Talk:Density matrix

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Initial comments

Would be helpful to tie this in to density operator as discusssed in quantum logic. CSTAR 23:03, 19 May 2004 (UTC)

Any objections to moving this page back to "Density matrix" where it belongs? --V79 16:35, 2005 Jun 15 (UTC)

Done --V79 13:08, August 7, 2005 (UTC)

The whole "C*-algebraic formulation of density states" section seems pretty unnecessary to me - right now, it's a lot of jargon that doesn't even make sense to a physicist. Also, a statement like "It is now generally accepted that the description of quantum mechanics in which all self-adjoint operators represent observables is untenable," definitely needs a reference. I will delete the section unless someone objects. --130.126.230.244 17:20, 27 January 2006 (UTC)

Done. If someone feels like rewriting the section to be more explanatory and less jargon-filled, go ahead. 130.126.230.244 16:22, 30 January 2006 (UTC)

The section has been replaced. It could be more explanatory, but as it stands is better than nothing. Those who find it useless may simply ignore it. Archelon 16:58, 5 May 2006 (UTC)

"It is now generally accepted" again! This is not good writing style. Was there a vote that I missed on that acceptance? ** However at last these hidden arguments against the "adjoint operator style" should be made clear.

Density matrix vs Density operator

Is it just me or does this article use “density matrix, $\rho$” for the “density operator, ${\hat {\rho }}$ ”? —Preceding unsigned comment added by Laser Lars (talk • contribs) 12:22, 10 September 2008 (UTC)

The terms "density matrix" and "density operator" are used interchangeably in everyday physics. Is there some technical difference? What is it? --Steve (talk) 20:10, 10 September 2008 (UTC)

Yes, the technical difference is whether or not one has chosen a basis. In Dirac notation an operator e.g. ${\hat {O}}=|\psi \rangle \langle \psi |$ can be represented by a matrix given a basis $\{|\phi _{k}\rangle \}$ by the elements $O_{mn}=\langle \phi _{m}|\psi \rangle \langle \psi |\phi _{n}\rangle$

OK, that makes sense. I just made an edit that hopefully corrects and clarifies this issue. --Steve (talk) 20:13, 12 September 2008 (UTC)

Actually, it's more complicated then that. For uncountably infinite dimensional systems, you can't write the operator as a matrix, even if you have a 'basis' .... so the term 'operator' covers a much more general case than the term 'matrix' does ... this article should actually be called Density Operator, NOT the specific case of density matrix Dr. Universe (talk) 23:58, 25 September 2010 (UTC)

There is such a thing as a matrix with infinitely or even uncountably-infinitely many rows. See Matrix (mathematics)#Infinite matrices. In my experience, people who say "matrix" in the context of "density matrix" are usually...but not always...thinking of finite dimensions. --Steve (talk) 19:05, 13 January 2011 (UTC)

The operator is the actual mathematical object, where the matrix is one of its many possible representations. Infact, in a different basis, the same operator is represented by a different matrix. This is true for finite and infinite dimensions. So I suggest to change the name of the article to "Density operator", and have "Density matrix" redirected to it. Oakwood (talk) 01:07, 17 October 2012 (UTC)

I have no preference between "density matrix" and "density operator" as the article title. I don't think there's anything wrong with "density matrix", for example Karl Blum chose Density Matrix, not Density Operator, for the title of his textbook on the subject. ...As long as the article explains the relation between the matrix and the operator, and I think it does. --Steve (talk) 12:08, 17 October 2012 (UTC)

on the dynamical equation

I think it's necesary to add to the article the dynamical equation of the density matrix: —The preceding unsigned comment was added by 201.232.162.150 (talk) 04:28, August 23, 2007 (UTC)

question on C* algebra formulation

so what's the problem with self adjoint operators as observables? in the finite dim case, the C* algebraic and SA operator formulations are the same. if one is to completely abandon the SA operator formulation, how does one come up with the C*-algebra of observables in the first place? As stated in the article, the GNS lets you recover the Hilbert space, is this not the state space you start with? Mct mht 01:41, 22 May 2006 (UTC)

For one thing, superselection rules. For another, one needs to tie in observables to geometrical structure and symmetry; Geometrical structure as in a "presheaf" kind of association between open sets and observables (in the Streater-Wightman axioms for instance. There's more to it than this of course, but this is the 2 minute summary I could most quickly think of.--CSTAR 02:03, 22 May 2006 (UTC)

here is a more elementary issue, but could also be a problem. as stated in article, if your observables are compact operators, then the states are precisely what's defined in the article. but if you wanna define a quantum operation in the Schrodinger picture, then it's a map between trace-class operators. but the dual map, between observables, is now between the full space of bounded operators. there seems to be some inconsistency there. Mct mht 00:50, 18 June 2006 (UTC)

