# Decoherence question

1. Apr 14, 2015

### zonde

I would like to understand why is decoherence needed in QM.
As I understand decoherence models irreversibility.
But if two quantum systems interact and resulting states are going in different directions they can't produce interference, right? Even if they do not interact with environment. So isn't it possible that isolated quantum system still undergoes irreversible evolution?

2. Apr 14, 2015

### Jilang

A couple of strategically placed mirrors could make the process reversible though. So is it really irreversible?

3. Apr 14, 2015

### atyy

Decoherence is simply a part of QM in any interpretation. It comes for free.

In Copenhagen, it is unknown whether there is such a thing as an isolated quantum system. The wave function is just a tool to make predictions for observations, which necessarily means an interaction between the apparatus and the quantum system.

In Bohmian Mechanics and MWI, an isolated quantum system undergoes reversible evolution.

Last edited: Apr 14, 2015
4. Apr 14, 2015

### zonde

But do not mirrors represent environment?
And secondly in order to get observable interference mirrors would have to be place very precisely and for many identical interactions. So FAPP that won't happen unless you try to do this (in experiment).

5. Apr 14, 2015

### Jilang

You asked if the process were reversible. The answer would be yes. Would it reverse spontaneously, not without the mirrors. Is there decoherence? I would say not.

6. Apr 14, 2015

### Staff: Mentor

It not a matter of being needed or not - its deducible from its basic axioms - no way around it.

What you MAY be asking is why its needed in interpretations and explaining collapse etc.

It isn't. In fact Ballentine specifically states it has no interpretational value at all and in his interpretation (the ensemble interpretation) it doesn't. However it has shed light on the so called measurement problem and because of that some interpretations take it on board eg Many Worlds, Consistent Histories, Bohmian Mechanics, Ignorance Ensemble to name a few. What it does is change how the measurement problem is viewed - now its not what causes collapse (collapse was never really part of QM anyway) but why do we get any outcomes at all; without going into what that means technically (it's the difference between an improper and a proper mixed state).

If you want to pursue it further here is THE book to get:
https://www.amazon.com/Decoherence-Classical-Transition-Frontiers-Collection/dp/3540357734

It approachable after a first course in QM.

Thanks
Bill

Last edited by a moderator: May 7, 2017
7. Apr 14, 2015

### zonde

Let's take light. Obviously there are light sources that won't produce interference in double slit experiment. But once you have increased coherence of light say with bandpass filter how do you loose it?

8. Apr 14, 2015

### zonde

Wikipedia says something similar:
"Specifically, decoherence does not attempt to explain the measurement problem. Rather, decoherence provides an explanation for the transition of the system to a mixture of states that seem to correspond to those states observers perceive."
But I don't get it.
Let's take example. Beam of polarized light goes through polarization beam splitter at an angle. Later two resulting beams are detected with photodiodes. If I look at two beams after beam splitter they represent two coherent states. When a beam hits detector I get signal as photons are absorbed by electrons. I don't really see where decoherence comes into the picture and what it explains here.

I don't want to go further. Basic understanding would be enough.

Last edited by a moderator: May 7, 2017
9. Apr 15, 2015

### Staff: Mentor

The photons are absorbed by the electrons. But what happens to the electrons? The absorption of the photons is an interaction, so we'd expect the electrons to end up in a state which is also a superposition; and the electrons are part of the detector, so instead of having a detector which either detected or didn't detect, we'd have a detector that is in a superposition of detected and didn't detect. But we know that real-world detectors don't act that way.

Decoherence explains how a macroscopic object can (and usually does, unless we do heroic things to prevent it) behave classically, even though it is made up of a very large number of particles each individually doing weird quantum things.

A good non-technical introduction, mentioned in many posts in this forum, is https://www.amazon.com/Where-Does-The-Weirdness-Mechanics/dp/0465067867

Last edited by a moderator: May 7, 2017
10. Apr 15, 2015

### f95toli

Don't get sidetracked by focusing too much on experiments in quantum optics. This is a very specific situation and it gets difficult to analyze because photons are so "weird".
If you want to understand how/why decoherence affects quantum systems you'd be better off looking at some other system. The easiest example is probably an electric RCL circuit:the oscillations of the LC circuit (which in QM terms is just a harmonic oscillator) are damped by the resistor. A full QM treatment of this which includes decoherence could e.g. be a hamiltonian consisting of a harmonic oscillator aa and a Lindbladian which accounts for the decay and is inversely proportional to the resistance.

