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

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?

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.

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)

• This can't be right. Firstly, I don't understand why you say that in my case I have $x^0 < y^0$. This ordering was not simply chosen on a whim; it is somehow fixed by the fact that $q_1$ and $q_2$ are in the forward light-cone, but the reason eludes me. Secondly, the fields appearing in the reduction formula eq 5-169 above are full Heisenberg-picture fields. I have a problem with your use of Wick's theorem in that they only apply to interaction-picture fields (where they evolve according to the free part of the Hamiltonian), and is applicable in perturbation theory exclusively. Commented Jul 9, 2014 at 19:39
• I actually never read Itzykson and I actually didn't look at your problem in specific, but this substitution is something that is done all the times (almost without thinking). You may have a point however when you say that this is done in the interaction picture: you gave me a little doubt now, but I would still say Wick's can be applied in any picture (could you tell me why you say/think it cannot?). I don't really see what is the problem when you say 'is applicable in perturbation theory exclusively'... plus the above expression is true at any order, it's not a perturbative expansion. Commented Jul 9, 2014 at 20:31
• for what concerns my statement "in your case you have $x^0<y^0$" I just assumed that because of the result you got. To be fair I don't understand the sentence " $q_1$ and $q_2$ are in the forward light cone": I mean what kind of relevant information is this giving me? Can't you infer from somewhere else that $x^0<y^0$? Commented Jul 9, 2014 at 20:34
• Oh of course you can! $A(q_2)$ is a product and so it exists in time after $A(q_1)$; the momentum $q_2$ is associated with the space-time variable $y$ while $q_1$ with $x$. Therefore $y^0>x^0$. Commented Jul 9, 2014 at 20:37
• I am actually pretty sure Wick's applies to your problem, and in general regardless of the picture considered!.The whole point of deleting terms in the expansion of the time-ordered product(all non-fully-contracted products)is that any expression of the kind $<0|\phi^\dagger...\phi|0>$ is zero because $\phi$ annihilates the vacuum.So $<0|\phi^\dagger=0=\phi|0>$. And this is true in any picture. Commented Jul 9, 2014 at 20:58