Ultraweak continuity takes care of this.--CSTAR 02:25, 18 June 2006 (UTC)

so it seems one needs to use the Heisenberg picture map, between observables, in general. it's probably a good idea to modify the article quantum operation a bit accordingly. Mct mht 02:26, 18 June 2006 (UTC)

Yikes No. That article is formulated in terms of the Schrodinger picture. The dual of a quantum operation (on trace class operators) is ultraweakly continuous.--CSTAR 02:36, 18 June 2006 (UTC)

Do we really want C* algebra in the wikipedia article. This seems quite technical and probably not of interest to many people. Maybe it needs its own page. — Preceding unsigned comment added by 109.153.172.124 (talk) 10:12, 2 November 2014 (UTC)

Comment on recent edit

CSTAR: ok, obviously they're not measures in the measure theoretic sense, but one can see the formal resemblence. and they are called quantum probability measures (was probably your contribution, if i have to guess) in their own right anyway. calling them just that makes the heuristic comment on mixed states as probability distribution on states (common in physics literature) more clear. Mct mht 16:57, 22 May 2006 (UTC)

merge

the articles pure state and mixed state can easily be merged with this one. Mct mht 05:20, 22 May 2006 (UTC)

Yep.--CSTAR 17:03, 22 May 2006 (UTC)

Support. Nick Mks 15:25, 8 June 2006 (UTC)

Entropy

I've made some additions to the article consisting of expansion & clarifications of exisiting material and also linking measurement with entropy. See what you think. --Michael C. Price ^talk 20:51, 23 June 2006 (UTC)

For von Neumann Entropy: I think Log should be to the base 2, not natural log? See Von_Neumann_entropy#Definition -Mark Moriarty —Preceding undated comment added 17:05, 28 April 2012 (UTC).

I switched Von Neumann Entropy to use ln instead of log_2. I found ln in lots of sources. It's possible that both definitions are in use. --Steve (talk) 20:23, 28 April 2012 (UTC)

Mixed State to pure state

It states in the context that "Therefore a pure state may be converted into a mixture by a measurement, but a proper mixture can never be converted into a pure state." So can anybody comment on how to produce a pure state? Reducing temperature to zero, or get Bosons under critical temperature, any other ways? — Preceding unsigned comment added by 173.20.49.9 (talk) 16:54, 16 July 2011 (UTC)

Put a beam of unpolarized light (mixed state) through a polarizer. Then you have polarized light (pure state)! Very easy. We're basically talking about the second law of thermodynamics here. Entropy can't go down. But wait, my freezer lowers the entropy of water by making it into ice! How does that work? Well, it decreases the entropy of the water, but increases the entropy in other places (outside the freezer and inside the power plant making the electricity). In the light example, you can't turn a beam of unpolarized light into polarized light without reducing the intensity. (By a factor of 2.) The entropy does not decrease if you take the lost light into account too. Whoever wrote that sentence was referring to the whole system, not a subsystem of a larger system. The latter can certainly be made into a pure state. I am adding some text to the article to clarify this point... --Steve (talk) 05:05, 17 July 2011 (UTC)

I disagree with the prescription, even though it is technically correct. The original explanation is clearly talking about pure ensemble states being converted into a mixed ensemble state by measurement. It might have been motivated by hindsight. If, instead, the measurement had been conducted on an indivisible particle / system, then a mixed state makes no sense; rather, a single particle can only be in a pure state, and merely changed from one into another by a measurement. It is particularly confusing because of the terminology being overloaded to mean something completely different. In this case, we tend to denote the pure state of a single particle, but specifically a linear combination of the eigenbasis of an incompatible measuring operator, as a "mixed state".

There is a theorem stating that a system with a Hamiltonian, be it classical or quantum, has an entropy that is never-increasing (luckily, it tends to be constant). The immediate consequence is that the entropy defined by density matrices would then be never-increasing, directly against the 2nd law of thermodynamics requiring the entropy to increase. You can find this under Feynman's set of lectures on statistical mechanics. Two things to note: 1) The entropy defined by density matrices for single particles stays constant under Schroedinger's time evolution and when measured (because it is always pure), in total agreement with Hamiltonian dynamics. 2) The same entropy defined by density matrices, but this time for ensembles, increases, in agreement with thermodynamics but against Hamiltonian dynamics.

There is another theorem, the very important Fluctuation-Dissipation theorem, of thermodynamics, that governs the entropy increasing process. Actually, from the random fluctuations, there is no reason why entropy cannot spontaneously decrease. Indeed, it actually does do that from time to time in regions. The theorem then says that the average entropy always increases.

I will be incorporating the ideas from here soon. Giving thought as to the implications on Hamiltonian dynamics, entropy, how single particle dynamics relates to the ensemble and whether it can be mathematically smoothed over, is returning headaches. 94.194.100.255 (talk) 03:07, 30 August 2012 (UTC)

"There is a theorem stating that a system with a Hamiltonian, be it classical or quantum, has an entropy that is never-increasing". I am awfully confused about the details of this theorem. The universe is a system with a Hamiltonian -- do you agree? The universe has an entropy that increases -- do you agree?

