Talk:Quantum state/Archive 2

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Archive 1

Archive 2

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Latest comment: 2 years ago1 comment1 person in discussion

This article was the subject of a Wiki Education Foundation-supported course assignment, between 20 August 2020 and 23 November 2020. Further details are available on the course page. Student editor(s): MAllison5.

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States are linear functionals, not vectors (or rays)

Latest comment: 7 years ago29 comments5 people in discussion

I am a bit surprised that the current article is going into pedantry on how states are rays not vectors, but then the formal technical definition is left to a one sentence section at the end. States simply are linear functionals on the algebra of observables. This actually translates nicely to something understandable and intuitive in normal English: "A state is a function that assigns an expectation value to each observable." I'd such that the article actually open with that somewhere in the first paragraph, before differentiating between pure and mixed states.T R 07:06, 5 April 2016 (UTC)

The definition (first sentence, not end)

A physical pure quantum state corresponds in the mathematical formulation of quantum mechanics to a ray in a Hilbert space over the complex numbers.

is more or less verbatim from Weinberg. I have heard of the definition you mention, but have not seen it clearly stated anywhere in the common literature. We could of course get rid of the terrible waste (pedantry) of devoting a complete sentence to distinguish rays from vectors. I don't mind either to replace the definition all together with yours, that I suspect has higher mathematical precision, something I usually strive for. Please go ahead. My edit was meant to ward off Chjoaygame's definition that quantum states are abstract objects (that refer to repeated practical replication of single quantum phenomena). YohanN7 (talk) 09:22, 5 April 2016 (UTC)

If we are digging into rays, let us say that a nonzero real multiple of a state is the same state as well. How about something like "state is a unit ray"? Petr Matas 19:38, 6 April 2016 (UTC)

Alternatively, we could say: "Pure state is represented by a non-zero vector. Non-zero complex multiples of the vector represent the same state." Petr Matas 22:34, 6 April 2016 (UTC)

The article is clear on this. Third sentence. Deemed as pedantry though. But I, as you, think that blue links can be accompanied by an explanatory sentence. It makes the lead more accessible. YohanN7 (talk) 09:48, 7 April 2016 (UTC)

Now, is there anything that prevents the current definition to be characterized as

"A state is a function that assigns an expectation value to each observable."?

Does not

{\hat {O}}\mapsto \langle {\hat {O}}\rangle _{\mathcal {R}}=\langle \Psi |{\hat {O}}|\Psi \rangle ,\quad \Psi \in {\mathcal {R}},

where

R

is a unit ray, express this? YohanN7 (talk) 09:48, 7 April 2016 (UTC)

After a helpful tutorial from a kind editor: A pure state is a unit eigenvector of a maximal observable. Dirac 1930, pp. 13–14: "We shall frequently have to consider the greatest possible number of independent compatible simultaneous observations being made on a system and shall, for brevity, call such a set of observations a maximum observation." I see that Cohen-Tannoudji et al. don't get their raw mixed states from ovens. They get them from "furnaces", and then sort them with "analyzers".<volume 1, p. 399.> On page 403: "We could recombine them to form one wave packet again by removing the screen (that is, by not performing the measurement) and by submitting them to a new field gradient, whose sign would be the opposite of the first one."

Also Newton (2002), p. 4:

After the measurement with the outcome

A

, a system originally in the state

Ψ

is left in the eigenstate P_$A$

Ψ

of

A

with the eigenvalue

A

, where P_$A$ is the projection on the eigenspace of

A

at

A

, i.e., the space spanned by the eigenvectors

Ψ A

(this is called the projection postulate).<Newton, R.G. (2002). Quantum Physics: a Text for Graduate students, Springer, New York, ISBN 0-387-95473-2.>

Chjoaygame (talk) 18:40, 11 April 2016 (UTC)Chjoaygame (talk) 19:32, 11 April 2016 (UTC)Chjoaygame (talk) 21:59, 11 April 2016 (UTC)

In the opinion of U. Fano (1957), 'Description of states in quantum mechanics by density matrix and operator techniques', Rev. Mod. Phys., 29(1): 74–93, pp. 75–76:

A pure state is characterized by the existence of an experiment that gives a result predictable with certainty when performed on a system in that state and in that state only. ...

