QM is Feynman path ensemble - is QFT Feynman field ensemble?

[vanhees71]

The isolated system however is given by the nucleus, the electrons and the (quantized) em. field. That's why also without an additional electromagnetic field the atom gets deexcited by spontaneous emission of photons (as you write yourself).

 

[Jarek 31]

Sure, from Feynman ensemble perspective.
But simultaneously there works also thermodynamical/statististical physicics and its perspective, e.g. having tendency to increase entropy for example by spreading energy like
"excited atom -> deexcited atom + photon"
... against QM unitarity maintaining von Neumann entropy.

I think the main problem with understanding of quantum mechanics is trying to see everything from single perspective, while physics is not that simple - there are multiple perspectives complementing each other.
We have wave-particle duality: while Feynman ensemble focuses on the wave part, Boltzmann ensemble is more focused on the particle part of the duality.

 

[vanhees71]

There is no wave-particle duality in modern quantum theory. The example with the atom is QFT at 0 temperature, i.e., there's no Boltzmann ensemble.

 

[Jarek 31]

So why we observe entropy growth in unitary quantum evolution?

 

[vanhees71]

Entropy stays constant under unitary time evolution.

 

[Jarek 31]

Exactly, while in physics around us it clearly grows.
You cannot escape statistical mechanics - mathematically universal principle of maximal entropy - and mathematically it is there in widely and succesfully used Boltzmann ensembles, even if you call it "Wick rotated Feynman".

 

[A. Neumaier]

The isolated system however is given by the nucleus, the electrons and the (quantized) em. field. That's why also without an additional electromagnetic field the atom gets deexcited by spontaneous emission of photons (as you write yourself).

Not really. Your isolated system has no excited states. The excited states of @Jarek31 are solutions of the time-independent Schrödinger equation without an external field! These do not decay but form the discrete part of the spectrum of the Hamiltonian. In order to get the excited atom to decay, the energy lost must be carried by an emitted photon. To make this possible, one has to add an interaction with the electromagnetic field!

 

[A. Neumaier]

Entropy stays constant under unitary time evolution.

Exactly, while in physics around us it clearly grows.

We observe this growth only in thermally isolated but mechanically nonisolated subsystems. These do not follow a unitary evolution but a dissipative one. In place of the Schrödinger equation one has a Lindblad equation, obtained in good approximation by contracting a larger isolated system (following unitary dynamics with constant entropy) to the system actually observed.

In most of observable physics, the system is not even approximately thermally isolated, and entropy has no reason to increase. It decreases in many chemical reactions observable at everyday temperatures; you only need to bother to do the calculations! Instead, the principle governing most of macroscopic physics is the decrease of free energy (usually Gibbs free energy, Helmholtz free energy, or enthalpy, depending on the boundary conditions).

 

[A. Neumaier]

Boltzmann ensembles, even if you call it "Wick rotated Feynman".

Boltzmann ensembles have nothing to do with "Wick rotated Feynman". A Wick rotation changes a physical system into a completely different system that bears hardly any relationship to the original one.

Boltzmann ensembles make sense only for weakly interacting collections of atoms. Already a crystal is very far from a Boltzmann ensemble.

 

[Jarek 31]

Boltzmann ensembles have nothing to do with "Wick rotated Feynman"

Except that formulas are the same, and we cannot escape mathematically universal principle of maximal entropy in Boltzmann ensemble - it is hidden in models with incomplete knowledge (not being The Wavefunction of The Universe).

There is no wave-particle duality in modern quantum theory.

Maybe let us take a look at "Imaging the atomic orbitals of carbon atomic chains with field-emission electron microscopy": https://journals.aps.org/prb/abstract/10.1103/PhysRevB.80.165404
They literally made photos of orbitals - striping electrons from single carbon atom, shape EM field to act as a lens, with matrix of detectors determine where in orbital electrons came from - getting electron densities as below (nicely seen s,p).
Doesn't it use both wave and corpuscular nature of electron?

 

[vanhees71]

Not really. Your isolated system has no excited states. The excited states of @Jarek31 are solutions of the time-independent Schrödinger equation without an external field! These do not decay but form the discrete part of the spectrum of the Hamiltonian. In order to get the excited atom to decay, the energy lost must be carried by an emitted photon. To make this possible, one has to add an interaction with the electromagnetic field!

