_{k,A}~:=~\sqrt{{\bf k}^2+m^2_A}~>0~. \tag{3-35}$$

Note in particular that we only integrate over non-negative $k^0\geq 0$ in the momentum space. At this point let us recall the statement below eq. (5-169) that the in and out momenta $q_i$ of particle $A$ is in the forward light cone $q^0_i\geq |{\bf q}_i| $ for $i=1,2$.

To warm-up consider a free complex scalar field $\varphi_A$ and its complex conjugate field $\varphi_A^{\dagger}$ for the particle $A$. They have Fourier expansions

$$ \varphi_A(x)~=~\int\! \widetilde{dk} \left[ a_A(k)e^{-ik\cdot x} + b_A^{\dagger}(k)e^{ik\cdot x}\right], \tag{3-78a} $$

$$ \varphi_A^{\dagger}(y)~=~\int\! \widetilde{d\ell} \left[ b_A(\ell)e^{-i\ell\cdot y} + a_A^{\dagger}(\ell)e^{i\ell\cdot y}\right]. \tag{3-78b} $$

It follows from eqs. (3-78) that

$$ \int_{\{x^0\}} \!d^3x~e^{-iq_1\cdot x}\stackrel{\leftrightarrow}{\partial^x_0} \varphi_A(x) ~\stackrel{q^0_1\geq |{\bf q}_1|}{=}~ib_A^{\dagger}(q_1),\tag{B1} $$ $$\int_{\{y^0\}} \!d^3y~e^{iq_2\cdot y}\stackrel{\leftrightarrow}{\partial^y_0} \varphi_A^{\dagger}(y) ~\stackrel{q^0_2\geq |{\bf q}_2|}{=}~-ib_A(q_2),\tag{B2}$$ cf. eq. (5-30). The creation and annihilation operators are time independent.

Now we are instead interested in an interacting complex scalar field. In this case, we define (time-dependent) asymptotic annihilation and creation operators via eqs. (B1) & (B2). For the general philosophy of the LSZ formalism, see also this & this related Phys.SE posts.

II) Let us now return to OP's question. The difference between the $T$-ordered product and the retarded commutator is

$$ T \varphi_A^{\dagger}(y)\varphi_A(x)~-~ \theta(y^0-x^0)[\varphi_A^{\dagger}(y),\varphi_A(x)] ~\stackrel{(3\text{-}87)}{=}~\varphi_A(x)\varphi_A^{\dagger}(y).\tag{D} $$

So OP's exercise is to show that (minus) the integrals on the rhs. of eq. (5-169) vanish if we replace the $T$-ordered product with the difference (D). We calculate:

