What are observables and how are they related to quantum decoherence and wavefunction collapse. I read this:

Observables - what are they?

but it was about the technical details on observables. Even some of the Internet links confused me...

  • 1
    $\begingroup$ Could you elaborate what doesn't satisfy you about the answer to the question you linked? $\endgroup$ – ACuriousMind Dec 14 '14 at 22:21
  • $\begingroup$ @ACuriousMind it doesn't say in detail what observables are, no info on how they are used, etc. $\endgroup$ – TanMath Dec 14 '14 at 22:23
  • $\begingroup$ Related: physics.stackexchange.com/q/43406/2451 and links therein. $\endgroup$ – Qmechanic Dec 14 '14 at 22:26
  • $\begingroup$ An observable is a physical quantity (or, a variable) that can be measured. Position, linear or angular momentum, energy, mass, charge, can be measured. The quantum theory associates to an observable $O$, an operator $\hat O$. So, I disagree with what is written in "Observables - what are they?". They say "the observable is denoted like $⟨\psi|\hat x|\psi⟩$". If $\hat x$ is the operator associated with the observable x, then $⟨ψ|\hat x|ψ⟩$ is the mean value of x in the state $| \psi >$. Well, the concept of operator is independent of the state, and is not equal with its average in some state. $\endgroup$ – Sofia Dec 14 '14 at 23:03
  • $\begingroup$ possible duplicate of What is an observer in quantum mechanics? $\endgroup$ – Brandon Enright Dec 15 '14 at 0:15

An observable basically is something you can measure. However compared to just any general measurement, an observable has one specific property: If you repeat the measurement without the system changing in between, you get the same result (this is obviously not true for all possible measurement procedures). Such measurement have indeed been experimentally realized; for example the Stern-Gerlach experiment to measure spin.

That you get the same result for repeated measurements means that after the measurement you have to assume that your system is correctly described by a state for which this measurement gives the observed result (a so-called eigenstate of the observable).

In classical mechanics, this is no problem because there all observables are assumed to have well-defined values all the time, and the measurement only reveals them. However in quantum mechanics, this assumption does not work; it can be shown that this assumption would contradict quantum mechanics.

So a quantum system can be in a quantum state where a given observable has no well-defined value and it is only possible to give probabilities about what value a measurement will give. On the other hand, after the measurement your quantum system has to be described using a state where the given observable does have a well-defined, predictable value. In other words, unless you already had an eigenstate of your observable, the state you have to describe your system with has changed. This change of state is what is sometimes referred to as "collapse of the wave function" or "reduction of the wave function".

Now one of the properties of this change is that it does not follow the same laws as the change of an unobserved system. On the other hand, a measurement is ultimately also just a physical interaction, and therefore one would expect them to follow the same laws as other physical processes. This mismatch is known as the measurement problem. How to solve the measurement problem is what interpretations of quantum mechanics are all about.

Now decoherence is the observation that as soon as your quantum system interacts with other systems (the measurement apparatus, or the environment), it becomes entangled with it, and therefore you cannot give the quantum system a wave function by itself. Instead you find that if you ignore the additional systems in your description, you have to describe the system as if it were a statistical mixture of wave functions. And it turns out it's the same statistical mixture that you also become when you measure your system. Some people claim that this already solves the measurement problem, but others disagree.

| cite | improve this answer | |
  • $\begingroup$ I need math in the answer please! $\endgroup$ – TanMath Dec 14 '14 at 23:03
  • $\begingroup$ Well, it's already past midnight here, so I won't add math today, and tomorrow I won't have time for it. But maybe on Tuesday I'll find some time to add something about the mathematical description. $\endgroup$ – celtschk Dec 14 '14 at 23:07
  • $\begingroup$ @TAbraham Read the answer, don't just apply the math operator to it. $\endgroup$ – Count Iblis Dec 14 '14 at 23:08
  • 1
    $\begingroup$ @celtschk: you mix things. The property that "If you repeat the measurement without the system changing in between, you get the same result" characterizes the classical measurement on a quantum system, not the observable. To use a unpleasant word, this is the collapse. Once an observable is measured for finding an eigenvalue of it, the wave-function collapses on the corresponding eigenstate. $\endgroup$ – Sofia Dec 14 '14 at 23:10
  • $\begingroup$ Sofia is correct. Repetition of a measurement in QM does not result in the same value for an observable in quantum mechanics. Strictly speaking it does not result in the same measurement in stochastic classical systems, either (which, thanks to the third law of thermodynamics are all of them). If you were to reduce the world to exactly reproducible observables the total number of these would be exactly zero. $\endgroup$ – CuriousOne Dec 15 '14 at 0:23

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