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##
 
Not an expert in QM. AFAIK, Schrödinger's equation is quite different from the classical wave equation. The former is an equation for the dynamics of the state of a (quantum?) system, the latter is an equation for the dynamics of a (classical) degree of freedom. As a matter of fact, Schrödinger's equation is first order in time derivatives, while the classical wave equation is second order. But, AFAIK, Schrödinger's equation is a wave equation; only its interpretation makes it non-classical...
I am not sure if this falls under classical physics or quantum physics or somewhere else (so feel free to put it in the right section), but is there any micro state of the universe one can think of which if evolved under the current laws of nature, inevitably results in outcomes such as a table levitating? That example is just a random one I decided to choose but I'm really asking about any event that would seem like a "miracle" to the ordinary person (i.e. any event that doesn't seem to...

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