I am also not sure how to understand your statement "a single particle can only be in a pure state". If I have a photon which is one-half of an EPR pair $(|00\rangle +|11\rangle )/{\sqrt {2}}$, would you describe it as a single particle? Would you describe it as being in a pure state? --Steve (talk) 21:24, 30 August 2012 (UTC)

Hi, surely there is some thing missing in the text and is bringing confusion to the reader. Confusion is arising because explanation started with the passing of vertical polarized through a circular polarizer as given below: "If we pass (|R\rangle +|L\rangle )/{\sqrt {2}} polarized light through a circular polarizer which allows either only |R\rangle polarized light, or only |L\rangle polarized light, intensity would be reduced by half in both cases. This may make it seem like half of the photons are in state |R\rangle and the other half in state |L\rangle ."

but, suddenly started asserting that photons are abosrbed by vertical linear polarizer as given below:

"But this is not correct: Both |R\rangle and |L\rangle photons are partly absorbed by a vertical linear polarizer, but the (|R\rangle +|L\rangle )/{\sqrt {2}} light will pass through that polarizer with no absorption whatsoever."

Kindly, address this issue. — Preceding unsigned comment added by Sharma yamijala (talk • contribs) 11:18, 14 February 2014 (UTC)

Recent edit on equivalence of ensembles

Quite briefly, I don't believe it. Is there a citation? The assumption is that ensembles are in 1-1 correspondence with operators of the form $A U$ where $A$ is non-negative trace class of trace $1$ and $U$ unitary.

it's not claimed that A is nonnegative. Mct mht 05:15, 29 June 2006 (UTC)

I still don't believe it,

it's also not claimed that A has trace 1. Mct mht 21:00, 29 June 2006 (UTC)

I still don't believe it. --CSTAR 21:07, 29 June 2006 (UTC)

Why? Ensembles have the structure of a convex set. One might naturally associate to an ensemble to a probability measure supported on the convex set of density operators, but from there to the characterization given in the most recent edit, is in my view unjustified. --CSTAR 04:56, 29 June 2006 (UTC)

This might be true for a very special kind of ensemble, in which the components are linearly independent.--CSTAR 05:04, 29 June 2006 (UTC)

let ρ be a mixed state. a square root factorization of ρ is of the from ρ = Σ v_i v_i ^*. from this we see that there is a one-to-one correspondence between such factors and ensembles describing ρ. combine this with the unitary freedom of square roots and that gives what's being claimed. Mct mht 05:11, 29 June 2006 (UTC)

As I said, I don't see what this has to do with ensembles. Please define what you mean by an ensemble. Could you provide a citation?

I think what you are trying to say is that if we have a convex representation

$\rho =\sum _{i}\lambda _{i}\tau _{i}$

where

$\tau _{i}$

are density states, then this is an ensemble (the "this" being the pair consisting of the mixture coefficients and the finite sequence of mixed states τ_i) and the failure of uniqueness is due to the multiplicity of such convex representations.

--CSTAR 14:31, 29 June 2006 (UTC)

yes, that's the, i believe pretty common in physical context, definition of an ensemble. if p_i is a discrete probability distribution, v_i a family of pure states, then {p_i, v_i}_i is called an ensemble. Mct mht 21:11, 29 June 2006 (UTC)

Well, then more generally we can write

$\rho =\int _{S}\tau d\mu (\tau )$

where μ is a probability measure on the compact convex space of states. We can regard the measure μ as an ensemble representing the state τ. The non-uniqueness follows from the failure of unique representation by probability measures, even probability measures which are supported on the set of extreme points. --CSTAR 21:21, 29 June 2006 (UTC)

yes, agreed. didn't you object to calling those things probability measures. :-) Mct mht 21:38, 29 June 2006 (UTC)

No. If I recall correctly, what I objected to here was calling a completely additive function on the orthocomplemented lattice of projections on a Hilbert space a probability measure. There is an affine bijection between such functions and density operators. A probability mesure on states as mentioned in my remark above (in this talk page) is a probability measure in the classical sense, that is, a set function on the σ-algebra of Borel sets.--CSTAR 21:48, 29 June 2006 (UTC)

In conclusion, then the non-uniqueness follows from the multiplicity of representations of the form

$\rho =\int _{S}\tau d\mu (\tau )$

by probability measures μ supported on the compact convex set of states S. This can be described as non-uniqueness of convex representations. What does this have to do with non-uniquess of factorizations (except in very special cases?) --CSTAR 00:45, 30 June 2006 (UTC)

in the formulation of my comments above, this is precisely the non uniqueness of square root factorization of positive semidefinite operators restated in this context. take for instance the finite dim case. given a mixed state described by the ensemble {v_i}, where the probabilities are absorbed into the states, for notational convenience. so ρ = Σ v_i v_i ^*. another ensemble {w_i} describes the same state iff there exists a unitary U such that [v_i] U = [w_i]. Mct mht 01:39, 30 June 2006 (UTC)

