Why are the quantum fields considered ontological?

[ilper]

TL;DR

The quantum fields stem from the quantum mechanical wavefunction, which in the standard QM is epistemological (amplitude of probability). How can they be then considered ontological?

Summary: The quantum fields stem from the quantum mechanical wavefunction, which in the standard QM is epistemological (amplitude of probability). How can they be then considered ontological?

The quantum fields stem from the quantum mechanical wavefunction, which in the standard QM is epistemological (amplitude of probability). How can they be then considered ontological?

 

[Demystifier]

The quantum fields stem from the quantum mechanical wavefunction

Do they? For instance, how does a real scalar field $\phi$ stem from a complex wave function $\psi$? Or how does non-commutativity of field operators stem from commutative wave functions?

 

[Morbert]

Summary: The quantum fields stem from the quantum mechanical wavefunction, which in the standard QM is epistemological (amplitude of probability). How can they be then considered ontological?

The analogy to a wave function would be a wave functional that reports amplitudes for field configurations. The wave functional would be epistemic, and the (measured) field configuration would be ontic.

This link might be useful: https://plato.stanford.edu/entries/quantum-field-theory/#Field

 

[ilper]

Do they? For instance, how does a real scalar field $\phi$ stem from a complex wave function $\psi$? Or how does non-commutativity of field operators stem from commutative wave functions?

Of course scalar and EM fields don't stem from WF. But the Dirac WF is epistemic as much as Schroedinger WF and it is it that is quantized and the electron-positron field is introduced. This field is widely considered ontological. [Though not connected with my question - In regard to non-commutativity IMO it is consequence from the oscillator decomposition of the field (e.g. field=sum/integral of oscillators) and birth/annihilation operators commutator follow from x,p commutators - e.g. from QM). ]

 

[ilper]

The analogy to a wave function would be a wave functional that reports amplitudes for field configurations. The wave functional would be epistemic, and the (measured) field configuration would be ontic.

This link might be useful: https://plato.stanford.edu/entries/quantum-field-theory/#Field

Well what about Dirac WF.

 

[ilper]

The analogy to a wave function would be a wave functional that reports amplitudes for field configurations. The wave functional would be epistemic, and the (measured) field configuration would be ontic.

This link might be useful: https://plato.stanford.edu/entries/quantum-field-theory/#Field

Thank you very much. This text is very useful.

 

[Morbert]

Thank you very much. This text is very useful.

As an aside, I've been looking at some literature and it seems the analogy I mentioned above results in the field interpretation inheriting some of the same problems as a particle interpretation ( see 6.3 here: https://arxiv.org/pdf/math-ph/0602036.pdf )

 

[Morbert]

Well what about Dirac WF.

I often work with the Dirac equation in a condensed matter context (though never in a high-energy context), and in that context the solutions serve the same purposes that solutions to the Schroedinger equation serve.

 

[Demystifier]

birth/annihilation operators commutator follow from x,p commutators - e.g. from QM

Can you explain how exactly the former follow from the latter?

 

[ilper]

The easiest way I recall is the article in Wiki - Quantum harmonic oscillator item 1.2. Ladder operator method.

 

[ilper]

The easeast way to Show this I recall is to go to Wiki -Quantum harmonic oscillator. Item 1.2 Ladder

Can you explain how exactly the former follow from the latter?

 

[Demystifier]

The easeast way to Show this I recall is to go to Wiki -Quantum harmonic oscillator. Item 1.2 Ladder

In this way you get one creation operator, but in quantum field theory you need an infinite number of creation operators. How do you get an infinite number of creation operators from $x$ and $p$?

 

[ilper]

his way you get one creation operator, but in quantum field theory you need an infinite number of creation operators. How do you get an infinite number of creation operators from $x$ and $p$?

I don't understand something maybe? Why one operator? Operators are functions of k -> a=a(k) and similar for birth operators, so they are continuous set from k = 0 to infinity, so the commutation is the same for every k [a(k),a+(k)]=1 (k=0, inf). You just use a different w -> k in the formula in wiki connecting a,a+ to x,p.
In case of Dirac though one needs anticommutators.

 

[Demystifier]

I don't understand something maybe? Why one operator? Operators are functions of k -> a=a(k) and similar for birth operators, so they are continuous set from k = 0 to infinity

From $x$ and $p$ you can construct only one $a$, given by the first formula at
https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator#Ladder_operator_methodThere is no dependence on $k$ in this formula.

 

[ilper]

Of course there is. w is the angular frequency. w=2πν and k=1/ν.

 

[ilper]

The analogy to a wave function would be a wave functional that reports amplitudes for field configurations. The wave functional would be epistemic, and the (measured) field configuration would be ontic.

I thought about this. IMO it brings nothing new. In the bosonic case the electromagnetic potential A (e.g. the field configuration) already gives an Amplitude for the presence of photons in x,t. From A one gets E and B and defines the energy in x.t. Deviding by E=h.ν he gets the number of photons in x.t. (e.g. the probability of QM)
It is peculiar that after quantization we loose where the photon is. It is everywhere. Because the particles are identified with single k (not a band around k) and is hence a sin wave. This way the quantized EM Looks more like probabilistic , more close to ordinary QM before the quntization.

 

[Demystifier]

Of course there is. w is the angular frequency. w=2πν and k=1/ν.

Your formula $k=1/\nu$ is true for photons, but not for a non-relativistic massive particle.

But that's not the point. The point is that $\omega$ (and hence $k$) in this wikipedia formula is a constant, not a variable. A more physical way to say this is the following. If you have only one particle with position $x$ and momentum $p$, then you have only one harmonic oscillator, while field theory is a theory of an infinite number of harmonic oscillators. Or if you still don't get it, then consider the Hamiltonian. In QM of a single particle in a harmonic potential the Hamiltonian is something like
$$H_{\rm QM}=\hbar\omega\left( a^{\dagger}a+\frac{1}{2}\right)$$
while in QFT the Hamiltonian is something like
$$H_{\rm QFT}=\int d^3k \, \hbar\omega(k)\left( a^{\dagger}(k)a(k)+\frac{1}{2}\right)$$
How would you get the integral $\int d^3k$ from QM?

 

[ilper]

Well, I don't know. Never thought about. So I suppose the 1 formulae is some idealization. As the particle is in a potential well, it's position is width of the well. So it can not be represented with just one p (k), but is a superposition Δp and you get integral. But if there is a free particle then k is sharp and the oscillator is just one. But if this is the case in QFT you also will be left with one k.

 

[Demystifier]

Well, I don't know. Never thought about. So I suppose the 1 formulae is some idealization. As the particle is in a potential well, it's position is width of the well. So it can not be represented with just one p (k), but is a superposition Δp and you get integral. But if there is a free particle then k is sharp and the oscillator is just one. But if this is the case in QFT you also will be left with one k.

It looks as if you don't know much about QFT. If you want to see how some version of QFT can be derived from QM of particles, I would suggest you to read some text on the theory of phonons (not photons) in solid state physics.