In Peskin and Schroeder the LSZ-formula is given as below where the states in the $S$-matrix element are fully interacting Heisenberg states.

$$\begin{array}{l}\prod_{1}^{n} \int d^{4} x_{i} e^{i p_{i} \cdot x_{i}} \prod_{1}^{m} \int d^{4} y_{j} e^{-i k_{j} \cdot y_{j}}\left\langle\Omega\left|T\left\{\phi\left(x_{1}\right) \cdots \phi\left(x_{n}\right) \phi\left(y_{1}\right) \cdots \phi\left(y_{m}\right)\right\}\right| \Omega\right\rangle \\ \underset{\begin{array}{c}\text { each } p_{i}^{0} \rightarrow+E_{\mathbf{p}_{i}} \\ \text { each } k_{j}^{0} \rightarrow+E_{\mathbf{k}_{j}}\end{array}}{\sim}\left(\prod_{i=1}^{n} \frac{\sqrt{Z} i}{p_{i}^{2}-m^{2}+i \epsilon}\right)\left(\prod_{j=1}^{m} \frac{\sqrt{Z} i}{k_{j}^{2}-m^{2}+i \epsilon}\right)\left\langle\mathbf{p}_{1} \cdots \mathbf{p}_{n}|S| \mathbf{k}_{1} \cdots \mathbf{k}_{m}\right\rangle.\end{array}\tag{7.42}$$

However on Wikipedia and in many other books it is given with in- and out states in the $S$-matrix.

$$\left.\left\langle p_{1}, \ldots, p_{n} \text { out }\right| q_{1}, \ldots, q_{m} \text { in }\right\rangle=\prod_{i=1}^{m}\left\{-\frac{i\left(p_{i}^{2}-m^{2}\right)}{(2 \pi)^{\frac{3}{2}} Z^{\frac{1}{2}}}\right\} \prod_{j=1}^{n}\left\{-\frac{i\left(q_{j}^{2}-m^{2}\right)}{(2 \pi)^{\frac{3}{2}} Z^{\frac{1}{2}}}\right\} \tilde{G}\left(p_{1}, \ldots, p_{n} ;-q_{1}, \ldots,-q_{m}\right).$$

My question would be why Peskin and Schroeder aren't using in and out states? Are in- and out-states not necessary ind deriving the LSZ-formula?

  • $\begingroup$ They are inverses of each other. $\endgroup$
    – Qmechanic
    Oct 15, 2022 at 4:25
  • $\begingroup$ @Qmechanic I understand that I can just multiply the propagators and Z factors to the other side and then using the asymptotic equal sign since then only the residuals coincide. However why is Peskin and Schroeder not using in and out states? I edited the question $\endgroup$
    – NicAG
    Oct 15, 2022 at 13:33
  • $\begingroup$ They are, see their eq. (4.71). $\endgroup$
    – Qmechanic
    Oct 15, 2022 at 13:43
  • $\begingroup$ Why do you think they are not using in and out states? In the RHS of their equation (7.42) you explicitly have $\langle \mathbf{p}_1,\dots ,\mathbf{p}_n|S|\mathbf{k}_1,\dots,{\mathbf{k}_m}\rangle$ which is an ${\cal S}$-matrix elemented computed between some in and out states. In particular, up to choices of notation, the two equations in your post are exactly the same. $\endgroup$
    – Gold
    Oct 15, 2022 at 13:56
  • $\begingroup$ To me the $p_i$ are just some states in the Heisenberg picture which do not evolve. And we want to compute there overlap $\langle{p_1,...,p_n}|k_1,...,k_m\rangle{}$. It does not even make sense to me to write the S matrix in between. I mean we just want their overlap. However in-states are states in the interaction picture with time-evolution and we need $S$ to time evolve the one in the other state $\endgroup$
    – NicAG
    Oct 15, 2022 at 14:06

2 Answers 2


The question is not really about LSZ, but rather about the relation between in/out states and free states and how the ${\cal S}$-matrix is defined. A good reference for this is Weinberg's The Quantum Theory of Fields Chapter 3 and this Phys.SE thread. The basic idea is this: the interacting theory has a Hilbert space ${\cal H}$ and a time evolution operator $U(t)$. In the same way the free theory has a Hilbert space ${\cal H}_0$ and a evolution operator $U_0(t)$.

