How can I derive Eq 9.5.11 in Scully's Quantum Optics

In summary: Is it possible to do that?In summary, the equation of motion for an operator ##(a^\dagger)^ma^nO_A## in the Heisenberg picture is given by$$\frac{d}{dt}[(a^\dagger)^ma^nO_A]=-\frac{i}{\hbar}[(a^\dagger)^ma^nO_A,H_F+H_A+H_{AF}]+\langle\frac{d}{dt}[(a^\dagger)^ma^n]\rangle_RO_A$$
  • #1
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Homework Statement
In the Scully's Quantum Optics textbook, section 9.5, an equation 9.5.11 is given to calculate the time derivative of the operator $$(a^\dagger) ^ma^nO_A$$. I try to derive this equation by myself, however, there are several problems.
Relevant Equations
Eq-9.1.1, Eq-9.3.5, Eq-9.5.5~10
Firstly, I don't know in which Picture this equation holds (if I hadn't missed some words in the previous text...). I think it may be the Heisenberg Picture. But if it is, the rest target is to prove $$\frac{i}{\hbar}[H_R+H_{FR},(a^\dagger) ^ma^nO_A]=\langle\frac{d}{dt}((a^\dagger) ^ma^n)\rangle_RO_A$$. Left side of this equation is something dependent on $\hat{b}$, but the right side is not. So how can I prove this equation?

(This is my first time to post a problem in Physics Forum, so if I did something wrong remind me please and I will correct.)
Thanks very much!
 
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  • #2
I think you have a lot of typos in the equation above it seems inconsistent. For starters, you seem to have ensemble average on the RHS, and just the pure operators on LHS. Please check and define all terms. Also, can you please present the question in a self-contained way, so that people would not need to check the book to understand what you mean?

Looking beyond this, what seems to be the problem? You have an operator, you have the Hamiltonian, you find the commutator of the former with the latter, and this is your time-derivative.

But you have already written it all that. So what is the question?
 
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  • #3
Cryo said:
I think you have a lot of typos in the equation above it seems inconsistent. For starters, you seem to have ensemble average on the RHS, and just the pure operators on LHS. Please check and define all terms. Also, can you please present the question in a self-contained way, so that people would not need to check the book to understand what you mean?

Looking beyond this, what seems to be the problem? You have an operator, you have the Hamiltonian, you find the commutator of the former with the latter, and this is your time-derivative.

But you have already written it all that. So what is the question?
Sorry for these mistakes. I will present this question in a self-contained way tomorrow or the day after tomorrow because the book is not at hand now. Thank you for your reply.
 
  • #4
The whole question is: (No typo this time I think...)
In section 9.5 of Scully's quantum optics, an atom in a damped cavity is researched. The Hamiltonian is
$$H=H_F+H_A+H_{AF}+H_R+H_{FR}$$
The subscript F represents Field, and A -> Atom, R ->reservoir and AF -> interactions between atom and field and so on.
$$H_F=\hbar\nu a^\dagger a$$
$$H_A=\frac{1}{2}\hbar\nu\sigma_z$$
$$H_{AF}=\hbar g(\sigma_+a+a^\dagger\sigma_-)$$
$$H_R=\sum_{k}\hbar\nu_kb^\dagger_k b_k$$
$$H_{FR}=\hbar\sum_k g_k(b^\dagger_ka+a^\dagger b_k)$$
where ##\sigma_z=|e\rangle\langle e|-|g\rangle\langle g|##, ##\sigma_-=|g\rangle\langle e|##,##\sigma_+=(\sigma_-)^\dagger##.
Then it says that the equation of motion for any operator of the form ##(a^\dagger)^ma^nO_A##, (where ##O_A## is an atomic operator) is given by
$$\frac{d}{dt}[(a^\dagger)^ma^nO_A]=-\frac{i}{\hbar}[(a^\dagger)^ma^nO_A,H_F+H_A+H_{AF}]+\langle\frac{d}{dt}[(a^\dagger)^ma^n]\rangle_RO_A$$
and the term ##\langle\frac{d}{dt}[(a^\dagger)^ma^n]\rangle_R##is given by an equation in the previous section:
$$\frac{d}{dt}\langle(a^\dagger)^ma^n\rangle_R=[i\nu(m-n)-\frac{\ell}{2}(m+n)]\langle(a^\dagger)^ma^n\rangle_R+\ell mn\bar{n}_{th}\langle(a^\dagger)^{m-1}a^{n-1}\rangle_R$$
However, this equation (above equation, in the previous section) is derived in Heisenberg picture using the Hamiltonian
$$H=H_0+H_1$$
$$H_0=\hbar\nu a^\dagger a+\sum_k\hbar\nu_kb^\dagger_kb_k$$
$$H_1=\hbar\sum_kg_k(b^\dagger_ka+a^\dagger b_k)$$

In current section (not the previous section), the picture is not specified. But form the form of the motion equation, I guess it is in interaction picture where ##H_0=H_R+H_{FR}##and##H_1=H_A+H_F+H_{AF}##. But I still can not derive this motion equation, and another question is that why one could apply the equation in the previous section (it is in the Heisenberg picture) to current section (in interaction picture)?

Besides, you may notice that the difference between ##\langle\frac{d}{dt}(a^\dagger)^ma^n\rangle_R## and ##\frac{d}{dt}\langle(a^\dagger)^ma^n\rangle_R##, but the original text is so.

PS. I try to find an "EDIT" button to modify my post directly, but I could not find such a button (using Chrome or Edge)...
 
  • #5
Cryo said:
I think you have a lot of typos in the equation above it seems inconsistent. For starters, you seem to have ensemble average on the RHS, and just the pure operators on LHS. Please check and define all terms. Also, can you please present the question in a self-contained way, so that people would not need to check the book to understand what you mean?

Looking beyond this, what seems to be the problem? You have an operator, you have the Hamiltonian, you find the commutator of the former with the latter, and this is your time-derivative.

But you have already written it all that. So what is the question?
Thank you. I have updated this post with more information. Now it is self-contained in some degree...
 

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