How Does Quantum Field Theory Preserve Causality?

[Jamister]

causality is a property that follows from QFT (or maybe even from the Dirac equation?)
the simplest example is a scalar field with Lagranagian $\mathcal {L}=\frac 1 2 \partial^\mu \phi \partial_\mu \phi - \frac 1 2 m^2 \phi^2$
The fact causality is preserved follows by that [ϕ(x),ϕ(y)]=0 outside the light cone. The common explanation of that is that ϕ(x) is a measurement operator and that two measurements with x,y outside are commuting and therefore do not effect each other.
but what is the $\phi(x)$ operator in QM? it does not correspond to any operator in QM.
It seems like this explanation of causality not complete.

 

[PeterDonis]

casualty

I think you mean "causality".

what is the $\phi(x)$ operator in QM?

By "QM" do you mean non-relativistic QM? If so, non-relativistic QM is derived from QFT in the non-relativistic approximation, so QFT's account of causality applies to non-relativistic QM as well.

It seems like this explanation of casualty not complete.

Sure it is. See above.

 

[Jamister]

I think you mean "causality".

thanks I fixed that

By "QM" do you mean non-relativistic QM? If so, non-relativistic QM is derived from QFT in the non-relativistic approximation, so QFT's account of causality applies to non-relativistic QM as well.

but that's not an explanation for my question.

 

[PeterDonis]

that's not an explanation for my question

Why not?

 

[Jamister]

Why not?

because $\phi(x)$ is not a measurement operator in QM, and I don't know what's the meaning of it as measurement operator in QFT. So all this explanation doesn't have any meaning to me. That's my question.

 

[PeterDonis]

I don't know what's the meaning of it as measurement operator in QFT

Have you looked at any introductory treatments of QFT that discuss this?

 

[Jamister]

Have you looked at any introductory treatments of QFT that discuss this?

yeah I did. They all have the same explanation which is not stratifying in my opinion.

 

[PeterDonis]

They all have the same explanation

I'm not asking about their explanation of causality. I'm asking about their explanation of what the operator $\phi(x)$ represents in QFT. Have you had trouble understanding their explanation of that? If so, which references and what did you have trouble with?

 

[Demystifier]

what is the $\phi(x)$ operator in QM?

In QM one works in the Hilbert space ${\cal H}^{(n)}$ of a fixed number $n$ of particles. QFT works in a larger Hilbert space
$${\cal H}=\bigoplus_{n=0}^{\infty} {\cal H}^{(n)}$$
that combines all possible numbers of particles. The operator $\phi(x)$ is an operator in ${\cal H}$ that cannot be presented as an operator in anyone of ${\cal H}^{(n)}$.

 

[vanhees71]

Well, of course you can also do QFT in non-relativistic QT. A bit misleadingly that's known as the "2nd-quantization formalism". That's even a very convenient choice when dealing with many-body systems of bosons or fermions, because the formalism takes care of the correct (anti-)symmetrization of the state kets.

In the case that you have a Hamiltonian such that particle number is conserved, the 1st-quantization and the 2nd-quantization formalism are exactly equivalent, since then the time evolution doesn't change the particle number, i.e., when you start in the 2nd-quantization formalism with a state of definite particle number, time evolution keeps you in $\mathcal{H}^{(n)}$. So all results from the 1st-quantization formalism are in this special case exactly the same as those from the 2nd-quantization formalism.

The distinction between relativistic and non-relativistic QFT is that in the relativistic case there's no interacting QFT where particle numbers are preserved. That has a deeper reason in the representation theory of the Poincare group together with the demand of formulating local (microcausal) QFTs with a stable ground state (Hamiltonian bounded from below). It forces you, already for free particles, to introduce creation operators in the mode decomposition of the quantized free fields (in front of the modes with negative frequency_) in addition to the annihilation operator (Feynman-Stueckelberg trick). The meaning of this is to avoid "particles running backward in time" (though this is ironically often said in popular-science writings about relativistic QFT).

That's why in relativistic QFT it's impossible to have the numbers of individual particle species to be conserved since the interaction terms in the Lagrangian/Hamiltonian have to be built from the complete field operators at one space-timepoint to preserve the microcausality condition for the Hamilton density. If particles and antiparticles are distinguishable and if there's a conserved charge (as in the Standard Model where the gauge structure demands to have several conserved charges) there's only charge conservation but not the conservation of particle numbers of individual particles. E.g. in QED you have the conserved electric charge, but you can create as many electron-positron pairs as you wish provided you have the energy to collied particles to do so. If the particles and antiparticles are indistinguishable (strictly neutral particles like e.g., photons) there's not even the possibility of heaving a conserved current for them. Then you can create and destroy the particles without any other restriction than energy, momentum, and anuglar-momentum conservation. That's why you can create and/or destroy easily as many photons as you wish. You don't even need any minimal energy for that, because photons are in addition massless!

In non-relativistic QT you the causality structure is much more robust. Here "actions at a distance" are a common thing, and you have spatially non-local interactions, like with the Coulomb interaction between two charged particles. In the 2nd quantization formalism the corresponding interaction term reads
$$\hat{V}=\frac{1}{2} \sum_{\sigma_1,\sigma_2} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}_1\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}_2 \frac{q_1 q_2}{4\pi |\vec{x}_1-\vec{x}_2|} \hat{\psi}_1^{\dagger}(\vec{x}_1,\sigma_1) \hat{\psi}_2^{\dagger}(\vec{x}_2,\sigma_2) \hat{\psi}_2(\vec{x}_2,\sigma_2)\hat{\psi}_1(\vec{x}_1,\sigma_1).$$
Since in this non-relativsitic case the field operators are pure annihilation operators (in the interaction picture) the total number of particles is conserved, i.e., you destroy two particles and create two particles. In physical terms there's only elastic Coulomb scattering no creation and annihilation processes included.