The assumption underlying the scattering problem is that for every free state $\Phi\in{\cal H}_0$ there is both an in state $\Psi^-$ and one out state $\Psi^+$ such that the interacting evolution of $\Psi^\pm$ as $t\to \pm \infty$ matches the free evolution of $\Phi$. In other words: $$\lim_{t\to \pm \infty}\|U(t)\Psi^\pm -U_0(t)\Phi\|=0\tag{1}.$$

We may rewrite this equation as $$\lim_{t\to \pm \infty}\|\Psi^\pm-U^\dagger(t)U_0(t)\Phi\|=0\tag{2}$$

which means that this condition is equivalent to $$\Psi^\pm = \Omega_\pm \Phi,\quad \Omega_\pm \equiv \lim_{t\to \pm\infty} U^\dagger(t)U_0(t)\tag{3}.$$

The operators $\Omega_\pm$ are called Moller operators and they map the free Hilbert space ${\cal H}_0$ onto a subset of ${\cal H}$ called the subspace of scattering states. If all you care about is the scattering problem, you may restrict your attention just to these states.

The important point here is that in/out scattering states are states in ${\cal H}$ while free states are states in ${\cal H}_0$. Up to this point I didn't use the Dirac notation just because writing norms $\|\|$ of kets gets a bit cluttered using it. To connect to your notation I now switch to the Dirac notation. In that context, the ${\cal S}$-matrix is defined to be the overlap between one in state and one out state:

$${\cal S}_{p_1,\dots,p_n;k_1,\dots, k_m}\equiv \langle{p_1,\dots, p_n;+}|k_1,\dots, k_m;-\rangle\tag{4}.$$

But now we recall that these states are defined in terms of free states by Moller operators: $$|k_1,\dots,k_m;-\rangle \equiv \Omega_-|k_1,\dots,k_m\rangle,\quad |p_1,\dots, p_n;+\rangle \equiv \Omega_+|p_1,\dots, p_n\rangle\tag{5}.$$

So the ${\cal S}$-matrix can be written in terms of free states as

$${\cal S}_{p_1,\dots,p_n;k_1,\dots, k_m}\equiv \langle p_1,\dots, p_n|\Omega_+^\dagger \Omega_-|k_1,\dots, k_m\rangle\tag{6}.$$

The operator appearing there is what we define to be the ${\cal S}$-operator, and it reads: $${\cal S}=\Omega_+^\dagger\Omega_- = \lim_{t\to \infty}\lim_{t'\to -\infty} e^{iH_0t}e^{-iH(t-t')}e^{-iH_0t'}\tag{7}$$

where we explicitly write out the evolution operators in exponential form.

So the answer to your original question is simply that the two formulas you have for the ${\cal S}$-matrix are the same because the RHS of (4) is equal to the RHS of (6) by the way the scattering problem is setup. One of the formulas is written in terms of the RHS of (4) and the other in terms of the RHS of (6), but once you identify them you see the two formulas are the same.

  • $\begingroup$ Thanks for the great answer. So to summarize. The S-matrix element is the expectation of the Moller operators in free states or just the overlap between in and out states. $\endgroup$
    – NicAG
    Oct 17, 2022 at 12:42
  • $\begingroup$ You're welcome, I'm glad it helped ! Yes, it is either the overlap between in/out states or the expectation value of the ${\cal S}$ operator built from Moller operators, ${\cal S}=\Omega_+^\dagger \Omega_-$ between the free states corresponding to the in/out states. $\endgroup$
    – Gold
    Oct 17, 2022 at 13:15

We have Heisenberg states in the far past and future of the fully interacting theory and want to take their overlap

$S=\left.\left\langle p_{1}, \ldots, p_{n} ; \text { out } (T\rightarrow \infty)\right| k_{1}, \ldots, k_{m} ; \text { in } (T\rightarrow - \infty)\right\rangle$

Going to the Schrödinger picture, we want the states to be asymptotically free ,i.e. they have the same time evolution at large times

$\lim_{t \rightarrow \infty} || U_0(t)\phi_0-U(t)\phi_{int}||=0$

The operator fulfilling this is $e^{iH_0t}e^{-iHt}$

Hence we can take the overlap of a two free states at the same time and

$S=\lim_{t \rightarrow \infty} \left\langle \phi_0(t=0)|e^{iHt} e^{-iH_0t} e^{iH_0t}e^{-iHt}|\phi_0(t=0) \right\rangle$

But isn't this just

$S=\lim_{t \rightarrow \infty} \left\langle \phi_0(t=0)|e^{iHt} e^{-iHt}|\phi_0(t=0) \right\rangle$

so we have gained nothing.

  • $\begingroup$ Hi @NicAG. What you are actually missing is that in this computation you have two limits. In any case, I have written a full answer, see if it helps you. $\endgroup$
    – Gold
    Oct 16, 2022 at 14:15

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