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 $p_i$: $$\rho = \sum_{i=1}^N p_i |\psi_i \rangle \langle \psi_i| \tag{1}\:.$$ 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.

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

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.

  • $\begingroup$ Shouldn't $q_j$ be distinguished because they are invariants of the unitary evolution in contrast to $p_j$. $\endgroup$ Commented Aug 11, 2016 at 12:52

4 Answers 4


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.


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!

  • $\begingroup$ 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. $\endgroup$ Commented Feb 12, 2014 at 9:10
  • 1
    $\begingroup$ 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. $\endgroup$
    – celtschk
    Commented Jun 22, 2014 at 22:00

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


I know that you said you're not interested in the case where the mixed state was obtained via a partial trace of a composite system, but there are certainly interesting things to say in that context as well, which are related to this discussion.

Consider this thought experiment: I give you two qubits, A and B. I've measured qubit A along the $z$-axis, but haven't told you the result of my measurement. Qubit B is entangled as a Bell pair with a third qubit which you don't have access to. Are the two qubits "equivalent"?

The answer depends on your interpretation of quantum mechanics. Someone who subscribes to a realist interpretation, an epistemic interpretation, and the many-worlds interpretation would all answer differently. (The many-worlds interpretation is usually classified as realist, but for the purpose of this question it's clearer to separate it off. Everyone would agree on the results of all physical experiments, but would only disagree on the right words to describe them.)

Ralph the realist, who subscribes to a (non-many-worlds) realist interpretation, would say that since I've measured qubit A along the $z$-axis, it's definitely in either the pure state $|\uparrow\rangle$ or the pure state $|\downarrow\rangle$ by the measurement postulate of quantum mechanics. You don't know which pure state, so you should describe the system with a maximally mixed density matrix, but the uncertainty is purely classical and simply reflects your classical ignorance of what I measured. Qubit B, on the other hand, is entangled with another qubit, so it's decribed by reduced density matrix, and the origin of the uncertainty in its state is fundamentally quantum-mechanical. Ralph would say that Qubit A is therefore in an unknown pure state while qubit B is in a mixed state. Philosophers of physics say that qubit A is a "proper mixture" because its probabilistic nature comes from classical ignorance, and qubit B is an "improper mixture" because it's described by a reduced density matrix and its probabilistic nature comes from quantum entanglement. By drawing this distinction, they implicitly claim that classical and quantum uncertainty are philosophically distinct, even if they cannot be distinguished empirically, as you point out.

Eva the epistemicist, who subscribes to an epistemic interpretation, would say that since there's no physical calculation or measurement that can distinguish classical and quantum uncertainty, there's no reason to consider them philosophically distinct, and that the alleged distinction between proper and improper mixtures doesn't actually exist. She would say that both qubits "really are" in the same maximally mixed state, not just that they are equivalent from your perspective. This view is appealing from a logical positivist standpoint, but it leads to the possibly counterintuitive implication that the purity of a physical system is subjective and depends on your "background knowledge": I would describe qubit A as being in a pure state, you would describe it as being in a mixed state, and we'd both be right.

Minerva the many-worlder, who subscribes to the many-worlds interpretation, would say that (assuming it wasn't already pointing parallel to the $z$-axis before the measurement) qubit A is in a mixed state - but that I am also in a mixed state because I measured it! Qubit A and I are together in a coherent superposition of my having measured "up" or "down" (although it will quicky decohere as the entanglement spreads further), and so qubit A and I are each individually in a mixed state. Minerva would agree with Ralph that there is a fundamental difference between classical and quantum uncertainty, but she would agree with Eva that both qubits are in the exact same mixed state. However, unlike Eva, who denies the existence of a distinction between proper and improper mixtures, Minerva would say that both qubits are in (identical) improper mixtures.

  • $\begingroup$ Umm, I'm a bit confused about Minerva. I understand that when you have measured spin A but you + spin A haven't yet interacted with Minerva, she would say that you + spin A constitute an entangled pure state w.r.t. her and thus the individual state of either you or spin A is a reduced density matrix for her. So for her, you + spin A is equivalent to spin B + the other spin. But as soon as you interact with her and tell her you know the spin of A, she'd herself have entangled with you + spin A so, at that point, she'd describe her lack of knowledge of spin of A as a proper mixture, no? $\endgroup$
    – user87745
    Commented Jun 3, 2020 at 17:43
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    $\begingroup$ @DvijD.C. Yeah that's a good point, I guess once once I "give you" the qubit, you presumably decohere right along with me. I'm not exactly sure how the philosophers who draw a distinction between proper and improper mixtures think about the many-worlds interpretation. (Maybe they don't.) $\endgroup$
    – tparker
    Commented Jun 4, 2020 at 2:22

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