This would imply that all convex representations of a state have the same cardinality.

that's not true. let me clarify. we don't assume [v_i] is full rank. same goes for [w_i]. when an ensemble v₁...v_m is such that m ≤ n, we simply append columns of zeros to get a square matrix [v_i 0]. as a (somewhat common) abuse of notation from linear algebra, above i have put [v_i]. all i am doing is stating a fact from linear algebra and relate to this context. Mct mht 03:13, 30 June 2006 (UTC)

Could you provide a citation for this fact?--CSTAR 03:17, 30 June 2006 (UTC)

PS BTW, in the example I suggested, no matter how much appending you do, that trick isn't going to work.--CSTAR 03:19, 30 June 2006 (UTC)

in all my comments, ensemble elements are pure states (vectors). i thought that was clear. that's not so in the e.g. you gave. Mct mht 03:30, 30 June 2006 (UTC)

But this isn't true. To take a trivial example; any state ρ is representable as a trivial convex combination of 1 times itself ρ= 1 × ρ (cardinbality 1) and if it's a non-extremal state, at least one other non-trivial convex combination of two other states (cardinality two).--CSTAR 02:20, 30 June 2006 (UTC)

BTW, in general the set of states ain't compact, no? Mct mht 03:14, 30 June 2006 (UTC)

appended after CTAR's reply below: i meant norm compact above. Mct mht 03:23, 30 June 2006 (UTC)

In the finite dimensional case, all TVS topologies are the same.--CSTAR 03:24, 30 June 2006 (UTC)

Yes, it's always compact in w* topology (even for C*-algebras) That's a result of fundamnetal importance. This follows from the Banach-ALouglu theorem.--CSTAR 03:17, 30 June 2006 (UTC)

Re:In all my comments, ensemble elements are pure states (vectors). i thought that was clear You thought that was clear? Um, it was not clear to me. Now with that caveat, the result you are claiming is equivalent to the following: Representations of the form

$\tau =\sum _{i}\lambda _{i}E_{i}$

where the $E_{i}$ are selfadjoint projections (that are however not assumed to be pairwise orthogonal) are uniquely determined up to unitaries. That is if

$\tau =\sum _{i}\lambda _{i}F_{i}$

then there is a unitary U such that

$E_{i}=UF_{i}U^{*}$

Frankly, I don't believe this is true. (If the projections are assumed pairwise orthogonal, of coure it's just the spectral theorem). This claim would imply that all probability measures on the compact convex set of states, supported on the extreme points, are "unitarily equivalent" provided they have the same center of mass. I would be surprised f that were true in anything but the abelian case. However, if you provide me with a reference, I'll believe it.--CSTAR 03:49, 30 June 2006 (UTC)

CSTAR, that is not at all equivalent to what was claimed! Mct mht 04:05, 30 June 2006 (UTC)

If a states is an extreme point, it is a rank 1 projection. therefore, if

$\tau =\sum _{i}\lambda _{i}A_{i}$

where the A_i are extreme points, then the A_i must be rank 1 projections.

Now I thought you just claimed that two representations as convex combinations of extreme poinys must be equivalent "unitarily"; What else could you possibly mean other than what I wrote? --CSTAR 04:15, 30 June 2006 (UTC)

Anyway, please provide a citation with whatever it is you are claiming. That will save everybody a lot of time.--CSTAR 04:16, 30 June 2006 (UTC)

CSTAR, i don't know, man, it's probably in most linear algebra books. It's pretty standard. For instance, Choi used this to show how the Kraus operators of a CP map are related by a unitary matrix. Again, the claim is simply the following: Let A = M M^* be a positive semidefinite matrix. Then N satisfies A = N N^* iff N = M U for some unitary U, where all matrices are square (If we drop the square assumption, then U need not be square, but it is isometric). Mct mht 04:53, 30 June 2006 (UTC)

OK that I believe and is easy to prove. What that has to do with representations of ensembles however, is not at all clear.--CSTAR 05:06, 30 June 2006 (UTC)

it is related exactly in the sense i describe before. if A is a state, the column vectors of M gives an ensemble describing the state, same goes for N (the probabilities can be recovered easily). if you take the unique positive square root, you get an ensemble of orthonormal states ... Mct mht 05:19, 30 June 2006 (UTC)

I guess we're back to square 1. My original question above remains as far as I can see, unanswered: What is the relation between "square root factorizations" and ensembles? Do you have a citation for this relation?

