# Decoherence - why does the fringe disappear?

1. Feb 4, 2006

### zekise

Recent double-slit experiments with massive molecules, such as fluoro-fullerene consisting of 108 atoms and atomic mass 1632, see heavy matterwave experiment, show that the interference fringe will disappear if the FF molecule emits a thermal photon, or collides with a gas particle, where the mass of the gas particle is immaterial. I am having a hard time understanding some of the explanations given for this phenomena in different places, which I think are as follows:

1- The emission or collision produces an entangled pair. Therefore, as shown in entanglement experiments (Scully, Walborn), the fringe gets obscured by the other entangled member, and can only be revealed if we capture the entangled particle, and do a coincidence selection. Original fringe cannot be revived.

2- Upon emission or collision, the wavefunction collapses and the molecule gets localized, and the partial wavefunction due to the other slit disappears, and so does the fringe. No need to explain this in terms of entanglement. Fringe cannot be revived.

3- The emission or collision passes which-path information to the emitted or collided particle. Although this information has not been measured, in principle it can be obtained. Thus the Principle of Complementarity dictates that the fringe would disappear as the which-path becomes known. Furthermore, if we erase this information, the fringe can be revived.

These are different and contradictory explanations. Any and all insights into this phenomena is highly appreciated. Thanks.

2. Feb 5, 2006

### vanesch

Staff Emeritus
Decoherence is best understood if one does NOT consider collapse (but takes on an MWI viewpoint - it is not for nothing that I push this view here ; it is for its explanatory power in exactly these situations).

The fundamental idea of decoherence is that, when a system 1 gets entangled with a system 2 (mostly through an interaction between both), the potential interference that was possible by observing system 1 alone, is now displaced to the overall system, and will only be observed when doing CORRELATION MEASUREMENTS acting on both systems.

Imagine that system 1 was in a state |a> + |b>, and that an experiment was designed to test the interference between |a> and |b> (in other words, an observable I1 that gives 1 for the eigenstate |a>+|b> and gives 0 for the eigenstate |a>-|b>). Clearly, if we look at system 1 with this observable (with this experiment), we will see interference (always a click, always result "1").
Even we consider the composed system of system1, and system2, as long as both are in a PRODUCT state (|a>+|b>) |u>, the observable I1 (which is now in fact, the observable I1 x 1 on the product hilbert space) will show "interference" (that is, always the same outcome).

But when we now have the systems entangle themselves, into
|a>|v> + |b>|w>, and we NOW apply the observable I1 x 1, we will NOT find interference anymore. We will find 50% "1" and 50% "0".
The reason for this is that the observable I1 operates ONLY on system 1, and that its results are hence determined by the reduced density matrix of system 1, which changed, due to the entanglement, from the pure state |a> + |b> into the mixture of 50% |a> and 50% |b>.

So, if you only look at system 1, IT LOOKS AS IF COLLAPSE OCCURED. It looks as if the state |a> + |b>, due to the entanglement with system 2, changed into a probabilistic mixture of 50% |a> and 50% |b>.
So did collapse occur or not (and hence, was system 2 a "measurement device" ?).
Unfortunately, no. Because if this were the case, we wouldn't be able TO RESTORE INTERFERENCE, by looking also at system 2, as is shown in several quantum erasure experiments. In order to explain THOSE results, one cannot accept that the interaction collapsed the wavefunction of system 1, but one needs the entire, entangled state
|a>|v> + |b>|w> and NOT the statistical mixture of 50% |a>|u> and 50% |b>|w>. So collapse didn't occur physically, it only APPEARED to be so when we only looked at system 1, because then the REDUCED density matrix is sufficient to explain all results (while the TOTAL density matrix is needed for the results on both systems, and all correlations - total density matrix which has not been changed from a pure state to a mixture - which is the density matrix form of collapse).

