# What is really that density matrix in QM?

• I
That would turn this thread into discussing every difference between Classical and Quantum Information theory.
The thread has already changed the original question which was about: What is really that density matrix in QM?
If the similarities go beyond a simple point of view, we should detect the physical consequences that result from them.
/Patrick

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And to measure in an eigenbasis of rho itself, where the two agree, is infeasible in all but the simplest situations. This answers @Morbert's query.
This basis is often an energy eigenbasis, which is why quantum chemists are interested in the von Neumann entropy as the Shannon entropy of the natural orbitals/modes of the system (It's a step in various investigations about the relationship between independent fermion entropy and wavefunction compactness, correlation energy, interaction energy functionals etc.)

PS I should clarify that by maximal knowledge I meant with respect to other measurement contexts as opposed to other possible preparations of the system.

DarMM
Gold Member
The thread has already changed the original question which was about: What is really that density matrix in QM?

/Patrick
Not for me, I was only considering entropy in so far as it related to distinguishing pure and mixed states. Even ignoring this you seemed to be discussing that quantum information sources cannot be modelled with Classical Information theory which I don't think anybody would dispute or has disputed.

A. Neumaier
2019 Award
This basis is often an energy eigenbasis
For your recipe to apply one needs an eigenbasis of rho. This can be an eigenbasis of energy only if the state is stationary. While this case is important it is still a very special state.

vanhees71 and Morbert
Not for me, I was only considering entropy in so far as it related to distinguishing pure and mixed states.
OK

The same density matrice can represent different mixed states, i.e the spectral decomposition of a density matrix is not unique (unless the state is pure) and there are several ways to achieve the same density matrix by mixing pure states.

/Patrick

DarMM
Gold Member
OK

The same density matrice can represent different mixed states, i.e the spectral decomposition of a density matrix is not unique (unless the state is pure) and there are several ways to achieve the same density matrix by mixing pure states.

/Patrick
A density matrix is a mixed state, as far as I've seen they are synonymous tetms. Though the decomposition into a sum of pure states, if it exists, is not unique as you said. Though again that has not been in question.

Though again that has not been in question.
Is the density matrix formulation is necessary to understand results from real experiments, or is it possible to do it without it ?
if entropy represents an intrinsic property of a physical system, the ambiguity associated with the representation carried by a density matrix does not make it an appropriate tool for talking about entropy.

/Patrick

atyy
Regardless of these issues with the interpretation of classical entropy are we agreed that mixed states are just general quantum states for two reasons:
1. Their state space is ##Tr\left(\mathcal{H}\right)## not ##\mathcal{L}^{1}\left(\mathcal{H}\right)##, thus they seem not to quantify classical ignorance of a pure state since they cannot be read as probability distributions over pure states

2. In QFT finite volume systems have no pure states.
Pure states are then just a special case where you have one totally predictable context, they don't constitute "the true state" of which one is ignorant. Such a totally predictable context seems to be absent in QFT, there is always some measurement uncertainty in QFT thus only mixed states.
I don't understand this, and will have to take your word for it, since this doesn't occur in elementary quantum mechanics (eg. it doesn't occur in anything Englert says). I have questions about the case in which pure states don't exist:

1) Does the Stinesprung theorem fail? Usually the Stinesprung theorem means that mixed states can be obtained as subsystems of pure states

2) If there are no pure states, how is time evolution defined? Is unitary evolution still fundamental? Isn't unitary evolution only expected for pure states or proper mixtures of pure states? For example, an improper mixture is not expected to undergo unitary evolution.

A. Neumaier
2019 Award
I don't understand this, and will have to take your word for it, since this doesn't occur in elementary quantum mechanics (eg. it doesn't occur in anything Englert says).

For example, an improper mixture is not expected to undergo unitary evolution.
Unitary evolution is reserved for truly isolated systems. Restricting unitary dynamics to an isolated subsystem preserves unitary evolution. If the isolation is imperfect the preservatives is imperfect. This is the usual situation.

DarMM
Gold Member
1) Does the Stinesprung theorem fail? Usually the Stinesprung theorem means that mixed states can be obtained as subsystems of pure states

2) If there are no pure states, how is time evolution defined? Is unitary evolution still fundamental? Isn't unitary evolution only expected for pure states or proper mixtures of pure states? For example, an improper mixture is not expected to undergo unitary evolution.
The distinction between proper and improper mixed states breaks down in quantum field theory. See Section 4 of this review of entanglement and open systems in QFT:
https://arxiv.org/abs/quant-ph/0001107
Time evolution of any finite volume system will be non-unitary as everything is an open system in QFT.

atyy
DarMM
Gold Member
if entropy represents an intrinsic property of a physical system, the ambiguity associated with the representation carried by a density matrix does not make it an appropriate tool for talking about entropy.
I don't understand, has anybody been discussing basing things on this decomposition?

atyy
The distinction between proper and improper mixed states breaks down in quantum field theory. See Section 4 of this review of entanglement and open systems in QFT:
https://arxiv.org/abs/quant-ph/0001107
Time evolution of any finite volume system will be non-unitary as everything is an open system in QFT.
Is it be possible to understand that as a case in which the state of every subsystem is an improper mixture?

