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Suppose a quantum system is initially at a state $\psi_0$ and that a measurement of an observable $f$ is performed. Immediately after the measurement, the system will be in a state that is an eigenvector of the operator $\hat f$ associated to $f$, the eigenvalue being the result of the measurement.

My question is the following: What if the candidate for this eigenvector does not represent a valid state? For example, the space of states of a 1D-system is $L^2(\mathbb{R})$ and there are operators acting on the space of all functions on $\mathbb{R}$ whose eigenvectors may not belong to $L^2(\mathbb{R})$. How does the wave function collapse to such an eigenvector?

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    $\begingroup$ Your situation would require a measuring device that has infinite precision, such as a perfect position or momentum measurement. This is impossible in the real world. We measure 'smeared' versions of $\hat{x}$, $\hat{p}$, etc. $\endgroup$ – knzhou Sep 9 '16 at 22:15
  • $\begingroup$ Give an example of a measurement which you would say produces an invalid quantum state. Then I will post an answer (or point out that this is a duplicate of a question I answered some time ago). $\endgroup$ – DanielSank Sep 9 '16 at 22:32
  • $\begingroup$ Here is a very related question: physics.stackexchange.com/questions/137122/… $\endgroup$ – DanielSank Sep 9 '16 at 22:32
  • $\begingroup$ @knzhou Indeed. Actually I was hoping someone could answer this question of mine on exactly the issue of smeared measurements. $\endgroup$ – DanielSank Sep 9 '16 at 22:33
  • $\begingroup$ @Qmechanic Why did you bold "My question"? $\endgroup$ – tparker Sep 10 '16 at 9:28
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Every observable is described by a self-adjoint operator $A : D(A) \to {\cal H}$, where $D(A)$ is a dense subspace of $\cal H$ and coincide with $\cal H$ if and only if the set $\sigma(A)\subset \mathbb R$ (the spectrum of $A$) of values which $A$ may attain is bounded.

The spectral theorem says that $A$ has an associated projection valued measure (PVM). That is a map associating every (Borel) subset $E\subset \sigma(A)$, for instance $E= [a,b]$ or a single point $E= \{\lambda\}$, with an orthogonal projector $P_E : \cal H \to \cal H$.

It turns put that the "formal egenvectors", like $\delta$ functions, are always associated the the continuous part of $\sigma(A)$, whereas the proper eigenvectors $\psi_\lambda$ are associated with the elements $\lambda$ of the point spectrum part of $A$, they are the proper eigenvalues $\lambda$ of $A$.

Regarding outcomes $E$ of the measurement procedure belonging to the continuous spectrum, what one actually measures is an interval $E= [a,b]$.

In this situation the postulate of collapse, known as von Neumann-Luders postulate, states that, if a pure state is represented by the normalized vector $\psi$ before the measurement of $A$ and the outcome of measurement is $E$, the post-measurement pure state is $$\psi_E = \frac{P_E \psi}{||P_E\psi||}\:.\tag{1}$$ The probability to obtain $E$ in the state $\psi$ if measuring $A$ is, in particular,
$$||P_E\psi||^2 \tag{2}$$

Remarks.

(1) This postulate concerns non-destructive idealized measurement processes. In the experimental practice with realistic instruments, the post measurement state is described by a quantum operation which is a more sophisticated mathematical tool extending the notion of PVM.

(2) von Neumann-Luders postulate includes the case of a measurement of a discrete value $\lambda$ which belong to the point spectrum, i.e., a proper eigenvalue. In the absence of degeneracy, $$P_{\{\lambda\}} = |\psi_\lambda \rangle \langle \psi_\lambda |\:.$$ and applying (1) and (2) you obtain the standard elementary results. If the eigenspace of $\lambda$ has dimension $d \leq +\infty$ and thus there is a Hilbert basis of eigenvectors $\{\psi_{\lambda k}\}$, more generally, $$P_{\{\lambda\}} = \sum_{k=1}^d|\psi_{\lambda k} \rangle \langle \psi_{\lambda k} |\:.$$

(3) von Neumann-Luders postulate can be extended to mixed states trivially. In this context it has a natural meaning in terms of conditional probability over the non Boolean quantum lattice of elementary events (see my answer here)

