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Density equations (light considered as reservioir)

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Robert_G
#1
Jun18-14, 11:23 PM
P: 35
Hi, there

I am reading the book called "Atom-Photon Interaction", the chapter of " Radiation considered as a Reservoir". My question is actually short, but I have to describe the background.

Following is the density equation which describes the interaction between the damped harmonic oscillator and the radiation.

[itex]\frac{d \sigma}{dt}=-\frac{\Gamma}{2}[a, b^\dagger b]_+ - \Gamma'[\sigma, b^\dagger b]_+-i(\omega_0+\Delta)[b^\dagger b, a]+\Gamma b \sigma b^\dagger + \Gamma'(b^\dagger \sigma b + b \sigma b^\dagger)[/itex].

Here, the ##\sigma## is the density operator for the harmonic oscillator, and ##b## (##b^\dagger##) is the annihilation (creation) operator of the harmonic oscillator, and all the properties of the radiation is contained in the paremeters ##\Gamma## and ##\Gamma'##. Now we want to see how the population evolves, and this is about the calculation ##\langle n| \cdot \cdot \cdot|n \rangle##. So we need to calculate the term ##\langle n|b \sigma b^\dagger|n \rangle##. The following is how I did it, and it actually can lead to the answer that printed in the book.

##\langle n| b \sigma b^\dagger|n \rangle=(b^\dagger |n\rangle)^\dagger \; \sigma \; b^\dagger|n \rangle##

Using ##b^\dagger |n \rangle = \sqrt{n+1}|n+1\rangle## can bring us

##(n+1)\sigma_{n+1,n+1}##

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My question is how about do it the other way.

##\langle n| b \sigma b^\dagger|n \rangle=\langle n | b \sigma (b^\dagger | n \rangle)##

##=\sqrt{n+1}\langle n | b \sigma|n+1\rangle##

Now, If I knew the commuter of ##[\sigma, b]## or, what's ##\sigma |n+1 \rangle##, I can go on with the calcuation, But I don't. Does anyone know how to do it in this way? Do not calculate from the left to right.




PS: It 's correct in the first way, right?
PPS: This is not a stupid question, I hope.
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Greg Bernhardt
#2
Jun26-14, 10:36 PM
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P: 9,489
I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?
rigetFrog
#3
Jul12-14, 11:54 PM
P: 109
What does the variable 'a' stand for?


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