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

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

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,853 answers , 20,624 comments
1,470 users with positive rep
501 active unimported users
More ...

Why Lorentz group for fields and Poincaré group for particles?

+ 8 like - 0 dislike
566 views

Wigner treatment associates to particles the irreps of the universal covering of the Poincaré group $$\mathbb{R}(1,3)\rtimes SL(2,\mathbb{C}).$$ Why don't we consider finite dimensional representations of this group? I understand we ask for unitarity when representing its action on states, so the representation cannot be finite dimensional. However we do consider finite dimensional representations of the Lorentz group $O(1,3)$ and associate them to fields. Why associate the Lorentz group to fields? Why do we look in this case for finite dimensional representations? What do we associate to unitary representations of the Lorentz group?

Thanks.

This post imported from StackExchange Physics at 2014-09-24 10:16 (UTC), posted by SE-user Issam Ibnouhsein
asked Sep 23, 2014 in Theoretical Physics by Issam Ibnouhsein (120 points) [ no revision ]
Related: physics.stackexchange.com/q/21801/2451 , physics.stackexchange.com/q/100844/2451 and links therein.

This post imported from StackExchange Physics at 2014-09-24 10:16 (UTC), posted by SE-user Qmechanic
About "fields", I think there is a difference between $\Phi_a$ and $\Phi_a(x)$. The former could be in a finite dimensional representation $R$ of the Lorentz group, $\Lambda(\Phi_a) = (\Lambda_R)_a^b \Phi_b$, while I don't see how the latter could be in a finite dimensional representation : it is because the Lorentz transformation acts on coordinates too, this part has to be represented by some differential operator $D_\Lambda$ : $\Lambda(\Phi_a(x)) = (\Lambda_R)_a^b \Phi_b(\Lambda^{-1}x) = (\Lambda_R)_a^b D_\Lambda (\Phi_b(x))$.

This post imported from StackExchange Physics at 2014-09-24 10:16 (UTC), posted by SE-user Trimok
For instance, in a unitary representation, we have $D_\Lambda = e^{-i \omega_{\mu\nu} L^{\mu\nu}}$, with $L_{\mu\nu} = -i(x_\mu \partial_\nu - x_\nu \partial_\mu )$. The space of functions (or operators in QFT) $\Phi_b(x)$, upon which the Lorentz transformation - and so the differential operator $D_\Lambda$ - acts, does not appear as finite-dimensional.

This post imported from StackExchange Physics at 2014-09-24 10:16 (UTC), posted by SE-user Trimok

A brief and hopefully correct and useful comment: Think of $\mathbb{R}^n$ as the coset space $ISO(n)/SO(n)$ (with similar modifications for Minkowsksi signature).  Similar ideas with Lie supergroups leads to superspace and superfields. 

1 Answer

+ 4 like - 0 dislike

Actually even the quantum fields support ("are") representations of Poincaré group acting on a suitable vector bundle whose sections are, in fact, the considered fields:

$$\phi^A(x) \to S^A_B(\Lambda) \phi^B (\Lambda x + t)$$

where $(\Lambda, t)$ is the generic element of the Poincaré group. Notice that $S$ sees only the Lorentz part of the Poincaré group and defines a vector representation on its own right.

When you fix an event in spacetime, i.e., you deal with a fiber of the vector bundle only, the Lorentz group part of the semidirect product  you wrote acts on that fiber,  by means of the representation

$$\phi^A(x) \to S^A_B(\Lambda) \phi^B (x)$$

leaving the fiber fixed. $S$ is a finite dimensional representation of the Lorentz group, since the fiber  has finite dimension as a vector space (the range of the index $A$ if finite). However this fiber does not admits a Hilbert space structure invariant under that representation, for this reason the representation is not unitary and it can be finite dimensional.

It is fundamental to notice that, in QFT, $\phi^A$ is also  an operator in the Hilbert space of the theory, and Poincaré group is a continuous symmetry of the physical system: It leaves  the  transition probabilities invariant. Essentially due to Wigner theorem, one has that this symmetry can be implemented unitarily  in the Hilbert space of the theory by means of a (strongly) continuous unitary representation  $U_{(\Lambda, t)}$.

It is natural tu assume that, under this unitary representation of Poincaré group acting in the Hilbert space, the quantum fields viewed as operators, transform covariantly with respect the other representation acting in the spacetime:

$$U_{(\Lambda,t)} \phi^A(x) U_{(\Lambda,t)}^\dagger = S^A_B(\Lambda) \phi^B (\Lambda x + t)\:.$$

answered Sep 26, 2014 by Valter Moretti (2,025 points) [ revision history ]
reshown Sep 26, 2014 by Valter Moretti

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$\hbar$ysicsOver$\varnothing$low
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).
To avoid this verification in future, please log in or register.




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

Your rights
...