Irreversibility of "Measurement" ?

[Widdekind]

If quantum mechanical "measurement" represents an irreversible alteration, to the Wave Function, of a quantum system (causing its "collapse" into a singular state)...

 

then doesn't QM measurement resemble Entropy, which irreversibly increases with time? What other physical processes, or quantities, are also "irreversible" ??

 

(Thanks in advance)

[Widdekind]

Are there any known examples, of interactions involving Wave Function Collapse, which conserve (interacting) particle number (or, is there always an "accompanying" annihilation / creation event, of a whole quanta of Wave Function) ?

[Sisyphus]

The entropy of a system always increases with time. Does the "measuredness" of a system always increase with time? I don't think so.

[swansont]

I think the point here is what measurement does to entropy. The number of states available to the system is larger, but the probability of being in that state is less than 1, such that normalized value is 1. Afterwards, you have 1 state. I don't see that the entropy has changed at all due to the wave function collapse.

 

The details of physically making the measurement will no doubt cause entropy to increase.

[Widdekind]

(Thanks for the responses.)

 

Typically, interacting Wave Functions do not "collapse", but become "entangled", entering a single state for all of the entangled particles. "Measurement" destroys this singular entangled state, causes all of the entangled Wave Functions to collapse, and, thereby, puts all of the particles into individually separate states. Would not that increase the system's entropy (making more states available to the ensemble) ?

 

Virtually every time two particles interact and aren't measured, they become entangled...

 

When two particles interact, there is no Wave Function Collapse. But, when a particle interacts with a 'measuring device', the measurement has a definite outcome: the particle's Wave Function "collapses" into a pure state, of whatever the device measures.

 

Marc Lange. An Introduction to the Philosophy of Physics, pp. 259,297.

[swansont]

(Thanks for the responses.)

 

Typically, interacting Wave Functions do not "collapse", but become "entangled", entering a single state for all of the entangled particles. "Measurement" destroys this singular entangled state, causes all of the entangled Wave Functions to collapse, and, thereby, puts all of the particles into individually separate states. Would not that increase the system's entropy (making more states available to the ensemble) ?

 

This is what I was referring to. What is the entropy of a|A> + b|B> (where the probabilities add to 1), as opposed to |A>, or |B>?

[Widdekind]

This is what I was referring to. What is the entropy of a|A> + b|B> (where the probabilities add to 1), as opposed to |A>, or |B>?

 

The calculation of entropy, from "counting microstates", weights every such state, by its probability, yes? If so, the entropy of a|A> + b|B> is constant, for all a,b s.t. a+b = constant.

 

Now, when Wave Functions (WFs), of quantum objects, interact "weakly" (when it's "business as usual" under "normal & ordinary" kinds of conditions, such that the SWE still applies, without WF collapse), their WFs become "entangled".

 

The system S is represented as having two eigenstates v₁ & v₂... The initial state of S is some superposition v of v₁ & v₂... If the radiation field [S] interacts with a quantum system M, such as an atom, then the evolution of the joint system S+M is governed by the Schrodinger Equation, and the resulting state will be a superposition [math]\Psi = c_1 (v_1 \otimes u_1) + c_2 (v_2 \otimes u_2)[/math]... Here, u₁ & u₂ are states of the atom... This is inconsistent with the state of S+M being either [math](v_1 \otimes u_1)[/math] or [math](v_2 \otimes u_2)[/math] [entangled states are not products of separate particle states]. Nevertheless, if an observer O now performs a measurement on the system M, this will project M into an eigenstate (u₁, say). Since the systems are correlated [entangled], S will also be projected into v₁.

 

R.I.G.Hughes. The Structure & Interpretation of Quantum Mechanics.

So, strongly seemingly, you're quite correct -- [math]\Psi = c_1 (v_1 \otimes u_1) + c_2 (v_2 \otimes u_2) \rightarrow v_1 \otimes u_1[/math] (or, [math]v_2 \otimes u_2[/math]) conserves entropy.

 

What if particle number is not conserved, in the interaction, which causes WFC ? Please ponder a standard double-slit experiment, in which a detector screen (D) is irradiated by an incident electron beam (e^-), in what amounts to a "reverse photoelectric effect":

 

[math]e^{-} + D \rightarrow e^{-} : D + \gamma[/math]

where the ejected photon is rapidly reabsorbed, within the detector (D), as thermal radiation [the incident electron beam heats up the detector screen]. At deepest root, this rather resembles the WF "localization", of when a formerly free electron becomes bound to a bare proton during Recombination in the ISM or IGM. The SWE is completely "probability preserving", so the irreversible act of particle creation / destruction, in which a whole new probability distribution is "born", or an old probability distribution "vanishes", cannot mathematically be described with said SWE. This is quite akin to WFC, which also represents a "discontinuous quantum jump", beyond the scope or purview of the SWE. So, could not the "quantum measurement problem", of what causes WFC, have something to do, with particle number non-conservation ? And, would not a particle number increase, in an interaction, increase the ensemble's entropy (a whole new particle, with new states available to it, in addition to those available to the previous particles...) ?

