# 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? – Floris 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. – Emilio Pisanty 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. – mohan_das 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. – Emilio Pisanty Jun 27 '14 at 11:44