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

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  How to prove the equivalence of two different difinitions of S-operator? i.e. $\Omega_+(\Omega_-)^\dagger= e^{i \alpha}(\Omega_-)^\dagger\Omega_+$

+ 2 like - 0 dislike
1383 views

I read there are two definitions about [S-operator](https://en.wikipedia.org/wiki/S-matrix#The_S-matrix):

The first one (e.g (8.49) in Greiner's Field Quantization) is:
$$S_{fi}\equiv \langle \Psi_p^{-}| \Psi_k^{+}\rangle$$
where $|\Psi_p^{-}\rangle$ is a state in Heisenberg picture which is $| p \rangle$ at $t=+\infty$ when you calculate the $|\Psi_p^{-}\rangle$ in Schrodinger picture , called out state. $| \Psi_k^{+}\rangle$  is a state in Heisenberg picture which is $| k \rangle$ at $t=-\infty$, called in state.

So$$S_{fi}\equiv \langle \Psi_p^{-}| \Psi_k^{+}\rangle= \langle p|(\Omega_-)^\dagger\Omega_+|k \rangle$$

In this case the S-operator $\hat S=(\Omega_-)^\dagger\Omega_+$,
where Møller operator 
$$\Omega_+ = \lim_{t\rightarrow -\infty} U^\dagger (t) U_0(t)$$
$$\Omega_- = \lim_{t\rightarrow +\infty} U^\dagger (t) U_0(t)$$
So $$S=U_I(\infty,-\infty)$$

Another definition (e.g (9.14) (9.17) (9.99) in Greiner's Field Quantization) is :
$$S_{fi}\equiv \langle \Psi_p^{-}| \Psi_k^{+}\rangle\equiv\langle \Psi_p^{-}| \hat S ^\prime |\Psi_k^{-}\rangle=\langle \Psi_p^{+}| \hat S ^\prime |\Psi_k^{+}\rangle$$
where S-operator
$\hat S ^\prime |\Psi_p^{-}\rangle =|\Psi_p^{+}\rangle$ that is $\hat S^\prime = \Omega_+(\Omega_-)^\dagger$.

It seems that these two definitions are differnt, but many textbook can derive the same dyson formula for these two S-operators. 
https://en.wikipedia.org/wiki/S-matrix#The_S-matrix

How to prove: $$\Omega_+(\Omega_-)^\dagger= e^{i \alpha}(\Omega_-)^\dagger\Omega_+$$

related to this question: http://physics.stackexchange.com/questions/105152/there-are-two-definitions-of-s-operator-or-s-matrix-in-quantum-field-theory-a

asked Apr 1, 2017 in Theoretical Physics by Alienware (185 points) [ revision history ]
edited Apr 1, 2017 by Alienware

I don't think the claimed equality holds. Probably different sources use slightly different definitions of the Moeller operators.

@ArnoldNeumaier The first definition is obiviously right. You can find the second definition still has a form of Dyson formula except a phase in many textbook, and authors have given a proof e.g. Greiner's Field Quantization  (9.14) (9.17) (9.99). Hatfield's QFT (7.43) (7.63)-(7.90) 

1 Answer

+ 1 like - 0 dislike

The first definition gives the  S-matrix in the interaction picture (a unitary operator in asymptotic space), the second formula gives the S-matrix in the Heisenberg, a unitary operator in Heisenberg space. These usually (e.g., if there is more than one channel) operate in distinct Hilbert spaces. Note that $\Omega_\pm$ go from the asymptotic Hilbert space to the Heisenberg Hilbert space (which are in general distinct). 

See Thirring's Course on Mathematical Physics, Vol. 3, Definition (3.4.23) (p.138 in the first edition).

answered Apr 3, 2017 by Arnold Neumaier (15,787 points) [ revision history ]
edited Apr 3, 2017 by Arnold Neumaier

So the S-operators in these two definition are two different operators. But are they only differ by a phase? Because I really found  in different textbook these two different operators have the same Dyson series.

@Alienware: In general they cannot be compared at all since they act on different spaces. To compare them one must identify the asymptotic Hilbert space in some way with the Heisenberg Hilbert space, and there is no canonical way to do so. To actually compute S-matrix elements to compare with cross sections or other collision data you need the interaction S-matrix. 

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$\hbar$ysicsOv$\varnothing$rflow
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
...