An experiment that yields a unique predetermined result for a system in a given pure state can be designed to act as a filter which leaves the system undisturbed, like a Nicol prism traversed by light of the pertinent linear polarization. ...

An experiment that characterizes uniquely a pure state, as indicated above, and thus provides a maximum of information about it is called a "complete" experiment. ...

A pure state can then be identified by specifying the complete experiment that characterizes it. Mathematically, one can construct a variety of Hermitian operators which have the given pure state as an eigenstate. ... Given such an operator, it proves possible to design, at least in principle, an experiment that constitutes a measurement of the corresponding observable. ...

I think Fano's paper is a standardly accepted rendition of orthodoxy. Therefore it is a fair candidate as a reliable source for Wikipedia.Chjoaygame (talk) 07:42, 12 April 2016 (UTC)

I think this non-mathematical characterization of a pure state fits well into the lead section. This will allow us to move the state vectors and Hilbert space into the section on pure states, away from the lead. Petr Matas 08:58, 13 April 2016 (UTC)

Random quotes from old references are not a good substitute for a precise definition that accords with how quantum states are perceived and used today. Weinberg, on the other hand, is a modern reference. Neither of the above "characterizations" of pure states are actual characterizations. Besides, they don't stand up to inspection, especially not when taken out of context.

Quantum mechanics gives predictions about physics that have never been proven wrong. But these predictions are based on precise definitions in the modern language of physics. Chjoaygame is of a very different opinion from mainstream physicists.

What a pure state is in an (idealized, non-existent) experimental setting does however belong in the article somewhere, but not as a primary definition.

The lead has in my opinion deteriorated from bad to worse with the last series of small edits, and now offers at least three mutually incompatible "characterizations" of what a pure state is. YohanN7 (talk) 11:03, 14 April 2016 (UTC)

I think that the primary definition is necessarily the mathematical one, which may not be comprehensible to the general public. As the lead section should try to target the widest audience possible, I propose the lead to use as little maths as possible, and move the precise definition (the current third to fifth paragraph, after some fixes) to the first section. Can we fix the different "characterizations" somehow, so that they become correct and compatible? Petr Matas 12:42, 14 April 2016 (UTC)

We can surely get both math and experiments out of the lead. It need not be long for this article. YohanN7 (talk) 09:53, 16 April 2016 (UTC)

Pondering ~~your~~ TR's formulation "A state is a function that assigns an expectation value to each observable", I am missing the statement that the expectation value is a statistical mean for some observables and the actual value of the observable with no uncertainty for others. But I think it is compatible with the current second paragraph. Petr Matas 12:42, 14 April 2016 (UTC)

It is not my formulation. The current second paragraph is problematic, so consistency with this is to me undefined. Now, my problem with the by TR proposed definition (the last section in the article is devoted to this by he way) is just a technical one, and about finding a good source. My definition yields linear functionals on the operator algebra per above. Given a linear functional, is there then a ray such that the linear functional is represented by the expression above? If this is so, the definitions are equivalent. (I probably would like to see the qualifier continuous to be perfectly happy. General linear functionals on infinite-dimensional spaces can be quite bizarre. This is why a good source is required, or at least an editor actually knowing this like presumably TR.) YohanN7 (talk) 09:53, 16 April 2016 (UTC)

From memory (which may be wrong) the classical reference for the definition of state as (continuous) linear functionals on the operator algebra's is Von Neumann's Mathematical Foundations of Quantum Mechanics and I am pretty certain that at the very least most modern sources on algebraic quantum field theory would cover it. You typically will not find it in most textbooks because the technical aspects of the definition require more advanced mathematics (functional analysis) than most books are willing to delve into. I'll try to see if I can find some modern sources.

