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Suppose my system, described by a separable Hilbert space $H$, is in the state $\Psi$ when I measure an observable that has only continuous spectrum. What is the state of the system after the measurement?

Say, for the sake of definiteness, that we are measuring the momentum of a particle whose state is $\Psi \in L^{2}(\mathbb{R}^{3})$. Will it "leave" the space $L^{2}(\mathbb{R}^{3})$ and "become" a wave?

If, instead, my measurement is only partly accurate and says that the momentum of the particle is in a set $\Delta =(a_x,b_x)\times(a_y,b_y)\times(a_z,b_z)$, will the measurement collapse the wave function into $P\Psi$ (where $P$ is the spectral projector of the momentum operator on the set $\Delta$)?

What is the correct description of the system after one such measure?

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Your intuition that

If, instead, my measurement is only partly accurate and says that the momentum of the particle is in a set $\Delta =(a_x,b_x)\times(a_y,b_y)\times(a_z,b_z)$, will the measurement collapse the wave function into $P\Psi$ (where $P$ is the spectral projector of the momentum operator on the set $\Delta$)?

is exactly correct. Measurements which project onto unphysical states, like momentum eigenstates, are not physically realizable; we can only project onto continuum states with a finite resolution, and the corresponding measurement projectors are the appropriate spectral projectors $P$. After the measurement, the state of the system is the projected vector $P\Psi$.

If you want a slightly more sophisticated approach, you can think of your detector as having a number of "pixels" each of which has a variable sensitivity $f_j(\mathbf p)$ to different momenta, and which is associated with a projection operator $$ P_j=\int\text d\mathbf p \, f_j(\mathbf p) \,|\mathbf p⟩⟨\mathbf p|. $$ The $f_j(\mathbf p)$ should be a partition of unity, in the sense that they're real, positive, and $\sum_j f_j(\mathbf p)$ should be identically 1. This then makes the $P_j$ into a POVM which obeys $$\sum_j P_j=\mathbb I.$$ After measuring on pixel $j$, the state of the system is $P_j\Psi$.

Why is this necessary? Your original scheme fits inside the new one, but it has discontinuities in the $f_j$. This means that you introduce spectral clipping into your state, with the result that $P\Psi$ is still not physical, and in general will have ~$\sin(x)/x$ ringings which will render it unphysical (though still integrable). Only if the $f_j$ are smooth and have compact support will the $P_j\Psi$ be completely physical. (Here, by 'physical state' I mean one which has finite $⟨\hat x^n⟩$ and $⟨\hat p^n⟩$ for all $n$.)

Obviously, this criterion precludes functions of the sort $f_j(\mathbf p)=\delta(\mathbf p)$, and projecting onto a pure momentum eigenstate is not a physical thing. If you did do that, the state does leave $L^2$ to become a (plane!) wave. This is not a particularly bothersome idea, because to really be able to pin down a function's wavelength to an exact real number, you need a physical procedure that assesses it at every position between $-\infty$ and $\infty$, which means that the unboundedness of the resulting state is not particularly surprising. (Obviously, an analogous version of this argument holds for "arbitrary momentum precision" and the "increasingly large spatial extent" necessary for the measurement and which obtains for the resulting post-projection state.)

To read more about the role of physical states as I've defined them and their relation to continuous spectra, the key concept to look up is rigged Hilbert spaces. A good reference on this, from a cursory reading, is

The role of the rigged Hilbert space in Quantum Mechanics. R. de la Madrid. Eur. J. Phys. 26 no. 2, 287-312 (2005). arXiv:quant-ph/0502053.

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  • $\begingroup$ Would you please explain to me in more detail this spectral clipping? The spectral projection $P$ described above is and orthogonal projection, and therefore a bounded operator. How could it render the state vector $\Psi$ nonintegrable? $\endgroup$ – Geno Whirl Aug 28 '14 at 15:47
  • $\begingroup$ 'Spectral clipping' refers to the introduction of discontinuities in a function's spectrum. The essential intuition is that this makes the function into a step function in momentum space, so its Fourier transform is like a sinc function (i.e. the transform of a step function). This is actually integrable (and I amended my answer) but it's still not physical, as it doesn't have any well-defined moments of position. Thus $P$ doesn't take $\Psi$ out of $L^2$, but it does take it out of the inner space $\Phi\subsetneq L^2$ of physical states. $\endgroup$ – Emilio Pisanty Aug 28 '14 at 16:12

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