4
$\begingroup$

Let's say that "decoherence" is that transition from a pure quantum state to a mixed state due to interactions with the environment. (A reasonable definition?)

How is that compatible with the Copenhagen interpretation of quantum mechanics -- specifically with the role of the observer?

By the above definition, observation is an example of decoherence; the observer in the Copemhagen interpertation is part of the environment and causes the wavefunction to collapse. However, decoherence can also happen by (often-unwanted) coupling of the quantum system to a heat bath. Does that mean that the heat bath is an "observer"? If not, what causes the apparent** wave function collapse when there is no human observer?

Having worked with crummy superconducting qubits, I'm aware that they can decohere very quickly and apparently without human observation. E.g. suppose I initialize a qubit in a pure state. If I measure it after 10 ns, the qubit still appears to be in a pure state. If I measure the qubit after 1ms, it appears to have decohered into a mixed state. (I could determine that by trying to perform some quantum operation on the qubit.) Since I did not measure or interfere with the qubit in the intervening time, it does not seem that I could have caused the decoherence ("collapse").

** I'm using the word "apparent" to mean "Appearing to the eye or mind (distinguished from, but not necessarily opposed to, true or real); seeming."

$\endgroup$
11
  • 1
    $\begingroup$ The "collapse" bit is when the observation picks out a single component of the mixed state. This is not the same as a pure state superposition decohering into a mixture. $\endgroup$ Commented Dec 4, 2014 at 16:00
  • $\begingroup$ Could you elaborate on the importance of the difference between the two? IIRC, a system in a "mixture" is (or can be) in one eigenstate of the hamiltonian; you just don't know which one state it is until you look. So in both cases, a superposition of states is reduced to a single state. The difference I see is that in one case you observe the system soon after it turns into one state but in the other case you wait a little while. $\endgroup$
    – lnmaurer
    Commented Dec 4, 2014 at 19:17
  • $\begingroup$ @Inmaurer - If you're talking about a mixed state generated by taking the reduced density matrix of a subsystem A of a larger pure state, I think it's potentially misleading to say that A "is" in one of the states, since the full state is the pure state of the whole system. I think it would be better to say the probability distribution for measurements of a subsystem A of a pure state is mathematically identical to the probability distribution you'd get for a distinct system A' that wasn't part of a larger entangled state, and which was in a mixed state that happened to match the reduced $\endgroup$
    – Hypnosifl
    Commented Dec 4, 2014 at 19:20
  • $\begingroup$ (continued) density matrix for A. Then you are free to assume that A' is definitely in one of the pure states that are part of its mixed state, and you just don't happen to know which one, but it's not so obvious that you can say the same for A. $\endgroup$
    – Hypnosifl
    Commented Dec 4, 2014 at 19:21
  • 1
    $\begingroup$ The so-called Copenhagen interpretation isn't even self consistent. If I were you I wouldn't worry about reconciling anything against it. Also, sorry to hear about the crummy qubits. May your $T_2$'s increase in future years. $\endgroup$
    – DanielSank
    Commented Jan 4, 2015 at 21:02

2 Answers 2

2
$\begingroup$

Let's say that "decoherence" is that transition from a pure quantum state to a mixed state due to interactions with the environment. (A reasonable definition?)

Mixed states are NOT decohered states, they are states where the phases of the wavefunctions are well defined, just not in an eigenstate that will give a unique eigen value at measurement.

How is that compatible with the Copenhagen interpretation of quantum mechanics -- specifically with the role of the observer?

In the quantum regime "observation" is interchangeable with "interaction", it is not necessary to have a human observer. An electron scattering off an atom "observes" the atom.

By the above definition, observation is an example of decoherence; the observer in the Copemhagen interpertation is part of the environment and causes the wavefunction to collapse.

The "collapse" language is a fancy way of saying that an instance was picked from a probability distribution, which is what the state function gives us where composed of pure states or mixed. The square of the state function gives us a probability distribution and an observation gives an instance in that distribution. It is as ridiculous as saying that when getting a five in a throw of the dice, the probability distribution of the dice throws has collapsed.

However, decoherence can also happen by (often-unwanted) coupling of the quantum system to a heat bath.

Now yes, a heat bath leads to decoherence. The easiest way to think of coherence and decoherence is in the density matrix formalism.

A density matrix is a matrix that describes a quantum system in a mixed state, a statistical ensemble of several quantum states. This should be contrasted with a single state vector that describes a quantum system in a pure state.

A density matrix has rows and columns of all the pure state functions comprising an ensemble of paticles. Off diagonal elements carry the phase information between two pure wave functions.

expect with densitymatrix

In theory, the whole universe could be described by one density matrix. h_bar though being a very small number and the macroscopic dimensions of a heat bath together with the enormous number of particles ( 10^23 per mole) reduce the density matrix to its diagonal elements. That is when a system of particles is decohered, when the phase information is lost.

Does that mean that the heat bath is an "observer"? If not, what causes the apparent** wave function collapse when there is no human observer?