That's why the MWI view takes it that this collapse NEVER occurs, and that all collapse is only apparent because we limited ourselves to a part of the system that got entangled with something else. Of course, once you get entangled with *the environment*, you are totally lost, and you'll probably never be able to do coincidence measurements that take all entanglements into account, to restore interference. One can call this a kind of practically irreversible entanglement, and this will result in you always obtain correct results through collapse for feasible experiments (because the result that would be wrong, namely the restoration of interference, is an experiment which is practically impossible to perform). As such, the *apparent* collapse in measurement has been explained, because measurement devices entangle the entire system to the environment, in an intractable way.

Yes, this is the "decoherence" view (part of the MWI view), and is in my opinion the cleanest one.

One has then to postulate some "special physics" for this process, which cannot be described by the usual unitary interaction operators, and one runs into trouble when one IS going to restore the fringes using the second system - as long as this is experimentally feasible.

That's the same idea as 1, but there's confusion about "erasing this information". "Erasing the information" comes down to performing a measurement that EXTRACTS THE COMPLEMENTARY INFORMATION. For instance, a measurement that will look at the (|u>+|v>) state of system 2, and will give you the right TAGGING to extract the subsample of a/b results that will show interference.

Quantum erasure has been too often presented as: when you "erase" the information on the remote system, magically, the interference fringes appear at the first system. That's NOT the case. You need the RESULT of the remote measurement which "erased" the which-path information (the a or b information, hence the u or v information) in order to find the SUBSAMPLE of the a/b data in which interference can be seen.

3. Feb 8, 2006

### pavsic

This is really a very clear explanation showing the merits of many worlds interpretation (or relative state interpretation) first proposed by Everett. Unfortunately Everett could not publish his long PhD thesis, but only a rather short paper. This has caused a lot of confusion and misunderstanding regarding the MWI.

In 2002 I discussed this with B. De Witt at a conference in Washington and he said that most problems that people find in MWI had already been clarified in Everett's PhD thesis. Had it been published right at the beginning (and not only with a great delay in a book), all such misunderstanding and confusion (and reluctance) concerning MWI would have been very probably avoided. Few people that criticize MWI have read his thesis.

However, I think that there are still issues that are not completely clarified in Everett's PhD thesis and subsequent works, including the modern ones. I discuss this in Part IV of my book The Landscape of Theoretical Physics: A Global View (Kluwer, 2001). A link to the contents of the book and sample pages can be found on my Home Page http://www-f1.ijs.si/~pavsic/ .
In my opinion quantum mechanics cannot be fully understood without employing some radical views that are discussed in my book. However, I do not claim to understand QM (Nobody really understands quantum mechanics" [Feynman]).

Matej Pavsic

Last edited: Feb 8, 2006
4. Feb 9, 2006

### setAI

I like the cut of your jib- vanesch!

"decoherence is just a matter of degree. There is never a moment after which an object's invisible counterparts cannot affect it any longer. It just gets too expensive to set up the apparatus that would demonstrate their existence. " ~David Deutsch

5. Feb 12, 2006

### zekise

Hi Patrick – many thanks for your wonderful explanation. I was delayed in getting back as I was for a while schlossed up with Schlosshauer. Not that I pretend to fathom the formalism.

A good answer generates more questions. So here they come if you don’t mind :

Question 4) What do you mean by the DIFFERENCE of two states |a> - |b> ? Is this simply a phase shift of the ‘b’ component of the wavefunction resulting in an anti-fringe when interfered?

Question 5) Is there such a thing as a “quantum coherent state” for a single particle? Now in the heavy molecule experiment, Zeilinger et al used a Talbot-Lau grating to collimate the FF beam. This would be a necessary step to observe interference. However, if the experiment can be practically done on a single molecule, would this be unnecessary? So is it correct to say that when the FF solid sublimates, and a single molecule flies at the interferometer, that it is already in a coherent state? That is, if the molecule is NOT entangled in any manner and therefore IS isolated, it will be in a coherent state?

Q6) Assume a single FF molecule is coherent and flies through the interferometer. Assume that we somehow are able to figure out (calculate) the sublimation ejection trajectory (without perturbation of the molecule) before the slits, and therefore calculate which slit the molecule will pass through - would this not put it in a mixed state and destroy the IP? Would this, if possible, be an example of non-entangled decoherence? Is there such a thing?

