- #1

Fredrik

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[tex]\frac 1 N\sum_{i=1}^N\delta_{kk_i}[/tex]

Now consider an ensemble of N identical systems. Because of the above, the operator [itex]f_k^{(N)}[/itex] defined by

[tex]f_k^{(N)}|a_{k_1}\rangle\cdots|a_{k_N}\rangle=\left(\frac 1 N\sum_{i=1}^N\delta_{kk_i}\right)|a_{k_1}\rangle\cdots|a_{k_N}\rangle[/tex]

where the [itex]|a_j\rangle[/itex] are eigenvectors of some observable A, is telling us what fraction of the systems in the ensemble are in state [itex]|a_k\rangle[/itex]. (The eigenvalue is equal to that fraction). It's also telling us what fraction of the results will be [itex]a_k[/itex] if we measure A once on each system in the ensemble (and it's prepared in this particular state). That fraction is equal to the eigenvalue, which is equal to the expectation value of [itex]f_k^{(N)}[/itex] in this state.

Now consider an ensemble such that all the systems are prepared in the state [itex]|s\rangle[/itex], which isn't an eigenvector of A. Does the last interpretation above still hold? I mean, is the expectation value of [itex]f_k^{(N)}[/itex] in an arbitrary state

[itex]|s^{(N)}\rangle=|s\rangle\cdots|s\rangle[/itex]

still equal to the fraction of measurements of A that will yield the result [itex]a_k[/itex]? Is it possible to justify this without using the axiom that the probability of measuring [itex]a_k[/itex] when the state is [itex]|s\rangle[/itex] is [itex]|\langle a_k|s\rangle|^2[/itex].