# The sign of coupling Hamiltonian in CQED

• I
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!

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.

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.