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Suppose we have two quantum mechanical systems with Hilbert spaces $H_1$ and $H_2$ respectively. I am trying to understand the difference between an entangled, pure, state compared to an entangled (mixed) state. If I am not mistaken, entangled pure states are merely unit vectors of $H_1 \otimes H_2$, whereas a mixed state $\rho$ is an ensemble: $$ \rho = \sum_i w_i |\psi_i \rangle \langle \psi_i |. $$

Here's where I am confused. If an entangled system has mixed state $\rho$, how should we interpret what is physically going on?

  • (Interpretation 1) The state of the system is literally a random variable with distribution $\rho$.
  • (Interpretation 2) The system is indeed in a pure state, but we don't know what pure state actually describes the system. Nevertheless the expected likelihood that the system is in state $| \psi_i \rangle$ is given by $w_i$.

For context, this question arose from comparing quantum mechanics to classical statistical mechanics. In the classical case, I think interpretation 2 is appropriate. Therefore, if quantum mechanics matches interpretation 1, then this would be a profound distinction from classical intuition.

Anthony D'Arienzo
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  • This may answer your question. – joseph h Dec 04 '20 at 05:59
  • @Drjh I saw that post. I don’t think I have an issue with the mathematics of mixed states. I am confused by what “statistical mixture” (in the language of that post) physically means. – Anthony D'Arienzo Dec 04 '20 at 07:16
  • The mixed states of photon polarization $p(+)=p(-)=.5$ and $p(\phi)=1/2/\pi$ where phi is the angle of the qubit state give the same mixed state, but are experimentally distinguishable by looking if full intensity is obtained for all polarizator angle, photon by photon. Hence is there another way to formalize mixed states ? – QuantumPotatoïd Jun 24 '23 at 18:26

2 Answers2

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(I stress that we are not discussing here mixed stats obtained by taking the partial trace over a subsystem.)

Interpretation 2 is problematic. There are infinitely many ways to write a mixed state as a convex combination (incoherent superposition) of pure states and there is no way to distinguish them from the experimental viewpoint, i.e, with measurements.

In other words there is no way to say, e.g., which are the pure quantum states the system is visiting during its apparently random evolution. We should choose them a priori.

Typical case: I incoherently superpose a pair of non orthogonal pure states and next I decompose the density matrix into its orthogonal eigestates. A posteriori, there is no experimental way to say how I have really produced the mixed state, if superposing the eigenstates or the original non orthogonal pure states.

In classical physics instead, in principle, we can determine the real states the statistical state is made of through accurate measurements. It is difficult but not impossible.

I think that a safe point of view is considering mixed states as generic quantum states and viewing pure states as special cases if them.

ADDENDUM. My idea is that a state of a quantum system is the full assignment of every probability of every outcome of every observable of that system.

That is the best information the quantum world permits us to know (excluding non-local/contextual realistic hidden variable theories).

Gleason's theorem proves that the said assignment is exactly a density matrix. (see my answer to this question Why is the application of probability in QM fundamentally different from application of probability in other areas?)

From this perspective the so called mixed states are more natural than pure states.

In this view pure states are states which cannot be "probabilistically" decomposed into other states. They are extremal elements in the space of the quantum probability measures.

As is well known they are one-to-one with the unit vectors up to phases of the Hilbert space. These are the familiar state vectors $|\psi\rangle$ of the Hilbert space.

However, the fact that someone (I in particular) finds more familiar pure states than mixed states is principally due to historical reasons, in my opinion, but it does not relies, in my view, upon strong physical reasons.

