Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Stone-von Neumann theorem & modern analog

  1. Aug 1, 2003 #1


    User Avatar
    Science Advisor
    Gold Member
    2015 Award
    Dearly Missed

    Several papers have appeared in the last few months suggesting a development in the quantum theory of general relativity which is analogous to the Stone-von Neumann theorem of 1931.

    The clearest and most representative of the bunch is probably the Okolow-Lewandowski paper dated February 2003:


    this thread is intended to describe the S-vN theorem and its importance in connecting Heisenberg and Schroedinger pictures and to draw the analogy with current work in quantum GR.

    [Lubos Motl kindly pointed out an error in the above link where I had written 0202... rather than 0302..., and I have corrected the typo.]

    I'll begin with a brief sketch of the S-vN theorem.
    Then, about 10 posts further along the thread, there will be some discussion of the Okolow-Lewandowski paper, the work of Sahlmann, and mention of a forthcoming paper by the four principal researchers involved in this development: Okolow-Lewandowski-Sahlmann-Thiemann. this last paper is referred to by O-L but is not yet in the preprint archive.

    Eric Weissteins MathWorld says this about the S-vN theorem:
    <<Stone-von Neumann theorem---A theorem which specifies the structure of the generic unitary representation of the Weyl relations and thus establishes the equivalence of Heisenberg's matrix mechanics and Schrödinger's wave mechanics formulations of quantum mechanics in Euclidean space.

    Neumann, J. von. "Die Eindeutigkeit der Schrödingerschen Operationen." Math. Ann. 104, 570-578, 1931.>>

    Weisstein omits to reference M.H.Stone, "Linear Transformations in Hilbert Space and Their Applications to Analysis," 1932

    Here is something Baez said about it:
    <<... the Stone-von Neumann theorem I know and love. The version I know says: all strongly continuous irreducible unitary representations of the Weyl relations for a finite-dimensional symplectic vector space are unitarily equivalent.

    This is the theorem that guarantees the equivalence between Schrodinger's and Heisenberg's approaches to quantum mechanics.

    Recall that the Weyl relations are the exponentiated version of Heisenberg's "canonical commutation relations". We exponentiate them to turn the nasty unbounded self-adjoint P's and Q's into nice one-parameter groups of unitary operators exp(isP) and exp(isQ), which are technically easier to deal with. The Stone-von Neumann theorem doesn't hold for infinite-dimensional
    symplectic vector spaces, which is why quantum field theory is trickier than quantum mechanics....>>


    Von Neumann was born in Budapest Hungary in 1903. His name was Janos. This is maybe the most famous theorem in Representation Theory---there is essentially only one irred representation of the Weyl algebra.

    Heisenberg formulated QM in terms that amounted to a matrix algebra. The important objects were operators called "observables". The Heisenberg picture did not have "wave functions". There was some technical trouble with unbounded operators, which were tamed by exponentiating them--the Weyl algebra resulted.

    The Schroedinger picture had wave functions from which you could build a Hilbert space of quantum states. You could translate the (Heisenberg) Weyl algebra into operators on that Hilbert space. But maybe there was some ambiguity about this! Maybe there were several essentially different ways of doing that?

    No, von Neumann showed that there is essentially only one way to set the Weyl algebra up in business operating on the Schroedinger wave functions or quantum state space. There was only one way to "represent" the algebra as operators on the Hilbert space. So even though matrices dont LOOK like wave functions the two were giving us the same picture of the world.

    Something interesting happened this year that was reminiscent of von Neumann. Jerzy Lewandowski drew the analogy. Hanno Sahlmann used the same exponentiation trick to tame some unbounded LQG observables to get an algebra Lewandowski called the "Sahlmann algebra" and proved a uniqueness theorem about its representations. the Sahlmann algebra is just big enough to contain the holonomies on a manifold and the (tamed) electric fluxes and all its reasonable representations are equivalent to one called the Ashtekar-Lewandowski representation. the theorem is on page 16 of

    (this is one that sA mentioned he came across back in February or so when it came out, maybe I should try to read it?)
    Last edited: Sep 27, 2003
  2. jcsd
  3. Aug 1, 2003 #2
    i hate to be demanding, but are you able to say more about the original stone-von neumann theorem? i know precious little about it.

    let me start with simple terminology: what does it mean for an operator to be bounded/unbounded?
  4. Aug 1, 2003 #3


    User Avatar
    Science Advisor
    Gold Member
    2015 Award
    Dearly Missed

    good lord Lethe in my little model of the universe you are the one who is supposed to be teaching me, not the other way round:wink:

    maybe you are just asking this for pedagogical reasons and could answer it yourself---if so feel free to take charge of exposition at any time, I just want to put Stone-von Neuman on the table for two reasons:

    A. it is an unmistakable monument (like the sphinx in the desrt) that shows why group (and algebra) representation theory is at the heart of Quantum stuff.

