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When we define the S-matrix, what are "in" and "out" states?

Physics Asked by A. Zerkof on May 8, 2021

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

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

so that

$$leftlangle vec{p}_1 cdots vec{p}_n ; text{out} mid
vec{q}_1 cdots vec{q}_m ; text{in}
rightrangle
=
leftlangle vec{p}_1 cdots vec{p}_n ; text{in}
mid S mid
vec{q}_1 cdots vec{q}_m ; text{in}
rightrangle.$$

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 $left| vec{p}_1 text{out} rightrangle$ describes a free particle with momentum $vec{p}_1$, and $left| vec{p}_1 text{in} rightrangle$ also describes a free particle with momentum $vec{p}_1$, then $left| vec{p}_1 text{out} rightrangle = left| vec{p}_1 text{in} rightrangle$ … 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.

One Answer

*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 $|p_1,inrangle=|p_1,outrangle$ (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
http://projecteuclid.org/euclid.cmp/1103839514

Correct answer by Arnold Neumaier on May 8, 2021

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