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Unbounded operators in non-relativistic QM of one spin-0 particle

  1. Apr 3, 2009 #1


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    What exactly are the axioms of non-relativistic QM of one spin-0 particle? The mathematical model we're working with is the Hilbert space [itex]L^2(\mathbb R^3)[/itex] (at least in one formulation of the theory). But then what? Do we postulate that observables are represented by self-adjoint operators? Do we say that a measurement of an operator [itex]A[/itex] on a system prepared in state [itex]|\psi\rangle[/itex] yields result an and leaves the system in the eigenstate [itex]|n\rangle[/itex] with probability [itex]|\langle n|\psi\rangle|^2[/itex]? Then how do we handle e.g. the position and momentum operators, which don't have eigenvectors?

    Can the problem of unbounded operators be solved without the concept of a "rigged Hilbert space"? Is it easy to solve when we do use a rigged Hilbert space? What is a rigged Hilbert space anyway?

    I think I brought this up a few years ago, but apparently I wasn't able to understand it even after discussing it. I think I will this time, because of what I've learned since then. Don't hold back on technical details. I want a complete answer, or the pieces that will help me figure it out for myself.
  2. jcsd
  3. Apr 3, 2009 #2
    I too would like a clarification on the subject of "rigged Hilbert space". Sometimes it seems like it is just a word people throw in to justify introducing non-normalizable eigenstates and treating them in a similar way as other eigenstates with the substitution [itex]\langle n|m\rangle =\delta_{nm}\rightarrow\langle x|x'\rangle=\delta(x-x')[/itex]. Is this just some trick or is there more to it.

    Sometimes people introduce boxes with periodic boundary conditions and let the size of those boxes go to infinity at the end.... is this more rigorous? Probably not ..?
  4. Apr 4, 2009 #3


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    More axiomatically, one can start with a complete normed algebra
    of operators (a Banach algebra), satisfying some extra axioms
    that make it into a C* algebra. Then construct a Hilbert space
    on which the elements of the algebra act as operators. This is
    called the "GNS" construction. The algebraic approach avoids
    some of the operator ambiguities that can arise with the
    "Hilbert space first" approach.

    Using Rigged Hilbert Space (RHS), aka "Gelfand Triple".
    (Personally I dislike both names, and prefer the more
    explicit "Gelfand Triple Space", though I think I'm alone in
    that usage.)

    Without an RHS, you've got to pay careful attention to the domains of
    operators. The general spectral theorem for s.a. operators on inf-dim
    Hilbert space is littered with domain stuff. But the whole point of
    the RHS idea is to avoid that stuff and provide a rigorous mathematical
    underpinning of Dirac's original bra-ket stuff that uses such improper

    Do you have a copy of Ballentine's QM textbook? It's one of
    the few that explain and emphasize how all the Dirac-style QM
    we know and love is really all being done in an RHS.
    (Ballentine also shows how some of the operators in non-rel
    QM arise by considering unitary representations of the
    Galilei group, which was another part of your question.)

    I just looked at the RHS Wiki page but it's very brief and doesn't
    tell you much. Although Ballentine describes RHS, it's only at an
    introductory level. There's an old book by Bohm & Gadella,
    "Dirac Kets, Gamow Vectors, and Gel'fand Triplets" which explains
    a bit more, but they too don't get into the mathematical guts.

    There's no way I can fit a complete technical answer in a Physics
    Forums post, but maybe I can get you started...

    The basic idea is to start with a Hilbert space "H" and then construct
    a family of subspaces. To do this, take the formula for your
    Hilbert space norm, and then modify it to make it harder for all
    states to have a finite norm. E.g., change the usual norm from
    \int dx \psi^*(x) \psi(x)
    to something like
    \int dx |x|^n \psi^*(x) \psi(x)
    Clearly, for n>0, only a subset of the original [itex]\psi[/itex]
    functions still have finite norm. It is therefore a "seminorm" (meaning
    that it's defined only a subset of H). This family of seminorms,
    indexed by n, define a family of progressing smaller and smaller
    subsets of the original Hilbert space H. It turns out that each such
    subspace is a linear space, and is dense in the next larger one.

    More generally, this construction comes under the heading of
    "Nuclear Space", with a corresponding family of "seminorms".
    The Wiki page for Nuclear Space has some more info.

