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I Difference between superposition and mixed state

  1. Apr 19, 2016 #1
    I get that a if we have complete information of the state of the system (i.e. all the possible knowledge we could have about it: the values its observables can take and their corresponding probabilities), then it is a pure state and can be represented by a vector (ket), ##\lvert\psi\rangle## in some Hilbert space. I also get that if we do not have complete information about the system, then we are not able to specify the exact state of the system and must therefore resort to an ensemble of pure states. There is a probability associated with each pure state in the ensemble which quantifies the probability that our system is in that specific state. Such systems are said to be in a mixed state and are described in terms of a density matrix which contains all the information about the possible pure states that it could be in and their associated probabilities.

    What confuses me is what happens when one brings the concept of superposition into the picture. The Schrodinger equation is linear and as such any linear combination of states is a solution to it. How does this differ from the mixed state case?

    Is it simply that one can project a pure state ##\lvert\psi\rangle## onto a particular eigenbasis of an operator (representing one of its observables), and as such, is represented as a linear combination of these eigenbasis vectors. Now, as we have complete information of the system, the coefficients of this expansion correspond to the probabilities that the state with be in a given eigenstate of the the associated operator.

    Physically, this corresponds to the fact that before we have made a measurement of the system, we cannot not a priori definitively say that the state is in one eigenstate or another of the given observable. Hence, we must consider the state to be in a superposition of all the possible eigenstates simultaneously, and once a measurement has been made, the system will "collapse" into a given eigenstate.

    If this is correct, then I can see how the mixed state formalism differs from the pure state formalism, but otherwise I'm left quite confused.
     
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  3. Apr 19, 2016 #2
    If you can only do a projective measurement onto only one basis, there is no observable difference between the mixture and the superposition of the basis vectors. This is generally not the case if you can do more projective measurements or implement an arbitrary POVM. If two states are described by different density operators, then you can always find a measurement(s) to distinguish them.
     
  4. Apr 20, 2016 #3
    Could one say that the difference is as follows:

    For a pure state one has the knowledge of all the possible configurations that the state could be in, for example, one knows that the spin of the system is either spin-up or spin-down. Quantum superposition arises here because, although we know all the possible spins that the system can have, we cannot say before measuring which spin it has and hence we must consider both configurations before measuring.

    For a mixed state we have a lack of knowledge of the exact state the system is actually in and so we must describe in terms of an ensemble of the possible states that it could be in, each with an associated probability. This is a classical superposition as one is forced to describe the system in terms of a superposition of all the possible states that it could be in, weighted by their corresponding probabilities. This is probability is purely classical as it arises due to a lack of knowledge about the state of the system, rather than the probabilities that arise due to the quantum nature of the system (which themselves arise due to the fact that the state must be described by a wave function and so one can only describe its observables in terms of the a set of values with associated probabilities even if one has a complete knowledge of the state).

    Would this be correct at all?
     
  5. Apr 20, 2016 #4

    blue_leaf77

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    I personally think that your description above is in general safe, except for
    I won't say that the density matrix for a mixed state is a superposition of the density matrix of pure states. The term superposition is usually reserved for states not for operators. The more correct term to use is, the density matrix for a mixed state is a convex sum of the density matrix of pure states.
     
  6. Apr 20, 2016 #5
    Agreed. Superposition of pure states ##-## states of complete knowledge ##-## gives another state of complete knowledge.
     
  7. Apr 20, 2016 #6
    Ok, thanks for pointing that out.

    Is it correct to say that the probability that arises in a pure state is due to the fact that the state of the quantum system is described in terms of a wave function and this inherently associates a distribution of possible values that the observables of the state can assume, each with a particular probability associated with it corresponding to how likely that particular value will be obtained upon measurement of that particular observable? By saying that we have complete information about the quantum state are we really saying that the information is as complete as it possibly can be, but this still leaves an inherent probability associated to observables - one can only have complete information about the spectrum of values that each possible observable of the system can take, but cannot predict a priori exactly which value each observable has until a measurement has been made, hence why we must consider the system to be in a superposition of the possible eigenstates of the particular observable?
     
  8. Apr 20, 2016 #7

    blue_leaf77

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    The probabilistic nature of the outcomes of a measurement is due to how physicists had agreed to interpret the implication of superposition postulates of quantum mechanics.
    That part of the sentence is unnecessary to introduce, if we know the system completely (because it's in a pure state), then we know it completely.
    For the rest of your post, I have got no problem.
     
  9. Apr 20, 2016 #8
    If you have a pure state, you know there is a measurement that always gives a definite outcome (measuring in the eigenbasis), and that's the most certainty quantum mechanics can give us.

    Whether the information is as complete as it possibly can be depends on the interpretation.
     