Q7) Now the FF molecule consists of 3,264 particles bound together by electric forces. I assume this entangles the components of the molecule with each other, as they are continuously interacting with one another. Why is it that this entanglement does not decohere the molecule, in the manner that an emitted thermal photon does - and result in IP loss?

Q8) If the FF emits an ionization electron instead of a thermal photon, I understand this electron will be entangled with the molecule. Is this always the case, or only if it is emitted in a certain manner? Now assuming the electron before ionization was entangled with the molecule, and remains entangled after emission, then what is now different that results in the IP disappearing?

Q9) Do all interactions result in a degree of entanglement? If this is the case a particle must always be in a state of perpetual entanglement with a huge number of other particles as force fields can travel quite a distance. Is it fair to say most particles are entangled with most other particles in the universe?

I will be polite, and take my seat. But there are more I need to ask – on degrees of freedom, superselection, disentanglement, reversal, and information. TIA

Last edited: Feb 12, 2006
6. Feb 13, 2006

### zekise

Help Help

bump

7. Feb 14, 2006

### vanesch

Staff Emeritus
Yes. I addressed this in another post on a parallel thread somewhere. It's a matter of choice of phases, initially, but you are entirely right that (whatever your phase choice), |a>+|b> and |a>-|b> will give you the fringe and anti-fringe (where you arbitrary can call one "fringe" and the other 'anti' of course).

There's a terminology issue here. I think you're mixing up the terminology for "coherent state", "pure state", entangled state, and mixture, no?
We can go into details if you want to. They will even be, at a certain point, interpretation-dependent.

When I look at your next question, it seems that what you call "coherent" state, seems to be a pure state that is not entangled, and that can give rise to interference when applied to a certain measurement. So I'll take it as such, knowing that I'm guessing here...

If we were to be able to know the ejection trajectory to such detail that we knew which slit it went through, then the measurement of the IP would simply not be such a measurement. Let's not forget that "interference" is such a measurement on a state, that we can consider the initial state as |a>+|b>, and that we apply a measurement that distinguishes |a>+|b> from |a>-|b> (fringe from anti-fringe). Interference shows when there is a high probability to get the first, and a low probability to have the second (or vice versa). If you now have an initial state which is NOT |a> + |b>, but simply |a>, then the IP will not show (the probabilities will be equal for the two outcomes). In all 2-slit interference, |a> stands for "beam through the first slit" and |b> stands for "beam through the second slit" (and the relative sign gives you the relative phase between both paths - which is up to a point conventional).

Because they all undergo the SAME interference measurement, in other words, we do an interference experiment on the TOTAL state, and not on some substate (usually, we loose the thermal photon, and we only look on "one leg" of the entangled pair (molecule-photon)).

Because the interference experiment didn't apply to the electron when it was emitted, while it took part in the experiment when it was still part of the molecule.

Here, we enter interpretation-dependent issues. In the von Neumann view (with projection), a measurement UNDOES entanglement (as the state is now an eigenvector of the measurement operator, and hence again in a product state).
In the MWI view, each measurement ENTANGLES the observer with the system, but as the observer is going to see only one branch, this entanglement becomes unobservable.
Interactions usually result in entanglement according to the Schroedinger equation (it doesn't have to be so, but there are chances that things go that way).

8. Feb 18, 2006

### zekise

Patrick - I am pretty sure you do not mean that if we only went through the motions of calculating the FF molecule ejection trajectory, then the system would somehow become a mixed state |a> or |b>.

Therefore it must be one of the following 2 cases (?):

1- The sublimated FF molecule is entangled with the rest of the FF solid (and is not coherent to itself). In this case, there will be no interference pattern observed because the molecule is entangled and the fringe is obscured by the anti-fringe. But in reality we do see a vivid interference pattern in matterwave experiments (without correlation selection).

2- There is no entanglement between the molecule and the rest of the FF solid (and the molecule is singularly coherent). In this case, in theory, we can obtain the recoil and calculate the trajectory and figure out |a> or |b>, and also get the IP. Surely it is not a matter of practicality. But, it would be incorrect to say, IMO, that the IP survives because in practice we cannot calculate the trajectory, even though in principle the information is there in the solid.