    Now, to proceed further, you need to know a couple of things about
    inf-dim vector spaces and their duals. First a Hilbert space H is
    isomorphic to its dual (i.e., isomorphic to the set of linear
    mappings from H to C). Then, if you restrict to a linear subspace
    of H, (let's call it [itex]\Omega_1[/itex], corresponding the
    case n=1 above), the dual of [itex]\Omega_1[/itex], which I'll
    denote as [itex]\Omega^*_1[/itex], is generally larger than H.
    I.e., we have [itex]\Omega_1 \subset H \subset \Omega^*_1[/itex].

    Note that the usual norm and inner product are ill-defined between
    vectors belonging to the dual space [itex]\Omega^*_n[/itex], but
    we still have well defined dual-pairing between a vector from
    [itex]\Omega^*_n[/itex] and a vector from [itex]\Omega_n[/itex].
    This is enough for Dirac-style quantum theory.

    Actually, I'm getting a bit ahead of myself. First, we should take an
    inductive limit [itex]n\to\infty[/itex] of the [itex]\Omega_n[/itex]
    spaces, which I'll denote simply as plain [itex]\Omega[/itex] without
    the subscript. This is the subspace of functions from H which vanish
    faster than any power of x.

    The "Rigged Hilbert Space", or "Gel'fand Triple", is the name given
    to the triplet of densely nested spaces:
    \Omega \subset H \subset \Omega^*
    The word "rigged" should be understood to mean "equipped and ready for
    action". (Even with this explanation I personally still think it's a
    poor name.)

    [Continued in next post because of "Database error"...]
    Last edited: Apr 4, 2009
  5. Apr 4, 2009 #4


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    [Continuation of previous post...]

    The master stroke now comes in that the so-called improper states of
    position, momentum, etc, in Dirac's bra-ket formalism correspond to
    vectors in [itex]\Omega^*[/itex]. It is possible to take a s.a.
    operator on [itex]\Omega[/itex], and extend to an operator on
    [itex]\Omega^*[/itex], where the extension is defined in terms of its
    action on elements of [itex]\Omega[/itex] via the dual-pairing.

    Taking this further, there is a generalization of the usual spectral
    theorem, called the Gelfand-Maurin Nuclear Spectral Theorem which
    shows that eigenvectors of A in the dual space [itex]\Omega^*[/itex]
    are "complete" in a generalized sense, even though they're not

    So although people "throw around" the phrase "rigged Hilbert space"
    it's actually very important to the mathematical underpinnings of QM,
    though perhaps less so if you just want to do Dirac-style basic
    calculations. The RHS *is* the arena for modern QM, rather than the
    simpler Hilbert space as widely believed. The RHS, with the G-M Nuclear
    Spectral Theorem, is a far more general mathematical foundation than
    the trick of "finite boxes", etc, that jensa asked about.

    There's also an old textbook by Maurin "General Eigenfunction Expansions..."
    which gives the rigorous proof (though not very clearly, imho). But I
    think I'll stop here and see what followup questions arise.
  6. Apr 4, 2009 #5
    This connection between unboundedness of operators and the non-normalizable eigenvectors is non-existing, and indicates some confusion.

    Example 1:

    If you consider the system defined by a Hamiltonian (this is the infinitely deep potential well)

    H = -\frac{\hbar^2}{2m}\partial_x^2 + \infty\;\chi_{]-\infty,0[\cup ]L,\infty[}(x)

    and solve it's eigenstates, you get a sequence of normalizable eigenvectors [itex]|\psi_1\rangle,|\psi_2\rangle,\ldots[/itex], and in this basis the Hamiltonian is

    H = \frac{\hbar^2\pi^2}{2mL^2}\left(\begin{array}{ccccc}
    1 & 0 & 0 & 0 & \cdots \\
    0 & 4 & 0 & 0 & \cdots \\
    0 & 0 & 9 & 0 & \cdots \\
    0 & 0 & 0 & 16 & \cdots \\
    \vdots & \vdots & \vdots & \vdots & \ddots \\

    This is an unbounded operator, but still it can be diagonalized in the Hilbert space in the standard sense.