The multitude of interactions in a heat bath and the macroscopic dimensions ensure decoherence, the loss of phases between state functions because they cannot be measured.

Having worked with crummy superconducting qubits, I'm aware that they can decohere very quickly and apparently without human observation. E.g. suppose I initialize a qubit in a pure state. If I measure it after 10 ns, the qubit still appears to be in a pure state. If I measure the qubit after 1ms, it appears to have decohered into a mixed state.

a mixed state is not decohered. still there exists a density matrix with non zero off diagonal elements

(I could determine that by trying to perform some quantum operation on the qubit.) Since I did not measure or interfere with the qubit in the intervening time, it does not seem that I could have caused the decoherence ("collapse").

All matter emits black body radiation according to the temperature, i.e. photons impinged on your qubit. Obviously some of them interacted and changed the state function describing the setup. On a large scale this is the heat bath.

$\endgroup$
11
  • $\begingroup$ "Mixed states are NOT decohered states, they are states where the phases of the wavefunctions are well defined, just not in an eigenstate that will give a unique eigen value at measurement" But there are also pure states that are not eigenstates--and if you decompose a pure state into a superposition of eigenstates, doesn't each of the eigenstates get assigned a "well-defined" phase? If so this isn't a good definition of "mixed state". I had thought the basic definition was just a statistical mixture of pure states, where each pure state can be assigned some classical probability. $\endgroup$
    – Hypnosifl
    Commented Dec 4, 2014 at 20:16
  • $\begingroup$ Yes, I'm also confused by your definition of a mixed state. Can you reconcile your definition with the one you linked to: "A mixed quantum state is a statistical ensemble of pure states". $\endgroup$
    – lnmaurer
    Commented Dec 4, 2014 at 22:35
  • $\begingroup$ "In the quantum regime 'observation' is interchangeable with 'interaction'". That basically answers my question. However, is your answer answer widely accepted? For example von Neumann (who came up with the density matrix formalism), does not appear to believe what you said. It looks like he believed that consciousness was required for observation: en.wikipedia.org/wiki/Von_Neumann%E2%80%93Wigner_interpretation Also, is your answer compatible with the Copenhagen interpertation? (You never explicitly answered that.) $\endgroup$
    – lnmaurer
    Commented Dec 4, 2014 at 22:41
  • $\begingroup$ "a mixed state is not decohered. still there exists a density matrix with non zero off diagonal elements" Why does it have to have non-zero off diagonal elements? (Heat baths have all non-diagonal elements as zero, right?) If there are non-diagonal elements after a qubit decoheres, how would I show that? $\endgroup$
    – lnmaurer
    Commented Dec 4, 2014 at 22:48
  • 1
    $\begingroup$ @annav, I agree that we're getting into philosophy here, but that's ultimately what I'm asking. Many introductory quantum mechanics books talk about the Copenhagen interpretation in simple terms ("observers", "observations", "measurements", etc.) without explaining what they are -- leading to the impression that human intervention is necessary. I like equating interaction with observation, but it leaves some things unanswered. E.g. two qubits can interact and remain in a superposition, so is that interaction not an observation? Anyhow, thank you for your answers. $\endgroup$
    – lnmaurer
    Commented Dec 5, 2014 at 19:51
1
$\begingroup$

I haven't studied this in any detail, so take what I say with a grain of salt. But from summaries I've read, like the essays on decoherence.de, I think that in the Copenhagen interpretation "collapse" would still have to be understood as conceptually distinct from decoherence. I think a way to see this would be by imagining an idealized Schroedinger's-cat type situation where a complex system could be kept entirely isolated from the external universe by some sort of perfect "box" (like an ideal square well with infinite potential barriers on each side), up until the moment when we on the outside choose to "look in the box" and measure what's inside. Then you can divide the system within the box into a combination of some small subsystem A + everything else, which is treated as the "environment" for the subsystem. In this case, even before we look inside and collapse the wavefunction, there can still be decoherence between the subsystem A and its environment within the box, but until the act of measurement, the entire system within the box must be modeled by us as being in a single pure state. In this case the effects of decoherence on probabilities would become apparent if we imagined "opening" the box and measuring only the state of subsystem A, which in the Copenhagen interpretation would be the first point at which any real "collapse" of the wavefunction occurs. From summaries I've read, measuring only one subsystem of a larger entangled system requires generating a "reduced density matrix" for subsystem A from the larger pure state of the entire system, and it is with this step that subsystem A is modeled as being in a mixed state. So the idea is that if there has been interactions leading to decoherence between subsystem A and its environment in the box before our opening the box and collapsing the wavefunction, the probabilities of different measurement results for A will look different than if there had not been decoherence between A and its environment prior to measurement (and there can also be degrees of decoherence rather than it being an all-or-nothing affair, from what I understand).