Re: the column vectors of M gives an ensemble describing the state, I'm assuming here A = M*M. What's the ensemble (understood as some convex combination of pure states) that gives the state A? Presumably, they're related to the columns or rows of M, but I'm sorry I don't see this.

Re: if you take the unique positive square root, you get an ensemble of orthonormal states ...

In fact, if you take the unique positive square root, you get a positive operator. The columns of this operator are not going to be orthogonal (except in some trivial cases). How do you get an ensemble of orthonormal states. Do you mean the spectral decomposition of A? Could you please be more specific, saying how you do this?

reply: say A has spectral resolution A = U Σ U^* = U Σ ^½ Σ ^½ U^*. Let M = U Σ ^½, then it's trivial to verify what i said. Mct mht 02:05, 1 July 2006 (UTC)

Just give me a citation (with the statement of the result about the relation between ensembles and factorizations) and I'll go away.--CSTAR 13:43, 30 June 2006 (UTC)

This applies to couple other related objects as well. Besides Kraus operators, purifications of a given mixed state are related in a similar way. Mct mht 05:31, 30 June 2006 (UTC)

What's the "this" that applies to other related objects?--CSTAR 13:43, 30 June 2006 (UTC)

CSTAR, i am done with this particular discussion. one can't get much more explicit than the explanations i've given. i am sorry you're not convinced. this is completely elementary and standard. Mct mht 02:05, 1 July 2006 (UTC)

I'm sorry I still don't believe what you said. In fact, to be honest, I don't understand what you said about the relation between ensembles and factorizations. I can't believe that it's standard; if indeed it were standard it would be easy to come up with a source in the form page X of Y which says Z. But, suit yourself. --CSTAR 04:03, 1 July 2006 (UTC)

PS If somewhat has understood mct mht's point about the relation between ensembles and factorizations and I'm just being dense, I'll listen. --CSTAR 04:07, 1 July 2006 (UTC)

Here's an answer

First of all definition of an ensemble: this is an indexed family of pairs

$p_{i},\mid \psi _{i}\rangle$

with

$p_{i}\geq 0,\quad \sum _{i}p_{i}=1$

and

$\mid \psi _{i}\rangle \,$

a unit ket vector. The corresponding density matrix is the convex combination of rank one projections

$\sum _{i}p_{i}\mid \psi _{i}\rangle \langle \psi _{i}\mid$

As Nielsen and Chuang point out, it is actually more convenient to absorb the probability coefficient into the state by replacing thekets with "renormalized" kets:

$\mid \psi _{i}\rangle \,$ by $\mid {\sqrt {p_{i}}}\psi _{i}\rangle$

Thus with this renormalized formulation, the corresponding density matrix is

$\sum _{i}\mid \psi _{i}\rangle \langle \psi _{i}\mid .$

Theorem. Two ensembles ψ, ψ' define the same density state iff there is a unitary matrix

$u_{ij}$

such that

$\mid \psi _{i}\rangle =\sum _{j}u_{ij}\mid \psi '_{j}\rangle$

This is Theorem 2.6 of Nielsen and Chuang. --CSTAR 04:43, 1 July 2006 (UTC)

OK, that seems clear. I'd rather not work with renormalised states, though, and re-express your final equation as:

$\mid \psi _{i}\rangle {\sqrt {p}}_{i}=\sum _{j}u_{ij}\mid \psi '_{j}\rangle {\sqrt {p_{j}^{'}}}$

--Michael C. Price ^talk 05:18, 1 July 2006 (UTC)

Are you going to replace the MM* stuff (which completely lost me) with the above crystal-clear construction?

I have another query; this line in the introduction seems false:

The description of states in Dirac notation by ket vectors is not sufficient to describe the effect of quantum operations, such as measurement, on a quantum mechanical ensemble.

Any mixture of kets can be analysed by treating each ket separately. I suggest deletion. The point (about the merit of density matrices in statistical analysis) seems adequately made by the surrounding text.--Michael C. Price ^talk 06:03, 1 July 2006 (UTC)

I trust your judgement on replacements and deletions (and yes I prefer your non-normalized form of Nielsen and Chuang Theorem 2.6.) I feel I've done enough damage for one day.--CSTAR 06:10, 1 July 2006 (UTC)

Question: Is it implicitly assumed that the states $\mid \psi _{i}\rangle$ in this definition are orthogonal (i.e. states in some orthonormal basis), as in the example with $\mid R\rangle$ and $\mid L\rangle$ polarization? If so, I think this should be stated explicitly. If not, it might be better to change the example to one with non-orthogonal states. 72.229.22.106 (talk) 21:50, 23 October 2019 (UTC)Anonymous User

Beating the dead horse

OK it may be standard, but here is the the explicit relationship:

By definition, a positive rank one operator on H is one of the form

$\mid \phi \rangle \langle \phi \mid$

Recall that an ensemble representing a density matrix is a sequence of positive rank one operators which add up to A.

