# Covariant commutation relation in Mandl and Shaw

Hi,

I am trying to study Quantum Field Theory by myself from Mandl and Shaw second edition and I am having trouble understanding the section on covariant commutation relations. I understand the idea that field at equal times at two different points commute because they cannot "communicate" with each other. This is where the equal time commutation relation comes in. However, when one takes field at two different "space-time" points, things might be different. I am having trouble following the Mathematics in Mandl and Shaw. I can write the delta function as (3.43),

Δ (x) = (-c/(2π)3) ∫(dk3 sin kx)/wk

I can also see how the equal time commutation relation comes from this. But how do they go from this to the four dimensional integral in k (3.45),
Δ (x) =(-i/(2π)3)∫(dk4 δ(k2 - μ2)ε(k0)e-ikx

Also, how can I see that the four dimensional integral is Lorentz invariant and how does Lorentz invariance implies that integral is zero for any space-like interval.

Any discussion will be much appreciated.

tiny-tim
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welcome to pf!

hi shahbaznihal! welcome to pf!
… how do they go from this to the four dimensional integral in k (3.45),
Δ (x) =(-i/(2π)3)∫(dk4 δ(k2 - μ2)ε(k0)e-ikx
(that's viewable online at http://books.google.co.uk/books?id=...M&hl=en#v=onepage&q=easily established&f=true )

have you tried substituting for the δ-function, and then integrating, as suggested in (3.47)?
Also, how can I see that the four dimensional integral is Lorentz invariant and how does Lorentz invariance implies that integral is zero for any space-like interval.
(3.45) is obviously lorentz covariant … every bit of it is lorentz covariant

and the standard way of proving it's zero is to choose the frame in which k = (ko,0,0,0) and calculate it there!

(you can't find such a frame if they can "communicate"! )

EDIT: there's a similar calculation, which you might find easier, at p.274 of Weinberg's "Quantum Theory of Fields, Vol. I", see http://books.google.co.uk/books?id=...epage&q=calculation of the propagator&f=false

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Thanks for the online reference of Mandl and Shaw.

So as a general principle can I say that if the integrand is Lorentz invariant, the integral will also be Lorentz invariant? It will be nice to actually see this proved. Any idea how I can proceed to proving it?

Yes, I tried doing that k0 integration but I cant get the original dk3 integral back. I checked other books in my library and found some useful references. I will try integrating it again and will come back.

I can prove the equal time commutation relation but I am having trouble seeing how the equal time commutation relation implies that commutation is zero for every space-like interval.

Bill_K
So as a general principle can I say that if the integrand is Lorentz invariant, the integral will also be Lorentz invariant?
Provided the domain of integration is also Lorentz invariant. In the present case the domain is all of k.
Yes, I tried doing that k0 integration but I cant get the original dk3 integral back.
δ(k02 - ωk2) = (1/2ωk)[δ(k0 + ωk) + δ(k0 - ωk)]
You get two very similar terms from this, one with k0 = ωk, the other with k0 = - ωk. They combine to give you sin.
I can prove the equal time commutation relation but I am having trouble seeing how the equal time commutation relation implies that commutation is zero for every space-like interval.
It's the Lorentz invariance that does it. If it's zero for one point outside the light cone, then by Lorentz invariance it's zero for every point outside.

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Okay, so I wrote down the original delta equation in terms of delta functions (please see attached picture), but I do not have common exponential to take out of the equation.

If I can just take the negative exponential common from the two factors I will get the equation. Kindly help me where I am going wrong, I can see that this should work Physically but I am able to see it Mathematically.

It's the Lorentz invariance that does it. If it's zero for one point outside the light cone, then by Lorentz invariance it's zero for every point outside.
So if we show that the commutation is zero at Δt = 0 (a space like event) then any other space like event will give you a zero commutation because Lorentz transformation preserves space-like events.

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I think I am only confused because the two integrals in k have different arguments in e (one is positive while other is negative) and I cannot find a way to show that they can be combined in single exponential with negative argument as it is written in the later equations. I believe it has something to do with integration running on all values of k. Is that the case?

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Guys, tiny-tim Bill_K.....

strangerep