Lindblad master equation

Von Neumann Picture§

Lindblad master equation is built upon von newmann picture which is somewhat just schrodinger picture but for the full density state operator, thus in somesense it is a more complete schordinger euqation. Thus, let's derive the von numann equation very quickly and easily. First of all, a general state can be written as a sum of pure state, which is

$$\rho = \sum_i p_i\ket{\psi_i}\bra{\psi_i}$$

Thus

$$\dot{\rho}=\sum_i p_i\left(\dot{\ket{\psi_i}}\bra{\psi_i} + \ket{\psi_i}\dot{\bra{\psi_i}}\right)$$

Where if we apply the Schrodinger equation now, here, we get

$$\begin{align*} \dot{\rho}&=\frac{1}{i\hbar}\sum_i p_i\left(\hat{H}\ket{\psi_i}\bra{\psi_i} - \ket{\psi_i}\bra{\psi_i}\hat{H}\right)\\ &=\frac{1}{i\hbar}\left(\hat{H}\sum_i p_i\ket{\psi_i}\bra{\psi_i} - \sum_i p_i\ket{\psi_i}\bra{\psi_i}\hat{H}\right)\\ &=\frac{1}{i\hbar}\left[\hat{H},\rho\right] \end{align*}$$

Or, although non standard, but personally feels more appropiete since we are comparing it to the schrodinger equation, we will write it as

$$i\hbar\dot{\rho}=\left[\hat{H},\rho\right]$$

Where the schrodinger equation is

$$i\hbar \dot{\ket{\psi}}=\hat{H}\ket{\psi}$$

The setup§

Now, to derive the master equation, we note that everything is derived in the interaction frame and then translate back to the lab frame as before with RWA. Thus, in the rotating from, our hamiltonian is

$$\hat{H} = \sum_{k,\lambda}g_{k,\lambda}\hat{A}_{k,\lambda}e^{i(\omega_{eg}-\omega_k)t}+g^*_{k,\lambda}\hat{A}_{k,\lambda}^\dagger e^{i(\omega_k-\omega_{eg})t}$$

Applying the Von Neumann equation,

$$i\hbar \dot{\rho}=\left[\hat{H},\rho\right]$$

Let's change it's form so that we can take approximation from there. By integration, we note that the density matrix can be evaluated as

$$i\hbar\rho(t)=i\hbar\rho_0+\int_0^t \left[\hat{H}(t'),\rho(t')\right]\,dt'$$

Thus, the Von neumann equation becomes

$$\begin{align*} i\hbar \dot{\rho}&=\left[\hat{H},\rho_0+\frac{1}{i\hbar}\int_0^t \left[\hat{H},\rho\right]\,dt\right]\\ \dot{\rho}&=\frac{1}{i\hbar}\left[\hat{H},\rho_0\right]-\frac{1}{\hbar^2}\int_0^t\left[\hat{H}(t), \left[\hat{H}(t'),\rho(t')\right]\right]\,dt'\\ \end{align*}$$

Born Approximation§

1. We assume The bath and the system interact weakly, such that their state can be factorized approximately

$$\rho(t) = \rho_S(t) \otimes \rho_B(t)$$

Where $S$ and $B$ stands for System and Bath. And $\rho_B(t)$ is approximatly constant

2. Also, as it is constant, we assume $\braket{B}=0$, in the terminology of density matrices, it is $\operatorname{Tr}\left({B\rho}\right)=0$.
3. In Born approximation, we also take Markov approximation

Thus, now, to make use of everything of here, let's take the trace of system $B$,

$$\begin{align*} \dot{\rho}_S(t)&=\frac{1}{i\hbar}\operatorname{Tr}_B\left[\hat{H},\rho_0\right]-\frac{1}{\hbar^2}\int_0^t\operatorname{Tr}_B\left[\hat{H}(t), \left[\hat{H}(t' ),\rho(t')\right]\right]\,dt'\\ \end{align*}$$

Then, because $H$ is bilinear upon $S$ and $B$, where in this case $S=\sigma$, $B=b$, $A=S\otimes B$, by assumption 2, the first order terms must vanish to 0. And we use assumption 3 onto that integral, Thus

$$\dot{\rho}_S=-\frac{1}{\hbar^2}\int_0^\infty\operatorname{Tr}_B\left[\hat{H}(t), \left[\hat{H}(t-\tau),\rho(t)\right]\right]\,d\tau$$

