I noticed that in Srednicki's derivation of the LSZ-formula the expression (chapter 5) for the creation (and also later for the annihilation) operator by the field operator:

$$a^\dagger(\mathbf{k}) = -i \int d^3x e^{ikx}\stackrel{\leftrightarrow}{\partial}_0 \phi(x)\tag{5.2}$$

is used although this expression is only valid for a free field theory whereas the LSZ-formula applies for interacting fields. He just introduces the derivation with

"Let us guess that this still works in the interacting theory."

The only difference that he makes with respect to the free theory is that the creation operators are time-dependent and then writing:

$$\begin{align}a_1^\dagger(+\infty) -& a_1^\dagger(-\infty)= \int_{-\infty}^{\infty} dt \partial_0 a_1^\dagger(t)\cr =& -i \int d^3k f_1(\mathbf{k})\int d^4x \,\partial_0(e^{ikx}\stackrel{\leftrightarrow}{\partial}_0 \phi(x))\end{align}\tag{5.10}$$


$$a_1^\dagger \equiv \int d^3k f_1(\mathbf{k}) a^\dagger(\mathbf{k})\tag{5.6}$$

with $f_1(\mathbf{k})$ describing the form of the wave packet. Further on in the chapter he apparently fixes this problem by requiring that

$$\langle p|\phi(0)| 0\rangle=1.\tag{5.25b}$$

I really would like to know how this condition makes the application of this free-field formula possible.

Actually I am tempted to consider $a_1^\dagger(\pm\infty)$ as asymptotic creation operators creating in- and out-states (as it is done in chapter 16 of Bjorken&Drell) and replace $\phi(x)$ in the expression (2) by $\frac{1}{\sqrt{Z}}\phi(x)$, (with $\sqrt{Z}=\langle p|\phi(0)| 0\rangle$ ) but this is actually not allowed because the replacement $\phi(x) \rightarrow \frac{1}{\sqrt{Z}}\phi(x)$ is only allowed on matrix element level.

For instance Edelhäuser & Knochel (other books as Peskin & Schroeder do that too, but I did not compare their computation to the Srednicki one carefully) sandwich the calculation between multi-particle states, so I wonder if Srednicki left simply out the sandwiching particle states in expression (2). But even then I feel uncomfortable as the asymptotic creation operators are supposed to be time-independent, however their time-dependence kind of is assumed in the use of the fundamental theorem of calculus.


2 Answers 2


OP asks good questions. Let us try to sketch the logic of the LSZ reduction formula.

  1. In the Heisenberg picture, a free real field $\hat{\varphi}(x)$ has a Fourier expansion $$ \hat{\varphi}(x)~=~\int \widetilde{dk}\left[\hat{a}({\bf k})e^{ik\cdot x}+ \hat{a}^{\dagger}({\bf k})e^{-ik\cdot x} \right], \tag{3.19} $$ where $k\cdot x= {\bf k}\cdot {\bf x}-\omega_{\bf k}t$ and $\omega_{\bf k}=\sqrt{{\bf k}^2+m^2}$. The annihilation and creation operators, $\hat{a}({\bf k})$ and $\hat{a}^{\dagger}({\bf k})$, are time-independent operators, i.e. they belong to the Schrödinger picture. Eq. (5.2) [or equivalently eq. (3.21)] can easily be derived from the Fourier expansion (3.19).

  2. The ideal free theory and its fields are envisioned to live in Fock Hilbert spaces ${\cal H}_{\rm in}$ and ${\cal H}_{\rm out}$.

  3. Now we are interested in an interacting field theory with an interacting field $\hat{\varphi}(x)$ that lives in a Hilbert space ${\cal H}$.

  4. NB: We emphasize that interactions are not turned off at asymptotic regions: the Lagrangian does not depend explicitly on spacetime!

  5. For an interacting field $\hat{\varphi}(x)$, one can in principle still perform a spatial Fourier expansion at each time $t$, however the spatial Fourier transform can no longer be identified with annihilation and creation operators. Instead, we shall define so-called (time-dependent) asymptotic creation and annihilation operators $\hat{a}^{\dagger}({\bf k},t)$ and $\hat{a}({\bf k},t)$ by the Klein-Gordon (KG) inner product of eq. (5.2) [and the Hermitian conjugate eq.]. The definition is extended to an asymptotic creation operator $\hat{a}^{\dagger}_1(t)\equiv\hat{a}^{\dagger}(f_1,t) $ for a Gaussian wave packet in eq. (5.6).