    Example 2:

    If you regularize the differential operator [itex]\partial^2_x[/itex] by making some cut-off in the Fourier-space, you obtain some pseudo-differential operator which will be approximately the same as the [itex]\partial_x^2[/itex] for wave packets containing only large wave lengths. So fix some large [itex]R[/itex] and set

    (H_R \hat{\psi})(p) = \frac{p^2}{2m} \chi_{[-R,R]}(p)\hat{\psi}(p),

    which is the same thing as

    (H_R \psi)(x) = \frac{1}{2m} \frac{1}{2\pi\hbar} \int\limits_{-R}^R\Big(
    \int\limits_{-\infty}^{\infty} p^2 \psi(x') e^{i(x-x')p/\hbar} dx'\Big) dp.

    This operator is bounded and [itex]\|H_R\| = \frac{R^2}{2m} < \infty[/itex]. However, it's eigenvectors are outside the Hilbert space [itex]L^2(\mathbb{R})[/itex].


    So it is possible to have an unbounded operator so that its eigenvectors are inside the Hilbert space, and it is possible to have a bounded operator so that its eigenvectors are outside the Hilbert space.
    Last edited: Apr 4, 2009
  7. Apr 4, 2009 #6
    If we ask a question that what is the probability for a momentum to be in an interval [itex][p_0-\Delta, p_0+\Delta][/itex], we get the answer from the expression

    \frac{1}{2\pi\hbar} \int\limits_{p_0-\Delta}^{p_0+\Delta} |\hat{\psi}(p)|^2 dp.

    I'm not convinced that it is useful to insist on being able to deal with probabilities of precise eigenstates. Experimentalists cannot measure such probabilities either.
  8. Apr 4, 2009 #7


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    Good answers, both of you. I appreciate that you are taking the time to explain these things to me. I need some time to think about the technically advanced parts of Strangerep's posts, so for now I'll just reply to Jostpuur. I'll reply to Strangerep later today, or tomorrow.

    Jostpuurs post #6 brings up one of the things I was thinking about when wrote the OP and when I was reading Strangerep's reply. The RHS stuff is interesting, and I definitely want to learn it, but I feel that as long as we're just talking about the non-relativistic quantum theory of one spinless particle, it should be possible to avoid the complications by stating the axioms of the theory of physics carefully, instead of changing the mathematical model (by replacing the Hilbert space by a Gelfand triple). What I mean by the "axioms of the theory of physics" are the statements that tell us how to interpret the mathematics as predictions of probabilities of possible results of experiments. Am I right about this, or do we absolutely need something like a RHS just to state the simplest possible quantum theory in a logically consistent way?

    Joostpur, I agree that your example 1 proves that it's possible for an unbounded operator on a Hilbert space to have eigenvectors. I didn't expect that. The Hilbert space in your example is [itex]L^2([0,L])[/itex], not [itex]L^2(\mathbb R^3)[/itex], but those two spaces are isomorphic (unless I have misunderstood that too), so this should mean that there's an unbounded operator on [itex]L^2(\mathbb R^3)[/itex] that has an eigenvector.

    I don't understand 100% of example 2, but I accept it as a convincing argument that it's possible for a bounded operator to fail to have eigenvectors. The part that's confusing me is that I don't see what the "eigenvectors" of HR are. I assume that they are some sort of distributions.
  9. Apr 4, 2009 #8


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    Confusion in terminology perhaps, but there is certainly a connection.

    Expressed more precisely, let me quote the Hellinger-Toeplitz
    theorem (from Lax, p377):

    "An operator M that is defined everywhere on a Hilbert space H and
    is its own adjoint, (Mx,y) = (x,My), is necessarily bounded."

    The proof is only a few lines.
    There follows a corollary (further down on p377):

    "It follows from this [...] that unbounded operators that are their
    own adjoints can be defined only on a subspace of the Hilbert space".

    The last bit about "in the Hilbert space" is incorrect. Let me
    simplify your example...

    Your eigenvectors can be written as infinite-length vectors:
    e_1 := (1,0,0,0,....)~~

    e_2 := (0,1,0,0,....)~~


    e_k := (0,0,...,0,1,0,0,...)~~
    Forgetting the constants, your Hamiltonian can be written as
    H_{nm} ~=~ n^2 ~ \delta_{nm}

    Now, I can construct a particular linear combination of the [itex]e_k[/itex]:
    \psi := \sum_{k=1}^\infty \frac{1}{k} ~ e_k
    whose squared norm is
    (\psi,\psi) ~=~ \sum_{k=1}^\infty \frac{1}{k^2} ~<~ \infty
    So [itex]\psi[/itex] is in the Hilbert space. Now consider
    \phi := H \psi ~=~ \sum_{k=1}^\infty k^2 ~ \frac{1}{k} ~ e_k
    ~=~ \sum_{k=1}^\infty k ~ e_k
    whose squared norm is
    (\phi,\phi) ~=~ \sum_{k=1}^\infty k^2 ~\to~ \infty
    and therefore [itex]\phi[/itex] is not in the Hilbert space.

