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At the start of chapter 5 of Mark Srednicki's lecture notes on quantum field theory we define an operator that creates a particle that is "localised in momentum space near $\mathbf {k_1}$, and localised in position space near the origin":

$$a_1^\dagger\equiv\int d^3k\text{ }f_1(\mathbf k)a^\dagger(\mathbf k) \tag{5.6},$$

in which:

$$f_1(\mathbf k)\propto \exp[-(\mathbf k-\mathbf {k_1})^2/4\sigma^2] \tag{5.7}.$$

I do not follow how this necessarily creates a wave packet with the required properties. I see that a related question has already been asked on the site, but the answer doesn't address what I'm asking. I understand that we want the particle to be localised in position space so that its asymptotic behaviour allows us to consider its interactions perturbatively, but what specifically about the above construction makes these particles "localised in momentum/position space"?

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  • $\begingroup$ Possibly linked. $\endgroup$ Nov 19, 2020 at 20:19
  • $\begingroup$ Are you cool with $\langle {\mathbf k}| a_1^\dagger |0\rangle$? $\endgroup$ Nov 19, 2020 at 20:24
  • $\begingroup$ @CosmasZachos I think $\langle \mathbf k|a_1^\dagger|0\rangle=(2\pi)^3\omega\delta^3(\mathbf k-\mathbf k')$? Which I realise now is written on the same page, but I do understand it. $\endgroup$
    – Charlie
    Nov 19, 2020 at 20:29

1 Answer 1

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This is not an essentially QFT question, but rather a question of quantum mechanics.

The point is simply

  1. that Gaussians minimize the position/momentum uncertainty (see the answer to this question)
  2. that Gaussians go to Gaussians under Fourier transform.

Fourier transform (5.7) to find a Gaussian (in position space) which should be localised (peaking at) not too far away from zero.

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    $\begingroup$ This helps a lot, I hadn't thought about those two points, thanks! $\endgroup$
    – Charlie
    Nov 19, 2020 at 23:28

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