    B. It shows that this year is historical: Hanno S just proved the Stone-von N theorem of quantum gravity.

    Another fantastic Hanno paper just came out:
    "The irreducibility of the Ashtekar-Landowski Representation"

    This had never been proven! It is the basic guts of LQG and people had just been assuming it was, like, probably irreducible and so this 12-page paper comes out in March 2003 that proves a theorem that it is after all. Buttresses the uniqueness result. Also the A-L rep (which we now know unique and irred) is based on a measure on the space of connections called the A-L measure mu-sub-AL, &mu;AL for which uniqueness theorems are coming from Hanno and others. It is exciting. Nature seems to have given the ticket to a certain algebra and a certain action on a certain hilbertspace. So we should have the 1931 Stone-von Neuman result on the table as a reminder.

    But you said BOUNDED! Well heck on a finite dim hilbertspace
    or any normed space like a Banach space any linear operator
    has a bound

    max ||Ax||/||x||

    Like polynomials of degree < n with some reasonable norm and the operator is differentiation?

    But if the space is infinite dim, which the mothers normally are, then differentiation is not bounded because the leading coefficient keeps getting multiplied by n. excuse the incoherence but I'm sure you know what I mean

    you could teach us hilbertspaces Lethe, Paul Halmos has an extraordinarily thin book on them that is a classic, I wonder if there is something on line like that

    I shall go check since really should refresh my memory
  5. Aug 1, 2003 #4


    User Avatar
    Science Advisor
    Gold Member
    2015 Award
    Dearly Missed

    invitation to selfAdjoint

    Ahah! it just occurred to me that the ideal person to
    review the hilbertspace basics and give a few civilized examples
    is selfAdjoint!

    He surely knows what a bounded operator is, if anyone here does. The Hitchhiker's Guide equivalent would be knowing where one's towell is---hope no one finds the reference confusing.

    The address for that 12-page irreducibility proof (of the Ashtekar-Lewandowski representation of quantum gravity) is:


    There is terminology on the first page of that which I wish someone here could explain. If selfAdjoint reviews basic Hilberthood maybe I will tackle this one: "...The pair (A,E) coordinatizes an infinite dimensional symplectic manifold (M,&sigma;) whose (strong) symplectic structure s is such that A and E are canonically conjugate..."
    Last edited: Aug 16, 2003
  6. Aug 1, 2003 #5


    User Avatar
    Staff Emeritus
    Gold Member
    Dearly Missed

    Ye Gods! I just discovered this thread and already I'm co-opted!

    Here's my duty on bounded operators: See Wikipedia and get back to me with any questions.

    Think of the simplest Hilbert space, the complex plane, and think of the simplest bounded set, the unit circle. Now a general operator takes every complex number into some other complex number, and some of them would leave the image of the circle inside some (maybe very big) circle in the plane. But it's possible to think of transformations that map the circle into some arc going out to infinity. That's unbounded. Now generalize to all bounded sets, and then to general Hilbert spaces and you have it.

    Obviously an unbounded operator doesn't preserve sup (supremum or least upper bound) the sup of the magnitudes of the complex numbers on the unit circle is of course 1, whereas there is no sup for the magnitude of the c.n.s on a line. So if you want sup (and you do for these algebras) you gotta have bounded.

    Now that that's out of the way, Marcus this is absolutely stunning, the parallelism between Stone-von Neumann and Sahlmann's theorem. The match is too close to be coincidence, I suspect a category explanation behind it.
    Last edited: Aug 1, 2003
  7. Aug 1, 2003 #6


    User Avatar
    Staff Emeritus
    Gold Member
    Dearly Missed

    Also on the vertical and horizontal, I went back to my Nash & Sen and reviewed connections on principle bundles, and as you say, the idea is simple. V is the subset of the tangent bundle T where the tangents are tangent to the fibers (i.e the group actions). And H is the orthogonal complement of V in T, so T is the direct sum of V and T. No prob.

