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Mixed state vs. pure state

  1. Sep 9, 2011 #1
    In quantum mechanics, regarding light (photon), how to tell that a wavefunction is in a pure state or mixed state?
    I am learning these stuffs for my first time.
    I have attempted to answer that question but I am not sure: a wavefunction can be wrtitten as a linear combination of linear independent vectors, i.e
    Where [itex]\Psi[/itex] is the wavefunction, [itex]\Psi[/itex][itex]_{1}[/itex] and [itex]\Psi_{2}[/itex] are unit vectors that form a basis, [itex]\alpha[/itex] and [itex]\beta[/itex] are constants.
    If this wavefunction is in a mixed state, the sum of [itex]\alpha[/itex] + [itex]\beta[/itex] = 1

    Am I thinking right?
  2. jcsd
  3. Sep 9, 2011 #2
    A mixed state cannot be represented by a state vector [itex]\psi[/itex].

    Fundamentally, a mixed state is when you have a classical mixture of pure states. So you have maybe half the states [itex]\psi_{1}[/itex] and the other half [itex]\psi_{2}[/itex]. This is not the same as [itex]\frac{1}{\sqrt{2}}(\psi_{1} + \psi_{2})[/itex] which is a pure state.

    If you are reading it for the first time, you should encounter something called density matrices soon which will deal with mixed states.
    Last edited: Sep 9, 2011
  4. Sep 9, 2011 #3
    Can you please suggest me any reading which gives more details. And thanks
  5. Sep 9, 2011 #4
    I'm not really sure about readings for mixed states. Griffiths, sadly, does not include mixed states in his book. Maybe someone else will reply in this thread with a good reference. I picked it up from here and there. I suppose Density Matrices on Wikipedia might not be a bad starting point.
  6. Sep 9, 2011 #5


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    McLaren is one hundred percent correct about mixed states and density matrices, and yet I think his answer may miss the original question. The density matrix is an advanced topic, used in quantum statistical mechanics for example to describe an ensemble of quantum systems in thermal equilibrium, where both quantum and classical uncertainties come into play.

    The OP says he is learning about quantum mechanics for the first time, and so I think it is more likely that when he uses the term "mixed state" he is referring to something else.
    Rather than "pure state," the term you're looking for is "eigenstate." An eigenstate is a state in which the value of a physical quantity is certain. You can have eigenstates of energy, eigenstates of momentum, and so on. If Ψ1 and Ψ2 are eigenstates, then like you say, they form a basis, and a more general state may be written as a linear combination of them.
    But actually what you want here is |α|2 + |β|2 = 1
  7. Sep 9, 2011 #6

    Ken G

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    I think some further clarification is needed, because we really have three concepts here-- eigenstates, pure states, and mixed states. An eigenstate is a special type of pure state, it is a pure state that has a definite value with respect to some observable. Then we say it is an eigenstate of that observable (often energy). Sometimes you will want to express a pure state as a linear combination of eigenstates, where the coefficients of the linear combination are complex numbers and their norms add up the way Bill_K said. But that has little to do with mixed states, because as McLaren_Rulez said, mixed states cannot be written as a single wave function or state vector.

    I agree that a "density matrix" is kind of an advanced way to think of a mixed state, so I offer a more intuitive meaning. A pure state is a (bizarre) complex linear combination of states, but a mixed state is a more everyday kind of "combination"-- not a formal algebraic combination, but more like a logical combination. A mixed state of A and B is like saying "the actual state is either A or B, I just don't know which." So it's a more common-sense combination of pure states, not a pure state itself, and we would associate probabilities with A and B instead of coefficients in a linear combination.

    If you want to think in terms of a linear combination, another way to think of a mixed state is a pure state like you wrote originally, with the added idea that the coefficients you would use in the combination have a random phase with respect to each other. By that I mean, the coefficients can still have a magnitude, but they are neither real nor complex-- they simply have an undefined or unspecified phase relationship. When you use them in calculations, you simply treat the phase difference between them as randomly distributed between 0 and 2pi, and let any calculation you do include a sum over all those equally-weighted possible phase relations.
  8. Sep 11, 2011 #7
    Thanks for your explanation. But now, let's take a physical example: "LIGHT". I've been reading that Light Polarization would help in the understanding of pure and mixed state. Suppose I have a beam of light which can be polarized or not. As I read, a polarized light (photon) is a pure state while an unpolarized one is mixed state.
    Is there any method (experimental or theoretical) that can be used to know weather that light is polarized (pure state) or unpolarized (mixed state)?
  9. Sep 11, 2011 #8

    Ken G

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    Yes there is, and yes that is a perfect example of the distinction. With light in a pure state of polarization, I can find a polarizer (perhaps circular or linear) that will pass 100% of the light that is in that state (say for a laser beam), and another polarizer that will transmit 0%. Other polarizers will produce a linear combination of those two results, depending on how the pure state expands onto the eigenstates of the polarizer. But for light in a mixed state, if it is perfectly mixed (so as far from a pure state as possible), then every polarizer, circular or linear at any angle, will transmit 50% of that light, in an ensemble average. The mixed state is a random collection of pure states, the pure state has all the photons in the same state.
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