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Question: In Quantum Mechanics, people use the word "pure state" for some states; however, what do they mean exactly ?

Thoughts: I mean, a state is a vector in our vector (Hilbert) space, so in that sense, any given vector can be written as a linear combination (superposition) of different vectors, so in that sense, no state is pure, i.e there is no a unique set of fundamental states that does not compose of anything.

I, first, thought that they might be talking about the eigenstates of some specific observable at hand, but, even in research papers, they mention a "pure" state without talking about any observable or anything else.Also, even if that was the case, an observable can have infinitely many different set of orthonormal basis when it has a degenerate eigenvalue, so that doesn't make sense also.

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  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/70436/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Jan 29, 2019 at 5:48
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    $\begingroup$ @onurcanbektas see here for more. Every element of the Hilbert space can be made into a state, but not every state can be made from an element of the Hilbert space. Those that can are called pure, and those that can't are called mixed. It's common practice in elementary contexts to ignore the latter and to blur the lines between the Hilbert space and the (pure) states which can be constructed from them, but at some point you need to step back and be more general. $\endgroup$
    – J. Murray
    Commented Jan 29, 2019 at 6:46
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    $\begingroup$ @onurcanbektas Strictly speaking it is not correct to distinguish pure and mixed states by whether they are written as vectors of the Hilbert space or not (as the other comments refer to, and as things are often presented naïvely in elementary QM). This is because the fact that a state can be written as a vector is representation dependent, where representation means the way of writing the algebra of quantum observables mathematically as linear operators acting on a Hilbert space. In fact, for every state there exists at least one representation in which it is a vector state. $\endgroup$
    – yuggib
    Commented Jan 29, 2019 at 7:44
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    $\begingroup$ There is a more proper definition of purity that is probabilistic, in terms of bayesian information carried by each state. The pure states are the ones carrying maximal information. More concretely, a pure state is one that cannot be written as a convex combination of states other than itself. That means that the information is encoded solely in such state; on the other hand the information of a mixed state is encoded already in two or more other states (the ones forming the convex combination), and thus it is not maximal. $\endgroup$
    – yuggib
    Commented Jan 29, 2019 at 7:49
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    $\begingroup$ Purity of a state is also strictly related to irreducibility of its vector representation. As I mentioned, for every state there exists a representation such that it becomes a vector state in such representation. This is called GNS representation. A state is pure if and only if the GNS representation is irreducible (this means that the only subspaces of the Hilbert space left invariant by the action of the observables are the trivial subspaces, the zero subspace and the whole Hilbert space) $\endgroup$
    – yuggib
    Commented Jan 29, 2019 at 8:01

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