When we descirbe the degenerate parametric oscillator or the Parametric Down-Conversion, we can use the hamitonian: $$\hat{H}=\hat H_0+\hat H_{int}+\hat H_{irr}$$ \begin{array}{l} \hat H_0=\hbar\omega \hat a^\dagger_1 \hat a_1+2\hbar\omega \hat a^\dagger_2 \hat a_2 \\ \hat H_{int}=i\hbar \frac{k}{2}({\hat a^\dagger_1}^2 \hat a_2 - {\hat a_1}^2 \hat a_2^\dagger)+i\hbar \gamma_2 (\hat a^\dagger_2 F e^{-2i\omega t}-\hat a_2 F^* e^{2i\omega t}) \\ \hat H_{irr} = i\hbar \gamma_1(\hat a_1^\dagger \hat B_1-\hat a_1 \hat B_1^\dagger)+i\hbar \gamma_2(\hat a_2^\dagger \hat B_2-\hat a_2 \hat B_2^\dagger) \end{array} Here, $\hat H_0$ is the energies of the signal and the pump fields, where $\hat a_1, \hat a_2$ is the annihilation operators for the signal field and the pump filed respectively. $\hat H_{int}$ is the nonlinear coupling between the signal and the pump field, $\hat H_{irr}$ is the irreversible interaction betweent cavity fields adn the reservoir.

In $\hat H_{int}$ and $\hat H_{irr}$, Why we use $i(\hat A - \hat A^\dagger)$ to construct an Hermite operator instead of $(\hat A+\hat A^\dagger)$ ?

But in the Jaynes–Cummings model, the Hamiltonian is: $$\hat H=\hat H_{filed}+\hat H_{atom}+\hat H_{int}$$ \begin{array}{l} \hat H_{filed}=\hbar\omega_c \hat a^\dagger \hat a\\ \hat H_{atom}=\hbar\omega_a \frac{\hat \sigma_z}{2} \\ \hat H_{int}=\frac{\hbar\Omega}{2}(\hat a \hat \sigma_+ +\hat a^\dagger \hat \sigma_-) \end{array} There is no $i$ in the $\hat H_{int}$.

So what is the reason for choosing $i(\hat A - \hat A^\dagger)$ or $(\hat A+\hat A^\dagger)$ ?

Thanks a lot.

  • $\begingroup$ I know it's a matter of notations, and depends on how you write the electric field, but I don't remember exactly what the difference is. I'll look it up and give a more detailed answer. $\endgroup$ – Milloupe Jun 26 '19 at 9:19


Ok, my memory was right, though not precise enough: it depends on the sign you choose for time evolution.

  • If you write forward time evolution as $\exp(i\omega t)$, then the electric field should be:

$$\hat{E} \propto \hat{a} e^{i \omega t} + \hat{a}^\dagger e^{-i \omega t}$$

  • But if you write forward time evolution as $\exp(-i\omega t)$, then the electric field should be:

$$\hat{E} \propto i\left(\hat{a} e^{-i \omega t} - \hat{a}^\dagger e^{i \omega t}\right)$$

All that matters is that the way you write it remains consistent all along, in order to make sure that the fields and other physical variables do follow Maxwell's equations (among others).

  • 1
    $\begingroup$ to be it seems you have gotten it backwards. When you declare your annihilation operator $\hat{a}(t)$ to evolve as $\hat{a} e^{-i\omega t}$, you can write the vector potential $\hat{A}$ as $\propto \hat{a}(t) e^{ikr} + \hat{a}^{\dagger}(t) e^{-ikr}$, you will see that the electric field $\hat{E} = -\partial \hat{A} / \partial t$ obtains a factor of $i$ in front. This is the usual exposition of quantization of EM field from Maxwell's equations (check Wikipedia or other sources). $\endgroup$ – wcc Jun 27 '19 at 14:04
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    $\begingroup$ Oops, exactly, I was only looking at the interaction hamiltonian, for which it is the opposite (because $H = -D \cdot E$). I corrected in my answer! $\endgroup$ – Milloupe Jun 27 '19 at 14:16

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