1. In the many statements of the QM postulates that I've seen, it says that if you measure an observable (such as position) with a continuous spectrum of eigenvalues, on a state such as(adsbygoogle = window.adsbygoogle || []).push({});

then the result will be one of the eigenvalues x, and the state vector will collapse to the eigenvector |x>.

However, for say the position operator, the eigenvectors are the dirac delta functions δ(x-x') and they do not represent physically realizable states (since they can't be normalized). So what does the state vector actually collapse to?

2. In practice there will always be experimental uncertainty Δ due to the resolution of your measuring apparatus, etc. So if you measure some position x, with an uncertainty +/- Δ, does the state vector still collapse to an eigenvector corresponding to someparticulareigenvalue within the [x-Δ, x+Δ] range? Or does it collapse perhaps to some (of infinitely many) states ψ(z) such that for z outside of the [x-Δ, x+Δ] range |ψ(z)|^2 = 0, but inside that range the probability amplitudes are equal for all z?

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# Collapse of state vector for continuous eigenvalues

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