Quantum optics is a reference to a c-number

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Quantum optics references a "c-number" electric field, indicating that it represents a classical value of a quantum mechanical quantity. The discussion clarifies that a c-number does not possess a corresponding Hamiltonian, as it simplifies the electric field to a constant external quantity, neglecting its degrees of freedom. Niles emphasizes that while a classical electric field has a defined Hamiltonian based on its energy density, treating it as a c-number leads to an approximation that omits these complexities.

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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.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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