I'm slowly working through Srednicki's QFT book and I had a question about section 3 (canonical quantization of scalar fields). At one point, he shows that the creation and annihilation operators ##a(\mathbf{k})## and ##a(\mathbf{k}')## are time-independent via the equation:(adsbygoogle = window.adsbygoogle || []).push({});

$$a(\mathbf{k}) =i\int{d^3x e^{-ikx}[i\partial_0 \varphi(x) +\omega\varphi(x)]}$$

I thought that was kind of odd given that the theory was supposed to be relativistic and one should expect to see time and space treated in an equal way. In one sense, we have the ##e^{ikx}## factor, where ##kx## is the dot product of the position and momentum four-vectors, so this part is at least nominally time dependent (except maybe it's constant, since it's a dot product?). On the other hand, tracking down where ##a(\mathbf{k})## came from in the first place, we have the solution to the KG equation, which Srednicki gives as the sum of plane waves weighted with arbitrary functions of the wave vector:

$$\varphi(\mathbf{x},t) =\int{ \frac{d^3k}{f(k)} [a(\mathbf{k})+b(\mathbf{k})]e^{i\mathbf{k} \cdot \mathbf{x}-i\omega t}}$$

My question is this: since the Klein Gordon equation is symmetric in time and space, why aren't the ##a(\mathbf{k})##'s functions of frequency as well? e.g., why don't we have:

$$\varphi(\mathbf{x},t) =\int{ \frac{d^4k}{f(k)} [a(\mathbf{k},\omega)+b(\mathbf{k},\omega)]

e^{i\mathbf{k} \cdot \mathbf{x}-i\omega t}}$$

Does this have something to do with the on shell condition? Srednicki mentions it briefly in this section but doesn't elucidate why it's important. Thanks for any help you can give!

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# I QFT operators time/space asymmetry?

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