If you want to describe spontaneous emission you have to include the quantized em. field. Here you have the atom (the nucleus, the electron, and the static Coulomb potential) as an open subsystem. That's why you can spontaneously emit one or more photons to deexcite the atom initially in an excited eigenstate of its energy. So indeed, you have to add the interaction with the em. field. Where is the contradiction?

 

[A. Neumaier]

If you want to describe spontaneous emission you have to include the quantized em. field. Here you have the atom (the nucleus, the electron, and the static Coulomb potential) as an open subsystem. That's why you can spontaneously emit one or more photons to deexcite the atom initially in an excited eigenstate of its energy. So indeed, you have to add the interaction with the em. field. Where is the contradiction?

In the fact that to define what an excited state is you need to consider a different isolated system - that without the e/m field.

 

[Jarek 31]

If only there was a simple straighforward way, like just using the maximal entropy principle - Boltzmann ensemble ...

Here they observe nice (n,l)-like quantization (distance and angular momentum) ... for classical objects with wave-particle duality: https://www.nature.com/articles/ncomms4219

 

[A. Neumaier]

Except that formulas are the same

Same formula does not imply same physics. The formula $x(t)=e^{-\alpha t}$ appears in very different domains of physics where it has very different meanings - except that something decays.

Doesn't it use both wave and corpuscular nature of electron?

The images of orbitals essentially depict a charge density - see my Theoretical Physics FAQ
Neither wave nor corpuscular nature plays a role. Though the Schrödinger equation is used to define the states in which the orbitals have the textbook form. But this is just harmonic analysis on the sphere...

hidden in models with incomplete knowledge

These are not described by the Schrödinger equation. I suggest that you read the book

• Calzetta and Hu, Nonequilibrium Quantum Field Theory

to see how quantum field theory covers incomplete knowledge. It is far from Boltzmann's qualitative picture, and agrees quantitatively with various experimentally accessible limits.

 

[Jarek 31]

Boltzmann distribution also appears in many places - universal maximal entropy principle says it is the safest assumption for incomplete knowledge situations (for fixed e.g. mean energy):

The principle of maximum entropy states that the probability distribution which best represents the current state of knowledge about a system is the one with largest entropy

Unless working with The Wavefunction of The Universe, quantum models work with incomplete knowledge situations - have hidden averaging over the unknowns.
While it can be expressed in many languages, there is always combinatorial domination of parameters maximizing entropy - you cannot escape that.

The photos of orbitals indeed nicely show e.g. s,p orbitals (consequence of wave nature resonating to standing wave described by stationary Schrödinger), but they are obtained by averaging over positions of single electrons there (corpuscular nature).
The wave-particle duality is at heart of physics, quantum mechanics - we can focus on one of these natures in some perspective/approximation, but should't forget about the compete picture.

 

[A. Neumaier]

for fixed e.g. mean energy

Fixed mean energy is very uncommon in experimental situations. Usually the temperature or the pressure is fixed, and entropy is not maximized.

 

[Jarek 31]

Mathematically Boltzmann ensemble is just principle of maximal entropy for weighted possibilities - also optimizing such mean weight.
These weights are usually energy, but e.g. for paths it should be integrated over time.

 

[A. Neumaier]

Mathematically Boltzmann ensemble is just principle of maximal entropy for weighted possibilities - also optimizing such mean weight.
These weights are usually energy, but e.g. for paths it should be integrated over time.

But what you get out depends on which expectations you assume to be given. Hence the principle is empty unless you make very strong assumptions, valid only under very restrictive conditions. Rather than toying around with your limited intuition, look first at how modern nonequilibrium thermodynamics is done! Then see whether you can add something substantial by creative modification. Starting at a point much older than 100 years is unlikely to lead you to something interesting and new...

 

[Jarek 31]

Starting at a point much older than 100 years is unlikely to lead you to something interesting and new...

Are you saying that they made combinatorics obsolete? (e.g. leading to maximal entropy principle saying what are the safest assumptions in incomplete knowledge situations).

If not, and in incomplete knowledge situations they get the same formulas, then maybe it is just new exciting dressing for long known universal mathematics.

 

[Heidi]

Can this connection be taken to field theories - can we see QFT as Feynman ensemble of fields?
With concrete field configurations corresponding to each Feynman diagram?

I am afraid that the answer is no.
In the quantum case, unlike in the classical case , the path integral is along virtual configurations not on concrete , observable ones.
in the classical case we can observe the Ising values on aset of spins on the x-axis after the Wick rotation, but with the quantum spin of a single particle, the path of the spin from t=0 to t=1 cannot be observed . if you try to look at them you get path information and you change the system.

