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Lorentz Invariance of Propagator for Complex Scalar Field

  • Thread starter eudo
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  • #1
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Homework Statement



Show that

[itex][\hat{\phi}(x_1),\hat{\phi}^\dagger(x_2)] = 0 [/itex]
for [itex](x_1 - x_2)^2 < 0 [/itex]

where [itex]\phi[/itex] is a complex scalar field

Homework Equations



[itex]\hat{\phi}=\int\frac{d^3 \mathbf{k}}{(2\pi)^3 \sqrt{2\omega}}[\hat{a}(k)e^{-ik\cdot x} + b^\dagger(k)e^{ik\cdot x}][/itex]

with

[itex][\hat{a}(k),\hat{a}^\dagger(k')]=(2\pi)^3\delta^3(\mathbf{k}-\mathbf{k'})[/itex]
[itex][\hat{b}(k),\hat{b}^\dagger(k')]=(2\pi)^3\delta^3(\mathbf{k}-\mathbf{k'})[/itex]

and all other commutation relations vanish.

The Attempt at a Solution


This is problem (7.3) in Aitchison and Hey's QFT book. I have a couple questions about the solutions that are posted online here. They get eventually to the difference of two integrals:

[itex]\int\frac{d^3\mathbf{k}}{(2\pi)^3 2\omega}e^{-ik\cdot (x_1-x_2)}-\int\frac{d^3\mathbf{k}}{(2\pi)^3 2\omega}e^{ik\cdot (x_1-x_2)}[/itex]

So far so good. Then they argue that for space-like intervals, x1-x2 can be transformed to -(x1-x2) by a Lorentz transformation, and thus the second term cancels the first.

Now, I agree that each of the integrals must be Lorentz invariant, so you are allowed to evaluate each one in whichever frame you want, but I don't see how one is allowed to transform just the x1-x2 vector without also transforming the k.

And a related question: In a different problem (5.6), when we come to a similar integral, they suggest transforming to a frame where t1=t2, and since the resulting integral is an odd function of k, it must be zero. And since it's Lorentz invariant, it must be zero for all space-like intervals. It seems to me that this argument works for this problem, too, so I'm not sure why they bring up this other argument about transforming x1-x2 to -(x1-x2).

Anyone want to help explain what I'm missing?

Thanks
 

Answers and Replies

  • #2
vanhees71
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You are integrating over [itex]\vec{k}[/itex]. Thanks to the invariance of [itex]\mathrm{d}^3 \vec{k}/\omega[/itex] where [itex]\omega=\sqrt{m^2+\vec{k}^2}[/itex] you know that the result is a Lorentz scalar [tex]F(x_1-x_2)[/tex]. That means that under Lorentz transformations
[tex]F'(x')=F(x)=F(\Lambda^{-1} x).[/tex]
Thus you only need to transform [itex]x=x_1-x_2[/itex].

It's only crucial to show that for a spacelike [itex]x[/itex] you can always find a Lorentz transformation such that [itex]\Lambda x=-x[/itex]. Since you can orient your coordinate system always such that [itex]x=(0,\xi,0,0)[/itex] you just need to find a Lorentz transformation which makes out of this [itex]\Lambda x=(0,-\xi,0,0)[/itex].
 
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  • #3
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Thanks for the reply. I guess I'm missing the last part of that equality [itex]F(x) = F(\Lambda^{-1}x)[/itex]. As a test, if I define a function

[itex]F(x)=k\cdot x[/itex]

where [itex]k[/itex] is some arbitrary 4-vector, then F(x) is a Lorentz scalar. If I have some transformation such that [itex]x'=\Lambda x[/itex], then

[itex]F'(x')=k'\cdot x'=k\cdot x=F(x)\not=F(\Lambda^{-1} x)[/itex]

I understand that in the original problem we're integrating over [itex]\vec{k}[/itex], but it still seems to me we need to transform the [itex]\vec{k}[/itex].
 
  • #4
vanhees71
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Argh. That's of course a typo. The correct rule is
[tex]F'(x')=F(x)=F(\Lambda^{-1} x'),[/tex]
because
[tex]x'=\Lambda x \; \Leftrightarrow \; x=\Lambda^{-1} x.[/tex]
 
  • #5
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Right after I posted that, I figured that's what you meant. And it finally struck me why we don't have to worry about the [itex]\vec{k}[/itex]. Thanks for your help.
 

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