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

Please welcome our new moderators!

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

122 submissions , 103 unreviewed
3,497 questions , 1,172 unanswered
4,543 answers , 19,337 comments
1,470 users with positive rep
407 active unimported users
More ...

Causality and Quantum Field Theory

+ 6 like - 0 dislike
17 views

I have a problem with proof of causality in Peskin & Schroeder, An Introduction to QFT, page 28. To avoid confusion I use three vectors notation, rewriting the Eq. (2.53) for $y=0$ as follows:

$[\phi(x,t),\phi(0,0)]=\int \frac{d^3p}{(2\pi)^3}\frac{1}{2\sqrt{p^2+m^2}}\left(e^{-i\mathrm{p}.\mathrm{x}-it\sqrt{p^2+m^2}}-e^{i\mathrm{p}.\mathrm{x}+it\sqrt{p^2+m^2}}\right)$

The book goes on about how the integrand being Lorentz invariant makes this integral zero for the x out of the light cone. But I (not being a special relativity expert) want to see it more rigorously:

after changing variables $p\to-p$ in the first term, the equation simplifies to:

$[\phi(x,t),\phi(0,0)]=\int \frac{d^3p}{(2\pi)^3}\frac{-2i}{2\sqrt{p^2+m^2}}e^{i\mathrm{p}.\mathrm{x}}\sin\left(t\sqrt{p^2+m^2}\right)$

using spherical coordinates:

$[\phi(x,t),\phi(0,0)]=\int \frac{dpd\phi d\theta p^2\sin\theta}{(2\pi)^3}\frac{-i}{\sqrt{p^2+m^2}}e^{ipx\cos\theta}\sin\left(t\sqrt{p^2+m^2}\right)\\ [\phi(x,t),\phi(0,0)]=\int_0^{\infty}\frac{dpp}{(2\pi)^2}\frac{-2i}{x\sqrt{p^2+m^2}}\sin (px)\sin\left(t\sqrt{p^2+m^2}\right)$

again after another change of variables $u=\sqrt{p^2+m^2}$,

$[\phi(x,t),\phi(0,0)]=\frac{-2i}{x}\int_m^{\infty}\frac{du}{(2\pi)^2}\sin (x\sqrt{u^2-m^2})\sin\left(tu\right)$

I cannot see how this integral should be zero for $x>t$ !!! Can somebody please explain this to me?

This post imported from StackExchange Physics at 2014-03-22 17:27 (UCT), posted by SE-user Blackie
asked Mar 3, 2013 in Theoretical Physics by Blackie (30 points) [ no revision ]

2 Answers

+ 1 like - 0 dislike

I'll address your point about why the integral is Lorentz invariant, as from the comments to cduston's answer, I think this is your sticking point:

You can see the relation between a manifestly Lorentz invariant form like this $$\int d^4p \frac{e^{-ipx}}{(p^2-m^2)} \ \ \ (1) $$ and the not-so-obviously Lorentz invariant form $$ \int d^3p \frac{1}{E_{{\bf{p}}}}{e^{-ipx}} \ \ \ (2)$$ by using the identity $$\frac{1}{(p^2-m^2)}= \frac{1}{2E_{{\bf{p}}}}\{\frac{1}{(E_{{\bf{p}}}+p_0)}-\frac{1}{(E_{{\bf{p}}}-p_0)}\} $$ Here $E_{{\bf{p}}} = \sqrt{{\bf{p}}^2+m^2}$ is the on-shell time component of the momentum four vector, and $p_0$ is the "generic" time component - not necessarily on-shell.

If you substitute this in (1) and do the $p_0$ integral using the appropriate contour, you'll get (2).

What's actually going on is explained in the discussion near equation (2.40), you're doing a 4 momentum integral, but just restricting it to the mass shell using a delta function. Restriction to a mass shell is a Lorentz invariant operation, so you're maintaining Lorentz invariance throughout (even though with the three momentum integral it doesn't look like it!).

This post imported from StackExchange Physics at 2014-03-22 17:27 (UCT), posted by SE-user twistor59
answered Mar 3, 2013 by twistor59 (2,490 points) [ no revision ]
+ 0 like - 0 dislike

In the text he says the two terms vanish under $(x-y)\rightarrow -(x-y)$. In other words, there is a Lorentz transformation which takes $(x-y)\rightarrow -(x-y)$ in the second term when the separation is spacelike ($(x-y)^2<0$ using the wrong sign...). Do that, and the commutator vanishes.

This post imported from StackExchange Physics at 2014-03-22 17:27 (UCT), posted by SE-user levitopher
answered Mar 3, 2013 by levitopher (160 points) [ no revision ]
I agree with your comment in the text says when the separation is space like one can do such a transformation and get zero. Provided that the term is Lorentz invariant. I cannot see how this term is Lorentz invariant. I can see how $\int d^3pf(p)/\sqrt{p^2+m^2}$ is Lorentz invariant but not a function like $\int d^3pf(p,x,t)/\sqrt{p^2+m^2}$. Could you please explain to me why $\int d^3p e^{-ip.x-it\sqrt{p^2+m^2}}/\sqrt{p^2+m^2}$ is Lorentz invariant?

This post imported from StackExchange Physics at 2014-03-22 17:27 (UCT), posted by SE-user Blackie
The thing in the exponent is just the Lorentz-invariant inner product between the 4-momentum $p_\mu = (\vec{p}, \sqrt{\vec{p}^2 + m^2})$ and the 4-position $x_\mu = (\vec{x}, t)$ (up to whatever index and sign convention they're using).

This post imported from StackExchange Physics at 2014-03-22 17:27 (UCT), posted by SE-user Michael Brown
The previous poster is correct, but PS also mentions this a few pages earlier in the their text. I don't have it on me right now but have a look at it.

This post imported from StackExchange Physics at 2014-03-22 17:27 (UCT), posted by SE-user levitopher
@Blackie Eq.(2.40) on page 23 shows it's Lorentz invariant.

This post imported from StackExchange Physics at 2014-03-22 17:27 (UCT), posted by SE-user luyuwuli

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
...