# A Optical Bistability and the Jaynes-Cummings Model

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1. Mar 13, 2016

### Raptor112

I understand that otical bistability only occurs in a specific parameter regime defined by the bimodial leaf, but I have read that bistability originates from highly non linear dynamics of the system. As we are dealing with a qubit in a cavity that is being driven so things become non-trivial, the system is described by the Jaynes-Cummings Model with an additional drive term, so what terms in the equations or what in particular is non-linear that gives bistabilty?

2. Mar 14, 2016

### A. Neumaier

Please give a reference where ''the system'' is defined and the equations are stated.

3. Mar 14, 2016

### Raptor112

In the pape: Bistability effect in the extreme strong coupling regime of the Jaynes-Cummings model by A Dombi et.al
"Optical bistability is a benchmark of nonlinear lightmatter
interaction....Fluctuations are due to the dissipative processes, whereas the switching(Bistability) originates from a highly nonlinear dynamics." From the same paper, "the system :

and the master equation:

4. Mar 14, 2016

### A. Neumaier

The nonlinearity is in the term of H containing the factor g. For g=0, the system is decomposes into a harmonic oscillator and a 2-level system, which are both exactly solvable. The interaction creates the bistability.

5. Mar 14, 2016

### Raptor112

But with the interaction term it is still possible to diagonalise the Hamiltonian:

$H =\omega_c\bigg(\hat{a}^\dagger\hat{a} +\frac{1}{2}\bigg) + \frac{\omega_q}{2}\hat{\sigma}_z + g_{c}(\hat{a}\hat{\sigma}_+ + \hat{a}^\dagger\hat{\sigma}_-)$

has energy eigen values:

$E_{|n, \pm\rangle} = \omega_q\bigg(n+\frac{1}{2}\bigg) \pm\frac{\sqrt{\Delta^2 + 4g^2(n+1)}}{2}$

6. Mar 14, 2016

### A. Neumaier

But only in the absence of interaction with the environment, which causes energy dissipation and hence activates the bistability under appropriate conditions. The resulting Lindblad equation (2) can no longer be diagonalized. The mean equation (4) becomes linear only if g=0 and inherits from the nonlinear Hamiltonian the nonlinear term $S^*\alpha+\alpha^*S$.

7. Mar 14, 2016

### f95toli

Note that they are talking about a driven system, the J-C Hamiltonian can give rise to very complicated dynamics once you add e.g. a coherent drive (Mollow triplets etc). This in combination with decoherence means that the behavior is far from non-trivial even if you are only using a "simple" Lindbladian to model the system. .

Note also that the J-C Hamiltonian is only valid in the RWA; if you are driving the system hard enough you need to use the full spin-boson model which is difficult to solve even with numerical methods.

8. Mar 15, 2016

### Raptor112

So interaction with the outside environment causes the bistability?

9. Mar 15, 2016

### f95toli

Not sure there is a single "cause" here. I would say it is the very large value of g that is the "cause", with the understanding that the effects would only appear if you are driving the system hard enough and there is the right amount of dissipation.

But again, be very careful when working with the J-C Hamiltonian in this regime, you can easily go outside the regime where it is valid and if you solving it numerically you can easily get numerical artifacts if you space is not large enough

10. Mar 15, 2016

### A. Neumaier

Bistability is present independent of the environment but becomes alive through the latter. This also holds classically: A classical particle in a double-well potential is a bistable system, but the effect of it is seen only if you add an external interaction.

11. Mar 15, 2016

### Raptor112

But I don't actually see whats non linear about the interaction term?

12. Mar 15, 2016

### A. Neumaier

Any interaction may reveal the bistability due to an appropriate nonlinearity of the isolated system.