[Widdekind]

I found this reference, which contradicts my suggestion:

 

The existence of measurements, in which "nothing happens" (Renninger-style measurement), where knowledge is gained by the absence of a detection, is also difficult to reconcile w/ the view that irreversible acts cause quantum jumps. In a Renninger-style measurement, there must always be the "possibility of an irreversible act" (a detector must actually be present in the null channel), but this detector does not click during the actual measurement. If we take seriously the notion, that irreversible acts collapse the Wave Function, Renninger measurements require us to believe, that the mere possibility of an irreversible act is sufficient to bring about a quantum jump. The fact that such "interactionless" measurements are possible, means that the wave function collapse cannot be identified with some specific random process occurring inside a measuring device.

 

Nick Herbert. Quantum Reality, pg. 191.

 

Does a "failed interaction", resulting in a "failed Localization", at one point in space, force that Wave Function to Localize elsewhere? Or, does a "failed Localization" here, merely mean, that the probability distribution is "expelled" from the "failure locality" (with the "missing probability" being "redistributed", back into the rest of the Wave Function, in a sort of Renormalization) ?

 

 

[Widdekind]

According to Wikipedia, when a Wave Function "fails" to interact with a detector device, then, not being absorbed or transmitted, the Wave Function reflects off the obstacle, producing a diffraction pattern. Such a situation is not [full] Wave Function collapse & Localization:

 

The Mott problem

 

The Mott problem concerns the paradox of reconciling the spherical wave function describing the emission of an alpha ray by a radioactive nucleus, with the linear tracks seen in a cloud chamber...

 

 

Renninger's negative-result experiment

 

In the Renninger formulation, the cloud chamber is replaced by a pair of hemispherical particle detectors, completely surrounding a radioactive atom at the center that is about to decay by emitting an alpha ray. For the purposes of the thought experiment, the detectors are assumed to be 100% efficient, so that the emitted alpha ray is always detected.

 

By consideration of the normal process of quantum measurement, it is clear that if one detector registers the decay, then the other will not: a single particle cannot be detected by both detectors. The core observation is that the non-observation of a particle on one of the shells is just as good a measurement as detecting it on the other.

 

The strength of the paradox can be heightened by considering the two hemispheres to be of different diameters; with the outer shell a good distance farther away. In this case, after the non-observation of the alpha ray on the inner shell, one is led to conclude that the (originally spherical) wave function has "collapsed" to a hemisphere shape, and (because the outer shell is distant) is still in the process of propagating to the outer shell, where it is guaranteed to eventually be detected.

 

In the standard quantum-mechanical formulation, the statement is that the wave-function has partially collapsed, and has taken on a hemispherical shape. The full collapse of the wave function, down to a single point, does not occur until it interacts with the outer hemisphere. The conundrum of this thought experiment lies in the idea that the wave function interacted with the inner shell, causing a partial collapse of the wave function, without actually triggering any of the detectors on the inner shell. This illustrates that wave function collapse can occur even in the absence of particle detection...

 

 

Diffraction

 

A true quantum-mechanical wave would diffract from the inner hemisphere, leaving a diffraction pattern to be observed on the outer hemisphere. This is not really an objection, but rather an affirmation that a partial collapse of the wave function has occurred. If a diffraction pattern were not observed, one would be forced to conclude that the particle had collapsed down to a ray, and stayed that way, as it passed the inner hemisphere; this is clearly at odds with standard quantum mechanics. Diffraction from the inner hemisphere is expected.

 

EDIT: When part of a Wave Function impinges upon, and reflects off from, some surface, that incident Wave Packet evolves through three broad states of development:

 

1. incoming wave packet: <p> towards surface, spatially varying phase
   
2. collision: <p> = 0, spatially constant phase & doubled probability density (incoming & outgoing waves "pile up", effectively compressing to half the original longitudinal length, as they cross through themselves)
   
3. outgoing wave packet: <p> away from surface, spatially varying phase
   

Now, bound-state electron orbitals, if they're Hydrogen-like, are stationary states, with no net momentum (<p> = 0), and, hence, spatially constant phases. Logically, therefore, it would seem that the probability for absorption, of the electron, into the obstacle surface, would be maximum during that second, "turn around", phase. If the absorption occurs, the Wave Function collapses & localizes, into an available bound-state orbital. Otherwise, the reflection process continues normally, according to the Schrodinger Wave Equation, with no discontinuous "quantum jump". In theory, that reflected wave could then, later, still interact with other obstacles into which it impinges.

 

For example, please ponder a beam splitting experiment, which "quantum splits" the Wave Functions of photons, along two separate pathways, where one such path is much longer than the other. Thus, each half wave packet reaches the nearer detector screen first. Each time that those half wave packets fail to interact, they reflect off, backwards, back towards the original source. Imagine that, during that collision & reflection process, another detector is inserted along the shorter path. If, before the other half wave packets, on the longer leg, reaches the far-off detector, the failed-to-interact reflected wave packet reaches the newly inserted detector, then the whole Wave Function could collapse, into said newly inserted detector, instead of that far-off detector, at the end of the longer leg (???).

Edited August 7, 2010 by Widdekind