    Hence, it's incorrect to say that this Hamiltonian is an
    operator on the entire Hilbert space. It's only a well-defined
    operator on a subspace of the Hilbert space.

    The rigged Hilbert space formalism was developed to make sense of
    this. The Hamiltonian *can* be diagonalized in a sense, but must
    be done in terms of generalized eigenvectors in a larger space
    of tempered distributions (the [itex]\Omega^*[/itex] from my
    earlier post).

    Your Hamiltonian in example 2 (Edit: in the limit [itex]R\to\infty[/itex] )
    is not well-defined on all of [itex]L^2(\mathbb{R})[/itex]. I.e., it's not
    an operator on all of [itex]L^2(\mathbb{R})[/itex].

    You seem to be defining
    the eigenvectors on a subset of [itex]L^2(\mathbb{R})[/itex] (with
    finite "R") and then assuming they remain well-defined when you
    take [itex]R\to\infty[/itex]. But the limit
    \lim_{R\to\infty} \|H_R\| ~=~ \lim_{R\to\infty} \frac{R^2}{2m}
    does not exist.

    Sure, plenty of people get along fine without understanding
    these subtleties. Dirac was one of them. He just knew intuitively
    that it was ok, somehow. Later, some mathematicians came along
    and made it more rigorous and respectable, using rigged Hilbert
    space and related concepts. And Fredrick's question was clearly
    asking about the mathematically precise stuff.
    Last edited: Apr 5, 2009
  10. Apr 4, 2009 #9
    I was already aware of the fact that unbounded operators are often[itex]{}^*[/itex] defined on some subsets of the original Hilbert space, although I thought it would not be necessary to get into that matter in my post. For example the domain of [itex]H=-\frac{\hbar^2}{2m}\partial_x^2[/itex] is

    D(\partial_x^2) = \big\{\psi\in L^2(\mathbb{R})\;|\; \int\limits_{-\infty}^{\infty} p^4 |\hat{\psi}(p)|^2 dp < \infty \big\},

    but this doesn't usually get mentioned in every post that is concerned with this Hamiltonian.

    I can admit that I didn't know this theorem.

    ([itex]{}^*[/itex]: Or let's say that I was under belief that this "often" the case, while not being aware of the fact that it is always the case, with self-adjoint operators.)

    I had not thought about this example carefully, and was not aware of the fact that the domain is not the full space, but now when I look my post, I don't think that I would have very explicitly claimed the domain to be the full space either.

    I don't agree on this. Let [itex](X,\mu)[/itex] be some measure space, and [itex]f\in L^{\infty}(X)[/itex] some measurable function. Then the formula

    \psi\mapsto M_f\psi,\quad (M_f\psi)(x) = f(x)\psi(x)

    defines a bounded operator [itex]M_f:L^2(X)\to L^2(X)[/itex], and [itex]\|M_f\|\leq \|f\|_{\infty}[/itex]. My example belongs to this class of operators.
    Last edited: Apr 5, 2009
  11. Apr 4, 2009 #10

    The harmonic oscillator is an example of operator which is defined on some subspace of [itex]L^2(\mathbb{R}^n)[/itex], and which has a sequence of eigenvectors whose span is dense in [itex]L^2(\mathbb{R}^n)[/itex].

    I'll continue with the example from my previous post, where [itex]M_f[/itex] was defined. Suppose for simplicity that the [itex]f[/itex] was also injective, so that it doesn't get same values at different locations. Now we ask whether [itex]\lambda[/itex] is an eigenvalue. If it is, then an equation

    f(x)\psi(x) =\lambda \psi(x)

    must be true for a.e. [itex]x\in X[/itex]. This cannot happen unless [itex]f(\overline{x})=\lambda[/itex] with some [itex]\overline{x}\in X[/itex], and unless [itex]\psi(x)=0[/itex] for a.e. [itex]x\neq \overline{x}[/itex]. So the eigenvectors would have to be [itex]\psi(x) = \chi_{\{\overline{x}\}}(x)[/itex]. If the measure [itex]\mu[/itex] is such measure that [itex]\mu(\{\overline{x}\})=0[/itex], then the eigenvector doesn't exist because it is zero. What happens in the non-rigorous formalism is that this kind of eigenvector is multiplied with an infinite constant so that it becomes non-zero.

