Unification between the Probability Density of the particle and the Energy Density of an electromagnetic wave

[Kartazion]

Hello.

Is there an equation or an approach in view of a unification between the wave function (amplitude of probability of presence, and volume density of probability of presence of the particle (represented by the square of its norm)), and the electromagnetic field?
 

[swansont]

the “dressed state” approach

Particle wave function has ground and excited states, with numbers of particles in each, and photon states have an occupation number. The photons and atoms can interact.

https://www.quora.com/What-are-dressed-states-in-Quantum-Optics

[Kartazion]

Apparently this proprotion is only valid for the photon, the photon being the expression of an electromagnetic wave at given wavelengths.

The probability density of the presence of photons is exactly proportional to the square of the amplitude of the electromagnetic wave and the square of the amplitude of the electromagnetic wave is equal to the energy density of the electromagnetic wave (see Poynting vector).

Lorentz long ago represented the electron as a small sphere charged with electricity, and of such dimensions that the inertia of the surrounding Coulomb field was equal to the mass of the electron. This sounded reasonable in Lorentz's day. The theorists of the time were already busy calculating the mass of the electron from the theory of the electromagnetic field. The energy of the radiation came from the kinetic energy lost by the particle under the effect of its own electromagnetic forces.
 

17 hours ago, swansont said:

the “dressed state” approach

Particle wave function has ground and excited states, with numbers of particles in each, and photon states have an occupation number. The photons and atoms can interact.

https://www.quora.com/What-are-dressed-states-in-Quantum-Optics

The dressed state approach is it valid for an electron? Indeed I only saw one relation with the photon.

[swansont]

32 minutes ago, Kartazion said:

Apparently this proprotion is only valid for the photon, the photon being the expression of an electromagnetic wave at given wavelengths.

Yes. It’s used for photon-atom systems

Quote

The probability density of the presence of photons is exactly proportional to the square of the amplitude of the electromagnetic wave and the square of the amplitude of the electromagnetic wave is equal to the energy density of the electromagnetic wave (see Poynting vector).

That’s the classical description.

Quote

Lorentz long ago represented the electron as a small sphere charged with electricity, and of such dimensions that the inertia of the surrounding Coulomb field was equal to the mass of the electron. This sounded reasonable in Lorentz's day. The theorists of the time were already busy calculating the mass of the electron from the theory of the electromagnetic field. The energy of the radiation came from the kinetic energy lost by the particle under the effect of its own electromagnetic forces.
 

That’s also classical. It gives incorrect results 

[Kartazion]

1 hour ago, swansont said:

That’s the classical description.

That’s also classical. It gives incorrect results 

I'll see if there is a solution in the second quantization.

The second quantization or canonical quantization, is a method of quantization of fields which consists, starting from a classical field such as the electromagnetic field, to consider it as a physical system and to replace the classical quantities (E, B) describing the state of the field by a quantum state and quantum physics observables. We naturally come to the conclusion that the energy of the field is quantized, each quantum representing a particle. wikipedia fr

Thank you swansont