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In the free theory [tex] \mid k \rangle=a^{\dag} (\vec{k}) \mid 0 \rangle [/tex]. Then in Srednicki chapt 5, he defines time-independent operator that he says in free theory creates a particle localized in momentum space about [tex] \vec{k_1} [/tex] as:

[tex] a^{\dag}_1 \equiv \int f_1 (\vec{k}) a^{\dag}(\vec{k}) \,d^3 x [/tex]

where,

[tex]f_1 (\vec{k}) \propto \exp{ -\frac{(\vec{k}-\vec{k_1})^2} {4\sigma^2} } [/tex]

I have a few questions about this:

Firstly why is [tex] \mid k \rangle=a^{\dag} (\vec{k}) \mid 0 \rangle [/tex], creating a one particle state only valid in the free theory?

Secondly what is the point in this time independent operator? Is it something along the lines of a particle you create shouldn't have absolutely definite momentum k, as would be the case using the first creation operator, it must have a smeared momentum range about some value, to account for then uncertainty relation?

I'm just finding it weird because in a previous QFT course I studied (not using the path integral approach), I never seen this relation, and everything just seemed to work fine with only [tex] \mid k \rangle=a^{\dag} (\vec{k}) \mid 0 \rangle [/tex].

Finally how does this time independent operator create a particle localized in position space near the origin?

thanks

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# Time dependent annihilation op

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