Valter Moretti
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  • Oh no, just an hour ago I was reading your answer to the question about the Noether theorem and was admiring your presentation. But here I could not disagree more. You are of course correct that certain statistical ensembles of quantum states (elements of the Hilbert space) are in principle experimentally indistinguishable. But that premise does not contradict the standard view that the Hilbert space is the space of ALL physical states of the system (which is what interpretation 2 relies on). – ReasonMeThis Dec 04 '20 at 09:26
  • In my view the set of the states is the whole set of the so-called mixed state, which are the true quantum states. In fact, from a more general point of view, states are probability measures on the non-Boolean lattice of the orthogonal projectors. Gleason's theorem proves that states are one-to-one with matrix densities. – Valter Moretti Dec 04 '20 at 09:32
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    However all that is matter of personal taste. The only physical fact is the one I pointed out: differently form classical physics one cannot say which are the "true" states (in the naive view the pure states) that really decompose a so-called mixed state. For that reason the intepretation (2) is in my view disputable. – Valter Moretti Dec 04 '20 at 09:34
  • There still needs to be a sound argument to demonstrate how your correct premise (indistinguishability of certain ensembles) implies your conclusion (that it's untenable to view the Hilbert space as the space of ALL physical states). If you have one, I will edit my answer to address it. – ReasonMeThis Dec 04 '20 at 09:45
  • I am not sure to understand your point. Are you saying that, when we describe the state of a quantum system in terms of a mixed state, actually there is a pure state that is the true state of the system hidden somewhere nd that in principle we could know? – Valter Moretti Dec 04 '20 at 09:49
  • My idea is that a state is the full assignment of every probability of every outcome of every observable. That is the best information the quantum world permits us to know. Gleason's theorem proves that the said assignment is exactly a density matrix. However there are beautiful density matrices which cannot be decomposed into other density matrices: these are the so-called pure states. They are one-to-one with the unit vectors up to phases of the Hilbert space. – Valter Moretti Dec 04 '20 at 09:58
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    The fact that someone is more familiar with pure states is principally due to historical reasons, but it does not relies, in my view, on strong physical reasons. – Valter Moretti Dec 04 '20 at 10:00
  • My point is that in order to argue, as you do, that interpretation 2 is untenable/problematic there would need to be a sound argument that it's untenable to view the Hilbert space as the space of ALL physical states. But what is this sound argument? Your correct premise (that certain ensembles are indistinguishable) does not by itself provide one. – ReasonMeThis Dec 04 '20 at 10:09
  • I added my argument. – Valter Moretti Dec 04 '20 at 10:11
  • @ReasonMeThis I recommend the first chapter of Asher Peres' book, where the formalism of mixed states is derived solely from a very simple Stern-Gerlach thought experiment. – Norbert Schuch Dec 04 '20 at 10:11
  • @ValterMoretti Thanks for adding your argument for untenability of interp. 2 to your answer, I love these foundational questions. I have added my response to my answer, where I hope I summarized your argument fairly (if I didn't, do let me know please!) I think you have a very interesting perspective on the foundations of QM. I'd be interested in learning which of the interpretations of QM this perspective most closely aligns with, and who are the main physicists/philosophers of physics currently advocating it. – ReasonMeThis Dec 05 '20 at 00:16
  • @ValterMoretti Could you explain why you made that preamble about not discussing mixed states resulting from partial traces? I'm intrigued as to why this matters. A mixed state simply encompasses epistemological ignorance. I'd go as far as saying that all mixed states are a result of some partial trace we're not aware of. – Tfovid Jan 01 '24 at 13:51
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Short answer: interpretation 2.

EDIT: the answer, and comments, by @ValterMoretti argue that interpretation 2 is untenable/problematic. However, that argument does not succeed, as I will discuss in the addendum below my original answer.

A few quick points:

  • for the question you're asking it doesn't matter whether or not it's a combined system
  • the density matrix formalism is used specifically to describe a statistical ensemble of quantum states, i.e. when the system is in a particular state but we don't know which one
  • the density matrix formalism does not negate the fact that the Hilbert space is the space of ALL possible states of the system
  • interpretation 1 is not just incorrect, it's hard or impossible to even make sense of it (how can the actual state of a physical system be a statistical ensemble of other states?)
  • this is not related to the question you asked, but your description of an entangled state is not right; an entangled state is not just any state of the combined system, it's a state that cannot be represented as a product.

ADDENDUM. The answer, and comments, by @ValterMoretti argue that interpretation 2 is untenable/problematic.

However, that argument does not succeed. Interpretation 2 says that the Hilbert space is the space of ALL physical states of the system (and therefore ensembles of physical states, represented by density matrices, are not themselves physical states OVER AND ABOVE the states in the Hilbert space). So what arguments are given to show that this view is untenable? At the time of my writing this, two points have been made by @ValterMoretti:

  1. Certain statistical ensembles of quantum states (states in the Hilbert space) are in principle experimentally indistinguishable.

I will call this the "main premise" for brevity.

My response: the main premise is of course true, but it does not establish the "desired conclusion", namely that it's untenable to view the Hilbert space as the space of ALL physical states. Examine carefully both the main premise and the desired conclusion, and you will see that the conclusion does not at all follow from the premise.

2.

My idea is that a state of a quantum system is the full assignment of every probability of every outcome of every observable of that system.

Gleason's theorem proves that the said assignment is exactly a density matrix.

My response: at best this is just another tenable interpretation. But the mere existence of an alternative tenable interpretation does nothing to demonstrate the untenability of interpretation 2.

I said "at best", but in reality I have deep philosophical reservations about the above quote. But this is a much longer conversation, and is irrelevant to the discussion of the tenability of interpretation 2.

ReasonMeThis
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