 

[Jarek 31]

But Feynman diagrams represent very concrete particle scenarios, e.g. electron-positron annihilation on the right below:

Such charged particles have E~1/r^2 electric field, what translates into rho~1/r^4 energy density - why can't we ask about such field/energy distribution (its ensemble) in scenario represented by a given Feynman diagram?
Shouldn't it be approximately rho~1/r^4 around a charged particle?

 

[A. Neumaier]

Are you saying that they made combinatorics obsolete?

Combinatorics only works in the simplest situations - independent particles. Once there are interactions one needs more advanced diagrammatic methods.

 

[A. Neumaier]

But Feynman diagrams represent very concrete particle scenarios, e.g. electron-positron annihilation

These are virtual scenarios only; see my Insight article https://www.physicsforums.com/insights/physics-virtual-particles/ and the related articles cited there in the leading section.

 

[Jarek 31]

When there are interactions, we need to weight these combinations e.g. with energy - use Boltzmann distribution e.g. as in Ising model.

Virtual particles are for more subtle scenarios like representing Coulomb interaction with exchange of virtual photons.
So what about more concrete scenarios (Feynman diagrams) like annihilation of electron + positron - is there e.g. energy density rho~1/r^4 before annihilation as for electric field of a charge?

 

[A. Neumaier]

When there are interactions, we need to weight these combinations e.g. with energy - use Boltzmann distribution e.g. as in Ising model.

I am not talking about simulations but about producing dynamics for mixed states (representing incomplete information).

So what about more concrete scenarios (Feynman diagrams) like annihilation of electron + positron

All Feynman diagrams refer to virtual particles only.

 

[Jarek 31]

For situations with incomplete knowledge, models need some hidden assumptions about the missing information - and the safest assumptions (combinatorially dominating) are maximizing entropy.
However, there are additional constraints e.g. from energy conservation - requiring to use weighted: Boltzmann distribution.
And this is about equilibrium like the ground state - dynamics is indeed more complicated - for example which diffusion is more natural: based on standard random walk (GRW, no localization) or maximizing entropy MERW ( https://en.wikipedia.org/wiki/Maximal_entropy_random_walk ) - leading to exactly quantum ground state stationary probability distribution from Boltzmann path ensemble (for M_ij = exp(-E_ij) called transfer matrix):

Regarding Feynman diagrams, so if electron+positron meet producing photon which flies away, what would be "virtual" about such scenario?

 

[vanhees71]

Fixed mean energy is very uncommon in experimental situations. Usually the temperature or the pressure is fixed, and entropy is not maximized.

Well, thermal equilibrium is not that uncommon, and that's maximum entropy under the constraints of the situation (given mean energy and conserved particle number or conserved charge(s) lead to grand canonical ensembles with temperature and chemical potential introduced as Lagrange multipliers).

 

[A. Neumaier]

Well, thermal equilibrium is not that uncommon, and that's maximum entropy under the constraints of the situation (given mean energy and conserved particle number or conserved charge(s) lead to grand canonical ensembles with temperature and chemical potential introduced as Lagrange multipliers).

I don't know of any experiments done at fixed mean energy; thus this seems to me a fictitious situation.

At fixed temperature and volume, the Helmholtz free energy is minimized as equilibrium is approached, not the entropy maximized.

 

[A. Neumaier]

For situations with incomplete knowledge, models need some hidden assumptions about the missing information - and the safest assumptions (combinatorially dominating) are maximizing entropy.
However, there are additional constraints e.g. from energy conservation - requiring to use weighted: Boltzmann distribution.
And this is about equilibrium like the ground state - dynamics is indeed more complicated - for example which diffusion is more natural: based on standard random walk (GRW, no localization) or maximizing entropy MERW ( https://en.wikipedia.org/wiki/Maximal_entropy_random_walk ) - leading to exactly quantum ground state stationary probability distribution from Boltzmann path ensemble (for M_ij = exp(-E_ij) called transfer matrix):

The hidden information that you may assume without running into troubles depends on the boundary conditions imposed. At constant temperature, a maximal entropy ensemble never gives consistent simulations.

 

[A. Neumaier]

Regarding Feynman diagrams, so if electron+positron meet producing photon which flies away, what would be "virtual" about such scenario?

The internal particle line in the tree diagram. All internal lines represent virtual processes only. In reality, there is no ''meeting'' of electrons and positrons.