I am trying to solve the following Master equation (also similar to damped quantum harmonic oscillator):

$$\frac{d\hat{\rho}}{dt} = \frac{\Gamma}{2}\left(2\hat{a}\hat{\rho}\hat{a}^{\dagger} - \hat{a}^{\dagger}\hat{a} \hat{\rho} - \hat{\rho}\hat{a}^{\dagger}\hat{a} \right)$$ with initial coherent state: $\hat{\rho}(0) = |\alpha\rangle\langle\alpha|$. My idea was to use Sudarshan function and Gilmore D-algebra to write out differential equation. First step is to decompose density matrix operator in coherent state basis: $$\hat{\rho}(t) = \int d^2 \beta\ P(\beta, t) |\beta\rangle \langle \beta|$$ and act with operators appearing in initial equation:

$$\hat{a}\hat{\rho}\hat{a}^{\dagger} = \int d^2 \beta\ |\beta|^2 P(\beta, t) |\beta\rangle \langle \beta|$$ $$\hat{a}^{\dagger}\hat{a} \hat{\rho} = \int d^2 \beta\ P(\beta,t)\beta\left(\beta^* + \frac{\partial}{\partial\beta} \right)|\beta\rangle \langle \beta| = \int d^2 \beta\ |\beta\rangle\langle\beta| \left(\beta^* - \frac{\partial}{\partial\beta} \right) \beta P(\beta,t)$$ $$\hat{\rho}\hat{a}^{\dagger}\hat{a} = \int d^2 \beta\ P(\beta,t) \beta^*\left(\beta + \frac{\partial}{\partial\beta^*} \right) |\beta\rangle \langle \beta| = \int d^2 \beta\ |\beta\rangle\langle\beta| \left(\beta - \frac{\partial}{\partial\beta^*} \right)\beta^*P(\beta,t) $$ Finally we get: $$\frac{\partial P(\beta, t)}{\partial t} = \frac{\Gamma}{2}\left(\beta \frac{\partial}{\partial \beta} + \beta^* \frac{\partial}{\partial \beta^* } +2\right) P(\beta, t)$$

I am pretty sure that the derivation is correct because the differential equation preserves the unit trace i.e. $$\frac{d}{dt}\text{Tr}\{\hat{\rho}(t)\} = \int d^2\beta\ \frac{\partial P(\beta,t)}{\partial t} = \frac{\Gamma}{2}\int d^2\beta\ \left(\frac{\partial}{\partial \beta}\beta P(\beta,t) + \frac{\partial}{\partial \beta^*}\beta^* P(\beta,t) \right) = 0$$ My idea now was to use exponentiation and write the almost final solution as: $$P(\beta, t) = \exp\left[t\frac{\Gamma}{2}\left(\beta \frac{\partial}{\partial \beta} + \beta^* \frac{\partial}{\partial \beta^*} +2\right)\right] \delta^{(2)}(\alpha - \beta)$$ and then from definition of the delta function $$\delta^{(2)}(\alpha - \beta) = \frac{1}{\pi^2}\int d^2\eta\ e^{-i\eta^*(\alpha^* - \beta^*)}e^{-i\eta(\alpha - \beta)}$$ I can write $$P(\beta,t) = e^{\Gamma t} \frac{1}{\pi^2} \int d^2\eta\ e^{-i\eta^*\alpha^2} e^{t\frac{\Gamma}{2} \beta^* \partial_{\beta^*}} e^{i\eta^* \beta^*} \times e^{-i\eta \alpha} e^{t\frac{\Gamma}{2} \beta \partial_{\beta}} e^{i\eta \beta} = e^{\Gamma t}\delta^{2}(\alpha - \beta e^{t\Gamma/2})$$ The last step can be found here. At the end I get $$\rho(t) = e^{\Gamma t}|\alpha e^{-t\Gamma/2}\rangle\langle \alpha e^{-t\Gamma/2}|$$ Problem is it does not preserve the trace as it grows to infinity. Any idea where I made a mistake?


1 Answer 1


Its an old question, I am sure OP might have figured it out. The issue with normalization is present because of incorrect conversion from $P[\beta,t]$ to $\rho(t)$.

The correct way to construct density matrix $\rho(t)$ from Glauber-Sudarshan function $P[\beta,t]$ is :

$P[\beta,t]=e_{}^{\Gamma t}\delta_{}^{2}(\alpha-\beta e_{}^{\frac{\Gamma t}{2}})=\delta_{}^{2}(\beta - \alpha e_{}^{-\frac{\Gamma t}{2}}) \Rightarrow \rho(t)=\int d_{2}\beta P[\beta,t]|\beta\rangle\langle \beta|=|\alpha e_{}^{-\frac{\Gamma t}{2}}\rangle\langle\alpha e_{}^{-\frac{\Gamma t}{2}}|$


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