Fix an orthonormal basis for H. The matrix of such an operator is

$\langle e_{i}\mid \phi \rangle \langle \phi \mid e_{j}\rangle =\phi _{i}{\overline {\phi _{j}}}$

Conversely any operator whose matrix has that form is rank 1. Now suppose

$A=\Psi \Psi ^{*}$

Thus

$A_{ij}=\sum _{k}\psi _{ik}{\overline {\psi _{jk}}}$

Consider the sequence of rank one operators with matrices

$\Phi _{ij}^{k}=\psi _{ik}{\overline {\psi _{jk}}}$

Then clearly

$A_{ij}=\sum _{k}\Phi _{ij}^{k}$

This is an ensemble representing A. Conversely revcersing the argument, one sees that any ensemble representing A has this form. --CSTAR 21:21, 1 July 2006 (UTC)

Spinors from Density Matrices

The page includes the usual method of obtaining the density matrix from spinors but does not show the less well known method of obtaining spinors from density matrices. I'll go ahead and add it in. If you haven't seen the method, see Julian Schwinger's "Quantum Kinematics and Dynamics" or www.DensityMatrix.com .CarlAB 01:16, 6 October 2006 (UTC)

Orders of density matrices

So far this article has ignored the idea of various orders of density matrices. Also, some of the particular expressions are in fact for reduced density operators where the true expressions for the reduced density *matrices* is not given.

WikiProject class rating

This article was automatically assessed because at least one WikiProject had rated the article as start, and the rating on other projects was brought up to start class. BetacommandBot 09:47, 10 November 2007 (UTC)

Non-negative

The article claims a density matrix is non-negative, which the link defines as all the elements are greater than zero. This is clearly not true of a density matrix. Also shouldn't there be a condition on the density matrix that keeps the coherences (off-diagonals) smaller than the populations (on-diagonal elements). Otherwise the matrix isn't physical?--J S Lundeen (talk) 23:20, 18 January 2008 (UTC)

The link was wrong. The way it's intended to be used here, "nonnegative" is synonymous with positive-semidefinite matrix. I corrected the link. --Steve (talk) 01:59, 19 January 2008 (UTC)

Help with references

The article says that the density matrix was first developed by von Neumann, Landau and Bloch in 1927, but Bloch's articles only go as far back as 1945 on ISI web of knowledge, and I can't find anything by von Neumann. Can someone please provide the references for where the density matrix first developed. —Preceding unsigned comment added by Dr. Universe (talk • contribs) 00:04, 26 September 2010 (UTC)

Mixed State

Mixed states redirects here, but I can't find any definition here of what a mixed state is, or an explanation of what a mixed state physically means. Could someone who understands it better than me either add one, or redirect mixed state to somewhere that explains it?--Physics is all gnomes (talk) 21:54, 26 December 2010 (UTC)

I added some text to try to explain it more simply, does it help? --Steve (talk) 06:22, 27 December 2010 (UTC)

Thanks, that was quick! I'm still somewhat confused... can a single photon be in a mixed state? Or does a mixed state just mean that the beam of light contains a range of photons in different pure states?

I'm not sure whether I understood the difference between a superposition and a mixed state - is it true or not that a superposition means that the photon is (in some way) in both quantum states at once until measurement, but in a mixed state, the photon is physically in one of the pure states but we just don't know which one it is? --Physics is all gnomes (talk) 11:13, 27 December 2010 (UTC)

I added more text, hope it helps: A bit about quantum superpositions, and a discussion of how you get mixed states.

I think most physicists think that mixed states are not a truly fundamental part of quantum physics or part of our universe, but rather a convenient tool to keep track of uncertainties that are there in various realistic situations. But I'm not sure that's universally agreed...starts getting into philosophy of quantum mechanics.... I'm not confident enough to write anything about that. :-) --Steve (talk) 16:29, 27 December 2010 (UTC)

Cool, I think I get it now :D One clarification: when you say "Unlike linearly polarized light, [unpolarized light] passes through a polarizer at any angle", do you mean that it all passes through or half of it passes through? Also, maybe mixed states should be mentioned in the lead by name - I realise now that this seems to be what parts of the second and third paragraph are talking about.

