Expectation Values, Einstein Equations, Bloch Equations

We can write the Master equation with the help of

$$\sigma_- \equiv \vert g\rangle \langle e\vert,\quad \sigma_+ \equiv \vert e\rangl... ...ert g\rangle\langle g\vert,\quad \sigma_+ \sigma_-=\vert e\rangle\langle e\vert$$

$$\leadsto\frac{d}{dt}{\rho}(t) = -i\frac{1}{2}\bar{\omega}_0[\vert e\rangle\langle e\vert-\vert g\rangle\langle g\vert,\rho]$$

$$- \frac{1}{2}\gamma_+\Big\{ \vert e\rangle\langle e\vert \rho + \rh... ...\vert -2 \vert g\rangle \langle e\vert{\rho}\vert e\rangle \langle g\vert\Big\}$$

$$- \frac{1}{2}\gamma\Big\{ \vert g\rangle\langle g\vert\rho + \rho \... ...\vert -2\vert e\rangle \langle g\vert{\rho}\vert g\rangle \langle e\vert\Big\}.$$ (130)

Taking matrix elements, we obtain

$$\frac{d}{dt} \langle e \vert\rho \vert e \rangle = -\gamma_+ \langle e\vert \rho \vert e\rangle + \gamma \langle g\vert{\rho}\vert g\rangle$$ (131)

$$\frac{d}{dt} \langle g \vert\rho \vert g \rangle = +\gamma_+ \langle e\vert \rho \vert e\rangle - \gamma \langle g\vert{\rho}\vert g\rangle$$ (132)

$$\frac{d}{dt} \langle e \vert\rho \vert g \rangle = \left(-i \bar{\omega}_0 -\frac{\gamma_++\gamma}{2} \right)\langle e \vert\rho \vert g \rangle$$ (133)

$$\frac{d}{dt} \langle g \vert\rho \vert e \rangle = \left(+i \bar{\omega}_0 -\frac{\gamma_++\gamma}{2}\right) \langle g \vert\rho \vert e \rangle .$$ (134)

The first two equations for the diagonal elements (which are linearly dependent because $\langle e \vert\rho \vert e \rangle + \langle g \vert\rho \vert g \rangle =1$) are called Einstein equations. We can re-write the four equations, subtracting the second from the first, as three equations,

$$\fbox{$ \begin{array}{rcl} \displaystyle \frac{d}{dt} \langle \si... ...a}_0 -\frac{\gamma_++\gamma}{2}\right) \langle \sigma_-\rangle. \end{array}$\ }$$ (135)

These equations are called Bloch equations. Introducing the relaxation time $T_1$ and the decoherence time $T_2$,

$${T_1}= \frac{1}{2} T_2 \equiv (\gamma_++\gamma)^{-1},$$ (136)

we can write

$$\frac{d}{dt} \langle \sigma_z \rangle = -\frac{1}{T_1}\left( \langle \sigma_z \rangle - \langle \sigma_z ... ...langle \sigma_z \rangle_{\infty} \equiv \frac{\gamma-\gamma_+}{\gamma+\gamma_+}$$

$$\frac{d}{dt} \langle \sigma_+\rangle = \left(+i \bar{\omega}_0 -\frac{1}{T_2}\right) \langle \sigma_+\rangle$$

$$\frac{d}{dt} \langle \sigma_-\rangle = \left(-i \bar{\omega}_0 -\frac{1}{T_2}\right) \langle \sigma_-\rangle.$$ (137)