Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

206 submissions , 164 unreviewed
5,103 questions , 2,249 unanswered
5,355 answers , 22,794 comments
1,470 users with positive rep
820 active unimported users
More ...

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

+ 6 like - 0 dislike
2462 views

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.

This post imported from StackExchange Physics at 2014-08-26 21:32 (UCT), posted by SE-user A. Zerkof
asked Oct 22, 2012 in Theoretical Physics by A. Zerkof (30 points) [ no revision ]
related: physics.stackexchange.com/q/41206/7924

This post imported from StackExchange Physics at 2014-08-26 21:32 (UCT), posted by SE-user Arnold Neumaier

1 Answer

+ 6 like - 0 dislike

*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.

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
http://projecteuclid.org/euclid.cmp/1103839514

This post imported from StackExchange Physics at 2014-08-26 21:32 (UCT), posted by SE-user Arnold Neumaier
answered Oct 22, 2012 by Arnold Neumaier (15,787 points) [ no revision ]
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.

This post imported from StackExchange Physics at 2014-08-26 21:32 (UCT), posted by SE-user A. Zerkof
@A.Zerkof: I added something to 4.

This post imported from StackExchange Physics at 2014-08-26 21:32 (UCT), posted by SE-user Arnold Neumaier
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]

This post imported from StackExchange Physics at 2014-08-26 21:32 (UCT), posted by SE-user A. Zerkof

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOve$\varnothing$flow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...