• 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
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Expectation values of interacting fields

+ 6 like - 0 dislike

I was motivated to ask this question by the equality claimed in equation 10.3.3 of Weinberg's volume 1 of QFT books.

My interpretation of that,

If $O_s$ is a quantum field of spin $s$, $\psi_s$ is the free field of spin $s$, $|p,\sigma\rangle$ is a one-particle state of some interacting theory and $|0\rangle$ is the vacuum state, then there should exist a constant $N$ such that,

$\langle 0|O_s|p,\sigma\rangle = \frac{N}{(2\pi)^3} \langle 0|\psi_s|p,\sigma\rangle$.

If I understand his equation 10.3.6 then that seems to say that the field $O_s$ will be said to have been "renormalized" if $N$ can be set to $1$.

So is one saying that all effect of interactions are absorbed into an overall factor at the level of matrix elements?

That sounds very surprising to me - and it seems that Weinberg claims that it follows merely from the fact that $O_s$ and $\psi_s$ have to transform under the same irreducible representation of the Poincare group.

I would be glad if someone can elaborate this point.

Also what happens if one replaces $|p,\sigma\rangle$ by multi-particle states (it's not clear to me as to what the complete set of labels is that will be required to index the continuum of multiparticle states, and clearly total momentum or/and the invariant mass is not enough).

This post has been migrated from (A51.SE)
asked Nov 19, 2011 in Theoretical Physics by Anirbit (585 points) [ no revision ]
Not all effects of interactions are absorbed into an overall factor at the matrix elements. First of all, interactions lead to a non trivial dynamics/evolutions of the field content (occupation numbers). It is renormalization that is fulfilled as renormalization of mass, charge, and the field coefficients.

This post has been migrated from (A51.SE)
It's not really surprising. Poincare invariance imposes strong constraints on the matrix elements. In case of this simple matrix element its form is completely determined by the symmetry. This is because all of the 1-particle states are related by Poincare transformations. More complicated matrix elements are also constrained by symmetry but less strongly and hence will differ from the free field case

This post has been migrated from (A51.SE)
@Squark Can you elaborate on how the Poincare invariance is fixing the matrix element to this simple form? Its not obvious that these matrix elements for an arbitrary complicatedly interacting field should have had anything to do with that of the free field.

This post has been migrated from (A51.SE)
There is no direct relation between the free field and the interacting field. The relation is that Poincare symmetry applies to both and as result this matrix element has the same form for both (since symmetry alone is sufficient to determine it modulo an overall factor).

This post has been migrated from (A51.SE)
@Squark I don't really get it. Since $O_s$ and $\psi_s$ both transform under the same representation of the Poincare group it only means that both are linear transforms of each other via some $U(\lambda,a)$ - a spin-s representation of the element of the Poincare group corresponding to a Lorent tranformation by $\Lambda$ and a space-time translation of $a$ - so i would have thought that matrix element for the interacting field is a sum of the corresponding matrix elements of the free field between the vacuum and to whatever $U(\lambda,a)$ takes the given 1-particle state to.

This post has been migrated from (A51.SE)
@Squark But the claim here seems to be stronger - as in the matrix elements are proportional and not just that the interacting matrix element is a sum over certain free field matrix elements - thats the point which I don't understand. How does the symmetry constrain to proportionality rather than each being a linear sum of terms like the other? ... somehow seems like that the irreducibility of the representation should allow for the Schur's Lemma to do something...

This post has been migrated from (A51.SE)

1 Answer

+ 1 like - 0 dislike

The answer about this equation in Weinberg's book is rather simple. You have to evaluate $\langle 0|{\cal O}_l(0)|q_1,s\rangle$ being $l$ the index due to the corresponding Lorentz transformation. This means that this matrix element should transform as an element of the Lorentz group. Weinberg's assumption is that the corresponding representation is irreducible. If this is true, being this a matrix element between the vacuum and a single particle state, the only factor that has such a property of transformation under Lorentz group is the one of the free particle state and that can be always singled out. What remains is the contribution of the interaction that amounts to a constant. This is just a requirenment of Lorentz invariance. Let us make some examples:

Scalar field:

$$\langle 0|{\cal O}_l(0)|q_1,s\rangle=\frac{1}{(2\pi)^\frac{3}{2}\sqrt{2E}}N_s$$

Spin-1/2 field:

$$\langle 0|{\cal O}_l(0)|q_1,s\rangle=\frac{1}{(2\pi)^\frac{3}{2}}u_l(q_1,s)N_f$$

and so on.

But this is a well-known theorem in quantun filed theory: It is named Lehmann-Symanzik-Zimmerman (LSZ) theorem that Weinberg puts forward in a rather discorsive way in his book at that stage. LSZ theorem permits one to pass from N-point functions to S-matrix elements and to obtain cross sections and decay rates. So, you can write down the partition function of your favourite theory, compute N-point functions for a given reaction and, from it, using LSZ theorem obtain the corresponding observable. Another good presentation of this reduction formula is given on Peskin and Schroeder staring from pag.222.

This post has been migrated from (A51.SE)
answered Nov 22, 2011 by JonLester (345 points) [ no revision ]

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