To answer your question: no, not all state can be represented thus by a vector. However, you can always find a density matrix representation. (I think that is due to some version of the Riesz representation theorem).T R 16:02, 16 April 2016 (UTC)

Here is a book (be well-known physicist Ludvig Faddeev) that introduces states in quantum mechanics in the way I meant: [1]

T R 22:53, 16 April 2016 (UTC)

Thanks for the reply. It is logically a most satisfactory state of affairs. States defined this way include both pure and mixed states (I didn't get that part until now). A state is pure if and only if there is a ray in Hilbert space such that..., etc. The references should be impeccable too. YohanN7 (talk) 10:38, 18 April 2016 (UTC)

Indeed, "general linear functionals on infinite-dimensional spaces can be quite bizarre". Specifically: no, not every positive, norm continuous linear functional (on the algebra of observables) corresponds to some density matrix (the latter case including matrices or rank 1, of course). Those that correspond are exactly those that are ultraweakly continuous, which is not a deep result but, basically, the definition of the ultraweak topology. Riesz representation theorem is for Hilbert spaces only; the space of operators is Banach, not Hilbert and, worse, not reflexive. Boris Tsirelson (talk) 14:59, 19 April 2016 (UTC)

As far as I remember, the question, whether all those functionals are "quantum states", or only ultraweakly continuous are, is a bit controversial. Just like the question, whether the "physical" Hilbert space must be separable, or not. If you start with a Hilbert space, you have density matrices and the ultraweak topology on the observables. Alternatively, if you start with an algebra of observables, then all functionals "are created equal"; choosing one, you get (via GNS) a Hilbert space that makes this one (and many others, but not all) ultraweakly continuous. Boris Tsirelson (talk) 15:07, 19 April 2016 (UTC)

Thanks for doting the i's on some of those results. I'll immediately admit that I am a bit hazy on most of the technical details. I don't think we would want to enter that level of detail in the lead (and probably not anywhere in this article.) The many reason I brought this characterization of states up is that it lends itself nicely to a less precise (but accurate) description in common english, which would be good for the first paragraph. A second useful property is that it combines mixed and pure states in a unified way. This means that this distinction can be pushed further down in the lead.T R 13:02, 22 April 2016 (UTC)

I agree, but how about replacing the expectation value with probability distribution: "A state assigns a probability distribution to each observable." I would also omit the words "is a function that" to avoid suggesting that this is the definition. There should be no problems with sourcing such a simple statement. Petr Matas 08:56, 25 April 2016 (UTC)

That would veer uncomfortably far away from the "linear functional on observables" definition, IMHO. (With some appropriate fine print that I do not recall) It is true that assigning an expectation value to all observables is equivalent to assigning probability distributions, but that seems a biggish step. I think it is probably better to clarify this in a follow on sentence to reduce information density.T R 09:14, 25 April 2016 (UTC)

An expectation value alone does not generally yield a probability distribution. Maybe it does so in the case at hand, and perhaps only for pure states, Don't know, I haven't dug into it. Why not wait until references are found? YohanN7 (talk) 09:31, 25 April 2016 (UTC)

To be clear: An assigned expectation value for one observable does emphatically not fix its probability distribution. "An assigned expectation value for all observables fixes the probability distribution for all observables" is a non-trivial statement that needs backing up. YohanN7 (talk) 09:38, 25 April 2016 (UTC)

I think that it is obvious that a state (pure or mixed) fixes the distributions regardless of the implication "all expectation values => all distributions" being true or false. We just need to source it. Petr Matas 10:39, 25 April 2016 (UTC)

Actually, most of this (including the equivalence of assigning expectation values/ distributions for all observables) is covered in Fadeev's book I cited above. Anybody feel like drafting a first paragraph? T R 11:00, 25 April 2016 (UTC)

No problem with the implication "all expectation values => all distributions". Having an observable and a measurable set of reals, we may introduce a new observable as the indicator function of the given set applied to the given observable (the so-called spectral projection). Its expectation is just the needed probability... Boris Tsirelson (talk) 20:00, 25 April 2016 (UTC)

Aha! In other words, the probability that the value of an observable a lies in the set X is equal to the expectation value of the observable [a ∈ X] (written using Iverson bracket and probably with a bit of confusion of an observable and its value). The probabilities for different sets X specify the probability distribution. This is quite straightforward. Now the TR's objection makes sense to me, although I would still prefer that the lead mentions just the distributions. I think that we do not need to separate the definition from its consequences here – we can make a simple and strong statement right away and put the details into another section. Now it is time for others to decide. Petr Matas 06:07, 26 April 2016 (UTC)

