A calculation about coherent state

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SUMMARY

The discussion centers on the coherent state in quantum optics, specifically the representation of the coherent state as ##|\alpha \exp(-i\omega_L t)\rangle##, where ##\alpha## is a real number. The average value of the electric field operator ##E(R)## in this state is given by ##\langle\alpha \exp(-i\omega_L t)|E(R)|\alpha \exp(-i\omega_L t)\rangle=\mathscr{E}_0 \cos(\omega_L t)##, with ##\mathscr{E}_0=2 \epsilon \sqrt{\frac{\hbar \omega_L}{2 \epsilon V}} \sqrt{\langle N \rangle}## and ##\langle N \rangle = \alpha^2##. The coherent state evolves over time due to the laser frequency, and the cosine term arises from the interaction of the two exponential components in the average value calculation.

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Robert_G
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Hi, there. The following is copied from a book " atom-photon interaction " by Prof. Claude Cohen-Tannoudji, Page 415.

If the laser mode is in a coherent state ##|\alpha \exp(-i\omega_L t)\rangle## with ##\alpha## being real, Then the average value

##\langle\alpha \exp(-i\omega_L t)|E(R)|\alpha \exp(-i\omega_L t)\rangle=\mathscr{E}_0 \cos(\omega_L t)##

with

##\mathscr{E}_0=2 \epsilon \sqrt{\frac{\hbar \omega_L}{2 \epsilon V}} \sqrt{\langle N \rangle}##

##\langle N \rangle = \alpha^2##

and

##E(R)=\sqrt{\frac{\hbar \omega_L}{2 \epsilon_0 V}}\epsilon_L(a+a^{\dagger})##

I do not understand it at all. I do know some thing about the coherent state.
such as

##a |\alpha\rangle = \alpha |\alpha\rangle##

and

##|\alpha\rangle = e^{-|\alpha|^2/2} \sum_n \frac{\alpha^n}{\sqrt{n!}}|n\rangle##

But I don't understand what's going on here.

(1) Why the coherent state is ##|\alpha \exp(-i\omega_L t)\rangle##?

(2) ##\langle N \rangle = \alpha^2## means ##\langle \alpha \exp(-i\omega_L t)| N |\alpha \exp(-i\omega_L t)\rangle = \alpha^2##?

(3) Where does the ##\cos(\omega_L t)## come from in the first equation?
 
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Robert_G said:
(1) Why the coherent state is ##|\alpha \exp(-i\omega_L t)\rangle##?
The exponential is just the time-evolution with the laser frequency.

(2) ##\langle N \rangle = \alpha^2## means ##\langle \alpha \exp(-i\omega_L t)| N |\alpha \exp(-i\omega_L t)\rangle = \alpha^2##?
Write N in terms of a and ##a^\dagger## and let them operate on the different sides, that should work.

(3) Where does the ##\cos(\omega_L t)## come from in the first equation?
It is composed of the two exponentials on the left side, but I didn't check in detail which part comes from what.
 
mfb said:
The exponential is just the time-evolution with the laser frequency.

Write N in terms of a and ##a^\dagger## and let them operate on the different sides, that should work.

It is composed of the two exponentials on the left side, but I didn't check in detail which part comes from what.

In a book called quantum optics by M.O.Scully, the eigen-state of the operator ##a## is written as ##|\alpha\rangle##, Here, in the book by Claude. the eigen-state is ##|\alpha \exp(-i\omega_L t)\rangle##. This is the same thing. The latter is just using the form of ##\rho e^{i \theta}## to represent the complex eigen-value of the ##a##.

am i right?
 
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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