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
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  What is the 2-point correlation function of the electron field in QED?

+ 7 like - 0 dislike
1829 views

The Feynman propagator for the free electron field is the Fourier transform w.r.t. $y$ of the time-ordered 2-point VEV $\left<0\right|\mathcal{T}[\hat\psi(x)\hat\psi(x+y)]\left|0\right>$, taking $\hbar=c=1$, $$\mathcal{F}[\left<0\right|\mathcal{T}[\hat\psi(x)\hat\psi(x+y)]\left|0\right>](k)= \frac{k\!\cdot\!\gamma+m_e}{k^2-m_e^2+\mathrm{i}\epsilon}.$$ In QED, after renormalization, we obtain, valid in the infrared, when $k^2-m_e^2$ is small (given in this neat form, representing the sum of many perturbative integrals, in, for example, Appelquist & Carrazone, Phys.Rev.D 11, 2856 (1975)), $$\mathcal{F}[\left<0\right|\mathcal{T}[\hat\psi(x)\hat\psi(x+y)]\left|0\right>](k)= \frac{k\!\cdot\!\gamma+m_e}{(k^2-m_e^2+\mathrm{i}\epsilon)^{1-{\alpha_{EM}}/{\pi}}},$$ where $\alpha_{EM}\approx 1/137$. This is often called an infraparticle propagator or a dressed particle propagator (I'm rehearsing this stuff, please bear with me, and feel free to comment on anything that seems to need some change). When we move to renormalization group methods, the coupling constant becomes a function of the renormalization scale. In this case, if we measure the Fourier transform of the 2-point VEV, using the operator $\mathcal{F}[\mathcal{T}[\hat\psi(x)\hat\psi(x+y)]](k)$, the measurement scale is determined by the wave-number $k$, so I take it we can write the 2-point VEV as $$\mathcal{F}[\left<0\right|\mathcal{T}[\hat\psi(x)\hat\psi(x+y)]\left|0\right>](k) {{?\atop =}\atop\ } \frac{k\!\cdot\!\gamma+m_e}{(k^2-m_e^2+\mathrm{i}\epsilon)^{1-{\alpha_R}(k)/{\pi}}},$$ where the running coupling constant $\alpha_R(k)$ is $\alpha_R(m_e)\approx 1/137$ when $k=m_e$, at about 0.5 MeV and $\alpha_R(m_Z)\approx 1/127$ when $k=m_Z$, the mass of the $Z$ particle, at about 90 GeV (see here).

My understanding is that in QED the function $\alpha_R(k)$ is an increasing function of $k$, even to the extent that there is a Landau pole at finite inverse length $m_L$, $\alpha_R(m_L)=\infty$, but that the high-energy behavior of QED has only been calculated perturbatively, so that an analytic form for $\alpha_R(k)$ is not known. I would like, however, to look at a good article, perhaps a review, that gives as closed a form for $\alpha_R(k)$ as is currently known, in as neat a form as possible (this is an implicit question that may be too much to ask of the literature, given that almost everyone has moved on to supersymmetry, noncommutative geometry, string theory, etc., etc., and QED is obviously not empirically useful at high energy).

I would also like better to understand the relationship between the running coupling constant formalism and the Källén–Lehmann formalism. Can we equate the two, at least approximately, $$\frac{k\!\cdot\!\gamma+m_e}{(k^2-m_e^2+\mathrm{i}\epsilon)^{1-{\alpha_R}(k)/{\pi}}} {{?\atop \approx}\atop\ } \int\limits_{m_e}^\infty \frac{(k\!\cdot\!\gamma+m)f(m^2)\mathrm{d}m^2}{k^2-m^2+\mathrm{i}\epsilon}.$$ Here, $f(m^2)\ge 0$ is undefined for $m>m_L$ if there is a Landau pole in QED, but I suppose $f(m^2)$ would not have any poles and would approach zero fast enough for at least the integral $\int\limits_0^\infty f(m^2)\mathrm{d}m^2$ to exist in a well-defined theory? We could equate the two exactly if we were talking about a scalar Feynman propagator, because we could solve for $\alpha_R(k)$, but the change from $k\!\cdot\!\gamma+m_e$ to $k\!\cdot\!\gamma+m$ doesn't look good because the two expressions have different effects for different components of the Dirac spinor.

Better understanding the 2-point VEVs would of course leave the $n$-point VEVs to think about. Uggh.

Finally, there are three explicit Questions here (marked by question marks!), but I would also be interested in any ruminations from this starting point that include references.

This post imported from StackExchange Physics at 2015-04-27 11:38 (UTC), posted by SE-user Peter Morgan
asked Oct 29, 2011 in Theoretical Physics by Peter Morgan (1,230 points) [ no revision ]
I heard the electron propagator is not gauge invariant and without being involved into calculations, its look does not say much.

This post imported from StackExchange Physics at 2015-04-27 11:38 (UTC), posted by SE-user Vladimir Kalitvianski
Thanks, Vladimir. Indeed, I asked a Question that is somewhat related six months ago, physics.stackexchange.com/questions/9373/…. Agreed, it's about the details, but I'd like to find a way to get some of the details understood in a clean enough way to be able to forget about them; so that, indeed, there would be some part of the time when the look would be enough. FWIW, I think I prefer the Källén–Lehmann formalism.

This post imported from StackExchange Physics at 2015-04-27 11:38 (UTC), posted by SE-user Peter Morgan

The Kallen-Lehmann formula will hold only if there is no Landau pole,

A Landau pole (if actually nonperturbatively present, which I doubt) would mean that QED doesn't exist, hence has no correlation functions.

Thus the Kallen-Lehmann formula will hold only if there is no Landau pole. In this case, your conjectured equation is probably an identity.

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$\hbar$ysicsOverf$\varnothing$ow
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
...