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Mixed states are defined as the statistical ensemble of pure states. Classically, I understand the word, "statistical" referring to a system with a large number of microscopic particles. So if I go with that then mixed states should only exist in a system which has a large number of quantum particles.

Is it possible that a single isolated quantum particle to be in mixed state?

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  • $\begingroup$ An example would be an arbitrary wave-function of a particle in an infinite square well. But it's not actually a mixed state, but a state that is a linear combination of stationary states. $\endgroup$ – philip_0008 Jun 13 '16 at 8:52
  • $\begingroup$ How do you measure the state of a system with just one particle? As soon as you have to add some instrument to do the measurement, your system is suddenly composed of many entangled particles. I'm not sure if your question makes any sense at all :D $\endgroup$ – Luaan Jun 13 '16 at 12:28
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A mixed state represents a lack of knowledge on the system (maybe caused by the observer, maybe more fundamental). It is a notion in my opinion closely related to the Bayesian interpretation of quantum theories (seen as non-commutative probability theories).

Quantum states are (non-commutative) probabilities, i.e. they encode all the information on expectation values of the non-commutative observables that constitute the system. Now it is mathematically possible to define a (partial) ordering of these states, and that can be interpreted as "how maximal" the Bayesian knowledge encoded by the state is.

In other words, an ordering $\leq$ is introduced on quantum states by means of the following: $\omega_1\leq \omega_2$ if $\omega_1-\omega_2$ is positive as an object of the dual space of observables (a dual element $x$ is positive if, given an observable $A\geq 0$, then $x(A)\geq 0$). With such definition, you see that a state $\omega$ that can be written as $\omega= \lambda\omega_1+(1-\lambda)\omega_2$ ($0<\lambda< 1$, and $\omega_1,\omega_2$ states) satisfies $\omega\leq \lambda \omega_1$, $\omega\leq (1-\lambda)\omega_2$.

I stress the interpretation: a state encodes a maximal (Bayesian) knowledge if, roughly speaking, it is never less or equal to another state with respect to the ordering $\leq$ defined above. More precisely, $\omega$ encodes maximal knowledge (is extremal) if $\omega\leq \omega_1$ implies $\omega_1=\lambda\omega$, $0\leq\lambda\leq 1$. In other words, it means that it is not possible to write $\omega$ as a combination of different probabilities with suitable statistical weights. This in turn explains why extremal states have "maximal" knowledge: the non-extremal $\lambda\omega_1+(1-\lambda)\omega_2$ encodes knowledge as every probability, but such knowledge can be derived from the one encoded already in $\omega_1$ and $\omega_2$, simply by linear combination.

The extremal states are called pure states, the ones like $\lambda\omega_1+(1-\lambda)\omega_2$ mixed states. As you can see, this notion does not depend on how many particles do you have, but on the (Bayesian) interpretation of a quantum theory as a (non-commutative) probability theory.

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Yes, it can. "Ensemble" may mean many experiments with one particle, not obligatory many particles at the same experiment.

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  • $\begingroup$ Yes, it does. QM results are always about ensembles, not about one point, which says nothing as a matter of fact. (Classical mechanics all the more!) $\endgroup$ – Vladimir Kalitvianski Jun 13 '16 at 10:53
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Yes, one particle can be in a mixed state when there is classical uncertainty present. Or at least it is the cleanest visualization one can make of a quantum state with classical uncertainty to think of it as mixed.

The classic example here is the state of the particle ("cat") in the Wigner Friend Thought Experiment as Wigner would calculate it if he had to before he measures his friend's state but after his friend has measured the particle.

Wigner's friend makes an observation of the particle, but Wigner does not know the result. So Wigner knows the particle has been "collapsed" (i.e. had its state measured) so that he knows it is in state $|0\rangle$ with some classical probability $p$ and state $|1\rangle$ with classical probability $1-p$.

So the particle's state, as Wigner would calculate it, is mixed, with density matrix $\rho= p\, |0\rangle\,\langle 0| + (1-p)\, |1\rangle\,\langle 1|$.

As in Vladimir Kalitvianski's answer, this is simply one way wherein a single particle can belong to a larger ensemble, not needfully present all at once. Another simple example would be a quantum optics experiment, where the laser used to source single photon states one at a time is mode hopping or otherwise having a bad hair day (as lasers are wont to do). Each mode of the laser leads to the output of a different, pure one-photon state, but it (classically) randomly chooses a different pure state each time it outputs the light state. Each one-photon state is now a mixture.

The Wigner Friend experiment is also a pithy and elegant illustration of quantum purification, for Wigner's mixed particle state is the reduced state derived from the pure state comprising Wigner, his Friend and the particle; the full pure state is written down in the Wigner Friend Wiki article.

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As stated in other answers, a single (non entangled) particle could be described by a mixed quantum state. But I would not say it can be in a mixed state. If there is any ontology to a quantum state, then it concerns pure states only. The density matrix merrily conflates classical and quantum probabilities because it is designed to represent our knowledge about a system, not what it is. Only for interpretations of QM (such as Quantum Bayesianism) which take the knowledge itself as the ontology is this distinction irrelevant.

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  • $\begingroup$ The local state of one entangled particle can never be a pure state, and the corresponding density matrix definitely does not confuse quantum and classical probabilities. $\endgroup$ – udrv Jun 13 '16 at 18:04
  • $\begingroup$ @udrv. The question is about a "single isolated quantum particle", not an entangled one. $\endgroup$ – Stéphane Rollandin Jun 13 '16 at 20:09
  • $\begingroup$ @udrv. "Confuse" is the wrong verb; I meant "conflate". In French "confondre" has both meanings. I edited my answer. $\endgroup$ – Stéphane Rollandin Jun 13 '16 at 20:17
  • $\begingroup$ Ok, conflate works. But a non-interacting entangled particle is still isolated. $\endgroup$ – udrv Jun 13 '16 at 20:43
  • $\begingroup$ @udrv. I see where you are coming from... I guess to me isolated implies non-entangled, but I see this can open a can of worms as far as ontology and interpretations are concerned. Ok, I edited my answer again. $\endgroup$ – Stéphane Rollandin Jun 13 '16 at 20:51

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