  6. We would like to identify the temporal asymptotic 1-particle states in the interacting theory with the corresponding 1-particle states in the free theory.

  • For this, we assume the existence of a translation-invariant vacuum state $| \Omega \rangle$ of the full theory, $$\hat{P}_{\mu}|\Omega\rangle~=~0, \qquad \langle\Omega|\Omega\rangle~=~1.$$ This implies that $$\langle \Omega | \partial_{\mu}\hat{\varphi}(x) | \Omega \rangle~=~i\langle \Omega | [\hat{\varphi}(x),\hat{P}_{\mu}] | \Omega \rangle~=~0.$$

  • We assume the existence of 1-particle states of the full theory $$\hat{\bf P}|{\bf p}\rangle~=~{\bf p}|{\bf p}\rangle, \qquad \hat{\bf H}|{\bf p}\rangle~=~\omega_{\bf p}|{\bf p}\rangle,\qquad \langle{\bf p}|{\bf p}^{\prime}\rangle ~=~(2\pi)^32\omega_{\bf p}~\delta^3({\bf p}\!-\!{\bf p}^{\prime}), $$ in the Heisenberg picture. (To avoid clutter of notation, let's assume only 1 species of particles for simplicity.) Let us also introduce a Gaussian wave packet $$|f_1\rangle ~\equiv~ \int \!d^3k~ f_1({\bf p})|{\bf p}\rangle. $$

  • We should get rid of tadpoles $$ \langle \Omega | \hat{\varphi}(x) | \Omega \rangle~=~0, \tag{5.25a} $$ which causes transitions between the vacuum and 1-particle states. Then $$ \langle \Omega | \hat{a}^{\dagger}({\bf k},t) | \Omega \rangle~=~0, \qquad \langle \Omega | \hat{a}^{\dagger}_1(t) | \Omega \rangle~=~0. $$

  • We should normalize 1-particle states
    $$ \langle {\bf p} | \hat{\varphi}(x) | \Omega \rangle~=~\sqrt{Z}e^{-ip\cdot x}, \tag{5.25b} $$ cf. the Källén-Lehmann spectral representation. (Refs. 1 & 2 furthermore assume that $Z=1$.)

  • One may show that $$ \hat{a}^{\dagger}({\bf p},\pm \infty)| \Omega \rangle ~=~ |{\bf p} \rangle, \qquad \hat{a}({\bf p},\pm \infty)| \Omega \rangle ~=~0, $$ cf. e.g. this Phys.SE answer by user1379857. (A possible relative phase is conventionally set equal to 1.) Note that the normalization is $\langle {\bf p} | {\bf k} \rangle = Z (2\pi)^3 2 \omega \delta^3 ({\bf p}-{\bf k}) $.

  • Similarly, for fixed total 3-momentum ${\bf p}$, multi-particle states are more energetic than the 1-particle state $| {\bf p} \rangle$, and hence automatically suppressed for asymptotic times due to the Riemann-Lebesque Lemma.

  1. Finally let us return to OP's main question. Eq. (5.10) is fine as it is because eq. (5.2) is valid for interacting fields. Alternatively, one can postpone the use of the fundamental theorem of calculus (FTC) to a later stage as follows: $$ \begin{align} \hat{a}^{\dagger}_1(\infty) &- \hat{a}^{\dagger}_1(-\infty)\cr ~=~&\sqrt{Z}\left[\hat{a}^{\dagger}_{1,{\rm out}}(\infty) - \hat{a}^{\dagger}_{1,{\rm in}}(-\infty)\right]\cr ~\stackrel{(5.2)}{=}&-i \sqrt{Z}\int\! d^3k f_1({\bf k}) \int\! d^3x\left[ e^{ik\cdot x}\stackrel{\leftrightarrow}{\partial}_0\hat{\varphi}_{\rm out}(\infty,{\bf x})- e^{ik\cdot x}\stackrel{\leftrightarrow}{\partial}_0\hat{\varphi}_{\rm in}(-\infty,{\bf x}) \right]\cr ~=~& -i \int\! d^3k f_1({\bf k}) \int\! d^3x\left[ e^{ik\cdot x}\stackrel{\leftrightarrow}{\partial}_0\hat{\varphi}(\infty,{\bf x})- e^{ik\cdot x}\stackrel{\leftrightarrow}{\partial}_0\hat{\varphi}(-\infty,{\bf x}) \right]\cr ~\stackrel{\rm FTC}{=}& -i \int\! d^3k f_1({\bf k}) \int\! d^4x~\partial_0\left[ e^{ik\cdot x}\stackrel{\leftrightarrow}{\partial}_0\hat{\varphi}(x) \right]\cr ~=~&\ldots. \end{align}\tag{5.10'} $$


