How to understand the electric-field operator?

[Jeffrey Yang]

I know the positive field operator E⁺ is actually an annihilation operator a while the negative field E^- is a creation operator a⁺.

I also learned that the absorption process can be represented as E^-E⁺, which should be the number of photons n accroding to the principle of ladder operator. Also, the emission is E⁺E^-, whose eigenvalu however is n+1.

How to understand this extra 1 here? Why emitted number is n+1 rather than n? Is that 1 the vacuum field? But should the vacuum field cannot released through emission?

Thanks a lot

 

[vanhees71]

The 1 is spontaneous emission. It's an effect of quantizing the electromagnetic field.

 

[Jeffrey Yang]

The 1 is spontaneous emission. It's an effect of quantizing the electromagnetic field.

Hi, thanks for your answer.
Well, what you are talking about make sense to me. However, it is always confusing to me that if the the 1 or 1/2 (symmetrized operator) is the spontaneous emission term, why we always ignore this term when we discuss about thermal radiation? Let's say thermal radiation is obviously a spontaneous emission however whose statistical occupation is described by the Boltzmann function $1/(exp(E/kT)-1)$ rather than $1/(exp(E/kT)-1)+1/2$ or just $1/2$ ?

 

[vanhees71]

The vacuum energy is by definition set to 0 (normal ordering of the Hamiltonian). The spontaneous-emission term is important to get the correct Bose distribution as the equilibrium distribution of the corresponding Boltzmann equation. The equilibrium condition is that you have as many photon production as absorption rates, and in the production part you have both the spontaneous as well as the induced part. Formally this reflects itself in the Bose-enhancement factors in the Boltzmann collision term (here collision of photons with the walls of the cavity). These precisely lead to the Bose-Einstein rather than the (relativistic) Maxwell distribution as the equilbrium distribution!

 

[Jeffrey Yang]

The vacuum energy is by definition set to 0 (normal ordering of the Hamiltonian). The spontaneous-emission term is important to get the correct Bose distribution as the equilibrium distribution of the corresponding Boltzmann equation. The equilibrium condition is that you have as many photon production as absorption rates, and in the production part you have both the spontaneous as well as the induced part. Formally this reflects itself in the Bose-enhancement factors in the Boltzmann collision term (here collision of photons with the walls of the cavity). These precisely lead to the Bose-Einstein rather than the (relativistic) Maxwell distribution as the equilbrium distribution!

Em..I'm not sure whether I understand your reply correctly. Here are my mind:

Let's say if now we are talking about the field in the space causing light emission of a certain thermalized source (heat body, semiconductor ...). As we said, the emission operator has an eigenvalue $n+1 = \frac{1}{exp(E/kT) - 1} + 1$. In the above equation, photon number $n$ has been replaced by the Bose-Einstein occupation number with a temperature $T$ (defied by the environment). Now, the term $\frac{1}{exp(E/kT) - 1}$ actually drives the stimulated emission while the term $1$ drives the spontaneous emission. If now, in the aspect of thermalized source, the excitation element such as dipole, also follow a Bose-Einstein occupation but under a temperature $T_{s}$ defined by the source itself. Finally, the overall probability of spontaneous emission is $1 \times \frac{1}{exp(E/kT_{s}) - 1} = \frac{1}{exp(E/kT_{s}) - 1}$, the probability of simulated emission is $\frac{1}{exp(E/kT_{c}) - 1} \times \frac{1}{exp(E/kT_{s}) - 1}$.

 

[vanhees71]

The Bose-Einstein distribution is not an eigenvalue of the photon-number operator but the phase-space distribution function of photons in thermal equilibrium. So your entire posting doesn't make any sense.

 

[Jeffrey Yang]

The Bose-Einstein distribution is not an eigenvalue of the photon-number operator but the phase-space distribution function of photons in thermal equilibrium. So your entire posting doesn't make any sense.

Well, maybe my express is not professional. What I intend to express is just because in thermal equilibrium, finally the ensemble averaged photon number $n$ should equal to the Bose-Einstein probability. What I'm thinking about is how these occupation probability is finally transferred to the emitted light considering the detailed physical meaning of spontaneous and stimulated emission term from the $\mathbf{E}^{+}\mathbf{E}^{-}$ operator