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 ...

  Why do quantum observables form an associative algebra in some contexts?

+ 2 like - 0 dislike
396 views

In elementary quantum mechanics, we learn that quantum observables are self-adjoint operators that act on the Hilbert space of states.

However, in more advanced context, we talk of local operators, which are supposed to be quantum observables supported locally, forms an algebra. 

First, there is the Weyl algebra appearing in deformation quantization. It is the universal enveloping algebra of the Heisenberg Lie algebra, and obtained by deforming the the space of classical observables.

Second, in CFT, we assume operator product expansion. By multiplying two fields, we obtain another.

I would be grateful to anybody who could clear things out for me.

asked Feb 20, 2022 in Theoretical Physics by WJL [ no revision ]

1 Answer

+ 1 like - 0 dislike

The usage of the term 'observable' in algebraic quantum field teory is more general than the textbook usage based on von Neumann measurements. it formalizes the common practice in quantum field theory to use this label for arbitrary operators with a common nuclear domain, which form an algebra.

Thsi is consistent with the fact that POVM measurements (generalizing von Neumann measurements) are unrelated to spectral theory and need neither self-adjointness nor even Hermiticity. For a discussion of the latter point, see my paper 'Quantum mechanics via quantum tomography'.

answered Feb 20, 2022 by Arnold Neumaier (15,787 points) [ revision history ]

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
...