Quantum optics is a reference to a c-number

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Quantum optics refers to a "c-number" electric field, indicating that it is treated as a classical value rather than a quantum operator. This classification implies that there is no corresponding Hamiltonian associated with the c-number, as it simplifies the system by ignoring the electric field's quantum degrees of freedom. While classical electric fields can have a well-defined Hamiltonian based on energy density, in this context, the field is considered a constant external quantity from a classical source. The discussion highlights the distinction between classical and quantum treatments of electric fields in quantum optics. Understanding this difference is crucial for accurately applying quantum mechanics principles to optical systems.
Niles
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quantum optics is a reference to a "c-number"

Hi

In my lecture notes on quantum optics is a reference to a "c-number" electric field, and it is stated that, since it is a c-number, there is no corresponding Hamiltonian. As far as I have understood, a c-number is a classical value of some (possibly quantum mechanical) quantity. However, how can this rule out that the c-number has a Hamiltonian? E.g., a classical electric field has a well-defined Hamiltonian given by its energy density.


Niles.
 
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Niles, This would be an approximation, in which the degrees of freedom of the electric field are ignored and the field is treated as a constant external quantity maintained by a classical source.
 
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