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In textbooks, it is sometimes written that a mixed state can be represented as mixture of $N$ (I assume here $N<+\infty$) quantum pure states $|\psi_i\rangle$ with classical probabilities $p_i$: $$\rho = \sum_{i=1}^N p_i |\psi_i \rangle \langle \psi_i|\quad (1)\:.$$ Above $p_i \in (0,1]$ and $\sum_i p_i =1$ and a do not necessarily assume that $\langle \psi_i|\psi_j\rangle =0$ if $i\neq j$ but I require that $\langle\psi_i |\psi_i\rangle =1$ so that $\rho \geq 0$ and $tr(\rho)=1$. (There is another procedure to obtain mixed states using a partial trace on a composite system, but I am not interested on this here).

I am not sure that it makes any sense to distinguish between classical probabilities embodied in the coefficients $p_i$ and quantum probabilities included in the pure states $|\psi_i\rangle$ representing the quantum part of the state. This is because, given $\rho$ as an operator, there is no way to uniquely extract the numbers $p_i$ and the states $|\psi_i\rangle$.

I mean, since $\rho = \rho^\dagger$ and $\rho$ is compact, it is always possible, for instance, to decompose it on a basis of its eigenvectors (and there are many different decompositions leading to the same $\rho$ whenever $\rho$ has degenerate eigenspaces). Using non orthogonal decompositions many other possibilities arise.

$$\rho = \sum_{j=1}^M q_j |\phi_j\rangle \langle \phi_j|\quad (2)$$

where again $q_j \in (0,1]$ and $\sum_j q_j =1$ and now $\langle \phi_i|\phi_j\rangle =\delta_{ij}$. I do not think there is a physical way to decide, a posteriori, through suitable measurements of observables if $\rho$ has been constructed as the incoherent superposition (1) or as the incoherent superposition (2). The mixed state has no memory of the procedure used to construct it.

To pass from (1) to (2) one has, in a sense, to mix (apparently) classical and quantum probabilities.

So I do not think that it is physically correct to associate a classical part and a quantum part to a mixed state, since there is no a unique physical way to extract them from it.

Perhaps my impression is simply based on a too naively theoretical interpretation of the formalism.

I would like to know your opinions about this issue.

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You can distinguish pure and mixed state by whether Tr$\rho^2=1$. Then anything beyond pure state belongs to classical probability. Because, by the postulates of quantum mechanics, a state is a pure state. Mixed state only comes from incomplete knowledge, then we use classical statistics. –  user26143 Feb 11 at 11:39
    
The problem is not to distinguish between pure and mixed states! I completely disagree on the fact that that there is an axiom in QM stating that states are pure states. That is just a naive and non-motivated interpretation. States are generalized measures on the lattice of orthogonal projectors. Then Gleason's theorem proves that they are all of density matrices. Pure states are nothing but the extremal elements of the convex body of the states. –  Valter Moretti Feb 11 at 11:42
    
First of all, I take the framework from Weinberg's lectures in quantum mechanics, p. 52, "The first postulate of quantum mechanics is that physical states can be representated as vectors in a sort of abstract space known as Hilbert space". The mixed state will be outside the Hilbert space. Later in p 68, "probabilities can enter in quantum mechanics not only because of the probabilistic nature of state vectors, but also (just as in classical mechanics) we may not know the state of system". I mean physically, the mixed state is just similar with classical statistics. –  user26143 Feb 11 at 11:53
    
Then if you want to distinguish pure or mixed state. You have use an ensemble description. Consider many silver atoms come from an oven and we do a Stern-Gerlach experiment, and we use a single-particle description, say, $|\psi \rangle$ vs $\sum_i p_i |\psi_i \rangle \langle \psi_i|$. The pure state will have a definite spin polarization direction but mixed state not. (Ref. Sakurai's modern quantum mechanics, rev edi, p174) However, we can also describe the silver atoms by a multiparticle state vector $|\psi_1 \rangle \bigotimes |\psi_2 \rangle \bigotimes \cdots \bigotimes |\psi_n \rangle $ –  user26143 Feb 11 at 11:53
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Well you could also quote Aristotle ;-), I could cite several other eminent authors assuming different axioms. I could think of states as positive functionals on the algebra of observables...Also in this case the distinction of pure and mixed states is done referring to the extremal property of the pure states. However It seems that you are not answering to my question: Is there a physical way, using measurements of observables, able to distinguish between two different preparations of mixed states though leading to the same matrix operator? –  Valter Moretti Feb 11 at 12:34

3 Answers 3

up vote 8 down vote accepted

Yes, the density matrix reconciles all quantum aspects of the probabilities with the classical aspect of the probabilities so that these two "parts" can no longer be separated in any invariant way.

As the OP states in the discussion, the same density matrix may be prepared in numerous ways. One of them may look more "classical" – e.g. the method following the simple diagonalization from equation 1 – and another one may look more quantum, depending on states that are not orthogonal and/or that interfere with each other – like equations 2.

But all predictions may be written in terms of the density matrix. For example, the probability that we will observe the property given by the projection operator $P_B$ is $$ {\rm Prob}_B = {\rm Tr}(\rho P_B) $$ So whatever procedure produced $P_B$ will always yield the same probabilities for anything.

Unlike other users, I do think that this observation by the OP has a nontrivial content, at least at the philosophical level. In a sense, it implies that the density matrix with its probabilistic interpretation should be interpreted exactly in the same way as the phase space distribution function in statistical physics – and the "quantum portion" of the probabilities inevitably arise out of this generalization because the matrices don't commute with each other.

