- #1
mike1000
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Lets assume there is an observable represented by the operator, ##\hat{A}##. It follow (I think) that the observed values and allowed states obeys an equation of the following form\begin{equation}\hat{A}|\psi^i\rangle=\lambda_f|\psi_f\rangle\end{equation} where $$|\psi^i\rangle = initial \ \ state \ \ vector$$ $$|\psi_f\rangle = final \ \ state \ \ vector$$My assumption is that the operator ##\hat{A}## operates on the state vector that exists immediately prior to the measurement.
Assume that the state vector, ## |\psi^i\rangle##, is in a physical state that is some superposition of the eigenvectors, ##\{ \psi_n\}##, of the operator ##\hat{A}##
\begin{equation} |\psi^i\rangle=\alpha_1|\psi_1\rangle + \alpha_2|\psi_2\rangle+\dots+\alpha_f|\psi_f\rangle+\dots+\alpha_n|\psi_n\rangle\end{equation}\begin{equation}\hat{A}|\psi^i\rangle=\alpha_1\hat{A}|\psi_1\rangle+\alpha_2\hat{A}|\psi_2\rangle+\dots+\alpha_f\hat{A}|\psi_f\rangle+\dots+\alpha_n\hat{A}|\psi_n\rangle= \lambda_f|\psi_f\rangle\end{equation}\begin{equation}=\alpha_1\lambda_1|\psi_1\rangle + \alpha_1\lambda_2|\psi_2\rangle+\dots+\alpha_f\lambda_f|\psi_f\rangle+\dots+\alpha_n\lambda_n|\psi_n\rangle = \lambda_f|\psi_f\rangle\end{equation}Which implies ##\alpha_{i\neq f}= 0## and ##\alpha_f =1 ##
Which shows that the original assumption, that the initial state was a superposition of basis states, is false.(Proof by contradiction)
Assume that the state vector, ## |\psi^i\rangle##, is in a physical state that is some superposition of the eigenvectors, ##\{ \psi_n\}##, of the operator ##\hat{A}##
\begin{equation} |\psi^i\rangle=\alpha_1|\psi_1\rangle + \alpha_2|\psi_2\rangle+\dots+\alpha_f|\psi_f\rangle+\dots+\alpha_n|\psi_n\rangle\end{equation}\begin{equation}\hat{A}|\psi^i\rangle=\alpha_1\hat{A}|\psi_1\rangle+\alpha_2\hat{A}|\psi_2\rangle+\dots+\alpha_f\hat{A}|\psi_f\rangle+\dots+\alpha_n\hat{A}|\psi_n\rangle= \lambda_f|\psi_f\rangle\end{equation}\begin{equation}=\alpha_1\lambda_1|\psi_1\rangle + \alpha_1\lambda_2|\psi_2\rangle+\dots+\alpha_f\lambda_f|\psi_f\rangle+\dots+\alpha_n\lambda_n|\psi_n\rangle = \lambda_f|\psi_f\rangle\end{equation}Which implies ##\alpha_{i\neq f}= 0## and ##\alpha_f =1 ##
Which shows that the original assumption, that the initial state was a superposition of basis states, is false.(Proof by contradiction)
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