Along these lines, the essay "How decoherence can solve the measurement problem" by H. Dieter Zeh (one of the founders of the current understanding of decoherence) on the site I linked to above says:

As an application, consider the particle track arising in a Wilson or bubble chamber, described by a succession of collapse events. All the little droplets (or bubbles in a bubble chamber) can be interpreted as macroscopic "pointers" (or documents). They can themselves be observed without being changed by means of "ideal measurements". In unitary description, the state of the apparently observed "particle" (its wave function) becomes entangled with all these pointer states in a way that describes a superposition of many different tracks, each one consisting of a number of droplets at correlated positions. ... Decoherence leads to the same local density matrix (for the combined system of droplets and "particle", which therefore seems to represent an ensemble of tracks. The correlations between the wave functions of different droplets as forming tracks were already known to Mott in the early days of quantum mechanics, but he did not yet take into account the subsequent and unavoidable process of decoherence of the droplet positions by their environment. Mott did not see the need to solve any measurement problem, as he had accepted the probability interpretation in terms of classical variables. In a global unitary quantum description, however, there is still just one global superposition of all "potential" tracks consisting of droplets, entangled with the particle wave function and the environment: a universal Schrödinger cat. Since one does not obtain an ensemble of potential states without a collapse, one cannot select one of its members by a mere increase of information.

For another possible example, in the experiment described here involving fullerene molecules being sent through a number of slits in a gas-filled environment, if we were to model the whole system as being isolated until the a position measurement of the fullerene after it has passed through the slits, I expect that--just as was found experimentally--we would predict different degrees of interference fringes in the spatial pattern of the fullerenes depending on the pressure of the gas, which affects the degree of decoherence that occurred when the fullerene was passing through the slits. With sufficient decoherence, I think the spatial distribution of the fullerenes upon the position measurements would be the same to the distribution we would have seen if we had simply placed detectors at each slit (and perhaps at points throughout the space between the slits), and modeled each detection-event as a collapse. If that's right, then in this sense decoherence would seem to mimic the effects of prior collapses, although in the Copenhagen interpretation you would still need to assume a final collapse after the fullerene had passed through all the slits in order to talk about a final probability distribution for the fullerene's position. But if the statistics for subsystem A in the ending measurement can be identical for the two cases "a number of individual collapse-inducing measurements during the time-span of the experiment, then an ending measurement" and "decoherence without collapse during the time-span of the experiment, then an ending measurement", that might be taken as a sort of intuitive reason to support interpretations in which all apparent collapse is really just a simplified way of modeling the effects of decoherence.

$\endgroup$
24
  • 1
    $\begingroup$ Why is your answer so long and so complicated? To the first question of Inmaurer you could have answered by "yes, correct". About the observer, human observer has no role, i.e. both decoherence and collapse (two DIFFERENT things) don't need a human, or a human apparatus, to be there and become aware of the result. $\endgroup$
    – Sofia
    Commented Dec 4, 2014 at 19:00
  • $\begingroup$ (continuation) Decoherence is caused by lack of isolation of the system from its environment, (which indeed can be a bath). It transforms a pure state to a mixture. The collapse selects from the mixture ONE component, that all. If the wave function occupies some volume, an absorbing block over part of this volume, already collapses the wave-function on the rest of the volume, without our apparatuses being there. $\endgroup$
    – Sofia
    Commented Dec 4, 2014 at 19:05
  • $\begingroup$ Sofia - If by "first question" you meant the parenthetical "(A reasonable definition?)", I guess it didn't really register, but I would say that if you model the subsystem and environment as a single quantum system as Zeh suggested, then decoherence only causes the subsystem's reduced density matrix to become a mixed state, while the state of the subsystem + environment is still modeled as a single pure state. And I'm not sure why you're making the point about "human observer" to me, I didn't say the final measurement which we model as a "collapse" needed to be done by a human. $\endgroup$
    – Hypnosifl
    Commented Dec 4, 2014 at 19:06
  • $\begingroup$ can we let Zeh aside? I read what Zeh said and what many, many others said (and none of them solved the problem - with some people I had exchange of letters, but when asked precise questions they ran from corner to corner - I prefer not to give names). Now, can you tell me simply which one is the SYSTEM and which one is the ENVIRONMENT? Do you still refer to Schrodinger's cat? This example was proposed just to exemplify that a macroscopic body cannot be in a quantum superposition of states. But for our talk it is not a good case to examine - too complicated. $\endgroup$
    – Sofia
    Commented Dec 4, 2014 at 20:17
  • $\begingroup$ @Sofia - I would have thought that all derivations of equations relating to decoherence ultimately start from a model similar to Zeh's, where you model both the system you're interested in and its environment (the division presumably being a matter of choice, depending only on what element's behavior you want to predict measurement outcomes for) as a single entangled system evolving according to the Schroedinger equation, and then show that after decohering interactions, the equations predict the reduced density matrix for the system will have greatly reduced off-diagonal terms. $\endgroup$
    – Hypnosifl
    Commented Dec 4, 2014 at 20:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.