6.3: Example - Oscillation in a fully mixing two state system
- Page ID
- 28776
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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.