6.3: Example - Oscillation in a fully mixing two state system

Consider the expectation value of a quantity $S$ in a system which has two non-degenerate energy eigenstates $|1 \rangle$ and $|2 \rangle$, and where the Hermitian operator $\hat{S}$ is defined by $\hat{S}|1\rangle = |2\rangle$, $\hat{S}|2 \rangle = |1 \rangle$.

The general state can be written:

$$|\phi \rangle = c_1 \text{ exp}(−iE_1t/\hbar )|1 \rangle + c_2 \text{ exp}(−iE_2t/\hbar )|2 \rangle \nonumber$$

if we assume real $c_1$, $c_2$ it follows that the expectation value $\langle \hat{S} \rangle$ will be:

$$\langle \hat{S} \rangle = \langle \phi | \hat{S} | \phi \rangle \\ = \left[ c_1e^{iE_1t/\hbar} \langle 1| + c_2e^{iE_2t/\hbar} \langle 2| \right] \left[ c^*_1 e^{−iE_1t/\hbar} |2 \rangle + c^*_2 e^{−iE_2t/\hbar} |1\rangle \right] \\ = c_1c_2 [e^{i\omega_{21}t} + e^{−i\omega_{21}t} ] \\ = 2c_1c_2 \cos(\omega_{21}t) \nonumber$$

Thus the expectation value of $\hat{S}$ oscillates in time at frequency $\omega_{21} = (E_2 − E_1)/\hbar$. This arises because $\hat{S}$ is not compatible with the hamiltonian, and hence does not define a constant of the motion.