# Particle picture in position space in quantum field theory

When I operate $$a^{\dagger}_k$$ on vacuum, $$|0\rangle$$, I get a particle created in momentum space with a 4-momentum equal to $$(\omega_k, \vec{k})$$ where $$\omega_k=\sqrt{m^2+\vec{k}^2}$$ here I'm only talking about Klein Gordon field.

Now what confuses me is that: is creation of particle in position space non-local? Since Fourier transform of Delta function is a constant so does it mean peak in momentum space (1 particle creation) leads to spread of the particle in $$M(1,3)$$?

I know causality is maintained in QFT by imposing the commutator of operators vanish for spacelike separation but does it answer this question:

If I'm able to somehow create particle at $$x=0,t=0$$ can it be shown by calculation that this particle cannot be observed outside the light cone with it's head at origin? What observable I have to use to observe the particle position at later time in QM it would simply be $$\hat{x}$$ but I haven't seen similar kind of operator in QFT?

Also you can see there is a link between my above two question but I cannot pinpoint it clearly.

• Do you have similar issues with the standard quantum-mechanical wave function which is also non-local? – Prof. Legolasov Mar 11 '20 at 6:42
• @Prof.Legolasov well QM is by definition non relativistic so nonlocality is not much of a issue there same as instantaneous change in gravitational force in newtonian mechanics – aitfel Mar 11 '20 at 9:10
• that is not really a good argument, since as you said yourself you understand that there is no technical contradiction between locality of observations and nonlocality of wavefunctions. – Prof. Legolasov Mar 11 '20 at 9:45
• Relativistic quantum mechanics is not by definition non-relativistic! It does require a Schrodinger equation, in particular the Dirac equation and the interacting Dirac equation. It does not use the Klein Gordon field, which is non-physical, and does not give rise to a locality condition, as is required in relativistic quantum mechanics. – Charles Francis Mar 11 '20 at 11:17
• @Prof.Legolasov I'm a bit confused since I'm not getting your point when I said locality of observation I meant it in context of QFT, not QM and as for non-locality of wavefunction it got no relation to lightcone. – aitfel Mar 11 '20 at 11:54

We introduce an interaction density, $$I(x)$$, and integrate the Schrodinger equation iteratively. This leads to the perturbation expansion, or Dyson expansion, which can be written $$U(t) \approx 1+ \sum_{n=1}^\infty \frac{(-1)^n}{n!}\int{d^4x} \cdots \int{d^4x} \mathscr{T} \{I(x_1) \cdots I(x_n) \}$$
Under Lorentz transformation, the order of interactions can be changed in the time-ordered product $$\mathscr{T}$$ whenever $$x_i - x_j$$ is space-like. Conversely, Lorentz transformation cannot change the calculation of probabilities under the condition that the initial and final kets are stable states of free particles, as in scattering experiments. The locality condition, or microcausality, follows immediately as a requirement on interaction operators. It states that for any $$x,y$$ such that $$x-y$$ is space-like, the commutator of the interaction densities at $$x$$ and $$y$$ vanishes, $$[I(y), I(x)] =0$$ We fulfil this condition by constructing the interactions densities from products of field operators for the particles, which obey the same condition for bosons, and the corresponding anti-commutation relation for fermions, which necessarily appear in pairs in the interaction density.
This leads us to Dirac field operators $$\psi(x)$$ which annihilate a particle or create an antiparticle at $$x$$ and its adjoint, which does the opposite, and the photon field operator $$A(x)$$ which creates or annihilates a photon at $$x$$. The interaction for qed describes the emission/absorption of a photon by an electron/positron, and also the creation and annihilation of particle antiparticle pairs. In general, of course, one has to integrate over $$x$$ because the interaction can take place anywhere, but if an electron is constrained at a point, then the interaction must take place at that point.