What gives?

(I think this is closely related to the issue raised in the other Decoherence thread.)

My own hunch is that if there are a large number of degrees of freedoms in the solid, then the information IS SIMPLY NOT THERE. As we reduce the degrees of freedoms, the information BECOMES THERE, but the molecule starts becoming entangled with the solid. Once entanglement starts to set in, we will start losing the IP.

If this is the case, then this points to three things. 1- there is no information without entanglement. 2- entanglement is not a binary yes-no thing and is graduated (or is it quantized?). And 3- Entanglement can be broken and destroyed in the Copenhagen (non-MWI) model, like when a member of an entangled pair collides with a system with large degrees of freedom (iow, decoheres and superselection results).

Last edited: Feb 19, 2006
9. Feb 19, 2006

### vanesch

Staff Emeritus
Ah, you mean, that by the simple act of working out the calculation, we'd change the state of the system ? No of course not

But you think already of "trajectory", I'm talking about "quantum state". The result of your most fine-grained calculation will be a *quantum state* of the ejected particle, and that can be a quantum state that "looks like a trajectory" (state |a> or state |b>, say, that go through slit 1 or slit 2), OR (most probably) it will be some kind of spherical wave state, which will be something of the kind |a> + |b>. If you say "trajectory" you take it already for granted that the quantum state will be of the |a> or |b> flavor, but that doesn't need to be so (I'd even say that it almost certainly will NOT be the case, by the very fact that interference is observed!).
It's a bit as if you are asking about the "precise trajectory calculation" of a particle in a momentum state. A pure momentum state (which could be the result of a precise calculation) is NOT to be seen as a "trajectory", but is a superposition of "position states" (trajectories, if you want to).

Yes, so that's probably NOT the case, as you point out.

No. IF we are able to distinguish between the two "recoil states", THEN we're back in your case 1. So you can only get an IP if the two different recoil states are, in fact, almost identical (see thread with mirror).

That the information is NOT there in the solid ! Not even in principle. That the quantum state of the solid is (almost) the same after emission |a> or |b> and that the (biggest part of the) state of the solid can be factored out.

YES! You've got it...

Yes. The only problem with this view is that you have to select arbitrarily when this thing is becoming "irreversible".

Concerning the fact that entanglement is gradual, yes, of course. Consider this:

Initial state (|a> + |b>) |c>

After interaction, |a>|c> could evolve into something like |a>(|d> + eps |k>), while |b>|c> could evolve into something like |b>(|d> + eps |l>), where |k> and |l> are orthogonal states.

We now have the final state:

|a>(|d>+eps |k>) + |b>(|d>+eps|l>) = (|a>+|b>) |d> + eps (|a>|k> + |b>|l>)

if eps is a small number, we've still essentially the product state, with a "tiny bit" of entanglement. That's what my "almost" statements referred to earlier.

If you do an interference experiment on |a>+|b>, you'll find a slightly decreased IP (because of the eps part). If you do a correlation eperiment to test entanglement between a/b and k/l, you'll find a very very small correlation.

10. Feb 22, 2006

### zekise

Very interesting and thanks for the excellent post. So we are saying:

A1) If the substrate consists of a large degrees of freedom (LDF), then the emitted (sublimated) molecule will NOT be entangled with it, and the substrate will not enter into a new state after emission. Therefore, there is no information in the substrate.

A2) Now consider the other side of the experiment, where the superposed molecule will hit the screen and get "absorbed". Now I understand the decoherence folks are saying (pls. correct if necessary) that if the screen consists of a large degrees of freedom, then the molecule will collapse, and statistically we will have a fringe, which implies that there will again be no entanglement with the screen. However, in this case, there is residual information left at the screen (aka the measurement) and the screen will enter a different state, and this will also be in principle irreversible ("indelible").

A3) So here we have an example of information transfer without entanglement. What is the nature of this information, and how is it different from entangled information? Is this information the energy of the molecule, its momentum, or its position or spin or what? What really bothers me here is that A1) and A2) are the time reverse of each other. But A2 does result in information transfer, and A1 does not.:uhh: What gives?

