I'm a bit confused by Weinberg's discussion of scattering. He defined the in and out states $|\Psi^{\pm}_\alpha\rangle$ with particle content $\alpha$ as states that transform under the Poincare group as a direct product of one particle states at $t\rightarrow\mp\infty$. He also shows that these states are energy eigenstates with energy $E_\alpha$.

He then says that these states are in the Heisenberg picture and hence are time independent but the appearance of the states at a time other that $t=0$ is given by the action of $U(t)$. First of all, since the states are in the Heisenberg picture, why are they evolving? I have kind of convinced myself that they're not actually time evolving, but I can't really explain it so I clearly don't really understand it, so I'd like some clarification on what's happening.

He then defines the S-matrix as $$S_{\beta\alpha} = \langle {\Psi_{\beta}}^{-} | \Psi_\alpha^+\rangle.\tag{3.2.1}$$

Then he introduces an operator $S$ such that $$\langle\Phi_\beta|S|\Phi_\alpha\rangle \equiv S_{\beta\alpha}\tag{3.2.4}$$ where $|\Phi_\alpha\rangle$ are eigenstates of the free particle Hamiltonian $H_0$. He also says that we can write $$S=\Omega^\dagger(\infty)\Omega(-\infty)\tag{3.2.5}$$ with $$\Omega(\tau)=e^{iH\tau}e^{-iH_0\tau}\tag{3.1.14}$$ which satisfies $$|\Psi^\pm_\alpha\rangle=\Omega(\mp\infty)|\Phi_\alpha\rangle.\tag{3.1.13}$$ However he says that this expression only makes sense when acting on wave packets since otherwise the states would only evolve up to a phase. The actual expression is $$\int d\alpha g(\alpha) |\Psi_\alpha^\pm\rangle =\Omega(\mp\infty)\int d\alpha g(\alpha) |\Phi_\alpha\rangle$$ where $g(\alpha)$ encodes the shape of the wave packet.

Now the obvious conclusion here is that it also only makes sense for $S$ to act on wave packets. But in his definition $$\langle\Phi_\beta|S|\Phi_\alpha\rangle \equiv S_{\beta\alpha}=\langle\Psi_\beta^-|\Psi_\alpha^+\rangle$$ the states $|\Phi_\alpha\rangle$ are energy eigenstates, and the only way I see the relation working is if we take $|\Psi^\pm_\alpha\rangle=\Omega(\mp\infty)|\Phi_\alpha\rangle$ to be an actual relation, which he's already said we can't do.

Also in every other field theory book I've seen the $S$ operator is also acting on eigenstates of the free Hamiltonian like here. So what's going?

  • $\begingroup$ $\uparrow$ Which page? $\endgroup$
    – Qmechanic
    Commented Sep 19, 2015 at 20:05
  • $\begingroup$ Pages 110, 111, 113, 114. The main issue at the end of my post takes place on page 114. @Qmechanic $\endgroup$
    – Okazaki
    Commented Sep 19, 2015 at 20:07

3 Answers 3


Let me clarify the confusion regarding the Heisenberg picture and time evolution. In Schrodinger picture, states evolve with respect to time. But in the Heisenberg picture states don't evolve with respect to time. In a coordinate system $(t,x^i)$, you choose let us say the state $|\psi(t=0)\rangle$ in the Schrodinger picture as your state $|\psi\rangle$ in the Heisenberg picture. But in another coordinate system $(x^{i},t^\prime=t+\tau)$ you have the state $|\psi(t^\prime=0)\rangle=|\psi(t=-\tau)\rangle$ as the state in the Heisenberg picture. i.e. In the Heisenberg picture, in one frame the state is $|\psi\rangle=|\psi(t=0)\rangle$, in a time translated frame with $t^\prime=t+\tau$, the state is $|\psi^\prime\rangle=|\psi(t=-\tau)\rangle$. Therefore if we apply a "time translation" operator on a state in the Heisenberg picture going from one frame to another, it is equivalent to doing a time evolution in the Schrodinger picture in a given frame. Weinberg clearly mentions that he is doing a time translation and not time evolution.

  • $\begingroup$ I see. I deleted my comment $\endgroup$ Commented Jul 1, 2021 at 18:08

Very important sentence can be found in section 3.1. It roughly goes as "to preserve manifest Lorentz invariance, state vectors don't change with time but instead describe whole spacetime history of the system". In usual Shrodinger quantum mechanics you have time dependent ket $| \psi (t) \rangle$ and time evolution operator acting as $e^{-i(t-t_0)H} | \psi (t_0) \rangle = | \psi (t) \rangle.$ In this approach $| \psi \rangle$ doesn't have $t$ label at all and $e^{itH}$ is not to be interpreted as time evolution operator but rather as time translation. Instead of taking as an input state at fixed time and giving back state at other time it acts on one possible history of the system producing other possible history where everything is shifted in time.

In states are those histories which in distant past match with the free states. So if you time shift both in state and free state to the future by an infinite amount they should become the same.

As for the question about domain of definition of $S$-matrix. It is true that the fundamental relation $$ \int \mathrm d \alpha g(\alpha) e^{-iE_{\alpha} t} | \alpha ^{\pm} \rangle \to \int \mathrm d \alpha g(\alpha) e^{-iE_{\alpha} t} | \alpha \rangle,$$ for $t \to \mp \infty$ is supposed to hold only for wavepackets. But this doesn't mean that we can't act with these $\Omega$ operators also on plane wave states.


Figured it out myself last night. I think this sorts it out but I'd love to hear other people's point of view as well!

Suppose we have states represented by wavepackets

$|\Phi_g\rangle=\int d\alpha g(\alpha) |\Phi_\alpha\rangle$ and $|\Psi^{\pm}_g\rangle=\int d\alpha g(\alpha) |\Psi^{\pm}_\alpha\rangle$

Then consider $\langle \Phi_g|S|\Phi_g\rangle$ which does make sense since the action of $S$ is on wave packets.

Expanding the states; $\langle \Phi_g|S|\Phi_g\rangle=\int d\beta d\alpha g^*(\beta) g(\alpha) \langle\Phi_\beta|S|\Phi_\alpha\rangle$

But from the relation

$\int d\alpha g(\alpha) |\Psi_\alpha^\pm\rangle =\Omega(\mp\infty)\int d\alpha g(\alpha) |\Phi_\alpha\rangle$

we also have

$\langle \Phi_g|S|\Phi_g\rangle=\langle \Phi_g|\Omega^\dagger(\infty)\Omega(-\infty)|\Phi_g\rangle=\int d\beta d\alpha g^*(\beta) g(\alpha) \langle\Psi_\beta^{-}|\Psi_\alpha^+\rangle $

Equating, we have $\int d\beta d\alpha g^*(\beta) g(\alpha) \langle\Phi_\beta|S|\Phi_\alpha\rangle= \int d\beta d\alpha g^*(\beta) g(\alpha) \langle\Psi_\beta^{-}|\Psi_\alpha^+\rangle$

Since this must be true for all smooth functions $g$ over the appropriate range, we have



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.