  1. M. Srednicki, QFT, 2007; eqs. (3.19-21) + chapter 5. A prepublication draft PDF file is available here.

  2. S. Coleman, QFT lecture notes, arXiv:1110.5013; p. 158-162.

  3. C. Itzykson & J.-B. Zuber, QFT, 1985; eq. (3-44) + p. 202-208.

  4. M.E. Peskin & D.V. Schroeder, An Intro to QFT, 1995; sections 7.1-7.2.

  • $\begingroup$ Nice! BTW I also liked the discussion of the same point in the book by Tom Banks. $\endgroup$
    – user21299
    Commented Nov 19, 2020 at 22:35
  • $\begingroup$ I very much liked the answer @Qmechanic but you say that the derivation of the LSZ reduction formula does not use the assumption that the interactions are turned off at asymptotic times $t\to\pm\infty$. But Itzykson & Zuber claims (lines just above Eq. no. 5-14a, page 202) that switching off interactions is required. What is your take on this? $\endgroup$ Commented Mar 9, 2021 at 14:52
  • $\begingroup$ The glib answer is that I&Z apparently didn't get the memo :) $\endgroup$
    – Qmechanic
    Commented Mar 9, 2021 at 16:10
  • $\begingroup$ Correction to the answer (v5). The name Lebesque should be Lebesgue. $\endgroup$
    – Qmechanic
    Commented Nov 1, 2021 at 19:12
  • $\begingroup$ @Qmechanic Can you suggest how this proof can be extended so that it is valid for arbitrary spin? $\endgroup$
    – Hiamph
    Commented Apr 7, 2023 at 17:05

The LSZ formula is built in the Heisenberg picture. The fields $\phi (x)$ are defined as a sum over creation $a^\dagger _p (t)$ and annihilation $a_p (t)$ operators as
$\phi (x) = \phi (\vec x, t) = \int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_p}} [a_p (t) exp(-ipx) + a^\dagger _p (t) exp(ipx)]$
$\eta_{\mu \nu} = diag(1, -1, -1, -1)$ metric tensor in Minkowski spacetime
$\omega_p = \sqrt{\vec p^2 + m^2}$
$[a_k, a^\dagger_p] = (2 \pi)^3 \delta^3(\vec p - \vec k)$ equal-time commutation relations
$a^\dagger_p |0\rangle = \frac{1}{\sqrt{2 \omega_p}} |\vec p\rangle$ creation of a particle with momentum $\vec p$

The fields are operators that create states at some particular time, however the creation and annihilation operators at time $t$ are in general different from those at some other time $t'$. An interacting Hamiltonian $H$ will rotate the basis of creation and annihilation operators. If $H$ is time-independent, we have
$a_p (t) = exp(iH(t - t_0)) a_p (t_0) exp(-iH(t - t_0))$
$a^\dagger _p (t) = exp(iH(t - t_0)) a^\dagger _p (t_0) exp(-iH(t - t_0))$
just as
$\phi (x) = exp(iH(t - t_0)) \phi (\vec x, t_0) exp(-iH(t - t_0))$
where $t_0$ is an arbitrary reference time where the interacting fields match onto the free fields.

As for the condition $\langle p|\phi (0)| 0 \rangle = 1$, you get this by setting $\phi (\vec x, t) = \phi (\vec 0, 0) = \phi (0)$ in the definition of the fields.


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