    If the Hamiltonian is defined in the Fourier-space by a multiplication

    (H\hat{\psi})(p) = \frac{p^2}{2m}\hat{\psi}(p)

    then in the non-rigorous formalism the eigenvectors are delta-functions [itex]\delta(p-p')[/itex], and in the spatial representation they are the plane waves [itex]e^{ixp/\hbar}[/itex]. If the Hamiltonian is made bounded by force by multiplying the operated function with [itex]\chi_{[-R,R]}(p)[/itex], then the same eigenvectors still work, but the eigenvalues are different. The eigenvalues are the same for [itex]-R\leq p\leq R[/itex], but go to zero for other [itex]p[/itex].
  12. Apr 5, 2009 #11


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    I don't see how this observation implies anything more than that H is unbounded, and we already knew that. :confused:

    Edit: I do now, thanks to Jostpuur. See my next post.

    It looks well-defined to me. It's just not injective on all of [itex]L^2(\mathbb R)[/itex] and I guess that means it's not self-adjoint on [itex]L^2(\mathbb R)[/itex].

    [tex]H_R\psi(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}dp\ e^{-ipx}\frac{p^2}{2m}\tilde\psi(x)\chi_{[-R,R]}(p)=\frac{1}{2\pi}\int_{-R}^{R}dp\ e^{-ipx}\frac{p^2}{2m}\int_{-\infty}^{\infty}dx'\ e^{ipx'}\psi(x')[/tex]

    We can take the expression on the right-hand side as the definition of HR, and if we do I think it's clear that this operator is well-defined. The middle expression implies that it's not injective. The integral doesn't depend on what values the Fourier transform [itex]\tilde\psi[/itex] has outside of the interval [-R,R]. (LOL, the tilde is invisible in itex mode).

    If I insert [itex]\psi(x')=e^{ikx'}[/itex] into the right-hand side of my equation above, I don't get a constant times [itex]e^{ikx}[/itex]. I get zero. But I might be doing something wrong.
    Last edited: Apr 5, 2009
  13. Apr 5, 2009 #12
    In my opinion strangerep wrote a relevant comment concerning my example 1, but made a mistake with the example 2.

    Recall that if an operator [itex]T:V\to V[/itex] between two norm spaces is unbounded, then it does not mean that [itex]\|T\psi\|=\infty[/itex] for some [itex]\psi\in V[/itex]. This would be contradictory with the implicit assumption [itex]T(V)\subset V[/itex]. Instead it means that there is a sequence of vectors [itex]\psi_1,\psi_2,\psi_3,\ldots\in V[/itex] such that [itex]\|\psi_n\|\leq 1[/itex] for all n, and

    \sup_{n\in\mathbb{N}} \|T\psi_n\| = \infty.

    When [itex]V[/itex] is a subset of some larger vector space in which the norms of vectors are infinite, and the image point [itex]T\psi[/itex] is defined with some formula in this larger vector space, it can happen that [itex]\|T\psi\|=\infty[/itex] for some [itex]\psi\in V[/itex]. In this case we don't obtain an operator [itex]T:V\to V[/itex], but instead an operator [itex]T:D(T)\to V[/itex] where

    D(T) = \{\psi\in V\;|\; \|T\psi\|<\infty\}

    is the domain of the operator. This is what happens in the example of the infinite potential well. The Hamiltonian is not defined on the entire Hilbert space [itex]L^2([0,L])[/itex], but only on some subspace. However, this subspace is dense in the Hilbert space.

