# Two-level system with spontaneous emission

I have been stuck with this elementary problem of two-level system including spontaneous decay. After solving by standard procedure the following pair of equations, when I plot the expressions of the ground state population $P_0 (=|C_0(t)|^2)$, excited state population $P_1 (=|C_1(t)|^2)$ and sum of the two, $P_0+P_1$ vs time $t$, I find that $P_0+P_1$ is not $1$, as if there is a loss of the system as a whole. It has got to be wrong.

I tried to solve the following coupled differential equations incorporating the spontaneous decay width, for a single two-level atom subjected to optical field:

$\dot{C_0(t)} = -\frac{i}{2} \Delta C_0(t) + \frac{i}{2}(\Omega-i\Gamma_{sp})C_1(t) \\ \dot{C_1(t)} = \frac{i}{2} (\Delta+i\Gamma_{sp}) C_1(t) + \frac{i}{2}(\Omega^*)C_0(t)$

Is it correct to incorporate spontaneous emission in the language of probability amplitude of a single atom like this?

• Without looking at the details... Did you account for the possibility that the incident photon was absorbed while the system was in the lower energy state? Jun 26 '14 at 11:58

To summarize them here, what you do is work in terms of the matrix $$\rho =\begin{pmatrix}\rho_{gg}&\rho_{ge}\\ \rho_{eg}& \rho_{ee}\end{pmatrix} \text{ which equals }|\psi⟩⟨\psi| =\begin{pmatrix}C_0\\ C_1\end{pmatrix}\begin{pmatrix}C_0^\ast& C_1^*\end{pmatrix} \text{ for a pure state}.$$ Driving the system with Rabi frequency $\Omega$ and detuning $\delta=\omega-\omega_0$, and allowing it to spontaneously emit at a rate $\gamma$ gives the optical Bloch equations: \begin{align} \frac{d \rho_{gg}}{dt} & = \gamma \rho_{ee}+\frac i2 (\Omega^* \rho_{eg}-\Omega \rho_{ge}), \\ \frac{d \rho_{ee}}{dt} & = -\gamma \rho_{ee}+\frac i2 (\Omega \rho_{eg}-\Omega^* \rho_{ge}), \\ \frac{d \rho_{ge}}{dt} & =-\left(\frac\gamma2+i\delta\right)\rho_{ge}+\frac i2\Omega^*(\rho_{ee}-\rho_{gg}), \\ \frac{d \rho_{eg}}{dt} & =-\left(\frac\gamma2+i\delta\right)\rho_{eg}+\frac i2\Omega^*(\rho_{gg}-\rho_{ee}). \end{align} It is these equations that you should be solving.
• The mixedness is measured easily by calculating the purity, $\text{Tr}(\rho^2)$. This is always between 0 and 1 and equals 1 iff the state is pure. It should be a good exercise to show that for nonzero gamma the purity is strictly decreasing in time unless you are in a completely mixed state. Jun 26 '14 at 18:15
• Yes, Right. That means the density matrix equations would not translate into the amplitude equations I wrote in my question. Am I correct by saying: when you wrote $|{\Psi}\rangle\langle{\Psi}|= (\begin{smallmatrix} C_0\\ C_1 \end{smallmatrix}) (\begin{smallmatrix} C_0^* & C_1^* \end{smallmatrix})$ it was for a pure state and can be translated to the amplitude equations (without decay term). Then suitably adding the decay terms, it becomes mixed and cannot directly be translated to amplitude equations. Jun 27 '14 at 2:27
• To be clear: open systems and mixed states cannot be represented by amplitude equations and require a density-matrix formalism. Density matrices of the form $\rho=|\psi⟩⟨\psi|$ are unique to pure states and cannot describe mixed states or open systems. Jun 27 '14 at 11:44