This is formalism is incidentally the starting point for Jaynes-Cummings Hamiltonian which is used to model cavity/circuit QED experiments.

11. Apr 15, 2015

### stevendaryl

Staff Emeritus
I think it may be just buried in the assumptions. You might say something like

If a system is in state $\psi$ and you measure an observable corresponding to an operator $O$, then you will get $\lambda$ with probability $|\langle \psi | \psi_\lambda \rangle|^2$, where $\psi_\lambda$ is the projection of $\psi$ onto the subspace where $O$ has eigenvalue $\lambda$.
That doesn't seem to have anything whatsoever to do with decoherence. However, you need to ask the question: What does it mean to say that we have performed a measurement of the observable corresponding to operator $O$? It seems to me that it means that you've set things up so that microscopic details (such as the value of some observable property of a particle) are amplfied to make a macroscopic difference (the presence or absence of a "click" on a Geiger counter, for instance). To really make sense of that, it seems that you need to know that macroscopic states are distinguishable, and don't interfere with each other the way that microscopic states do. And something like decoherence is involved in making macroscopic states distinguishable (so they "classical" for many purposes).

Of course, you can ignore the question of what it means to measure something, and just develop the theory under the assumption that observables are measurable.

12. Apr 15, 2015

### zonde

As I understand decoherence does not solve problem of definite outcomes. Only disappearance of interference.
Speaking about states I believe that it's not exactly detector that has two states of detected and not detected but electrons instead. And the click state is represented by electrical pulse that have left detector and does not dissipate in environment (as we would not be able to make a record of that click).

How do you know? In quantum world there are no superposition states, there are only superposition of states. If you say that in classical world you could see superposition state if there would be one because it's different that quantum world ... well it would seem quite strange reasoning. But I suppose it's related to cats.

13. Apr 15, 2015

### StevieTNZ

I always think that why we don't see a 'classical' system in superposition is because 'collapse of the wave function' has occurred.

14. Apr 15, 2015

### zonde

And what it is that we don't see? How does "'classical' system in superposition" look like?

15. Apr 15, 2015

### Staff: Mentor

QUOTE="zonde, post: 5076477, member: 129046"]As I understand decoherence does not solve problem of definite outcomes. Only disappearance of interference.
[/quote]
Yes, but that's enough to explain the lack of observed macroscopic weirdness (except in experiments where we have gone to great lengths to suppress interaction with the environment), and that's all that most people claim for decoherence. We still end up not knowing what the detector detected until we look, but we know that we have one or the other result with no interference between them.

Before the discovery of decoherence, there was nothing in theory to explain why Schrodinger's cat would be either in a dead state or a live state, but never in a superposition of the two before someone opened the box. The best answer most interpretations could come up with was:
That was a less than completely satisfying answer because there's nothing in the theory that explains why some interactions have to collapse the wave function and others don't. Yes, it's true that the wave function collapse happens at the boundary between the quantum and classical worlds, but because "classical" was defined as where the wave function collapse happens, that answer explained little. What made the cat "classical"? Suppose we ran the experiment with an oyster? Or a bacterium? Or just the detector and the cyanide vial?

Decoherence basically says that the ordinary unitary evolution of a quantum system interacting with its environment will produce results that are experimentally indistinguishable from collapse, but without requiring an arbitrary division into classical and quantum and a near-magical non-unitary collapse at the boundary.

16. Apr 15, 2015

### zonde

Can you give reference for experiment where we have gone to great lengths to suppress interaction with the environment and observe macroscopic weirdness?

So from this I take that decoherence is attempt to make QM unequivocal. So it has more to do with consistency of theory than some specific observations.

17. Apr 15, 2015

### Staff: Mentor

18. Apr 15, 2015

### zonde

Photons are not weird (at least not weirder than matter). People who try to model photons the same way as matter particles are weird.

19. Apr 16, 2015

### f95toli

The fact that you are not already aware of these experiments sort of proves my point. Both pop-science and intro courses to QM tend to focus way too much on e..g the double slit experiment and largely ignore just about every interesting experiment that has been done since Aspects work on the Bell inequalities. It is of course true that this is a key experiment in QM, but it is ALSO true that is a very difficult experiment to understand in detail . Most of the recent work in QM (over the past 30 years or) has focused on other types of systems, e.g. trapped ions, atoms and artificial two-level systems such as solid state qubits (quantum dots, superconducting circuits etc). This latter development has to a large extent been what has motivated the development of a relatively simple framework for handling decoherence; this is simply because decoherence is much more noticeable (and important) in these systems, the underlying mechanisms are much easier to understand than in optics and the coherence can also be dramatically increased by careful design (choice of material etc) so understanding it is of considerably practical importance.