    I don't think that I said anything wrong in my example 1 though. Despite the fact that the Hamiltonian is not defined in the entire Hilbert space, the Hamiltonian has a sequence of orthogonal eigenvectors, whose span is dense in the Hilbert space, so the Hamiltonian is pretty diagonalizable there.
    Last edited: Apr 5, 2009
  14. Apr 5, 2009 #13


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    D'oh. This is what I missed. Thanks. [itex]\|H\psi\|[/itex] must be finite when the codomain of H is a Hilbert space, so Strangrep's calculation does prove that H can't be defined on all of [itex]L^2([0,L])[/itex]. It's a proof by contradiction:

    Assume that [itex]H=-(\hbar/2m)d^2/dx^2[/itex] is a linear operator from [itex]L^2([0,L])[/itex] into [itex]L^2([0,L])[/itex]. Then it's defined on the specific [itex]\psi[/itex] that Strangrep defined, but that [itex]\psi[/itex] satisfies [itex]\|H\psi\|=\infty[/itex], and that contradicts the assumption that the range of H is a subset of [itex]L^2([0,L])[/itex].
  15. Apr 5, 2009 #14


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    I intended (but neglected to say) "in the limit [itex]R\to\infty[/itex]".
    I've now edited my earlier post to fix this. Sorry for my lack of care.
  16. Apr 5, 2009 #15


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    I have heard about it (e.g. in a reply you wrote to me months ago), but I haven't studied it yet. It's on page 250 of the functional analysis book I bought a while ago (Conway), and I have read very little of that book so far. It's going to take a while before I get there, so maybe we can skip the details (especially proofs), and just talk about what the point is. What is the point? I thought it had something to do with relativity and causality.

    Edit: I think I see the point you were going for. I was asking about how to introduce observables into the theory, and you just meant that this is one way to define them. Is it a better way than to define the observables as self-adjoint operators on a separable Hilbert space?

    Does the C*-algebra/GNS approach have anything to do with the RHS concept, or are the two unrelated? (They seem unrelated to me).

    Spectral theorem...page 262...and that's probably not the most general one. It's about normal operators (A*A=AA*). I think it would be easier for me to learn the RHS stuff than to get through a whole book of functional analysis. (I think I know what I need to know about measures, integration and distributions).

    Unfortunately no. It's one of several books that I'm thinking about buying, but I haven't done it yet. I checked it out after reading Demystifier's thread about it (I noticed you did too), and I think it looks great. Fortunately, the relevant pages are available at Google books.

    I'm trying to find a specific set of statments that can be said to define the theory. I'm sure there are many different sets of statements that do the job (in the sense that each set is logically consistent and makes the same predictions about the results of experiments as some other set). I'd like to see the simplest set of statements that can define the theory, and also the set of statements that's the easiest to generalize to the relativistic case.

    I chose to ask specifically about non-relativistic QM of one spin-0 particle because it's the simplest of all relevant quantum theories, and I felt that it should be possible to define it in a pretty simple way. The traditional way (which is kind of sloppy) is to postulate among other things that states are represented by the rays of a (separable) Hilbert space (or specifically [itex]L^2(\mathbb R^3)[/itex]) and that the time evolution of a state is given by the Schrödinger equation. I think I would prefer to drop the explicit stuff about the Schrödinger equation, and instead postulate something about inertial observers and unitary representations of the Galilei group. This would give us both the Schrödinger equation and a definition of the Hamiltonian, the momentum operators and the spin operators (and probably the position operator too, but I haven't fully understood that part...something about central charges of the Lie algebra).

    I'm also interested in how the axioms must be changed when we go from non-relativistic to special relativistic quantum mechanics, and finally to general relativistic quantum mechanics. (But we can ignore that last one in this thread :smile:).

    There are some parts of of Ballentine's explanation where I feel that he dumbs it down a bit too much, but I think I understand what he should have said instead, so it's not a problem. :smile: His explanation, combined with yours, is very helpful actually.

    Hm, this part sounds familiar. I read the part about tempered distributions in Streater and Wightman recently, but I didn't try to understand every word. They defined a space of test functions that vanish faster than any power of x, and defined a tempered distribution to be a member of its dual space. The part I didn't understand was the exact definition of "vanish faster than any power of x". I'm going to read that part again, and see if I can understand it.

    Is the bottom line here that the members of H* are distributions with H (square integrable functions) as the test function space, and that the members of [itex]\Omega^*[/itex] are tempered distributions? Hm, what you said to Jostpuur in #8 looks like a "yes" to that question.

    I just realized that there's one small difference. The members of [itex]L^2(\mathbb R^3)[/itex] are not all infinitely differentiable, and test functions are usually assumed to be.

    It seems a bit strange and complicated to define a sequence [itex]\Omega_n[/itex] instead of defining [itex]\Omega[/itex] right away, but then I didn't understand S & W on a first read, and they seem to go straight for [itex]\Omega[/itex] (if I remember correctly). Maybe that's why I didn't understand them.