  10. Apr 20, 2016 #9
    So by saying that a system is in a pure state if we have complete knowledge of its quantum state, in as much as we can describe it in terms of a state vector which contains complete information of the statistics of each observable attributed to the system?!

    I think what confuses me the most at the moment is what a superposition of quantum states is actually describing (I get that mathematically it is due to the linearity of the Schrödinger equation), and how it differs from the case of a mixed state? Is it simply that even though we have completed knowledge of the possible eigenstates and eigenvalues of a given operator (corresponding to a given observable), we will in general, not know exactly which eigenstate (of the operator) that the state is in before measurement and hence we must consider it to be in a superposition of the set of eigenstates of the observable?
    In the mixed state case, is it then that we simply don't know for definite which state the system is in, however we may know the set of possible states that it might be in and the associated probabilities. In this case we don't have superpositions of states because the state of the system is not well defined (since it is described by a distribution of pure states) and hence we have to resort to using a density matrix to describe it?!
     
  11. Apr 20, 2016 #10
    Yes, yes and yes.

    I'm not sure if this will help or not, but you can think of two states in superposition as an altogether qualitatively different state. For example, a superposition of a spin "up" and "down" state is a spin state "pointing" in a different direction. If you don't measure the spin in that direction, you will get random results whose probabilities are given by the Born rule. These probabilities are not due to our ignorance of the quantum state.
     
  12. Apr 20, 2016 #11

    blue_leaf77

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    When dealing with a pure state, it's not right to ask the question "which eigenstate (of the operator) that the state is in before measurement?". Before any measurement was done on this pure state system, you know that the system is in this very pure state which turns out to be a superposition of eigenstates of the observable you are going to measure. Before this measurement, the system is not in either eigenstates of the observable, it's in a superposition of them.
    When asked to distinguish the difference between mixed and pure states (which might as well be a superposition of other states), I like to interpret the state of a quantum system as a result of preparation procedure. Long story short, a pure state is a result of a measurement procedure which is known to be able to produce a certain quantum state all the time, i.e. the chance that this procedure will give out this particular state is 100%. In contrast, a system in a mixed state is produced by a preparation procedure whose output covers more than one possible pure states with a certain probability of occurrence associated to them, that's why we are blind as to which pure state we will end up measuring when the system is in a mixed state. This probability is what you termed classical probability. I don't know the related history how it got its name, but I think it makes sense that this probability is called classical because it does represent the physical chance of a process to output what it can output - it's not related to the fundamental probabilistic nature of quantum mechanics. By analogy it's like the probability of getting the face or tail when you toss a coin in the air.
    Yes.
     
  13. Apr 21, 2016 #12

    DrDu

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    The difference between mixed and pure state is not an absolute one but depends on the observables you are controlling. For example, if you consider a particle in a box of volume 2V in a pure state, then if you only have under control the variables within a sub-volume of volume V then the particle will be described as a mixed state. E.g. it may be a mixture to observe one or zero particles in V. In fact, this is how mixed states often are obtained in practice, i.e. you consider a system and it's surrounding. You assume that the system + surrounding are in a pure state, but if you only control the system and can't gather information about the surrounding, the system alone will be in a mixed state.
     
  14. Apr 21, 2016 #13
    So would one say that as a pure state is one which contains complete information of the statistical distributions of each observable, the state vector must necessarily be in a superposition of a given set of eigenvectors (if we project onto a basis generated by a particular operator) since it must contain all possible information one can possibly know about this observable. When we measure the observable the state of the system will collapse into a particular eigenstate of the observable (in this sense do we lose some information from the system?)
     
  15. Apr 21, 2016 #14
    Would it be correct to say that in the case of a pure state, probabilities arise because the observables of the state are intrinsically non-deterministic (although their evolution is deterministic in the case of closed or isolated systems where they evolve according to the Schrödinger equation). In the case of mixed states, there is also a probability that arises due to the fact that we have a lack of knowledge about which state the system is actually in and thus we must consider it in terms of an ensemble of pure states, i.e. the state of the system is not well-defined. This is a so-called classical probability since it is not intrinsic to the observables of the system and hence the non-deterministic nature of which state the system is actually in arises solely due to our lack of complete information about the system, i.e. this non-determinism would disappear in the case where we have complete knowledge of the state of the system (its state is well-defined), however, the probabilistic nature of its observables would remain (this is a fundamental property of nature and not due to any hidden variables that we lack knowledge of).
     
  16. Apr 21, 2016 #15

    blue_leaf77

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    Yes.
    Yes.
    Yes.
    Yes.
     
  17. Apr 21, 2016 #16
    Ok cool, thanks for your help. I think I understand it now?!
     
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