• 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.


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


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,054 questions , 2,207 unanswered
5,347 answers , 22,726 comments
1,470 users with positive rep
818 active unimported users
More ...

  On the transformation rule of 'in and out‘ states

+ 2 like - 0 dislike

In p. 108 (as shown below) of Weinberg's field theory book 1, he said that the transformation rule of non-interacting particles is (3.1.1). Because of  the definition of the in and out states, they transform as (3.1.1) . So, they transform as the same way.

But in p. 116 (as shown below) , he said: "For any proper orthochronous Lorentz transformation....... as in (3.1.1) on both 'in' and 'out' states. " 

My question is, wherein I was mistaken in the first paragraph? Thank you for your patience!



asked Dec 14, 2017 in Theoretical Physics by shi_zonghua (15 points) [ no revision ]

I guess that the 'in and out' states transform as the same rule, but in different reference frame.  The inner product must be calculated in the same reference frame. But I still find other inconsistencies, and still need your opinion.

Weinberg states both times that they transform the same way, and that is correct. The in and out states are specified in the same frame, and when you change the frame both change according (3.1.1) to get their equivalent description in the new frame. Thus everything is consistent. 

Thank you, Mr. Neumaier ! But I think I  hadn't been able to show my question clearly.   The conclusion  that in and out states transform as the same way seems to merely need  the definition of in and out states. It need not the restriction on the form of the interreaction! 

So I guess there must be some mistakes on the understanding of the definition, but where is it?

1 Answer

+ 0 like - 0 dislike

I try to give an answer, welcome corrections!

When we write a state, we must notice the reference frame which the state lies in, because the form of state is diffrent in diffrent frame. Now I give 3 frames $O,O', O''$. Their correlations are $t'=t-\tau'$,  $t''=t-\tau''$, with $\tau'=-\infty$, $\tau''=+\infty$. Let's specify that the collision happens in $t=0$, in $O$ frame's view. 

When we say $\Psi^\pm$ are tranform as free particles, we actually mean $\Psi^+$ transforms as free particles just in frame $O'$, and  $\Psi^-$ transforms as free particles just in frame $O''$. 

Turn now to the inner product between in and out states $(\Psi^-, \Psi^+)$. Noticing that the product must be calculated in the same frame, we specify the frame is $O$. Now, we do a transformation $T$, and to see how the inner product changes. Please note, in frame $O$, neither of $\Psi^\pm$ transforms as free particles. However, we can use time translation to take $\Psi^+$ to frame $O'$, and act on it with $U_0(T)$, then take back to frame $O$. 

So, under transformation $T$,    $(\Psi^-, \Psi^+)$ changes into

$$(exp(+iH\infty)U_0(T)exp(-iH\infty)\Psi^-, exp(-iH\infty)U_0(T)exp(+iH\infty)\Psi^+)$$

Obviously, they are not equal, unless there are some restrictions on $H$. 

answered Dec 18, 2017 by shi_zonghua (15 points) [ revision history ]
edited Dec 20, 2017 by shi_zonghua

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:
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