    Do we need to go over symplectic?
  8. Aug 1, 2003 #7


    User Avatar
    Science Advisor
    Gold Member
    2015 Award
    Dearly Missed

    replying to both your posts, I would like to go over symplectic

    it is another place where the parallelism is strong
    to follow the parallelism I will tell it very roughly
    and approximately:

    In the Ashtekar "new variables" version of classical UNquantized
    General Relativity there is a pair of conjugate variables
    (A,E)-----the connection and the triad
    analogous to position and momentum

    and then that gradually changes to
    "holonomies and fluxes"
    as the quantization proceeds
    and there is a conjugate pair of operators

    And in the most recent step the flux operators get
    exponentiated to make them bounded and then somehow
    put back into the algebra which now consists of
    holonomies and (exponentiated) fluxes

    the thing is that all the way along there is this pair
    of variables or operators or observables, this conjugate
    pair which always retains something of the look and feel
    of position and momentum in the phase space of mechanics

    And the 2n dimensional phase space of mechanics is
    the prototype symplectic manifold, or? so we should
    get the basics out in the open for that

    Your other post said:
    <<...this is absolutely stunning, the parallelism between Stone-von Neumann and Sahlmann's theorem. The match is too close to be coincidence, I suspect a category explanation behind it.>>
    I share your enthusiasm, and a sense that the quality of the mathematics is high, though I have not speculated at all as yet about where the parallelism might come from. Stunning is the word for it. Glad its happening now when we can watch by internet.
    Last edited: Aug 1, 2003
  9. Aug 2, 2003 #8
    well i hate to burst your bubble, but while it may seem i know quite a bit, i am in fact just an aspiring student, struggling with my math homework. there are a lot of areas where my lack of knowledge shows, and analysis is one of the worst.

    ok, the definition of a bounded operator seems quite simple. it is one for which max ||Ax|| is finite, for ||x||=1. right?

    ok, but wikipedia seemed to be saying that bounded is synonomous with linear and continuous. is this so? this seems obviously true to me.

    they also said that in finite dimensions linearity implies continuity (or was it the other way around, no i think i ve got it right), but not necessarily in infinite dimensions. i m having trouble imagining a mapping that is linear but not continuous. would such a mapping be unbounded?

    some reading on spr lead me to some talk of unbounded operators. it sounded like they were saying that lie generators tend to by unbounded, while the exponentials thereof tend to be bounded. are they refering to elements of the lie group?

    ok, that s all for now.
  10. Aug 2, 2003 #9


    User Avatar
    Staff Emeritus
    Gold Member
    Dearly Missed

    Wikipedia is saying linear and continuous is necessarilty bounded. So if you wanted to prove some operator was bounded it would be sufficient (but not necessary) to show that it was linear and continuous. Consequently these linear continuous, guaranteed bounded, operators are important in algebras of functions.
  11. Aug 2, 2003 #10


    User Avatar
    Science Advisor
    Gold Member
    2015 Award
    Dearly Missed


    selfAdjoint I think what you said is right and modulo minor differences in ways of talking we are all three on the same wavelength (except I havent read Wikipedia about this and you two have)

    If you recommended specific articles in Wiki and gave me a link or a keyword or something I'd probably check them out but for now it sounds like there is a workable agreement

    Somebody should describe "symplectic" perhaps. Or? Conjugate variables (Qs and Ps) have always mystified me. Why does nature want us to have symplectic manifolds? Do I run the risk of destroying this thread by asking such an outlandishly naive question? Should we not try to discuss "symplectic" and just grind ahead with trying to understand these papers approximately and as best we can?

    Well Lethe you impress me as having a lot of gumption, as an aspiring student.
    BTW Chroot and Hurkyl know stuff that is applicable to this
    business. Rutwig, if he would appear, knows everything including about operator algebras and representations probably. Glad you are hear and wish some others would help out too.

    selfAdjoint already pretty much answered but i will go over this
    in sort of heavyhanded fashion, for practice

    1. we are are talking normed linear spaces
    a norm is subadditive ("triangle inequality") and compatible with scalar multiplication ("homogeneity") and ||x|| = 0 iff x = 0

    2. a normed linear space is just a vectorspace with a norm which is endowed with the norm topology----all that means is that continuous functions are defined using the norm exactly as you expect.

    3. a Banach space, in case anyone asks, is a normed linear space which is "complete" meaning cauchy sequences have limits---it doesnt have unexpected gaps where there should be something and there isnt,

    4. Damn, just got a call and must leave,
    Yes if A is linear map X --> Y between two normed linear spaces
    then these three statements are equivalent
    a. A is continuous at one point
    b. A is continuous at all points
    c. A is bounded

    and bounded means just what you (selfAdjoint, Lethe) have
    said it means, must go but will get back later

    we dont have to be this ceremonious! we can just go ahead
    with Sahlmann's papers or with the earlier A-L papers which are
    very good too---Sahlmann just made a continuation