I was reading Bloch sphere which cryptically suggests "The interior points are called the mixed states."... but doesn't explain further. Do you know how a mixed state would be represented on a Bloch sphere? I'm guessing that your unpolarized light example would be right in the centre of the sphere... But how would you represent light that had, say 80% probability of being |R>, and 20% probability of being |L>. Would the Bloch sphere (or perhaps Poincare sphere) then tell you what would happen if put that light through a linear polarizer?--Physics is all gnomes (talk) 20:50, 27 December 2010 (UTC)

Those are good suggestions, I tried again. Yes, unpolarized light is the center of the Bloch sphere. Light that had 80% probability of being |R>, and 20% probability of being |L> would be on the line between R and L, four-fifths of the way towards R, if memory serves. Suppose you put light through a linear polarizer that allowed (1,0,0) on the sphere to pass 100%, and (-1,0,0) on the sphere is blocked 100%. I bet the intensity fraction passing through for light at (x,y,z) on the sphere (or inside the sphere) is x/2+0.5. But that's just a guess. If you can find the book "Density Matrix Theory and Applications" it has a good simple discussion of the Bloch sphere interior as I recall. :-) --Steve (talk) 22:50, 27 December 2010 (UTC)

Thanks for explaining it to me! This seems quite a good way a writing an article - a knowledgable person repeatedly rewriting it until the beginner finally understands :) There is one sentence that's still a bit unclear "By contrast, an example of a mixed state would be $(|\psi _{1}\rangle +e^{i\theta }|\psi _{2}\rangle )/{\sqrt {2}}$, where $\theta$ is a randomly-varying real number." What does it randomly vary with? - between photons? I thought at first it meant randomly-varying with time but that doesn't seem to fit.--Physics is all gnomes (talk) 14:29, 28 December 2010 (UTC)

"By contrast, an example of a mixed state would be $(|\psi _{1}\rangle +e^{i\theta }|\psi _{2}\rangle )/{\sqrt {2}}$, where $\theta$ is a randomly-varying real number."

I'm not sure about this, but shouldn't that read "an example of an ensemble of mixed states" or something similar? Illustrates the basic idea that mixed states come from lack of knowledge (statistical probabilities). $(|\psi _{1}\rangle +e^{i\theta }|\psi _{2}\rangle )/{\sqrt {2}}$ by itself is a pure state, I think? 2.25.206.58 (talk) 20:22, 8 April 2011 (UTC)

How about "where $\theta$ is a randomly-varying real number, changing from photon to photon"? I don't like "an ensemble of mixed states", because an ensemble of mixed states is always another mixed state :-) --Steve (talk) 20:29, 9 April 2011 (UTC)

Yes, that sounds like a good idea :) 137.73.4.92 (talk) 14:04, 7 May 2011 (UTC)

Btw.: a state is pure if and only if $tr(\rho ^{2})=1$! so how does this artificial distinction of a mixed state and a superposition make sense? If I calculate $tr(\rho ^{2})$ for the suggested "pure" state I arrive at $tr(\rho ^{2})=1/2$! to put it in another way: a pure state is a normalized, positive, weakly continuous functional on the kinematical algebra which is represented by the scalar product of a unique vector in Hilbert Space so how can we represent $\left|\psi _{1}\right\rangle +\left|\psi _{2}\right\rangle$ by a unique vector? —Preceding unsigned comment added by 130.75.25.219 (talk) 14:00, 2 May 2011 (UTC)

When you say "the suggested "pure" state", are you referring to the state $(|\psi _{1}\rangle +|\psi _{2}\rangle )/{\sqrt {2}}$? If so, how did you calculate $tr(\rho ^{2})=1/2$? I calculated $tr(\rho ^{2})=1$ as expected. I think you may have done the calculation wrong...could you please give more details? --Steve (talk) 06:17, 3 May 2011 (UTC)

Accessible intro

The Manual of Style says that an article should have an accessible introduction. This lead, particularly the first sentence, failed this requirement to an almost comical degree. I have rearranged the text and streamlined it to improve readability without changing any of the content. However, someone should take this even further and replace some of the technical terminology in the last paragraph by a more accessible discussion. RockMagnetist (talk) 16:31, 5 April 2012 (UTC)

Löwdin paper

I would suggest to include at least a reference to the Löwdin paper that explains density matrices and natural orbitals. It has helped me a lot in my PhD work. Here it is: http://prola.aps.org/abstract/PR/v97/i6/p1474_1 Jocasa (talk) 11:42, 5 June 2012 (UTC)

Unfortunately it's not useful to most readers because it's behind a paywall. Longitude2 (talk) 10:53, 17 February 2021 (UTC)

A density matrix represents a mixed state (first line)

I don't see why this is necessary as density matrices apply just as well to pure states: If I have the superposition, $|\psi >={\frac {1}{\sqrt {2}}}(|1>+|2>)$, I can write

$\rho =|\psi ><\psi |=1/2(|1><1|+|1><2|+|2><1|+|2><2|)$

which in the $|1>,|2>$ basis gives me the matrix: $\rho ={\begin{bmatrix}1/2&1/2\\1/2&1/2\end{bmatrix}}.$ which is fully valid and something we do in quantum optics all the time.

