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Pure States vs. Mixed States?

  1. Oct 2, 2011 #1


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    I’m having a little difficulty understanding the difference between pure states and mixed states.

    Given a physical system and its associated Hilbert Space, is each member (or ray) of the Hilbert Space always a pure state?

    Is every linear combination (sum or integral) of pure states in the Hilbert Space always a pure state?

    Thank you in advance.
  2. jcsd
  3. Oct 2, 2011 #2
    I am strongly inclined to say "yes", since that is probably the answer you are looking for (*). That what physicists usually refer to as "Hilbert space" is the amount of all possible physical states for an object. These states are "pure states" in the language of quantum mechanics, and so are their linear combination (I guess I don't have to tell you that one, you're the mathematician ). "Mixed states" are an additional layer of complexity, that is usually only treated in statistical physics lecture.

    It is widely accepted and well known to most that the superposition of two QM states does not have a classical analogy, and that in particularly the superposition of say two location-eigenstates does not mean that the object is at one of the locations with some probability, but that it actually is at both locations at once (**). However, in classical physics you can have the case that there is an object which can be in either location but you don't know which, and can only give probabilities for it. A state "with probability x at A and with probability y at B" is not really a physical state (it's merely a lack of information), but can of course still be a helpful construction. Now, this case of "mixing states" also an equivalent in quantum mechanics, and that is then mixed states. Over a finite-dimensional Hilbert space, those mixed states are represented by a matrix, rather than a tuple (as in case of the pure states); you may encounter the term "density matrix".

    (*) Now the caveat: I'm just a dumb computational physicist, so I am no expert on this. But I was under the impression that "Hilbert space" merely mean a vector space with a norm. I am not sure that you couldn't consider the mixed states also forming a vector space with a norm, and therefore also being elements of a Hilbert space. I have not encountered anyone speaking of them in that manner, which is why I believe "yes" is the answer you are looking for. But I am not sure if the answer is rigid.
    (**) Caveat 2 (less severe): That is the mainstream view. There are serious people working on different interpretations of quantum mechanics.
  4. Oct 2, 2011 #3
    A mixed state is represented by a linear combination of any number of pure states, so to answer your question, no they are not the same.

    My professor explained it like this:
    A pure state (in classical or quantum mechanics) is as the name implies, pure. It contains all possible available information for a given state. Classically this is an exact value, such as the particle is located 'here', quantum mechanically this means you know where the particle's wavefunction resides and how far its probability density is spread out.

    A mixed state can be viewed in one of two ways. Either as a ensemble of various pure states, all spread out so that you can't be sure which pure state you'll pick up when you try. or an equivalently mixed up single state where you don't have all information that is possible. There is some 'ignorance' in the nature of a mixed state.
  5. Oct 2, 2011 #4

    Ken G

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    But a superposition of pure states is still a pure state, because the coefficients in the superposition are complex numbers with not only magnitude, but phase. This means you have "coherences" between the states you are superimposing, and that is what preserves the purity of the state (and also allows for interference between the contributing states as you follow the future time evolution of the entire pure state). A mixed state is essentially a weighted combination of pure states, where the weights refer to the probability of each state being actualized in a measurement, but the weights have no phase (or random phase interrelations, if you prefer), so the presence of multiple states does not lead to interference.

    To give you an example, in the two-slit experiment, if the apparatus does not establish which slit the particle went through, then the emergent wave function is a pure state that is a superposition of states that go through either slit (and create an interference pattern). If the apparatus does establish which slit the particle went through, but we are not privy to that information, then the emergent wave function we would use is a mixture of the states that involve going through each slit (and this does not create a two-slit interference pattern because the pieces of the mixture don't interfere with each other). The way this might be said in common parlance is, in the first case, the "particle went through both slits" (or, I prefer saying the issue of which slit is simply not meaningfully defined by that experiment), and in the second case, "the particle went through one slit or the other we just don't know which."

    Whether or not there is such a thing as a mixed state that does not represent information that has been established but we are not privy to, is a matter of interpretation of quantum mechanics. For example, if we have an ensemble of identically prepared particles, and they encounter an apparatus that we would use a mixed state to describe the outcome, that might mean the particles in the ensemble "really are" in one state or the other, and our lack of information is just around which one is which (which wavefunction that is, not which particle-- the particles are usually identical and indistinguishable), or it might mean that every particle is itself "really" in a mixed state, and no further information really exists. The former is more commonly taught, as it is the "Copenhagen" approach, but the latter has many proponents as well (it is the "many worlds" approach).
  6. Oct 3, 2011 #5
    In my understanding, a linear combination of pure states is a pure state again (basically just a change of basis in the hilbert space). To represent mixed states, you need to use density matrices, which are not vectors in Hilbert space but operators.
  7. Oct 3, 2011 #6


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    It's the density matrix of a "mixed state" that is a linear combination of the corresponding "pure density matrices" - not the "state vector" itself, that's probably the confusion here. The linear coefficients are simply the real valued "probabilities" that the system is in the particular pure state.


  8. Oct 3, 2011 #7
    Thanks Fredrik, I was starting to get really confused there, but it was just my terminology.
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