    That sounds interesting, but it's not even in my book. :smile:

    I'm definitely going to have to learn the details then. I really appreciate your effort in this thread. I still don't get it completely, but I'm getting closer.
    Last edited: Apr 5, 2009
  17. Apr 5, 2009 #16


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    Hmmm. That's a rather large question. It's not specifically about
    relativity and causality, but more about constructing a quantum theory
    starting from an algebra of observables, instead of starting from a
    Hilbert space.

    You can read the axioms for C*-algebras for yourself, but briefly,
    they're a subclass of Banach *-algebras, which in turn are *-algebras
    with a norm defined on every element satisfying certain extra axioms
    (e.g., the norm is submultiplicative -- See the top of Wiki Banach
    algebra page for what that means).

    One then considers linear functionals in the dual space of this normed
    algebra to arrive at a way of mapping observables to numbers, thus
    getting a quantum theory.

    When starting from a Heisenberg algebra, one usually employs the
    regularized (Weyl) form of the canonical commutation relations to
    banish the pathological behaviour that caused the need for RHS in the
    other formalism. The GNS construction basically means being given a
    vacuum vector (and its linear multiples, i.e., a 1D linear space), and
    multiplying it by all the elements of the algebra to generate a full
    Hilbert space (of course, I'm skipping lots of technicalities here).

    Of which book? (I don't have Conway.)

    Lax does the same, and stops short of talking about RHS and the
    Gelfand-Maurin generalization. (Looking at the index, I couldn't
    even find "nuclear spaces" mentioned.) Reed & Simon Vol-1 talk
    about nuclear stuff, but in the context of tempered distributions.
    That's why I asked a question here a while ago about proofs of the
    G-M theorem. Eventually I got hold of Maurin's book, difficult
    though it is.

    Specific examples of the [itex]\Omega, \Omega^*[/itex] are indeed
    (respectively) the Schwartz space of test functions, and its dual space
    of tempered distributions. The important thing about functional
    analysis is that abstracts away from specific spaces to general
    properties of inf-dim linear spaces, abstracting a lot of messy
    integration stuff by expressing things as linear operators instead.
    Although I too found functional analysis quite challenging and
    bewildering initially, I've now come to prefer it immensely and I only
    drop back to explicit integral stuff when considering specific examples.
    Functional analysis is an essential tool for the mathematical
    physicist, imho.

    Yes, so far. It boils down to: (1) pick a algebra of observables (actually
    their universal enveloping algebra), and (2) find all unitary irreducible
    representations of this algebra. (3) Construct tensor-product spaces thereof.
    The details fill many books of course. Weinberg takes this approach
    (more-or-less) in his volumes.

    Ah, the position operator (and localization) can get tricky. It's not
    too bad for the Galilei case (Ballentine covers it), but constructing a
    relativistic position operator is still controversial and problematic.

    No, H is self-dual. The test function space (Schwartz space) is an
    example of my [itex]\Omega[/itex] space (i.e., a dense subspace of H).

    Yes to that part.

    This (and the sequence of progressively stricter norms I mentioned
    originally) are just a rigorous way to define and generalize the notion
    of "...functions vanishing faster than any power of x...".

    The Wiki page on "nuclear space" has a bit more, though it doesn't
    mention the G-M theorem. Try to find Maurin's book if you can.
    (or maybe vol-4 in the series by Gelfand & Vilenkin -- I couldn't
    obtain the latter, but many authors reference it).

    Edit: I just remembered... there's some old ICTP lecture
    notes by Maurin on this stuff, available as:


    It covers a lot of the theorems, but skips the lengthy proofs.
    Last edited: Apr 5, 2009
  18. Apr 5, 2009 #17


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    I don't see why it's necessary to go beyond Hilbert space. Rather than defining a position operator, we could define projection operators with eigenvalue 1 if the particle is in some particular volume V, and 0 otherwise; heuristically, these would be

    [tex]P_{x\in V} \equiv \int_V d^3\!x\,|x\rangle\langle x|[/tex]

    Similarly for a volume in momentum space. Then, instead of defining a hamiltonian whose action could take a state out of the Hilbert space, we could define a unitary time evolution operator.
  19. Apr 6, 2009 #18

    George Jones

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    This is a very interesting thread in which I would like to participate actively, and from which I would like to learn, but I'm too busy with work for the next week or two to do the necessary reading and thinking.
    Even for a non-relativistic particle in a box, there is a lot a technical "grit". The position operator doesn't have any eigenstates that live in the Hilbert space of states, it only has distributional eigenstates. Also, the momentum operator is unbounded, and thus, by the Hellinger-Toeplitz theorem (as already posted by strangerep), the momentum operator cannot act on all the states in the Hilbert space of states.

