View Full Version : Some questions about Decoherence

kevin1981

2013-Sep-28, 03:10 PM

Hi there,

I have read a lot about how, wave function collapse, has nothing to do we our instruments and that it is an intrinsic part of nature. But, i was reading about decoherence yesterday and it implied, at least to me, that the above phrase is not quite true.

It makes sense to me that the information is dissipated through the millions of particles that make up our instruments rather than the particle suddenly being in one certain place, acting all spooky.

Is decoherence a tested theory or just another interpretation of quantum mechanics ? How likely is it, that decoherence is right ?

Cheers

Shaula

2013-Sep-28, 04:05 PM

Decoherence is an interpretation. So is wavefunction collapse. The underlying maths is the same, the model is the same (because it only predicts measurements we make, not what is 'actually' going on).

Decoherence makes it far easier to understand things like weak measurements, delayed choice eraser experiments and so on because it doesn't have collapse in. It is certainly a more useful interpretation for actually using the models and the maths because it is closer to the way the model works. That said it blurs the lines between classical and quantum in a way some people find unsatisfactory, the Copenhagen interpretation with a collapse marking the end of quantum behaviour fits what we see better for some. Decpherence also leads to a bit of a problem for some in that when you take the universe as a whole there are no external sinks and so you end up with a universe in superposition. I am not sure there is an exact mechanism within the collapse interpretations to solve this but I believe it is seen as a more tractable problem.

Decoherence is not a function of our instruments per se - it is a function of anything that can act as an information sink.

kevin1981

2013-Sep-28, 06:04 PM

Decoherence is an interpretation. So is wavefunction collapse.

I see. So, when we make an observation/interaction, that changes the nature of the quantum system. If i remember correctly, the wave function is a mathematical tool to calculate the probability amplitude of where a particle may be. Is that correct ?

If so, then thinking of it physically collapsing, is wrong. Though, i have only just realized that ! So, instead of thinking that it collapses, would it be right to think that once an interaction happens the coherent state changes. It is just that, the wave function has gone because now the particle is located in a certain position (relatively).

(because it only predicts measurements we make, not what is 'actually' going on).

That is a helpful sentence.

Shaula

2013-Sep-28, 07:43 PM

The wavefunction is a mathematical tool that encodes all possible states of a system as a superposition of wave-like mathematical functions. Each of these waves has a phase and it is the interaction of these phases that affects the relative probabilities of each state. When you make a measurement you extract information from this bundle of wavefunctions. After doing this you have a constraint on which ones were possible/probable. This extraction of information affects the relative phases causing previously in phase waves (coherent waves) to become incoherent and start to interfere with each other. This means that future states (which evolve from the constrained wavefunction) cannot have certain values because that would have been inconsistent with the measurement taken.

As for the wavefunction vanishing - this is why I mentioned weak measurements. It is possible to know something about a quantum state without knowing everything you can (which is collapsing the wavefunction in Copenhagen terms). This is where decoherence as an interpretation shines. You can partially collapse the wavefunction, or conditionally collapse it in ways that are hard to imagine if you picture collapse in the simpler Copenhagen terms. If you think of the wavefunction as just a bunch of waves which you can tinker with the phases of complicated concepts like weak measurements or eraser experiments are suddenly not quite so counter intuitive.

So the wavefunction is never really gone in the decoherence picture, instead it is just very, very constrained by measurements. But the thing with quantum behaviour is that as soon as you move away from the point in time you took those measurements quantum behaviour creeps back in.

kevin1981

2013-Oct-03, 08:24 PM

Thanks for answering Shaula. I always read what you write even if i do not answer, i am very busy with life at the moment !

When looking at my keyboard, it looks like a solid object. So, are all the particles that make up the keyboard in one place rather than in a superposition, due to them interacting with each other and the environment? Are the electrons still in a cloud of probability or are they in certain positions ?

Cheers

Shaula

2013-Oct-03, 08:38 PM

Electrons, atoms - all in a superposition. However they are in a constrained superposition - thanks to their interactions with the locale all states are not equally probable. If you think about it there is no experiment that you are doing right now that specifies where any individual atom is. You are not extracting that level of information from a desk by just interacting with it in the way you are. So as long as the overall desk object acts in accordance with your observations of it the atoms and electrons in it are relatively free to do what they like - which is the essence of a superposition in QM. If there is no constraint forcing a system to be in one particular state then it can be in many. As you extract more information (make finer measurements) the potential states the system can be in get more and more constrained.

kevin1981

2013-Oct-04, 04:47 PM

Okay thanks. That sounds a lot like decoherence. Am i right in thinking, decoherence would suggest that, everything is in a superposition but that information gets dissipated through the surrounding environment, so i have no way of seeing the superposition of the atoms in my keyboard?

kevin1981

2013-Oct-04, 04:57 PM

I thought that the more particles there are the less there is superposition. But, what you are explaining is that, the less particles there are and the more information we want to know about single, isolated particles, the less they are in superposition. If we want to know where a particle is then knowing that information forces the other states to disappear. So it is not so much interaction that causes wave function collapse but more the leaking of information from the superpositioned system. Is that right ?

