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Help clarify the probability of pure states and mixture of states

  1. Jul 14, 2012 #1

    KFC

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    Hi there,
    I am reading several books on quantum mechanics and they mention some concepts on coherently and incoherently superpose a bunch of wave functions. As my understanding, if there are series of wave functions, which satisfying a certain distribution, the coherent way to calculate the probability is to sum all the wave functions first and then take the |.|^2 to find the probability. For incoherent way, we are adding the probablity for each wave function, is that right? Thanks.
     
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  3. Jul 14, 2012 #2
    There is no way to describe an incoherent system with wave functions. You must use the density matrix formalism for this. The QM book by Cohen-Tanudji does a nice job of explaining this stuff in the chapter on density matricies.
     
  4. Jul 15, 2012 #3

    tom.stoer

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    Suppose you have a (infinite dimensional) Hilbert space H and a pure state |ψ> i.e. a one-dimensional subspace of H. Now you can introduce a projector |ψ><ψ|.

    Suppose you have a basis |n> of your Hilbert space; every basis vector defines such a one-dim. subspace, i.e. a projector |n><n|.

    Generalizing this to mixed states means to introduce a so-called density operator

    [tex]\rho = \sum_n p_n\,|n\rangle\langle n|[/tex]

    as a sum over projectors |n><n| and "classical" probabilities pn.

    To calculate an expectation value of an obervable A in this generalized state ρ one uses

    [tex]\langle A \rangle_\rho = \text{tr}(\rho A) = \sum_n \langle n| \rho A|n\rangle = \sum_n p_n\,\langle n| A|n\rangle = \sum_n p_n\,\langle A \rangle_n[/tex]

    It is interesting to see that <A>ρ is a weighted average of the individual expectation values An for each pure state n.

    For a detailed introduction please have a look at textbooks.
     
  5. Jul 23, 2012 #4

    KFC

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    hi all,
    I read the Cohen-Tanudji's text, now I get a bit more on the pure and mixed states. To learn more on the difference, I would like to verify my understanding

    1) So far, as my understanding, pure states are always the eigenstates of the operator of measurable, right?

    2) in some text, they mentioned that according to the definition of density operator, if the density matrix is a diagonal matrix with all zeros on the diagonal element except for one with unity, the corresponding state is pure state. So what about the superposition of all eigenstates? Will that also give a pure state?

    3) in some texts, they set one example to illustrate the mix states as follows: let says we only know that the initial state of the system is a mixture of two states ψ1 and ψ2 with probabilities p1 and p2 respectively. To measure a physical quantity A, the probability of getting an outcome an (an eigenvalue of A, is given as Tr(ρ An)

    what confusing me is when they mention the states ψ1 and ψ2, do they mean the eigenstates or any state expanded by the eigenstates? Why?
     
  6. Jul 23, 2012 #5

    tom.stoer

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    I think this is something we should discuss:
    Look at my example regarding the general state |ψ> and a projector |ψ><ψ|. This is a valid density density matrix with pψ=1 and all other p's=0. All what you have to do is to use a basis transformation such that this state vector |ψ> becomes a basis vector. In this new basis all other p's are zero, therefore the density matrix becomes a projector on |ψ>, and this is in general an indication for a pure state: a pure state is described by a density matrix which is a projector,. i.e. ρ² = ρ. In an approproiate basis that means that the density matrix is diagonal with only one non-zero element, which in my case corresponds to the state |ψ> .

    This works for an arbitrary superposition of states. Use basis vectors |n>, construct an arbitary superposition |ψ>, make a basis transformation such that |ψ> becomes a basis vector, voila
     
  7. Jul 25, 2012 #6
    While this is true (from any pure state it is easy to construct an Hermitian operator, i.e. a measurable operator which has the pure state as an eigenstate), thinking about eigenstates of operators is probably not useful, only confusing.

    By eigenstate, you mean a (unit) vector satifying O|v> = e|v> (eigenvector) or an eigenfunction satisfying a similar relation. Any unit vector, including superpositions of other vectors, represents pure states. If you think in terms of wavefunctions, any wavefunction represents a pure state. I write "represents" instead of "are" here, since a wavefunction, a vector, and a density operator are different mathematical objects which all may represent the same quantum mechanical state.

    This was already anwered above. I will just add that the technical for a density operator to represent a pure state is that is rank one. That is, all eigenvalues but one must be zero (and that must be one since all eigenvalues must sum to one for a density operator).

    They mean any pure states, orthogonal or not. If they are eigenstates of some particular operator is irrelevant in this context. The take-home point here is that a mixed state can always be viewed as a probability distribution over various pure states (this is called an ensemble). The reason the mixed state is specified by a density operator rather than such a probability distribution is that many different ensembles give rise to the same density operator. From the density operator, the probability of any measurement result can be calculated, and it will be identical to what you would calculate using one of the ensembles.
     
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