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 features!

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

123 submissions , 104 unreviewed
3,547 questions , 1,198 unanswered
4,552 answers , 19,366 comments
1,470 users with positive rep
411 active unimported users
More ...

Unitary quantum field theory

+ 7 like - 0 dislike
768 views

What do physicists mean when they refer to a quantum field theory being unitary? Does this mean that all the symmetry groups of the theory act via unitary representations? I would appreciate if one could provide some references where the definition of a unitary QFT could be found. Especially in the case where there might not be a special direction singled out as "time".

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user Daniel
asked May 24, 2013 in Theoretical Physics by UnknownToSE (500 points) [ no revision ]
retagged Mar 23, 2014
No, that's a different story (unitary reps of symmetry groups). It means that all states $|n\rangle$ in the theory have positive norm, i.e. $\langle n|n \rangle > 0$, but the full answer is more involved and should involve a discussion of reflection positivity.

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user Vibert
...(and also the S-matrix and the Froissart bound, of course).

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user Vibert
In a quantum theory (quantum mechanics or quantum field theory), unitarity means conservation of probability (or conservation of information), that is, if a state $|\psi>$ evolves in a state $|\psi'>$, you will have $<\psi|\psi> = <\psi'|\psi'>$. This means that the operator which transforms $|\psi>$ into $|\psi'>$ must be unitary. Unitarity is mandatory for the probabilistic coherence of the quantum theory.

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user Trimok

2 Answers

+ 5 like - 0 dislike

To expand on @user26374's answer a little, the phrase "A QFT is unitary" comes from the requirement that the $S$-matrix is unitary, i.e. $S S^\dagger = S^\dagger S = 1$ which is equivalent to the statement that sum of probabilities is 1. Unitarity implies several serious constraints on how a QFT can be formulated. For example, unitarity implies the Froissart bound, $\sigma \leq s \log s$ ($\sigma$ is the total cross-section and $s$ is the center of mass energy). It also implies that the propagator for a field must go no faster than $\frac{1}{p^2}$ at large $p^2$.

Unitarity is discussed in Weinberg Vol. 1.

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user Prahar
answered Jun 27, 2013 by prahar21 (520 points) [ no revision ]
+ 0 like - 1 dislike

Analogically, the unitarity of the theory of field $\psi $ in the most cases means, that field $\psi$ must transforms by the irreducible unitary representations of the Poincare group with mass $m$, momentum $p^{\mu}$ and spin $s$ (or with chirality $\lambda $ and momentum $p^{\mu}$). The unitarity requirement, of course, leads from QM postulate that the density of propability for particle is $\psi^{+}\psi$.

But if we don't connect fields with particles, we can analyze the non-unitary representations for fields.

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user PhysiXxx
answered Oct 3, 2013 by PhysiXxx (45 points) [ no revision ]
Most voted comments show all comments
The finite-dimensional irreps of the Poincare group which you refer to in your first paragraph are not unitary.

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user user1504
@user1504. Do we talk about representation by generators $M_{\mu \nu}, P_{\mu}$ with Casimir operators $P_{\mu}P^{\mu}, W_{\mu}W^{\mu}$ (for massive case)? If yes, I think that you're wrong.

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user PhysiXxx
@user1504 . Why did you assume that irreps are finite-dimensional?

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user PhysiXxx
I assumed that because you said that the field transforms by unitary representations. I usually understand this phrase to refer to the irreps in which the classical fields take their values. If this is not what you mean, then your 2nd paragraph seems broken to me.

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user user1504
@user1504 . I wrote the first paragraph in sense that if some field $\psi$, determined in Minkowski space-time, an own transformation can be represented as $$ \psi ' = e^{ia^{\mu}\hat {P}_{\mu} + \frac{i}{2}\omega^{\mu \nu}\hat {M}_{\mu \nu}}\psi , $$ where $\hat {M}_{\mu \nu}, \hat {P}^{\nu}$ are unitary generators which satisfy Poincare algebra and $a^{\mu}, \omega^{\mu \nu}$ are the local coordinates of the group of representation, it realizes the unitary representation of the Poincare group with spin $s$, mass $m$ and impulse $p^{\mu}$, or with helicity $\lambda$ and impulse $p^{\mu}$.

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user PhysiXxx
Most recent comments show all comments
And the representation, of course, is unitary

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user PhysiXxx
@user1504. So, why did you downvote?

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user PhysiXxx

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).
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
...