But I understand decoherence theory says the pointer states that will emerge are superselected for stability or some other criteria ("Quantum Darwinism" as per Zurek). So they are not arbitrary (indeterminate). And this is why classicisity is not arbitrary.

A4) What is it about the "large degrees of freedom" that inhibits entanglement (and can cause disentanglement)? Is it the density matrix quickly losing its off-diagonal entries when there are more degrees of freedom? How many degrees of freedom does a single isolated atom have, and what does the density matrix look like, and why does the off-diagonals NOT approach zero?

A5) Now is it fair to say that when an entangled particle meets a system with large degrees of freedom (LDF), it will disentangle and classicisity will emerge? If it did not disentangle, then the particle will put the system into a superposed entanglement, and the entanglement will spread throughout the system and the environment. Soon you will be seeing things vanishing before your own eyes! (Schro's cat). Then the universe would become one big gigantic entangled mess, and wierd things would be happening, such as particles refusing to collide, and there would be no ability to sustain life or consciousness. So we are saved by this LDF. But I do not understand how this LDF causes disentanglement.

bruce2g, thanks for the article on Penrose. I wonder if there have been any results from this proposed experiment? I am not sure of the significance of demonstrating a macroscopic superposition. I am not familiar with Penrose's theses (except that he claims a new sort of physical phenomena, and that the brain is loaded with that, resulting in consciousness). I think someone wrote a paper that there can't be anything coherent in the brain for too long (heh heh, no pun intended), and it will decohere, and the brain is a classical device. I think decoherence theory has really taken the steam out of the mystics and the consciousness people.

Last edited: Feb 22, 2006
11. Feb 23, 2006

### vanesch

Staff Emeritus
The point is not so much the large number of degrees of freedom, but the coherent interaction of the particle with the entire system, which ends up in the same state after the interaction or not. As a simplistic classical example, the difference between, say, an elastic collision (where the colliding objects have no "souvenir" of their collision) vs an inelastic collision (where some internal degrees of freedom got changed due to the collision, and can hence serve as "memory" for a collision).
I'd even say that having a lot of degrees of freedom would *increase* the chance of the inelastic collision ; so it is only in special cases that this doesn't happen (such as with a mirror). The mirror is rather the exception, where no "remnant" is left. Most stuff doesn't act as a perfect mirror, and some remnant is left.

Well, decoherence doesn't make much sense outside of an MWI perspective. And then things don't collapse, they just entangle! The screen just *entangles* with the molecule:

initial:
(|moleculepos1> + |moleculepos2> +...) |virginscreen>

becomes:
(|moleculepos1>|screenflash1> + |moleculepos2>|screenflash2>+...)

As we'll get entangled up also in this way, we'll only observe one of the branches.

Yes, it is quite hopeless (although in principle possible) to evolve again backwards the final state back to the initial state, where the screen information is "erased" (every branch gets back to the |virginscreen> state). At this point, you can just as well call this "a collapse" because of the quasi-impossibility to merge again the different terms, and forget about all the others. (at least, that's how one can introduce a "practical" collapse in the MWI view - IMO that's how decoherence is to be seen).

No, on the contrary: it is because the screen DID get entangled with the molecule state that it got the information. And it is because WE got entangled with the screen that we knew about it. (in the MWI - decoherence view).

What decoherence says is that the Schmidt decomposition in the observer x rest of universe basis will result in classically-looking states for each term. But you still have to pick out one of them.

In the decoherence view, this is rather irreversible entanglement, and not disentanglement! But IF YOU APPLY A COLLAPSE, of course ,this irreversible entanglement (resulting in different terms) followed by the picking out of one term (collapse) will then result in a product state (= non-entangled system). In fact, more degrees of freedom you have, more chance you have that this happens. It is only in special circumstances that all those degrees of freedom act coherently (such as is the case with a mirror), and then you do not have entanglement.

That depends entirely on the interactions that are going on.

