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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