Talk:Density matrix

message: 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) 
I completely agree: the time evolution of a density matrix is, in general, more general that the von Neumann equation. And this time evolution simply “generates“ density matrices/operators (or mixed states) from pure states. Just adding “Lindblad equation“ to the further references list, is not sufficient, I think. Although I know, of course, that there are lots of discussions which is the “correct” (or “most general form”) of such an equation of motion. DieHenkels (talk) 10:30, 8 December 2023 (UTC)
thread_title:  on the dynamical equation 

message: 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)
replies: text: 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)

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

text: 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)
thread_title:  question on C* algebra formulation 

message: 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)
thread_title:  Comment on recent edit 

message: the articles pure state and mixed state can easily be merged with this one. Mct mht 05:20, 22 May 2006 (UTC)
replies: text: Yep.--CSTAR 17:03, 22 May 2006 (UTC)

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

message: 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)
replies: text: 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)
thread_title:  Entropy 

message: 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^k_{i j} = \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)
thread_title:  Beating the dead horse 

message: 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)
thread_title:  Spinors from Density Matrices 

message: 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)
thread_title:  WikiProject class rating

message: 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)
replies: text: 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)
thread_title: Non-negative

message: 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)
thread_title:  Help with references 

message: 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)
thread_title:  Accessible intro 

message: 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)
thread_title:  Löwdin paper 

message: 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)
thread_title:  A density matrix represents a mixed state (first line) 

message: 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 \rang \lang \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 \rang \lang \psi|+\frac{1}{2}|\psi \rang \lang \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)
replies: text: Thanks! I endorse. --Steve (talk) 00:20, 25 July 2018 (UTC)
replies: text: 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)
thread_title:  A formula relating Shannon and Von Neumann entropies, and why it is incorrect. 

message: 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)
thread_title:  Removed incorrect criterion for pure states 

message: 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)
replies: text: , 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)

text: 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)
thread_title:  I will add: diagonal matrix 

message: 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)
thread_title:  Where is the matrix??? 

message: For explaining 'projection', which is essential to the motivation, this article links only to projection operator (linear algebra). So does Measurement in quantum mechanics, and Quantum mechanics doesn't yet link anywhere. I believe readers need more explanation since it's so intimately related to measurement and so often used in QM. And to define, for example projective measurement.
Q: Is it advisable to make a new article Projection (quantum mechanics) or Projective measurement (quantum mechanics)? Seems like a linkable section of one of these QM articles would work well. Thoughts? --MadeOfAtoms (talk) 23:01, 19 May 2021 (UTC)
replies: text: I don't think such an article would be advisable. Projective measurements should be explained in context, in the article Measurement in quantum mechanics. Well, it is already there, in the section Measurement in quantum mechanics#Projective measurement, it's just not well-explained. By the way, thanks for finding the mistake in the lead. Tercer (talk) 08:59, 20 May 2021 (UTC)
replies: text: Thanks, sounds like good advice. Kudos on your clear adjustment to the lead. Please smooth out any other clunky attempts I make to explain more for non-expert readers. MadeOfAtoms (talk) 07:49, 21 May 2021 (UTC)
thread_title:  Link or explanation for 'projector' in quantum mechanics  

message: It's lacking an explanation here. The two references date back to 1963 and 1972, is this "now"? or the "now" more recent? would it be possible to give an example?  — Preceding unsigned comment added by 89.3.212.183 (talk) 01:23, 9 December 2022 (UTC)
thread_title:  "It is now generally accepted that the description of quantum mechanics in which all self-adjoint operators represent observables is untenable.[24][25] "