Thanks (sorry, couldnt be bothered to rewrite the page during exam period, might try during the summer) — Preceding unsigned comment added by 142.150.226.247 (talk) 17:12, 26 March 2013 (UTC)

CONTRADICTION
The very first sentence of this topic contradicts the section on 'Pure and mixed states'. One says the density matirx describes mixed states; the other says it describes pure states or mixed. — Preceding unsigned comment added by Stephiefaulkner (talk • contribs) 08:50, 10 January 2017 (UTC)

It may be better to alter the first sentence to say that the density matrix describes a state prior to measurement, which in practice is always a mixed state, since measurement can never be guaranteed to exactly align with the singularly pure aspect (orientation) of the state prepared.

A formula relating Shannon and Von Neumann entropies, and why it is incorrect.

The formula which I have deleted from the article is the following: $S(\rho )=H(p_{i})+\sum _{i}p_{i}S(\rho _{i})$

Where $\rho =\sum _{i}p_{i}\rho _{i}$

This formula is clearly incorrect. Consider the case of a pure state: $\rho =|\psi \rangle \langle \psi |$. The Von Neumann entropy is zero for this case (you can regard this density matrix to have an eigenvalue of 1, with the rest zero). On the other hand looking at the right hand side we can obtain a non zero result by expanding (for example): $\rho ={\frac {1}{2}}|\psi \rangle \langle \psi |+{\frac {1}{2}}|\psi \rangle \langle \psi |$ Such that the Shannon entopy on the r.h.s is $log_{2}(2)=1$, and the Von Neumann ones on the r.h.s are still zero. --Nomadbl (talk) 15:19, 24 July 2018 (UTC)

Thanks! I endorse. --Steve (talk) 00:20, 25 July 2018 (UTC)

The formula is correct. See, for example, Theorem 11.8 of Nielsen and Chuang's Quantum Computation and Quantum Information. The counter-example given above is not valid, because the right-hand side is not actually an expansion into multiple different pure states. The qualification that the $\rho _{i}$ must have support on orthogonal subspaces is quite important. XOR'easter (talk) 01:29, 3 September 2019 (UTC)

Removed incorrect criterion for pure states

There are all kinds of criteria under which a density matrix $\rho$ represents a pure state. The one currently used ($\rho ^{2}=\rho$), however, I believe is wrong, which is the reason I removed it. My argument is as follows: Let ${\vec {v}},{\vec {w}}$ be two mutually orthogonal, normed vectors. The operator $P({\vec {x}}):=\langle {\vec {v}}\mid {\vec {x}}\rangle {\vec {v}}+\langle {\vec {w}}\mid {\vec {x}}\rangle {\vec {w}}$ is an orthogonal projection on the plane spanned by ${\vec {v}},{\vec {w}}$. It cannot be written as a pure state (in the form $\mid \phi \rangle \langle \phi \mid$) but it is idempotent. Thus, $P$ is a counter-example. — Preceding unsigned comment added by 217.95.160.47 (talk • contribs)

$P$ is not of trace 1, so this is no counter-example. The problem with the original text comes from the word "equivalently", which attaches to the earlier sentence about rank one. — Preceding unsigned comment added by 217.95.163.80 (talk) 12:20, 9 September 2019 (UTC)

I will add: diagonal matrix

I'd like to add something like the following to the lead:

The density matrix for an experimental system becomes a diagonal matrix when a measurement happens, and it is even possible to see the matrix gradually becoming diagonal as the system naturally decoheres due to connection with the environment without a measurement happening.

I will give this a week for feedback before I edit the article, since I am not an expert in the field, and this might include a mistake or two. David Spector (talk) 14:38, 29 November 2019 (UTC)

David spector, something like that could be a good addition, though I'd be wary of the "see the matrix gradually becoming diagonal" phrasing — "see" and "observe" are, after all, loaded words in quantum mechanics. XOR'easter (talk) 17:13, 2 December 2019 (UTC)

True. How about, "we know the matrix is gradually becoming diagonal"? We do know that there is a transition period between the existence of a quantum state and its replacement by a classical state (the diagonal matrix). Collapse doesn't happen in a femtosecond, right? David Spector (talk) 19:07, 2 December 2019 (UTC)

Well, a diagonal density matrix isn't really a "classical state"; it's a quantum state that is diagonal in the given basis (and would generally have coherences with respect to a rotated basis). I'll go back and read some standard references like Schlosshauer; maybe that'll suggest some phrasing that's really solid. XOR'easter (talk) 15:46, 3 December 2019 (UTC)

Since a week has elapsed, I will edit the article as I indicated. But if an expert visits here, please feel free to improve what I've added. David Spector (talk) 12:02, 7 December 2019 (UTC)

Where is the matrix???

An article on a matrix, but there is no matrix to be found!!! Poor again. — Preceding unsigned comment added by Koitus~nlwiki (talk • contribs)

The "Definition" section shows how to write a density matrix in Dirac notation, and the figure caption gives two examples written as square arrays of numbers. XOR'easter (talk) 16:32, 2 November 2020 (UTC)