It depends whether these DoF interact coherently or not. If they do NOT interact coherently, you will get irreversible entanglement, which can also be seen (after projection) as disentanglement (although it was in fact entanglement from a decoherence PoV). If they interact coherently (as do the electrons in a mirror), you can avoid extra entanglement, and hence preserve the *original* interference which was a proof of the simple entanglement.

Yes, that's in fact the real decoherence-MWI view. And once YOU get entangled too, you'll only observe "one branch" (one term). It is then up to you to decide whether or not you want to drag along the entire wavefunction (and keep the entanglement) or whether you will limit your attention only to the term you are observing (= collapse), at which point you arbitrarily decided to "throw away" the other terms, and hence disentangle the entire circus.

No, on the contrary (that's the MWI view). Because of this entangled mess, we WON'T often be seeing quantum interference effects and we'd have the impression to live in only one branch, which looks fairly classical.
So there is no *real* disentanglement in this view (as you say, we're living in a huge entangled mess) but things are so hopelessly entangled with eachother that almost ALL OBSERVATIONAL CONSEQUENCES (which are quantum interference and correlation results) are gone. Which is exactly what we observe ! Remember that we cannot observe entanglement, we can only observe *interference*. And entanglement KILLS interference of the subsystems, to make it only obervable of the overall system if you do a carefull correlation study. Now, if we get entangled with the entire universe, we won't ever be able to do all the necessary measurements to find out the overall correlation ; we'll always be limited to measurements on the subsystem, and as such, NOT find any correlation (= NO proof of quantum interference, and hence "local" entanglement).

cheers,
Patrick.

12. Feb 23, 2006

### zekise

A2) Now I am a bit confused - in this case (A2), the experiment is just a standard double-hole matterwave experiment and it shows (self) interference on the screen, due to wavefunction components from each hole. So would this not indicate that there is NO entanglement between the molecule and the screen, because if it were entangled we should get a gaussian? (MWI really confuses me, so let us first discuss the decoherence view in the non-MWI interpretation, if that is OK with you.) Since we see an interference pattern emerge from statistical registrations of the molecules on the screen, then there cannot be entanglement between the molecule and the screen. And there is some information passed to the screen such as momentum transfer (or in the case of a photon, it would be the absorbtion of the photon energy). If the screen was a detector array, then it could say, "molecule #n was registered at (x, y)".

A2.1) So if indeed there is no entanglement in this non-MWI view, then what caused that? Collisions between the molecule and an isolated gas atom has shown to produce entanglement. But collision between molecule and the screen in a vacuum has shown to produce interference, so presumably there is no molecule-screen entanglement.

A4) Now imagine in the same expirement we place in front of the (macro) screen (which is really an array detector) an array of suspended and isolated atoms (micro), one atom for each cell. Then if the molecule first collides with one of these atoms, we WILL get entanglement, and lets say the molecule (but not the atom) then registers itself on the screen array detector. We will not observe interference anymore, and instead will see a bell-shaped clumping statistically. So again, a (micro) "screen" made of isolated atoms will show gaussian, while a regular (macro) screen will show interference. What is the qualitative difference?

A5) Now if it is the case of no entanglement with a regular macro screen, then would this not indicate that it is possible to break entanglement? Imagine the molecule first colliding with an isolated atom on its way to the screen, and getting entangled with this atom. The molecule then proceeds to hit the screen and essentially becomes part of the screen. We will observe a gaussian statistically. Now, since the screen did not generate an entanglement as per A2) with the molecule, then it would be reasonable to assume that the entanglement of the molecule with the isolated atom would NOT be able to survive, and will get broken or decohered. Otherwise you would have an entanglement between the atom and the macro screen (where the molecule is now "integral" to the screen), which as per A2) was shown not to be possible.

I also believe this is what Scully has shown. The 2nd photon is not entangled with the detector that absorbed the 1st photon. There is no mention of this.

So in this non-MWI view, it is possible to disentangle, and the universe would be generally disentangled, because most of the universe acts like a decohering macro screen.

A6) BTW, in addition to lack of interference, can't we use violation of the Bell inequality for proof of entanglement between 2 particles?

regards

Last edited: Feb 23, 2006