Of course they are weird: they are intrinsically quantum mechanical. They are also much more difficult to handle mathematically than e..g a two-level system since you can't even write down a SE. This makes them extremely counter-intuitive.
If you want to get a better understanding of decoherence it is therefore much better to start with a classical system (e.g. a electrical or mechanical resonator or a two-level system) that behaves "classically" when it interacts with the environment, but can be made to exhibit QM effects when sufficiently isolated ( by cooling to down to very low temperatures etc). Some useful models for these can even by analyzed with just some simple algebra (decaying Rabi oscillations in a two-level system coupled to a heat bath)

20. Apr 16, 2015

### lucas_

When you see chairs and tables. They are collapsed. Yet when you see atoms and molecules. They are not collapsed. So how do you reconcile the two? Can we say the position eigenstates are collapsed while the Hamiltonians are uncollapsed.. but how could the atoms/molecules in position collapsed eigenvalues still respond to the environmental hamiltonians?

21. Apr 16, 2015

### StevieTNZ

I think you are reading more into my, I admit short, response. I speech marked 'classical' and refer to a 'classical' system as Newtonian Mechanics would (although in principle these are also quantum systems). I don't believe collapse happens at the boundary of classical and quantum worlds; there is no division in quantum theory. I was merely stating that because we see everyday items in definite states that 'collapse of the wave function' (however it may occur, and I say collapse to say 'so we see a definite result [it does not mean the system is not still in superposition and we are only seeing one of its states]).

22. Apr 16, 2015

### Staff: Mentor

When you see chairs and tables and other macroscopic objects, they are not "collapsed" - their state is a mixture that can only be described with a density matrix, and it is neither a pure state nor an eigenstate of any physically interesting operator.

The macroscopic object has an enormous number of internal degrees of freedom and therefore an enormous number of accessible states, and it moves at random among them. It just so happens that the overwhelming majority of those states are ones in which the object is located within a few atomic diameters of one position, so that's where we'll find it.

As an analogy, you could think about a volume of gas in a container. It contains some enormous number of molecules, all doing their own quantum mechanical thing, collapsing their wave functions into position eigenstates (Could I ask the specialists here to please withhold their maledictions for a moment?) when they interact with their neighbors and then evolving into new states in which their position is no longer certain. Despite all that quantum mechanical weirdness, on average the number of molecules interacting with the walls with container per unit time will be constant, so the pressure on the walls will be constant and consistent with the ideal gas law $pV=nRT$. You would not, however, conclude that the gas is in a single pure state that is an eigenstate of some operator that gives you the pressure; instead you would conclude that any measurement of the pressure is going to yield a value that is very close to the expectation value.

Last edited: Apr 16, 2015
23. Apr 16, 2015

### Staff: Mentor

Seeing and not collapsed doesn't make any sense since seeing is an observation.

Thanks
Bill

24. Apr 16, 2015

### atyy

In the Copenhagen interpretation, whenever you see something, the wave function will collapse. The tables and chairs and the atoms are equally collapsed. It turns out that in many cases, if one wishes to include the environment in one's description, decoherence causes apparent collapse of the atoms to the energy eigenstate. When you make the observation, then the atom collapses to an energy eigenstate.

Schlosshauer, http://arxiv.org/abs/quant-ph/0312059 (p14): When the interaction with the environment is weak and $\hat{H}_{S}$ dominates the evolution of the system (that is, when the environment is “slow” in the above sense), a case that frequently occurs in the microscopic domain, pointer states will arise that are energy eigenstates of $\hat{H}_{S}$ (Paz and Zurek, 1999)

Paz and Zurek, http://arxiv.org/abs/quant-ph/9811026: We investigate decoherence in the limit where the interaction with the environment is weak and the evolution is dominated by the self Hamiltonian of the system. We show that in this case quantized eigenstates of energy emerge as pointer states selected through the predictability sieve.

25. Apr 16, 2015

### lucas_

"When the interaction with the environment is weak".. is this referring to isolated quantum system or everyday object?

If the object is already in eigenstates of position, is it still available to be perturbed such that the pointer states will change to energy eigenstates?

In systems (irregardless of decomposition) where the pointer states are energy eigenstates. What would be the effect of different values of the pointer states of energy eigenstates from the environment on the system.. would this make the electrons become higher in orbital or would it just contribute to more vibrational degree of freedom in the molecules (making it hotter for instance)?