Another way to phrase the same interpretation: In classical physics, everyone agrees that we may have an incomplete knowledge about a physical system and use the phase space probability distribution to quantify that. Now, if we also agree that probabilities of different, mutually excluding states (eigenstates of the density matrix) may be calculated as eigenvalues of the density matrix, and if we assume that there is a smooth formula for probabilities of some properties, then it also follows that even pure states – whose density matrices have eigenvalues $1,0,0,0,\dots$ – must imply probabilistic predictions for most quantities. Except for observables' or matrices' nonzero commutator, the interference-related quantum probabilities are no different and no "weirder" than the classical probabilities related to the incomplete knowledge.

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I will provide an answer but from a different perspective, and hopefully convince you that there is information in a density matrix which has no classical counterpart. Furthermore this can hence be considered a quantum component, and it can be shown that this information is stored as the eigenvectors of $\rho$.

I will give an example of how this manifests. The Fisher Information $I(\theta)$ is a statistic from classical probability theory which characterises how quickly one can learn about a parameter $\theta$ which characterises a probability distribution $p(\theta)$.

Specifically the variance of an unbiased classical estimator $\hat{\theta}$ respects the Cramer Rao bound $$\mathrm{var}(\hat{\theta})\geq \frac{1}{I(\theta)}$$

The additivity of information means that if you sample the distribution $n$ times, collecting measurements each time the expected error $\Delta \theta_c = \sqrt{\mathrm{var}(\hat{\theta})}$ of any estimator goes like $$\Delta \theta_c \propto \frac1{\sqrt{n}}$$

This is recognised in the scaling of the standard deviation $\sigma$ in things like central limit theorem.

We can define a quantum analogue, to the fisher information $J(\theta)$ which satisfies an analogus bound, known as the Quantum Cramer Rao bound.

However it is found that by permitting entanglement between classically independent sampling events, the bound is much better. And after having collected a dataset of $n$ measurments, the best possible quantum estimator is bound only by the error $$\Delta \theta_q \propto \frac1{n}$$.

This shows that a general quantum state $\rho$ can definately support statistics which a classical probability distribution cannot.

The quantum Fisher information of a density matrix which depends on a parameter $\theta$ $$\rho(\theta) = \sum_i p_i(\theta) |\psi_i(\theta)\rangle\langle\psi_i(\theta)|$$ can be seen to seperate into several contributions, one of which is the classical Fisher information of the spectrum $p_i(\theta)$, another of which is a Fubini-Study like term which accounts for the information stored in the basis $|\psi_i(\theta)\rangle$. The possibility of (super-classical) quantum scaling depends entirely on the existence of this quantum term.

Alternatively stated, in terms of the behaviour of the Fisher information statistic and its quantum analogues, a density matrix $\rho$ supports non classical behaviour only if the basis set $|\psi_i(\theta)\rangle$ contains information relevant to the measurment, and in this sense, information stored in this way may be considered non-classical.


Useful stuff

If you are interested in some of the topics discussed here see this good review for an explanation. http://arxiv.org/pdf/1102.2318v1.pdf

This for an accessible but mathematical explanation of the QFI. http://arxiv.org/pdf/0804.2981.pdf

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Let's look at a famous, concrete example: Perfectly unpolarized light.

Alice creates unpolarized light by randomly (incoherently) mixing left-circular-polarized light with an equal intensity of right-circular-polarized light.

Bob creates unpolarized light by randomly (incoherently) mixing vertically-polarized light with an equal intensity of horizontally-polarized light.

There is no measurement that will tell you which light is Alice's and which is Bob's.

Are Alice's light fundamentally the same as Bob's light, or are they different kinds of light that are impossible to tell apart?

Well, one shouldn't make too much of these kinds of questions. But if I had to choose, I would say that they are different kinds of light, because the classical incoherent mixing process leaves a trail of information out there that is sufficient to tell the two beams apart (even though I may not have that information right now in practice).

For example, maybe Alice and Bob are each combining two different laser beams with slightly-different (and randomly fluctuating) frequencies. (This is a legitimate way to incoherently add two light beams in practice.) If I don't have a very fancy spectrometer, I can describe all my possible measurements by saying that these are unpolarized beams. But if I do have a fast and high-resolution spectrometer, I can figure out which beam is Alice's and which is Bob's.

This is an example of a broader truth: Classical probabilities are more situation-dependent than quantum probabilities. Specifically: If two people each think that a particle is in a pure state, they will always agree on what state it is in, and therefore they will agree on the probability distribution for any possible measurement of that particle. But if two people each think that a particle is in a mixed state, then they will often disagree on what mixed state it is in, because they may have different auxiliary knowledge, which leads them to assign different classical probabilities. (For example, maybe the particle is one of an EPR pair, and its twin has been measured, but only one of the observers knows the measurement result.)

But, given a state of "my knowledge right now", there is no way to draw a line between classical probabilities and quantum probabilities---and no reason to!

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However, considering the state prepared by, say Bob, even it is not perfectly unpolarized it is described by some density matrix $\rho$ "close" to $I/2$. No matter its specific form, you can always write it as an incoherent superposition of pure states different from those used by Bob, corresponding to a different experimental preparation. Once the state has been constructed you cannot distinguish between the two preparations by measuring any observable on the state. That is my point. –  Valter Moretti Feb 12 at 9:10
    
A third way to prepare the same $\rho$ is to generate a polarization-entangled two-photon state and send one of the photons. If you retain the other photon, you can turn the sent-out photon into either Alice's or Bob's ensemble by measuring the other photon in either the horizontal/vertical or circular polarization basis. Note that the actual measurement statistics of the sent-out photon won't depend on which way (or even whether) you measured the second photon. –  celtschk Jun 22 at 22:00

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