Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Difference between Mixed State and Superposition

  1. Jun 26, 2008 #1
    What is the difference between a mixed state and a superposition?
  2. jcsd
  3. Jun 26, 2008 #2
    From my modest experience, there is no real technical difference. Perhaps when you're talking about the net effects / measurables etc - you refer to it as a superposition; while you're talking about the constituent states etc - you refer to it as a mixed state?
  4. Jun 26, 2008 #3


    User Avatar

    If I understand you correctly, by superposition, you mean a pure state (which is the opposite of mixed). A mixed state is sometimes also called a classical superposition.

    For example, consider a photon which could be either right-handed or left-handed (equiv. circularly polarized in either clockwise or anticlockwise direction).

    A superposition of these states would give you a photon with maybe a 50/50 chance of either polarization being measured, and would be linearly polarized, so it would pass through a polarizing filter at the correct orientation with high probability.

    A mixed state might also have 50/50 chance of being either left or right handed, but this is a classical probability, so it is definitely in one state or the other, and would consequently have about a 50/50 chance of passing through a polarizer of any orientation.

    Alternatively, with the double slit experiment, you need a superposition of the states of passing through each slit to get the standard interference pattern. If you measure which slit the electron goes through, but ignore the result, then you still have a 50/50 chance of going through either slit. However, this will be a classical probability not a quantum superposition, so it is a mixed state and you won't see the interference pattern.

    You can get a mixed state from a pure one simply by looking at a part of a larger system. For the double slit experiment above, if you include the state of the measuring equipment and anything it has interacted with, it will still be in a pure state.
  5. Jun 26, 2008 #4
    I understand them to basically be the same in structure, but the difference is the context in which they are used. I always see the term mixed state associated with the idea of a linear combination of projectors where as a superposition is a term that is used for any linear combination in ANY Hilbert space.

    As an example, let [tex]\psi_1, \psi_2[/tex] be elements in a Hilbert space.
    [tex]\psi=\lambda_1\psi_1+\lambda_2\psi_2[/tex] is a superposition of two vectors but is generally still regarded as a pure state. Technically the density operator
    [tex]|\psi\rangle\langle\psi|[/tex] is a pure state.
    Where as the density operator
    is a mixed state.
  6. Jun 27, 2008 #5


    User Avatar

    Those expressions agree with what I was saying. Taking linear combinations of state vectors is quantum superposition, but taking combinations of density operators is the same of classical superposition.

    Also, if the two vectors are normalized, you must have [itex]|\lambda_1|^2+|\lambda_2|^2=1[/itex] in the first expression, but [itex]\lambda_1+\lambda_2=1[/itex], and [itex]\lambda_1,\lambda_2\ge 0[/itex] in the second one.

    In the second case, there is no way to distinguish between that state and the case where it is either in the definite state [itex]\psi_1[/itex] or in state [itex]\psi_2[/itex], and where you just don't know which one due to lack of knowledge. So, no quantum interference between the states occurs.
    Mathematically, they are both just different types of linear combinations, but they have a very different meaning physically.
    Last edited: Jun 27, 2008
  7. Jun 30, 2008 #6
    How do you express a mixed state using vectors?
  8. Jul 1, 2008 #7
    In general a mixed state (or statistical mixture) writes with the density matrix as follows :
    [tex] \hat{\rho}=\sum_{\alpha} p_{\alpha} \left| \psi_p \rangle \langle \psi_p \right| [/tex]
    Where [tex] p_{\alpha} [/tex] is the probability for the system to be in the state [tex] \left| \psi_{\alpha} \rangle [/tex] (that is not necessarily an eigen state). These [tex] p_{\alpha}[/tex] can be derived from expermimental data or by a priori assumptions.

    The only cases for which a mixed state is equivalent to a superposition is when there exist a state [tex] \left|\psi_{\alpha_0} \rangle[/tex] whose [tex] p_{\alpha_0} =1 [/tex] (therefore we have [tex] p_{\alpha} =0 [/tex] for [tex] \alpha \: \neq \: \alpha_0 [/tex] ).

    In this case you get a pure state and the density matrix writes :

    [tex] \hat{\rho} = \left| \psi_{\alpha_0} \rangle \langle \psi_{\alpha_0} \right| [/tex]

    EDIT : To answer straight forwardly to the question asked, you can't express a mixed state using only the ket notation.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Difference between Mixed State and Superposition
  1. Superposition state (Replies: 3)