Inexact measurement and wavefunction collapse

Question

As is usually said, measurement of an observable $q$ leads to collapse of wavefunction to an eigenstate of the corresponding operator $\hat q$. That is, now the wavefunction in $q$ representation is $\psi(q)=\delta(q-q_0)$ where $q_0$ is result of measurement.

Now, in reality measurements are never exact, so the wavefunction should not have that sharp peak. Instead it should be broadened depending on precision of measurement. In this case, can we still introduce an operator, an eigenstate of which would the new wavefunction appear? Is it useful in any way? Or does the new wavefunction depend too much on the way it was measured so that each instrument would have its own operator? How would such opeartor look e.g. for a single-slit experiment?

Answer 1

First of all, rigorously speaking, as the spectrum of the position operator $X$ on $L^2(R)$ is purely continuous, the spectral measure $P_E$ is labeled by Borel sets $E\subset R$, so, in particular $E$ can be an interval $[a,b]$. In position representation: $$\left(P_E \psi\right)(x) = \chi_E(x) \psi(x)\quad \forall x \in R$$ where $\chi_E(z)=1$ if $z\in E$, or $\chi_E(z)=0$ otherwise. When measuring the position of the particle, if the precision of the instrument is $2\delta>0$, so that one cannot distinguish anything inside the interval $[x_0-\delta, x_0+\delta]$, the wavefunction immediately after the measurement is, up to normalization, $$\chi_{[x_0-\delta, x_0+\delta]}\psi$$ provided $\psi$ was the wavefuntion before the measurement and the found position was $x_0$ (taking the precision of the instrument into account).

This is nothing but a particular case of the von Neumann - Luders axiom on quantum measurement that includes both observables with point spectrum and continuous spectrum. In the second case the notion of eigenvector cannot apply and, however, it is by no means necessary. Just the notion of spectral measure associated with a self-adjoint operator is sufficient.

The fact that real instruments for observables with continuous spectrum are truly described by that axiom, even taking the precision into account as done before, is however questionable due to many practical reasons. It is more plausible that, in real (non-destructive) experiments of position the wavefunction after the measurement is obtained from the incoming one through a so called quantum operation (http://en.wikipedia.org/wiki/Quantum_operation).

Answer 2

With imprecise measurements you need to use the more general statistical quantum mechanics rather than simple pure-state quantum mechanics.

Specifically, rather than ending up with a pure quantum state, after an imprecise measurement you instead have a "dirty" state (mixed quantum state) which is blurred by uncertainty in the classical sense. This mixture of quantum state with classical uncertainty cannot be represented by a single quantum state, but can be represented by a density matrix.

(In my opinion, it's a good idea anyway to use statistical methods even when considering exact measurements. That way you can include things like environmental decoherence.)