I Quantization of the electromagnetic field

Konte
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quantization, electromagnetic field
Hi everyone,

It is about the quantization of the electromagnetic field. The expression of field E and B are defined with:
-the annihilation a- and creation a+ operators, and the frequency ω.
So my question is: how does these fields must be expressed if they where "static"? I mean, how the electrostatic and magnetostatic fields must be expressed in quantized version?

Thank you.
Konte
 
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Konte said:
Summary:: quantization, electromagnetic field

Hi everyone,

It is about the quantization of the electromagnetic field. The expression of field E and B are defined with:
-the annihilation a- and creation a+ operators, and the frequency ω.
So my question is: how does these fields must be expressed if they where "static"? I mean, how the electrostatic and magnetostatic fields must be expressed in quantized version?

Thank you.
Konte
In quantum theory a "situation" is described by the state, not by the operators describing observables.
 
Here's a rough sketch of how this problem is solved, using a scalar field instead of a vector field for simplicity.

We start with the KG Hamiltonian with a spatially-dependent source ##J(\mathbf{x})##:
$$H = \int d^{3}x ~\frac{1}{2}\pi^2 + \frac{1}{2}(\nabla \phi)^2 + \frac{1}{2}m^2\phi^2 + J\phi.$$ Rewrite the Hamiltonian in terms of a new field ##\psi## related to ##\phi## via
$$\phi = \psi + K(\mathbf{x}),$$ where ##K## is some fixed function. If we choose ##K## such that
$$-\nabla^2 K + m^2 K +J = 0,$$then (after integrating out surface terms) the Hamiltonian will simplify to
$$H = \int d^{3}x ~\frac{1}{2}\pi^2 + \frac{1}{2}(\nabla \psi)^2 + \frac{1}{2}m^2\psi^2 + ...,$$ where the ommited terms depend only on ##K##. The important point is that now the linear term is gone and we can expand via creation-annihilation operators as if the field were free.

The groundstate of the field in the presence of ##J##, which I will call ##|\Omega\rangle##, is defined by $$a_\mathbf{k} |\Omega\rangle = 0$$ where ##a_\mathbf{k}## is the annihilation operator associated with the field ##\psi.## Now ##a## is related to the annihilation operators of the original field ##\phi##, which I will call ##b##, by some relation along the lines of $$a_\mathbf{k} = b_\mathbf{k} - f(\mathbf{k})$$ where ##f(\mathbf{k})## is some c-number. You can figure out what ##f(\mathbf{k})## should be exactly from the second expression above if you feel so inclined (I don't). In any case the definition of the groundstate ##|\Omega\rangle## can now be rewritten as
$$b_\mathbf{k} |\Omega\rangle = f(\mathbf{k})|\Omega\rangle,$$ which means that ##|\Omega\rangle## is some coherent state of the free theory.
 
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HomogenousCow said:
Here's a rough sketch of how this problem is solved, using a scalar field instead of a vector field for simplicity.

We start with the KG Hamiltonian with a spatially-dependent source ##J(\mathbf{x})##:
$$H = \int d^{3}x ~\frac{1}{2}\pi^2 + \frac{1}{2}(\nabla \phi)^2 + \frac{1}{2}m^2\phi^2 + J\phi.$$ Rewrite the Hamiltonian in terms of a new field ##\psi## related to ##\phi## via
$$\phi = \psi + K(\mathbf{x}),$$ where ##K## is some fixed function. If we choose ##K## such that
$$-\nabla^2 K + m^2 K +J = 0,$$then (after integrating out surface terms) the Hamiltonian will simplify to
$$H = \int d^{3}x ~\frac{1}{2}\pi^2 + \frac{1}{2}(\nabla \psi)^2 + \frac{1}{2}m^2\psi^2 + ...,$$ where the ommited terms depend only on ##K##. The important point is that now the linear term is gone and we can expand via creation-annihilation operators as if the field were free.

The groundstate of the field in the presence of ##J##, which I will call ##|\Omega\rangle##, is defined by $$a_\mathbf{k} |\Omega\rangle = 0$$ where ##a_\mathbf{k}## is the annihilation operator associated with the field ##\psi.## Now ##a## is related to the annihilation operators of the original field ##\phi##, which I will call ##b##, by some relation along the lines of $$a_\mathbf{k} = b_\mathbf{k} - f(\mathbf{k})$$ where ##f(\mathbf{k})## is some c-number. You can figure out what ##f(\mathbf{k})## should be exactly from the second expression above if you feel so inclined (I don't). In any case the definition of the groundstate ##|\Omega\rangle## can now be rewritten as
$$b_\mathbf{k} |\Omega\rangle = f(\mathbf{k})|\Omega\rangle,$$ which means that ##|\Omega\rangle## is some coherent state of the free theory.
I have a little question that answer can surely help me to understand more: how do we express the interaction term (in the Schrodinger equation) with electrostatic field in the "Stark effect" case when we treat the field as quantized field ? In semi-quantum case, it is only expressed as the product of dipolar moment with classical electrostatic field : ##\mu . E##
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!

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