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
5,082 questions , 2,232 unanswered
5,355 answers , 22,793 comments
1,470 users with positive rep
820 active unimported users
More ...

  What is an observer in QFT?

+ 10 like - 0 dislike
3137 views

In non-relativistic quantum mechanics, an observer can be roughly describe as a system with wavefunction $\vert \psi^O \rangle$ which, upon interaction with another system $\vert \psi^S\rangle$ (in some way that measures the observable $\hat A$) evolves into the following system

$$\vert \psi^O \rangle \otimes\vert \psi^S \rangle \to \sum_\alpha a_\alpha \vert \psi^O_\alpha \rangle \otimes \vert \phi_\alpha \rangle $$

with $\hat A \vert \phi_\alpha \rangle = A_\alpha \vert \phi_\alpha \rangle$ and $a_\alpha = \langle \phi_\alpha\vert \psi^S \rangle$ the probability of measuring the system in the state $\alpha$. $\vert \psi^O_\alpha \rangle$ is the way the observer will be when it has interacted with the system in the state. From the "point of view" of the observing system, the state will be

$$\vert \psi^O_\alpha \rangle \otimes \vert \phi_\alpha \rangle$$

for some $\alpha$.

The basic example works fairly well because the two systems can be decomposed in two fairly distinct rays of the Hilbert space. But in the case of a quantum field theory, how does one define an observer? Any "realistic" object (especially for interactive QFTs) will likely be a sum of every state of the Fock space of the theory, hence I do not think it is trivial to separate the system and the observer into a product of two wavefunctionals.

Is there a simple way of defining observers in QFT? Perhaps by only considering wavefunctionals on compact regions of space? I can't really think of anything that really delves into the matter so I don't have a clue.

This post imported from StackExchange Physics at 2017-10-11 16:32 (UTC), posted by SE-user Slereah
asked May 3, 2017 in Theoretical Physics by Slereah (540 points) [ no revision ]
I like to think of "observer/system" separation in the context of boundary formalism, where quantum fields live on the compact bulk region of spacetime bounded by a 3-surface where boundary states live. These states describe the interaction with the outside "observer", though in this picture the term "observer" completely loses its original meaning.

This post imported from StackExchange Physics at 2017-10-11 16:32 (UTC), posted by SE-user Solenodon Paradoxus
Nima Arkani-Hamed speaks very eloquently on the general question of observers in quantum field theory and quantum gravity. See for example pirsa.org/displayFlash.php?id=10080010

This post imported from StackExchange Physics at 2017-10-11 16:32 (UTC), posted by SE-user Bruce Greetham

2 Answers

+ 2 like - 0 dislike

As long as only scattering experiments are involved, the observer prepares the in-state at time $-\infty$ and takes a measurement on the out-state at time $+\infty$. In this setting, the observer is completely outside the QFT formalism. 

A correct account of an observer in relativistic QFT would have to model it as a very massive (many-particle) part of the quantum field localized in some region, in the spirit of the nonrelativistic treatment by in the work by Allahverdyan et al. reviewed by me here.  I haven't seen anything like this for the relativistic case. 

On the other hand, papers by Peres and Terno (e.g., https://arxiv.org/abs/quant-ph/0212023) discuss relativistic quantum mechanics (not QFT) for multiple observers in different Lorentz frames.

answered Nov 27, 2017 by Arnold Neumaier (15,787 points) [ revision history ]

I generally agree with Arnold's opinion and comments, and I would like just underline that "the observer" makes sure that the incident particles have the necessary energy/momentum and other properties, that the target is also has the necessary properties, that the collisions happen in a deep vacuum (no other obstacles), that the measuring devices detect the scattering products correctly, etc., etc. The observer makes a huge preparatory work, accompanying work, and result processing work. In other words, he makes a permanent work keeping the necessary and sufficient conditions for a given experiment. Only with all that we may be sure of the initial and final states, withing the experimental uncertainties established with the observer. As you can see, "observer" includes experimentalists, theorists, staff and stuff, all working to make sure that a (thus simplified) QFT is applicable to the studied processes.

+ 2 like - 0 dislike

In the book

the author amplifies two points (right at the beginning, first page of the preface on page "v" in volume 1):

  1. The relevance of the Peierls bracket for the spacetime-covariant formulation of QFT;
  2. its implication for a good theory of observers and measurement in QFT, which he attributes to Bohr-Rosenfeld 1933.

There is no doubt about the relevance of the Peierls bracket: This is the covariant form of the Poisson bracket (explained in detail in "Mathematical QFT - 8. Phase space"); and the positive frequency part of its integral kernel is nothing but the vacuum 2-point function (explained in "Mathematical QFT - 9. Propagators").

Chapters 7 and 8 of DeWitt's book (volume 1) mean to lay out a theory of measurement and observers in QFT based on this.  I don't feel quite qualified to review this here, but if you are interested, I would suggest to take a look.

answered Nov 30, 2017 by Urs Schreiber (6,095 points) [ revision history ]
edited Nov 30, 2017 by Urs Schreiber

Does the Peierls bracket exist for interacting systems too? In particular, does the statement about 2-point functions still apply? It sounds very interesting.

The positive frequency part of the Peierls bracket is the vacuum 2-point function of a free (linear) quantum field theory corresponding to the associated linear classical theory. Did you mean this? I don't know of any nonlinear version of this statement, and your PhysicsForums article is too big to see quickly the precise statement you intend here.

The Peierls bracket exist generally, also for interacting field theories. I recommend Khavkine 14.

Answering whether it's relation to the vacuum 2-point function goes beyond perturbation theory would seem to require non-perturbative theory which does not exist at this point.

Unfortunately, the book by deWitt has no significant contribution to the original question; no definition of a model observer is given. Chapters 7 and 8 of volume 1 are about classical measurement and heuristic quantum measurement of a primitive form. All problems are swept under the many-worlds carpet where no measurement is ever done - since there is no mechanism for splitting the worlds, and without splitting there are no definite measurement results. (See also my critique of Everett's many-worlds = relative state interpretation.)

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$\hbar$y$\varnothing$icsOverflow
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
...