# Maxwell-Bloch equations and operator-products

1. Aug 5, 2012

### Niles

Hi

I have a question regarding the Maxwell-Bloch equations, which I read about in a dissertation. I've looked through many papers online, but no one offers an answer. It is a very specific question, which is based on the first 2-3 lines on page 13 of this thesis: http://mediatum2.ub.tum.de/download/652711/652711.pdf

It is regarding the statement above equation (2.23a): The author states that
$$\left\langle {o^\dagger} \right\rangle = \left\langle {o} \right\rangle^*$$
which I agree 100% with. Now the author uses this relation to go from an equation for $\left\langle {a} \right\rangle$ to an equation for $\left\langle {a^\dagger a} \right\rangle$. However this is what I don't understand, because
$$\left\langle a \right\rangle \left\langle a \right\rangle ^* = \left\langle a \right\rangle \left\langle {a^\dagger } \right\rangle = \left\langle {a^\dagger } \right\rangle \left\langle a \right\rangle \ne \left\langle {a^\dagger a} \right\rangle$$
Does anyone know what the assumption is for making this factorization? I would be very happy to recieve some feedback.

Best,
Niles.

2. Aug 5, 2012

### Cthugha

Isn't this a consequence of investigating the steady state photon number of a coherent field?

If I get you right, you are interested in some expectation value:
$\langle \alpha|\hat{a}^\dagger \hat{a}|\alpha\rangle$

Couldn't you now just throw in the identity operator
$\sum_{n=1}^\infty |n\rangle \langle n|$
between the two operators and get the result you want or did I misunderstand you completely?

Last edited: Aug 5, 2012
3. Aug 5, 2012

### Niles

Thanks for replying. If we are looking at a cavity-QED system consisting of atoms + cavity, shouldn't the expectation value be taken over the combined states instead of the field states?

I don't think you have misunderstood me. Basically I am trying to figure out why the author of the dissertation is allowed to go from (2.22a) --> (2.23a) and (2.22b) --> (2.23b) by using the relation in my OP. I also thought that it may be because it is steady-state, but I don't see why the relation should be valid in steady-state.

Thanks for helping out.

Best,
Niles.

4. Aug 6, 2012

### Cthugha

No, I do not think that is really a cavity-"QED" treatment, but rather the classical approach to it. Sometimes the field is modeled as classical (or as a coherent state which is an eigenstate of the photon annihilation operator and therefore behaves very classical), but the operator formalism is kept for convenience.

Did you check reference 3 of the thesis? It is available on the ArXiv as well and is a bit more detailed:
http://arxiv.org/abs/quant-ph/0304015

5. Aug 6, 2012

### Niles

Ah, I see. I didn't know that, thanks.

Thanks for the link. I just checked it out, and he does explain the factorization in the case where the two operators act on different subsystems (page 4). But in our case the two operators work on the same system.

I found another paper where they use that trick without accounting for it: http://arxiv.org/abs/1105.2373. On page 2 they derive the Maxwell-Bloch equations, and find the steady-state solutions. So I guess they solve for <σ -> in (3) and then find <σ z> in (4). This they use in (2) to find <a> and thus <a><a>*. But <a><a>* = |<a>|2 I don't get, and it is getting very frustrating.

$$\sum\limits_n {\left\langle {\alpha |a^\dagger \left| n \right\rangle \left\langle n \right|a|\alpha } \right\rangle } = \sum\limits_n {\left\langle {n|a\left| \alpha \right\rangle ^* \left\langle n \right|a|\alpha } \right\rangle }$$
Can you throw me a hint to what to do from here?

Best,
Niles.

6. Aug 6, 2012

### Cthugha

Remember that coherent states are eigenstates of the annihilation operator, so that
$\hat{a}|\alpha\rangle=\alpha|\alpha\rangle$

7. Aug 6, 2012

### Niles

Thanks, so it becomes
$$\sum\limits_n {\left\langle {n|\alpha } \right\rangle ^* \alpha ^* \alpha \left\langle {n|\alpha } \right\rangle } = \sum\limits_n {\left\langle {\alpha |n} \right\rangle \left\langle {n|\alpha } \right\rangle \alpha ^* \alpha } = \alpha ^* \alpha$$
So since $\alpha$ is the amplitude, I guess this is why they say it gives us the intensity. But do you think that this is is simply the relation that both papers use? Personally I don't, but reading papers like the ones above are difficult for me as they don't really explain the details, and the worst thing is that the authors never bother replying when I ask them.

Best,
Niles.

8. Aug 6, 2012

### Cthugha

It is not really what they use, but the a-bit-quantum version of it. In a nutshell if I get the thesis you mentioned right, they just say they use a classical approach as an approximation which basically means that you can factorize the needed terms per definiton. Using coherent states just allows to keep the quantum treatment without having the odd formalism of using operators for a classical treatment.

That happens sometimes. People move around or leave university and at some time the mail addresses may even become obsolete. Contacting authors of papers which are older than the duration of a PhD thesis cycle often does not work.

edit: post #1000, yay :)

9. Aug 6, 2012

### Niles

The paper I linked to before (http://arxiv.org/pdf/1105.2373v2.pdf), do you agree with me that the approach one has to take in order to obtain (7) is to solve for $\sigma_-$ in (3), insert in (4) and obtain an expression for $\sigma_z$ and then insert in (2) and find $a$, which is then conjugated in order to find $<a^2>$? I tried doing it, but it became *very* tedious, but I can't see how it can be done otherwise.

You say that
So since the Maxwell-Bloch equations are semi-classical, I guess it is 100% valid to make the factorization I questioned in my OP, even though the operatores work on the same subsystem?

Congratulations with 1000 posts.

10. Aug 6, 2012

### Cthugha

Yes, I would go about it the same way, but it sure will get lengthy. Maybe that is a task Mathematica can handle well.

Yes, typically in such approaches light is assumed to be treated classically which allows factorization.

11. Aug 6, 2012

### Niles

Thanks, it is very kind of you to take the time to help.

Best,
Niles.