    LETHE: I happened to see the following post of yours about the discontinuity of the derivative. The question is to selfAdjoint but I will respond anyway by giving an example showing discontinuity at zero
    {sin(Mx)/M} restricted to a finite interval like [-1,1]

    as M gets large this sequence of functions converges to zero
    but the image sequence under deriv is {cos(Mx)} and does not
    converge to zero
    Last edited: Aug 11, 2003
  12. Aug 11, 2003 #11
    i believe that the derivative operator on L2(R) for example, is unbounded. does that mean that it is not linear and continuous? well obviously it s linear, so is it then simply not continuous? that seems weird to me
  13. Aug 11, 2003 #12
    I want to put my tiny little bit of knowledge regarding bounded operators in here. They operate over a linear vector space, but are not themselves linear, i.e. we can't add two bounded operators and get another bounded operator. But we can in general multiply them and get one.
  14. Aug 11, 2003 #13


    User Avatar
    Science Advisor
    Gold Member
    2015 Award
    Dearly Missed

    Much depends on the assumptions, for example there is a
    classic thin book by Lynn Loomis "Abstract Harmonic Analysis" which has
    a brief summary of Banach spaces at the beginning and on page 15 he says:

    "If X is a normed linear space the set B(X) of all bounded linear transformations of X into itself is not only a normed linear space but is also an algebra...."

    Normed linear spaces are very useful and the def is simple:
    it is just a vectorspace with a norm function that satisfies 3 natural conditions

    ||x|| = 0 iff x = 0
    (triangle ineq.) ||x + y|| =< ||x|| + ||y||
    (homogeneity) || cx|| = |c| ||x||

    Lots of things are normed linear spaces----the continuous functions on something, with the "supremum" norm, and lots more. To get fancier, you start asking that the linear space be "complete". the complex plane with one point removed is not topologically "complete" because there are sequences which you know are convergent but dont have anything to converge to---because of that gap.

    a normed linear space which is "complete" is called a Banach space

    An inner product space is called a Hilbert space if it is "complete"----taking limits is the essential thing in analysis.
    Last edited: Aug 16, 2003
  15. Aug 16, 2003 #14


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You just haven't been exposed to the right class of pathological functions. :smile:

    For simplicity, I will consider just the interval [0, 1].

    Consider the sequence of functions

    fn(x) = sin(nx) / n

    clearly, this sequence converges to the zero function. However:

    fn'(x) = cos(nx)

    This clearly does not converge to the derivative of the zero function (it doesn't converge at all!), so the derivative operator cannot be continuous.
  16. Aug 17, 2003 #15


    User Avatar
    Science Advisor

    Marcus already gave that example.
  17. Aug 25, 2003 #16


    User Avatar
    Science Advisor
    Gold Member
    2015 Award
    Dearly Missed

    this thread may have stuff to keep track of
  18. Aug 25, 2003 #17


    User Avatar
    Science Advisor
    Gold Member
    2015 Award
    Dearly Missed

  19. Aug 29, 2003 #18
    Let's look at the particle/wave ratio from a different point of view.

    Math looks at the continuum as infinitely many points with no gaps (R members).

    By QM there is a XOR ratio between particles and waves.

    Let us say that particle=point(localized element) and wave=continuous line(non-localized element).

    We know that any quantum element is an association between these opposites (or complementary) properties.

    It means that any mathematical language that tries to define a QM model, must be built on these associations, without forcing one property to be express in terms of the other.

    But this is what Math does, by forcing a continuous line to be expressed in terms of infinitely many points with no gaps.

    So, as I see it, we need some mathematical language that is built on a new perception of the continuum concept, which is not expressed by localized elements like points.

    If we follow this idea than particle/wave ratio = point/line ratio.

    It means that there is a XOR ratio between LINES to POINTS.

    XOR ratio between LINES to POINTS:
    0(LINE) 0(POINT) -> 0-(No information) -> no conclusion.
    0(LINE) 1(POINT) -> 1-(Clear Particle-like information) -> conclusions on points.
    1(LINE) 0(POINT) -> 1-(Clear Wave-like information) -> conclusions on lines.
    1(LINE) 1(POINT) -> 0-(No clear information) -> no conclusion.

    More detailed information about this approach, you can find here:

  20. Aug 29, 2003 #19
    Doron: i think you are off topic for this thread.

    marcus: can you recommend a text that covers stone-von neumann (the classical version, not the modern), it s not in any of my QM textbooks.
  21. Aug 29, 2003 #20


    User Avatar
    Science Advisor
    Gold Member
    2015 Award
    Dearly Missed

    Lethe, great to hear from you! I have to go out at the moment
    but will look for stuff (if possible on line) when I get back.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Stone-von Neumann theorem & modern analog
  1. Modern Algebra? (Replies: 6)

  2. Von Neumann analysis (Replies: 1)