The sign of coupling Hamiltonian in CQED

[thariya]

Hi all,

I've always regarded the coupling Hamiltonian for a bosonic cavity mode coupled to a two-level fermionic gain medium chromophore to be of the form,

$$H_{coupling}=\hbar g(\sigma_{10}+\sigma_{01})(b+b^{\dagger})$$,

where $b$ and $b^{\dagger}$ and annihilation and creation operators for the bosonic cavity mode and $\sigma_{ij}$ are the raising and lowering operators for a two level atom. $g$ is the coupling constant.

Using the rotating wave approximation, this sometimes is simplified to,

$$H_{coupling}=\hbar g(\sigma_{10}b+\sigma_{01}b^{\dagger})$$.

I've recently come across some texts that seems to use an alternate form(under the rotating wave approximation) to describe (what I perceive is) the exact same system,

$$H_{coupling}=\hbar g(\sigma_{10}b-\sigma_{01}b^{\dagger})$$,

the main difference being the negative sign. Would you be able to explain why this difference occurs and what the significance of the negative sign is?

Thanks!

 

[vancouver_water]

It depends on your convention for quantizing the EM field. You either get terms like $$\boldsymbol{E} \sim \boldsymbol{\epsilon_k}b_k+\boldsymbol{\epsilon_k}^*b_k^\dagger$$ or $$\boldsymbol{E} \sim i(\boldsymbol{\epsilon_k}b_k -\boldsymbol{\epsilon_k}^*b_k^\dagger).$$

Then when you add in the dipole-field interaction you end up with extra minus signs.

 

[thariya]

It depends on your convention for quantizing the EM field. You either get terms like $$\boldsymbol{E} \sim \boldsymbol{\epsilon_k}b_k+\boldsymbol{\epsilon_k}^*b_k^\dagger$$ or $$\boldsymbol{E} \sim i(\boldsymbol{\epsilon_k}b_k -\boldsymbol{\epsilon_k}^*b_k^\dagger).$$

Then when you add in the dipole-field interaction you end up with extra minus signs.

Thank you very much for the reply! Do the conventions vary based on the redefinition of the field operators $b_k^\dagger$ and $b_k$? If what I think is correct, in one convention, the $b_k^\dagger,b_k$ is proportional to the quantized vector potential, while in the other, it's the quantized generalized momentum. Am I right here?

Thanks.

 

[vancouver_water]

There is no redefinition of the field operators. Before the quantization it is the same as taking the E or A field as the real or imaginary part of a complex function.

 

[thariya]

There is no redefinition of the field operators. Before the quantization it is the same as taking the E or A field as the real or imaginary part of a complex function.

Thanks for that. I wrote down the derivation and I see what you mean. It basically depends on whether the cosine or sine waveform is used to describe the oscillating field.