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

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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.

This page titled 6.3: Example - Oscillation in a fully mixing two state system is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Graeme Ackland via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.