$$\begin{align}\int \!d^4x~ d^4y~ & e^{i(q_2\cdot y-q_1\cdot x)} (\square_x+m_A^2)(\square_y+m_A^2) \langle p_2,\text{out}|\underbrace{ \varphi_A(x) \varphi_A^{\dagger}(y)}_{\text{difference}} |p_1 ,\text{in}\rangle \cr ~\stackrel{\begin{matrix}\text{Spatial int.}\\ \text{by parts}\end{matrix}}{=}&~\int \!d^4x~ d^4y~ e^{i(q_2\cdot y-q_1\cdot x)} ((\partial^x_0)^2 + (q^0_1)^2)((\partial^y_0)^2 + (q^0_2)^2) \cr &\times~\langle p_2,\text{out}| \varphi_A(x) \varphi_A^{\dagger}(y) |p_1,\text{in} \rangle \cr ~\stackrel{\text{p. 206}}{=}&~\int \!d^4x~d^4y~ \partial^x_0\partial^y_0\langle p_2,\text{out}|\left( e^{-iq_1\cdot x}\stackrel{\leftrightarrow}{\partial^x_0} \varphi_A(x)\right) \left(e^{iq_2\cdot y}\stackrel{\leftrightarrow}{\partial^y_0} \varphi_A^{\dagger}(y)\right)|p_1 ,\text{in}\rangle \cr ~=~&\left[\left[\langle p_2,\text{out}| \left( \int \!d^3x~e^{-iq_1\cdot x}\stackrel{\leftrightarrow}{\partial^x_0} \varphi_A(x)\right) \right.\right. \cr &\left.\left. \times~\left(\int \!d^3y~e^{iq_2\cdot y}\stackrel{\leftrightarrow}{\partial^y_0} \varphi_A^{\dagger}(y) \right) |p_1 ,\text{in}\rangle\right]_{x^0=-\infty}^{x^0=\infty}\right]_{y^0=-\infty}^{y^0=\infty} \cr ~\stackrel{(B)}{=}~&\left[\left[\langle p_2,\text{out}|b_A^{\dagger}(q_1)b_A(q_2) |p_1,\text{in} \rangle\right]_{x^0=-\infty}^{x^0=\infty}\right]_{y^0=-\infty}^{y^0=\infty} \cr ~=~&Z\langle p_2,\text{out}| \left(b_{\text{out},A}^{\dagger}(q_1)b_{\text{out},A}(q_2) +b_{\text{in},A}^{\dagger}(q_1)b_{\text{in},A}(q_2) \right. \cr &\left. -b_{\text{out},A}^{\dagger}(q_1)b_{\text{in},A}(q_2) -b_{\text{in},A}^{\dagger}(q_1)b_{\text{out},A}(q_2) \right)|p_1 ,\text{in}\rangle \cr ~=~&-Z\langle p_2,\text{out}|b_{\text{in},A}^{\dagger}(q_1)b_{\text{out},A}(q_2) |p_1,\text{in} \rangle~\stackrel{\text{p. 204}}{=}~0 ,\end{align} \tag{C}$$

where the last equality in eq. (C) follows from an argument similar to the paragraph in Ref. 1 above eq. (5-21): Well-separated particle states of different species are stable, so that their in- and out-states can be identified.

References:

1. C. Itzykson & J.-B. Zuber, QFT, 1985.

Answer 2

You obtain this by Wick's Theorem, which can be stated as $$T\{\phi_1\phi_2...\phi_n\}=N\{\phi_1\phi_2...\phi_n+\sum\text{all possible contractions of }\phi_1\phi_2...\phi_n\}$$ where N is the normal ordering operator which puts all the daggered fields on the left ( for example $N\{\phi\phi^\dagger\phi\phi\phi^\dagger\}=\phi^\dagger\phi^\dagger\phi\phi\phi$); the contraction is defined below.

In your specific case $$T\{\phi^\dagger(y)\phi(x)\}=N\{\phi^\dagger(y)\phi(x)+contraction\{\phi^\dagger(y),\phi(x)\}\}$$

and the contraction is defined as it follows

if $x^0>y^0$ $$contraction\{\phi(x),\phi(y)\}=[\phi(x)^+,\phi(y)^-]$$ if $x^0<y^0$ $$contraction\{\phi(x),\phi(y)\}=[\phi(y)^+,\phi(x)^-]$$ where $\phi^+$ and $\phi^-$ are the positive and negative frequency parts of $\phi$ so that $\phi=\phi^++\phi^-$, ($\phi^+={\phi^-}^\dagger$); the contraction definition can be rewritten in short as

$$\theta(y^0-x^0)[\phi(y)^+,\phi(x)^-]+\theta(x^0-y^0)[\phi(x)^+,\phi(y)^-]$$ in fact note that $\theta(y^0-x^0)=0$ if $y^0<x^0$ and $\theta(x^0-y^0)=0$ if $y^0>x^0$.

In your case you have $x^0<y^0$, so only the first term remains. The normal ordered first term is not present anymore because it gives a zero expectation value. Therefore $$\langle 0|T \{\phi^\dagger(y) \phi(x)\} |0\rangle=\langle 0|\theta(y^0-x^0)[\phi^\dagger(y),\,\phi(x)]\,|0\rangle$$

You can easily show that the contraction of two fields is actually given by the Feynman propagator.

To show Wick's Theorem holds for a given number of fields is just a matter of writing down explicitly the time ordered product.

A clear explanation of Wick's theorem can be found in any QFT book (look Peskin's Introduction to Quantum Field Theory, page 88; your particular case is equation 4.37)