I found a very good reference,

Varadarajan, V. S. (2004). Supersymmetry for Mathematicians: An Introduction. AMS. ISBN 978-0-8218-3574-6. {{cite book}}: Invalid |ref=harv (help),

if anyone feels ready to go to work. YohanN7 (talk) 12:50, 24 January 2017 (UTC)

Conflicting definitions of a ray

Latest comment: 8 years ago7 comments3 people in discussion

There are two conflicting definitions of a ray: The article Projective Hilbert space says that vectors belonging to the same ray may differ in amplitude, whereas Wigner's theorem#Rays and ray space says that they may not. Which one is correct? Petr Matas 03:08, 30 April 2016 (UTC)

Really? Looking at "Wigner's theorem#Rays and ray space" I see "ray" and "unit ray" respectively. Boris Tsirelson (talk) 05:15, 30 April 2016 (UTC)

Yes. Per Wigner's..., all elements (vectors) of a ray are of the same magnitude, which may differ from one; unit ray is a special case, with elements of the unit (i.e. again of the same) magnitude. Petr Matas 01:58, 1 May 2016 (UTC)

Disentangling the definitions, one has that the unit rays of Wigner's theorem#Rays and ray space (second definition) are precisely the rays of Projective Hilbert space. (Unit rays of Wigner's theorem#Rays and ray space (first definition) are in an obvious one-to-one correspondence with those of the second definition, hence of those of Projective Hilbert space.) YohanN7 (talk) 10:57, 4 May 2016 (UTC)

IMHO, you cannot say that the unit rays (B) from the second definition are precisely the rays (A) from the first, although the difference is only formal. The spaces of A and B are isomorphic, but not identical. This is important when you treat an element of the projective space as a set of vectors of the Hilbert space (equivalence class is a set). Furthemore, using B you have to use the modifier "unit", but using A you don't. Being an encyclopedia, we should keep our terminology consistent. I think that the word "ray" in all quantum-physics-related articles should link to the article Projective Hilbert space, which should mention both definitions and explain the difference. Petr Matas 06:37, 5 May 2016 (UTC)

Treated as a mathematical structure, the projective Hilbert space has many equivalent definitions and isomorphic implementations. If a context requires to distinguish between them, then it is not quite about the projective Hilbert space, but about something more specific: specific subsets of a Hilbert space, etc. Boris Tsirelson (talk) 08:53, 5 May 2016 (UTC)

By (A) and (B) you probably refer to different things than I did in my post. I comparen

S

from

\Psi \approx \Phi \Leftrightarrow \Psi =z\Phi ,\quad z\in \mathbb {C} \smallsetminus \{0\},

S={\mathcal {H}}/\approx ,

(this is the set of unit rays (second definition) in Wigner's theorem) to

S' =

"the set of equivalence classes of vectors

v

in

H

, with

v\neq 0

, for the relation

\sim

given by

v\sim w

when

v=\lambda w

for some non-zero complex number

\lambda

" (from Projective Hilbert space).

These things are equal – as sets, i.e.

S = S'

. (Well, almost. It seems like the (set containing the) null-vector becomes an element of

S

. This should probably be fixed.) In my terminology, the elements of

S

are exactly the equivalence classes, i.e. subsets of

H

.

I agree that the term "unit ray" is unfortunate. It comes from the first definition of

S

(in Wigner's theorem). YohanN7 (talk) 09:21, 6 May 2016 (UTC)

Schrödinger equation

Latest comment: 8 years ago14 comments3 people in discussion

The current formulation about the role of the Schrödinger equation is not very general. States are defined too generally, just any ray in any complex Hilbert space. The former formulation, while not very well put, is true in QFT as well as QM 101. States are built from solutions of the relevant Schrödinger equation. Then why take away Weinberg as a reference? YohanN7 (talk) 09:44, 26 May 2016 (UTC)

"Schrödinger equation describes how the state changes with time" (the summary of this diff) — true in the Schrödinger picture (often used in QM), but false in the Heisenberg picture (often used in QFT). Boris Tsirelson (talk) 11:09, 26 May 2016 (UTC)

