I Quantization of the electromagnetic field

[Konte]

TL;DR

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

 

[HomogenousCow]

This thread asked a similar question:
https://www.physicsforums.com/threads/can-a-static-em-field-consist-of-photons.1009201/#post-6566300

 

[vanhees71]

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.

 

[Konte]

This thread asked a similar question:
https://www.physicsforums.com/threads/can-a-static-em-field-consist-of-photons.1009201/#post-6566300

Thank you, very interresting.

 

[HomogenousCow]

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.

 

[Konte]

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 : $\mu . E$