The following lecture from University of Oxford contains an explanation of the constancy of probability distributions for all observables when a system is in a stationary state: http://www.youtube.com/watch?v=0yvX4jhzblY#t=15m35s. However, the derivation of the vanishing amplitude does not actually use the fact that the state is stationary, leading to the conclusion that all observables have fixed probability distributions for all states, stationary or otherwise. Clearly this is false, but where is the error in the maths?(adsbygoogle = window.adsbygoogle || []).push({});

For a system in state [itex]\psi[/itex], the probability amplitude [itex]A[/itex] of measuring the value of an observable [itex]Q[/itex] as [itex]q_n[/itex] (the [itex]n[/itex]th eigenvalue of [itex]Q[/itex]) is [itex]\langle q_n|\psi\rangle[/itex]. The time derivative of this amplitude goes as:

\begin{align}

i\hbar\frac{\partial A}{\partial t}

= i\hbar\frac{\partial \langle q_n|}{\partial t}|\psi\rangle

+ i\hbar\langle q_n|\frac{\partial |\psi\rangle}{\partial t}

\end{align}

From the Schrödinger equation applied to [itex]|\psi\rangle[/itex], [itex]i\hbar\frac{\partial |\psi\rangle}{\partial t} = H|\psi\rangle[/itex], and from its conjugate form applied to [itex]\langle q_n|[/itex], [itex]-i\hbar\frac{\partial \langle q_n|}{\partial t} = \langle q_n|H[/itex]. So

\begin{align*}

i\hbar\frac{\partial A}{\partial t}

&= -\langle q_n|H|\psi\rangle + \langle q_n|H|\psi\rangle \\

&= 0

\end{align*}

This has concluded that for any state [itex]\psi[/itex], the probability amplitude for an observable [itex]Q[/itex] is constant in time. Clearly this is false, but which step in the derivation was dodgy?

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# All states are stationary, all observables are constant.

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