Where it is not obvious here, but the first markov approximation ( take $t'=t$ on the density matrix ) is justified because we already assumed that the time we are interested is below the decay timescale, which is very small. And on the second markov approximation ( substituting $\tau=t-t'$ and then making the upper limit $\infty$ ) is justified because there will be the correlation function $\operatorname{Tr}_B\left(H(t)H(t-\tau)\rho\right)$ appearing on all 4 terms after exanding the commutation, and the coorelation should drops to zero at large time differences, also below the decay time scale.

The Lindblad Master Equation§

The rest of the math are the same as previous derivation for spontanuos emission, thus it will be omitted here. The conclusion after transforming back to the schrodinger frame are

$$\dot{\rho}_S = -\frac{i}{\hbar}\left[\hat{H}_S,\rho_S\right]+\gamma\mathcal{D}_{\hat{\sigma}^-}(\rho_S)\qquad\mathcal{D}_L(\rho)=L\rho L^\dagger - \frac{1}{2}\{L^\dagger L,\rho\}$$

Notice, everything now is in the system frame, the bath is simplied to just giving a way to decay and we didn't need to deal with interaction terms from here.

Adjoint Lindblad Master Equation§

With

$$\dot{\braket{O}} = \operatorname{Tr}\left(O\dot{\rho}\right)$$

We direclty use the Lindblad Master Equation on it. The first term gives,

$$\begin{align*} -\frac{i}{\hbar}\operatorname{Tr}\left(O\left[\hat{H},\rho\right]\right) &= -\frac{i}{\hbar}\operatorname{Tr}\left(O\hat{H}\rho-O\rho\hat{H}\right) \\ &= -\frac{i}{\hbar}\operatorname{Tr}\left(O\hat{H}\rho - \hat{H}O\rho\right) \\ &= -\frac{i}{\hbar}\operatorname{Tr}\left(\left[O,\hat{H}\right]\rho\right) \\ \end{align*}$$

The dissipation term gives,

$$\begin{align*} \gamma\operatorname{Tr}\left(O\mathcal{D}_L(\rho)\right) &= \gamma\operatorname{Tr}\left(OL\rho L^\dagger-\frac{1}{2}OL^\dagger L\rho-\frac{1}{2}O\rho L^\dagger L \right) \\ &= \gamma\operatorname{Tr}\left(L^\dagger OL\rho - \frac{1}{2}OL^\dagger L\rho - \frac{1}{2}L^\dagger L O\rho \right) \\ &=\gamma\operatorname{Tr}\left(\mathcal{D}_L^\dagger(O)\rho\right)\qquad\mathcal{D}_L^\dagger(O)=L^\dagger O L - \frac{1}{2}\{L^\dagger L,O\} \end{align*}$$

Thusm easily we can see

$$\dot{\braket{O}}=\operatorname{Tr}\left(\left[-\frac{i}{\hbar}\left[O,\hat{H}\right]+\gamma\mathcal{D}_L^\dagger(O)\right]\rho\right)$$

Now, if we change to the Heisenberg frame, the expectation value should not change, but who stays constant have changed. Now $\rho$ stays constant and the operator changes, so in Heisenberg frame, we get

$$\dot{O}=-\frac{i}{\hbar}\left[O,\hat{H}\right]+\gamma\mathcal{D}^\dagger_L(O)$$

This is the adjoint equation.

Purcell effect ( again )§

From the equationes, we note $\braket{\sigma^+\sigma^-}=P_e$ , we can get

$$\dot{P}_e=ig\braket{a\sigma^+-a^\dagger\sigma^-}-\gamma P_e$$

Then we assume

1. a weak excitation ( $\sigma_z = -1$ )
2. adiabatic elimination, $\dot{a} = 0$
3. weak correlation limit between atom and field $\braket{a\sigma}=\braket{a}\braket{\sigma}$
4. $\kappa \gg g$

$$\dot{P}_e=-\left(\gamma + \frac{4g^2}{\kappa}\right) P_e$$

When with the same assumption but $\kappa \ll g$, we get

$$\dot{P}_e=-\frac{\gamma}{2} P_e$$