I have seen the scattering matrix defined using initial ("in") and final ("out") eigenstates of the free hamiltonian, with

$$\left| \vec{p}_1 \cdots \vec{p}_n \; \text{out} \right\rangle = S^{-1} \left| \vec{p}_1 \cdots \vec{p}_n \; \text{in} \right\rangle$$

so that

$$\left\langle \vec{p}_1 \cdots \vec{p}_n \; \text{out} \mid \vec{q}_1 \cdots \vec{q}_m \; \text{in} \right\rangle = \left\langle \vec{p}_1 \cdots \vec{p}_n \; \text{in} \mid S \mid \vec{q}_1 \cdots \vec{q}_m \; \text{in} \right\rangle.$$

1) What are "in" and "out" states?

2) Are they Fock states?

3) In Schrödinger or Heisenberg or interaction representation?

4) How are they related? (I believe that I see what they handwavily represent physically, but not formally.)

My main issue is that, if "in" and "out" states are one-particle eigenstates of the free hamiltonian, i.e. if $\left| \vec{p}_1 \text{out} \right\rangle$ describes a free particle with momentum $\vec{p}_1$, and $\left| \vec{p}_1 \text{in} \right\rangle$ also describes a free particle with momentum $\vec{p}_1$, then $\left| \vec{p}_1 \text{out} \right\rangle = \left| \vec{p}_1 \text{in} \right\rangle$ ... which is false. Still, books (some at least) describe these "in" and "out" states like that.

Moreover, I have seen (e.g. in Wikipedia, but also on this answer) that the scattering matrix is a map between two different Fock states, and I don't understand that.

5) Do states of the interacting system live in the same Fock space that asymptotic free states?

6) And if not, where do they live?

Understandable references would be appreciated.


1 Answer 1


*1/2. In and out states (of massive theories) are joint energy-momentum eigenstates (spanning asymptotic in and out Fock spaces) of asymptotic, free Hamiltonians (and momentum operators) associated with the bound states of a theory.

These Hamiltonians are not identical with the Hamiltonian defining the finite-time dynamics of the theory; in simple cases (ordinary quantum mechanics without bound states) they are just the Hamiltonians obtained by discarding the interaction terms. (For a rigorous discussion of this well understood situation see Chapter 3 in Volume 3 of Thirring's treatise on mathematical physics.)

*3. The representation (Schrödinger or Heisenberg or interaction) doesn't change the meaning of the states; it just changes where the dynamics is recorded.

*4. In and out states are related by the S-matrix, through the formula in your original question. For a single particle in an external potential (which is equivalent to two particles with a translation-invariant interaction, viewed in the rest frame of their center of mass), this is usually handled via the Lippman-Schwinger equation, treated in most textbooks.

In translation invariant theories, the relation $|p_1,in\rangle=|p_1,out\rangle$ (which you believe to be false) is in fact true, as single bound states do not scatter. The S-matrix is the identity on (dressed) single-particle states of a translation invartiant theory. Things get interesting when there are at least two particles around. Since only the total momentum is conserved, there is typically an exchange of momentum, and the amount is determined by the S-matrix. (The classical analogue is the change of direction when playing a golf ball across an uneven lawn - in the analogy the unevenness would be due to the influence of the second particle.)

*5. In a relativistic quantum field theory, the asymptotic Fock spaces are not equivalent to the Hilbert space in which the dynamics happens. The latter is never a Fock space (which means that the commutation relations are realized in an inequivalent manner). This is called Haag's theorem, and is the main reason for the UV divergences in perturbative QFT, where one tries to ignore this fact. See, e.g.,
Haag's theorem and practical QFT computations
Renormalization scheme independence of beta function

*6. The asymptotic spaces are obtained by a limiting procedure from the space where the finite-time dynamics happens, via Haag-Ruelle theory. In the nonrelativistic case, there is a somewhat less technical construction due to Sandhas

  • $\begingroup$ Thanks ! I have lots of things to read, but a first more question : for (4), I am tempted to say that n free particles with momenta $p_1$ ... $p_n$ long before or long after interaction are the same thing, so that $\left| p_1, p_2, \cdots p_n \; \text{in} \right\rangle = \left| p_1, p_2, \cdots p_n \; \text{out} \right\rangle$, thus leading to $S = \mathbb{I}$ without additional terms ... This is probably false, and I fail to see why it is. Perhaps after understanding what you said it will be more clear. $\endgroup$
    – A. Zerkof
    Oct 22, 2012 at 20:01
  • $\begingroup$ @A.Zerkof: I added something to 4. $\endgroup$ Oct 22, 2012 at 20:28
  • $\begingroup$ Another time, thanks for your answer. I'll allow myself one more naive question : is it correct to say that we begin form asymptotic Fock states $\left|p_1, p_2, \cdots \right\rangle$ and somehow map them to interacting states in an Hilbert space where they evolve (with some evolution operator $U$), and "after" the interaction, we map them back to Fock states $\left|p_1, p_2, \cdots \right\rangle$, so that, at the end, we have something like $B U A \left|p_1, p_2, \cdots \right\rangle$, with $B$ and $A$ the operators that map between asymptotic and interacting spaces. [cf. next comment] $\endgroup$
    – A. Zerkof
    Nov 12, 2012 at 14:07
  • $\begingroup$ @A.Zerkof: The asymptotic plane-wave states are already smeared out over infinite volume and are "noninteracting" in the sense that the scattering is a subleading correction to their behavior. This is why the Fock space is already in the theory. $\endgroup$
    – Ron Maimon
    Aug 29, 2014 at 17:14
  • 1
    $\begingroup$ @ArnordNeumaier Accroding to Bjorken and Drell, if I understand correctly, they show that in and out states are eigenstates of the full Hamiltonian of the interacting theory. They show that in and out states are eigenstates of $P^mu$ of the full theory. Peskin and Schroeder, too, tend to say something similar. But I'm throughly confused. $\endgroup$
    – SRS
    Jun 1, 2017 at 4:24

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