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  • $\begingroup$ Great, this is in line with that I was thinking about! So $\psi$ simpy gets mapped into $P_E\psi$, and it formally needn't be the case that, for example, $P_E\psi \in D(A)$, right? Or yes and this follows from how the projection valued measure is constructed? $\endgroup$ – darko Sep 10 '16 at 18:16
  • $\begingroup$ Yes. However if $\psi \in D(A)$ also $P_E \psi \in D(A)$. $\endgroup$ – Valter Moretti Sep 10 '16 at 18:17
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Hermitian operators corresponding to physical observables act on the Hilbert space of physically valid states. It's clear from the definition of an eigenvector that for any vector space $\mathcal{H}$ and linear map $f: \mathcal{H} \to \mathcal{H}$, the eigenvectors must lie in $\mathcal{H}$. Therefore, the eigenvectors for any physical observable will be physically valid states, and your issue can't come up.

For example, since position eigenkets $| x \rangle$ and momentum eigenkets $| p \rangle$ do not lie in the $L^2(\mathbb{R})$ Hilbert space (although they do lie in a more general "rigged Hilbert space"), the position and momentum operators technically aren't physical observables - only operators that are slightly smeared in position or momentum space are. Strange but true. They're still extremely useful mathematical idealizations though. Physically, this just means that no real measurement could ever have infinite precision.

In practice, this is almost never an issue, because all the usual formulas of quantum mechanics are true "in the distributional sense" - they're true if you multiply both sides by a smooth "envelope" function and then integrate. Or you can often discretize your Hilbert space into a large but finite set of points, in which case everything's well-behaved (this is what's almost always done in computational physics).

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I do not really understand the question fully, but I am guessing it means how the wave function is collapsed when the particle is not observed.

I am not sure if we know exactly how/why the wave function collapses on observation. I will try, may be you will find the answer somewhere in it.

Let us take an example of double slit experiment where observing the electrons/photons causes them to abandon the wave nature and start behaving like particles. i.e. the interference pattern disappears.

Now, why the measurement causes a switch from wave nature to particle behavior? I will try explaining it with one specific example scenario. This example can be extended to other scenarios with necessary adjustments

Collapse of a wave into an observed particle -

Suppose the entity moving through double slit is an electron (in wave form). Let us observe it by shining a laser on it. If we were to observe the electron with the laser, then the laser has to be reflected off the electron. Right? But, in order to reflect the laser, the electron has to turn into a particle. Why? Because a wave can not reflect off a wave. A wave (laser) can only reflect off a particle. Therefore, in order to reflect the laser, the electron must turn into a particle; otherwise, the observation is just not possible using a laser. Therefore, if there is a measurement via laser, the electron must behave like a particle, not a wave. This is collapse of the wave into a particle that was observed.

Collapse of a wave into a particle that was not observed -

Suppose you shine the laser on one slit and the electron passes through the other slit. Even in this case, before measurement, the electron wave was passing through both the slits. By shining the laser on part of the electron wave, we force the electron wave to turn into a particle, but due to wave density distribution, the particle is materialized through the other slit and does not reflect the laser. So, in this case also, the laser causes the electron to turn into a particle without being reflected off it. This is collapse of the wave into a particle that was NOT observed. Same concept can be applied to measurement via other type of detectors.

Why the Collapse of a wave becomes mandatory?

Why the laser is reflected off the electron in the first place? The actual reason for this can be the fact that the specific laser frequency that detects the electron can not coexist with electron wave at the same point of space time. So, when they meet, one has to collapse. Laser being more fundamental wave, it is the electron wave (less fundamental wave) that gives way and collapses. When collapsed, it becomes a particle and may reflect the laser depending upon where the particle is materialized. This is why when we try to observe the electrons in either of the slits, it causes interference pattern to disappear irrespective of whether the electron was seen or not. This is an example description of wave collapse in classical sense.

Quantum eraser – No need to go back in time

Past the slits, the part waves that passed through each slit interfere with one another even though the two parts remain connected. The connection now is not as smooth as it was before the slits and so, the non-smooth connection causes the whole wave to behave as if there were two different waves, and so, it behaves as if there is interference.

When the laser is shined past the slits, the wave nature collapses as described in previous sections and the electron turns to particle and interference is gone again, so the pattern disappears even if we shine the laser past the slits. Therefore, there is no need to go back in time and there is no erase of anything. It is just a collapse happening past the slits and the particle materializes as if it passed through one of the slits.

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  • $\begingroup$ I don't think this is what the OP meant. $\endgroup$ – tparker Sep 10 '16 at 7:30

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