Shaula

2013-Oct-04, 10:38 PM

The concepts are kind of linked. The properties of the table are derived from the atoms that make it up. This very fact means that any interaction with anything constrains the superposition of every other atom in it. The way to think of it is, I think, like this: If we have a sack and fill it with spheres. These can in theory be made of any material. Assume that they are quantum and that there are only 3 materials they can be made of - stone, water, helium (call them a, b, c). So we throw the 3 spheres into our sack and because we don't know what any of them are made of they are in a superposition. We have (a OR b OR c) (a OR b OR c) (a OR b OR c). There are 24 possible configurations ( aaa, aab, aac, aba, abb, abc, aca, acb, acc, baa, bab, bac, bba, bbb, bbc, caa, cab, cac, cba, cbb, cbc, cca, ccb, ccc ).

Now this sack is left outside. It is interacting with the environment, and from that things can be seen - the sack doesn't float away, for example. So the combination ccc (all helium) is strongly unfavourable because otherwise it would have interacted differently with the outside world. We can up the amount of interaction. It rains, suddenly and a lot. The place floods and the sack bobs away. That strongly disfavours aaa (all stone) if the sack doesn't immediately sink. So each and every interaction that effectively tells us something about the contents of the sack reduces the possible combinations that could be in there. Of course in quantum terms this has a very definite effect - it doesn't just tell us something about the pre-determined state of the spheres in the sack, it actually causes a change in something quite fundamental.

I find this sort of thing hard to convert into words well - but essentially the more a system is interacting with the environment the more chances there are for information to leak out irreversibly (this is the key point - the information has to be removed from the system and dissipated into a general environment). When that happens the number of states the system could be in (the amount of superposition) are reduced because these interactions constitute a measurement that constrains the system.

So it is not actually some hard and fast number of particles. It is simply how easy it is for information to escape. More particles makes this easier in almost every case simply because there are more things for the outside to interact with. This kind of hints at the fundamental weakness of the decoherence interpretation - how you define a system and an information sink. This is, to be honest, largely arbitrary. Or at least difficult to actually pin down precisely. It is easy for an experiment, where you have the instruments and the objects of study. In the real world it gets more complex.

kevin1981

2013-Oct-04, 10:56 PM

I find this sort of thing hard to convert into words well - but essentially the more a system is interacting with the environment the more chances there are for information to leak out irreversibly (this is the key point - the information has to be removed from the system and dissipated into a general environment). When that happens the number of states the system could be in (the amount of superposition) are reduced because these interactions constitute a measurement that constrains the system.

That is sort of what i said :) I do understand what you have wrote and i feel i have gained some more knowledge. How you guys (physicists) remember so much information is beyond me..

Can we pin point exactly where an electron is if we know absolutely nothing about it's momentum ? The same question goes for an atom as well.

Shaula

2013-Oct-05, 04:41 AM

Can we pin point exactly where an electron is if we know absolutely nothing about it's momentum ? The same question goes for an atom as well.

For both the equation is the Heisenberg uncertainty principle - so dX.dP >= h/2pi which implies that as the uncertainty in P tends to infinity the uncertainty in position (X) tends to zero. In experimental practise it is hard to reduce our uncertainty in P to actually zero (things like how the electron was produced, how it has interacted since tend to constrain P and so give a finite, if large, size to dP)

As for how physicists remember this - well, like everyone else (and like I keep telling people not to rely on!) we make up little stories and pictures in our head that make sense to us. Then we try to communicate them and realise as we go so that they unreliable to reason with! I always wonder how people remember things like sports teams histories (I'm lucky if I can name one member of a team I just watched play), or navigate (I get lost in small industrial parks). I guess it is just what kinds of data you find easier to store in your head!

caveman1917

2013-Oct-05, 04:14 PM

This is, to be honest, largely arbitrary. Or at least difficult to actually pin down precisely.

In as far as we are talking about an actual observer in that environment watching what happens the definition is easy enough. The environment is all the degrees of freedom that are not being tracked by that observer, or in other words those that are random to him.

The difficulty is to generalize this to a situation where you don't have an observer, which is arguably the goal of the decoherence approach - to do away with the special position of the observer. There was an interesting paper linked to by loglo in that thread where KenG and i discussed the many-worlds interpretation. It argues for the environment to constitute everything outside the closed null surface along the past/future lightcones of the events of start and end of the experiment as that would constitute the necessary one-way information membrane (somewhat like irretrievably losing information to a black hole - ignoring hawking radiation here).

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