Yes. The real question may be how general the article should be. QM 101 or more general? I don't know, which is why I posted this thread. I mean, if we plan to include states as linear functionals on operator algebras, then we certainly go beyond QM 101 (and QM 201 as well). I m o, we should. The former formulation was "not wrong" and vague enough to allow for later specification in the body of the article. The present formulation is also "not wrong" and is more or less a postulate of QM 101, but it commits to a limited scope. YohanN7 (talk) 11:51, 26 May 2016 (UTC)

The article, as of now, contains Subsection 1.2 "Schrödinger picture vs. Heisenberg picture". Boris Tsirelson (talk) 14:04, 26 May 2016 (UTC)

The former formulation already spoke about the time evolution of a state, which IMO implied the Schrödinger picture. I suggest introducing the two pictures early on, because the time evolution is an important aspect of the state. Is it true that "Heisenberg's state" equals to "Schrödinger's state at t=0"? If yes, then why do we need the Schrödinger equation in the definition of the former, but not of the latter? Isn't the solution of the Schrödinger equation a function of time? The Schrödinger's state depends on time, but it is not a function of time, just a snapshot of the system from a particular time. The Heisenberg's state is completely independent of time, so it is not its function either. Petr Matas 17:59, 26 May 2016 (UTC)

Solutions of the Schrödinger equation are in a one-to-one unitary correspondence with "Schrödinger's states at t=0"; in this sense, we have here two implementations of the same Hilbert space; which one to use, is a matter of taste, convenience etc. The former are more natural for Heisenberg picture, the latter for Schrödinger picture.

"The Schrödinger's state depends on time, but it is not a function of time" — sorry, I do not understand. For me, generally, "depends on time" and "is a function of time" means the same. And for you? Boris Tsirelson (talk) 18:10, 26 May 2016 (UTC)

In practice yes, but having y = f(t), I would say that f ∈ (functions from ℝ to ℝ) is a function, but y ∈ ℝ is not – it is a variable, whose dependency on t ∈ ℝ is given by the function f. I wanted to point out the difference between the snapshot from a chosen time (an example from classical physics: the state of a particle is given by its position and velocity vectors) and the complete history (like a worldline). Petr Matas 21:27, 26 May 2016 (UTC)

Ah, yes, I see. But only after your explanation... Boris Tsirelson (talk) 21:35, 26 May 2016 (UTC)

Weinberg went away together with the removal of the statement, that a state corresponds to a ray, sourced with it. The current text still implies this correspondence, but only indirectly, to avoid the conflict with the statement that a state corresponds to a vector. Please reintroduce the reference if you find it appropriate. Petr Matas 17:59, 26 May 2016 (UTC)

There is no conflict since the correspondence between physical states and rays is one-to-one and the correspondence between physical states and vectors is one-to-many. I.e. different correspondences, of which the first is conceptually the important one (and correct picture to have) one, and the latter is (sloppily, but it would be too tedious for anyone to carry along disclaimers about multiplicative factors) used most in practice. The Weinberg reference is good (to have somewhere at least) because it is by many considered as the best book on QFT around. It has also been said that he wrote the QFT series mostly for his physicist colleagues to explain to them what QFT is.

The former formulation,

Knowledge of the quantum state and the (deterministic) rules for its evolution in time exhausts all that can be predicted about the behavior of the system

did not really commit to some time dependent Schrödinger equation (e.g. not to the non-relativistic time dependent Schrödinger equation of particle in external potential). Which picture (Schrödinger, Heisenberg, interaction picture) is here a minor issue.