If so, wouldn't the state of the total system still be pure, so that unitary evolution still applies to the total system, and governs the evolution of the subsystems?

DarMM
Gold Member
Is it be possible to understand that as a case in which the state of every subsystem is an improper mixture?

If so, wouldn't the state of the total system still be pure, so that unitary evolution still applies to the total system, and governs the evolution of the subsystems?
The total system would be the entire universe. This remains an open question in QFT whether there are global pure states. Complications involved are:
1. QED coloumb fields might always be mixed. This is the mathematically most intractable problem.

2. Such a pure state has no operational meaning. The theorised global purification might be a state over an algebra of self-adjoint operators that includes Wigner's friend type "observing macroscopic systems to the atomic scale" type observables which almost certainly lie outside the observable algebra. Thus over the true algebra of observables the state will still be mixed.

3. Poorly understood complications from QFT in curved backgrounds, e.g. the information loss problem.

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vanhees71
Gold Member
2019 Award
It is time invariant hence stationary but leads to completely wrong predictions for thermal q-expectations such as the internal energy.
How do you come to this conclusion? It depends on the system!

Of course the MEM doesn't tell you what's the relevant information for a given system. That you have to determine yourself. The MEM is a very general principle. It's almost inevitable to make "objective" guesses based on the available information about the system.

Now 2 issues are usually raised against it, also in this thread.

(1) Choice of the "relevant observables" to be constraint in maximizing the entropy

Your criticism is quite common: Given only the constraint that you want a stationary state, constrains the possible choice of "relevant information" to be either averages of conserved quantities (like for energy in the canonical and grand-canonical ensemble; in the latter case you also give the average on one or seveal conserved charge-like quantities) or strict constraints of conservation laws (energy is strictly in an on macroscopic scales very small interval, as for the microcanonical ensemble).

That usually the Gibbs ensembles are preferred, i.e., giving at least constraints on ##U=\langle \hat{H} \rangle## and not some on any other function of ##f(\hat{H})##, is due to the fact that one considers the thermodynamical state as a small system coupled (weakly) to some reservoir, defining the canonical and grand-canonical ensembles (depending on whether you allow for exchange of conserved charges or not). Treating than the closed system large (reservoir+system)-system as a microcanonical ensemble (where it doesn't matter, which function ##f(\hat{H})## you look at), you are inevitably let to the usual canonical or grand-canonical ensemble with ##\langle \hat{H}_{\text{sys}} \rangle## is the relevant constraint. In the thermodynamic limit higher cumulants don't play a role. For small systems it can be necessary to take such higher-order constraints into account. The same argument holds if you have other relevant conserved quantities to take into account: In the typical "reservoir situation" for macroscopic thermodynamical systems the additive conserved quantities are the relevant observables and usually not more general functions thereof. For details, see

https://journals.aps.org/rmp/abstract/10.1103/RevModPhys.85.1115

and references therein.

This view is also solidified by the usual dynamical arguments using non-equlibrium descriptions of open quantum systems (master equations, transport equations, etc.): For short-range interactions in the collision term and truncating the BBGKY hierarchy (or the corresponding analogs for more detailed quantum descriptions; as the general Schwinger-Dyson hierarchy of QFT) at the lowest order leads to the standard equilibrium distributions (Bose-Einstein, Fermi-Dirac, Maxwell-Boltzmann), corresponding to the choice of additive conserved quantities in the MEM as discussed above.

(2) Choice of the information measure

This is the question, which type of entropy to use. The Shannon-Jaynes one in the physics context refers to the classical Boltzmann-Gibbs entropy. It's well-known that this doesn't work with lang-range forces present (electromagnetic, but that's pretty harmless, because in many-body systems you usually have Debye screening which comes to the rescue, but there's no such thing for gravity, and that's important for structure formation in the universe and our very existence). Here other (non-additive) entropy forms like Renyi or Tsallis entropies may be the better choice. As far as I know, there's however no generally valid dynamical argument as in the Boltzmann-Gibbs case.

vanhees71
Gold Member
2019 Award
OK

The same density matrice can represent different mixed states, i.e the spectral decomposition of a density matrix is not unique (unless the state is pure) and there are several ways to achieve the same density matrix by mixing pure states.