The question remains. Do we want to be general enough to include QFT states? In this case, "the relevant Schrödinger equation" refers to the free (one-particle) equation, which may be e.g the free Dirac equation. Usually plane-wave solutions are used, but any complete set will do, to build Fock space. ~~There is no (afaik) time-dependent Schrödinger-type equation relating states from one time to another.~~ Edit: If there is a Lagrangian density available, then there is an Euler-Lagrange equation of motion, but this is pre-quantization. The best one can do is to compute the S-matrix. YohanN7 (talk) 08:59, 27 May 2016 (UTC)

I see my above paragraph reads quite wrong. The strikethrough text should be something like "There is no (afaik) time-dependent closed form standard Schrödinger-looking equation relating states from one finite time to another." There is an equation of course, but it is pretty formal (and can be expanded in terms of an infinite sum of of creation and annihilation operators), with an easy, but very formal solution, the time-evolution operator. Sorry to bring this up all together. It is out of context. YohanN7 (talk) 09:23, 28 May 2016 (UTC)

I have made a new attempt with a ray while trying to maximize accessibility and returned Weinberg. Concerning the picture, isn't the state independent of time in Heisenberg picture? That is why I think that speaking about time evolution of the state implies the Schrödinger picture. But I agree that we should cover (at least) both pictures. Petr Matas 21:14, 29 May 2016 (UTC)

It reads fine to me.

The time evolution (QFT) is most easily obtained in the interaction picture. In it, the equation is again formally a Schrödinger equation. But all this is maybe not necessary to have in the lead. I made a tweak of the first paragraph that avoids the issue all together. What you think about that formulation? The downside is that the Schrödinger equation is gone altogether from the lead. It could be worked back in into later paragraphs.

I believe that once we are happy with the content of the lead, it should be cut up into pieces with most of it moved into the body of the article. YohanN7 (talk) 09:17, 30 May 2016 (UTC)

I tried to improve the language, but I think that your formulation is usable. Petr Matas 14:28, 30 May 2016 (UTC)

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Spin

Latest comment: 8 years ago3 comments2 people in discussion

"which" tag

There is a {{which}} template in the Spin section, but I think that it can be safely removed. "A quantum system" means "any quantum system", therefore in my eyes it does not make sense to ask which system. Petr Matas 09:09, 8 September 2016 (UTC)

Pinging 70.247.163.128. Petr Matas 09:13, 8 September 2016 (UTC)

Error in article?

Latest comment: 1 year ago8 comments4 people in discussion

I think there is a severe problem with the part of this article between the phrases "A more complicated case is given" and "the particles are arbitrarily far apart."

The problem is with the premise "superposition of joint spin states for two particles with spin 1/2".

In particular, the problem is that a requirement that the two particles be entangled, that is, that they share a single quantum state, is omitted.

The problem can be seen in the case of two particles with spin 1/2 that are not entangled. For any two such particles, this section is incorrect. For example, both particles can be measured with their spins in the down direction. David Spector (talk) 13:18, 14 May 2020 (UTC)

The article is perfectly correct. The state in question is the singlet, and it is defined in an equation. The properties that are mentioned are properties of the singlet, not of any quantum state. It wouldn't hurt to mention that this state is entangled, though. Tercer (talk) 15:35, 14 May 2020 (UTC)

I went to edit this myself, but then realised that the entire lead is a catastrophe and needs to be rewritten anyway. Tercer (talk) 15:39, 14 May 2020 (UTC)

Yes, that bit of the lead is technically correct, but the lead overall is a disorganized, overgrown mess. It's hard to tell how even to start fixing it; quite possibly, the best thing to do would be to ignore it and start an entirely new write-up in a sandbox. XOR'easter (talk) 18:18, 17 May 2020 (UTC)

OK, I tried my hand at shortening the introduction. I think it's better than it was before (which is of course not the same thing as being good outright). XOR'easter (talk) 19:38, 17 May 2020 (UTC)

Thanks for your work! It did improve, although I concur that it's still not good. It is still too long, and I'm skeptical of putting equations in a lead. Most of the material should me merged into the body of the article. I'll get to it eventually, but I can't make promises as I'm rather busy IRL. Tercer (talk) 21:06, 17 May 2020 (UTC)

Yes, I can imagine times when putting an equation in the lead would make sense, but this isn't really one of them — why that vector instead of any of the other equally famous examples? This is a busy time off-wiki for me, too, and not a week/month when I should be starting big new projects here, and planning out a new organization for this article (which might be what's necessary) sounds at least moderately big. XOR'easter (talk) 14:55, 18 May 2020 (UTC)

I waded right in here. Please review the new lead, but perhaps reply on the following topic to avoid getting too deep. Johnjbarton (talk) 03:23, 16 June 2023 (UTC)