/Patrick
No! The statistical operator uniquely determines the state of the system, and its spectral decomposition is unique, if you use a complete set of compatible observables, including the statistical operator. It's a self-adjoint operator!

It's of course right that mixing pure states is not unique, but why should it be?

No! The statistical operator uniquely determines the state of the system, and its spectral decomposition is unique, if you use a complete set of compatible observables, including the statistical operator. It's a self-adjoint operator!

It's of course right that mixing pure states is not unique, but why should it be?
here

/Patrick

atyy
The total system would be the entire universe. This remains an open question in QFT whether there are global pure states. Complications involved are:
1. QED coloumb fields might always be mixed. This is the mathematically most intractable problem.

2. Such a pure state has no operational meaning. The theorised global purification might be a state over an algebra of slef-adjoint operators that includes Wigner's friend type "observing macroscopic systems to the atomic scale" type observables which almost certainly lie outside the observable algebra. Thus over the true algebra of observables the state will still be mixed.

3. Poorly understood complications from QFT in curved backgrounds, e.g. the information loss problem.
Is this related: https://arxiv.org/abs/1406.7304 ?

Is it really true then that there is no unitary time evolution in QFT? For measurement, I can buy that there are no mixed states. But I find it hard to buy that there is no unitary evolution. Really? Then there will be complications with the information loss problem, as you say.

DarMM
Gold Member
Is this related: https://arxiv.org/abs/1406.7304 ?

Is it really true then that there is no unitary time evolution in QFT? For measurement, I can buy that there are no mixed states. But I find it hard to buy that there is no unitary evolution. Really? Then there will be complications with the information loss problem, as you say.
It's probably true, see remark 15 (p.31) in this paper:
https://arxiv.org/abs/1412.5945

Auto-Didact and atyy
atyy
It's probably true, see remark 15 (p.31) in this paper:
https://arxiv.org/abs/1412.5945
But how about for flat spacetime? In the case of no no pure states in flat spacetime, can we have unitary time evolution?

DarMM
Gold Member
That's an open question due to issues with infrared representations in QED.

atyy
That's an open question due to issues with infrared representations in QED.
Well, maybe QED doesn't exist - is this also expected to be a problem in say Yang Mills? Do we have pure states and unitary time evolution in Yang Mills?

vanhees71
Gold Member
2019 Award
Sure, that's why I said, if you use a complete set of compatible observables including ##\hat{\rho}##. As any operator, a given ##\hat{\rho}## is of course unique. It doesn't depend on it's representation in terms of a complete orthnormal system (CONS), i.e.,
$$\hat{\rho} = \sum_{ij} |i \rangle \langle i|\hat{\rho}|j \rangle \langle j|.$$
If now you have a ##\hat{\rho}## with degnerate eigenvalues ##p_j##, there are of course arbitrarily many CONS. Let ##|i,\alpha \rangle## be one such CONS, but in all of them you get
$$\hat{\rho}=\sum_{i} p_i \sum_{\alpha} |i,\alpha \rangle \langle i,\alpha |.$$
The inner sum is just the projector to the uniquely defined "degenerate" eigenspace to the degenerate eigenvalue ##p_i##. These projectors are independent of the chosen degenerate CONS since obviously for another CONS ##\widetilde{|i,\alpha \rangle}##
$$\hat{P}_i=\sum_{\alpha} |i,\alpha\rangle \langle i,\alpha| = \widetilde{|i,\alpha \rangle}\widetilde{\langle i,\alpha|}.$$

DarMM
Gold Member
Well, maybe QED doesn't exist - is this also expected to be a problem in say Yang Mills? Do we have pure states and unitary time evolution in Yang Mills?
Yes for Yang-Mills in flat spacetime we should have global pure states.

As for QED not existing I've always found the arguments for this very weak. It's based on the existence of a perturbative Landau pole, but the Gross-Neveu model has a Landau pole perturbatively, but still exists as a well-defined QFT in the non-perturbative sense.

However all of this would still only be in the idealisation of flat space. In curved space for any theory there simply isn't unitary time evolution simply because that's not how time evolution can be modelled for field theories in general, it is to be replaced with the notion of algebraic automorphisms.

dextercioby, atyy and vanhees71
DarMM
Gold Member
I should say I'm slowly in the process of gathering all this info in a coherent form. What nonperturbative QFT in curved spacetime is actually like renders many debates about QM pointless or seriously recasts the issues and I think it would be useful for others to know.

Auto-Didact and vanhees71
atyy