    This is an example of something much more general. If two self-adjoint operators satisfy a canonical commutation relation, then it is easy to show that at least one of the operators must be unbounded.
    I think you're referring to the fact that (unlike the case for the Poincare group) non-relativistic quantum mechanics deals with representations of a central extension of the Galilean group, not with representations of the Galilean group. This is related to mass in non-relativistic quantum mechanics. Ballentine never uses the term "central extension," but, unlike most (all?) standard quantum mechanics texts, he does give a non-rigorous version. See: page 73, Multiples of identity (c); page 76; pages 80-81.
    I think that you would like chapter 9, Generalized Functions, form the book Fourier Analysis and Its Applications by Gerald B. Folland.
    I think that it's just a matter of taste whether one uses Hilbert spaces or rigged Hilbert spaces as a rigourous basis for quantum mechanics. For example, Reed and Simon write (in v1 of their infamous work):

    "We must emphasize that we regard the spectral theorem as sufficient for any argument where a nonrigorous approach might rely on Dirac notation; thus, we only recommend the abstract rigged space approach to readers with a strong emotional attachment to the Dirac formalism."

    I also think that the reason for the popularity of the Hilbert space approach is historical.

    In the early 1930s, before the work of Schwartz and Gelfand on distributions and Gelfand triples, von Neumann came up a rigorous Hilbert space formalism for quantum theory.

    I think if a rigorous rigged Hilbert space version of quantum theory had come along before the rigorous Hilbert space version of quantum theory, then the Hilbert space version might today be even less well-known than the rigged Hilbert space version actually is. Students would now be hearing vague mutterings about "making things rigourous with Gelfand triples," instead of hearing vague mutterings about "making things rigourous with Hilbert spaces."
    Last edited: Apr 6, 2009
  20. Apr 6, 2009 #19
    In my understanding, the reason why standard Hilbert space formalism is not suitable for QM is rather simple. Let's say I want to define an eigenfunction of the momentum operator. In the position space such an eigenfunction has the form (I work in 1D for simplicity)

    [tex]\psi(x) = N \exp(ipx)[/tex]

    where [tex]N [/tex] is a normalization factor. This wavefunction must be normalized to unity, which gives

    [tex] 1 =\int \limits_V |\psi(x)|^2 dx = N^2V[/tex]

    where [tex]V[/tex] is the "volume of space", which is, of course, infinite. This means that the normalization factor is virtually zero

    [tex] N = 1/\sqrt{V}[/tex]

    So, the value of the wavefunction at each space point is virtually zero too, and it can't belong to the Hilbert space. But the wavefunction is not EXACTLY zero, because its normalization integral is equal to 1. So, here we are dealing with resolving uncertain expressions like "zero x infinity".

    As far as I know, there is a branch of mathematics called "non-standard analysis", which tries to assign a definite meaning to such "virtually zero" or "virtually infinite" quantities and to define mathematical operations with them. I guess that using methods of non-standard analysis in quantum mechanics could be an alternative solution for the "improper states" in QM (instead of the rigged Hilbert space formalism). Did anyone hear about applying non-standard analysis to QM?
  21. Apr 6, 2009 #20


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    I'm still a bit confused by distributions and tempered distributions. Let's see if we can sort this out.

    We define D to be the set of all [itex]C^\infty[/itex] functions from [itex]\mathbb R^n[/itex] to [itex]\mathbb C[/itex] with compact support. (This D isn't used in the construction of a rigged Hilbert space. I'm defining it just for completeness). We say that [itex]\phi_n\rightarrow\phi[/itex] if there's a compact set K that contains the supports of all the [itex]\phi_n[/itex], and every [itex]D^\alpha\phi_n[/itex] converges uniformly on [itex]\mathbb R^n[/itex] to [itex]D^\alpha\phi[/itex]. Here we're using the notation


    [tex]D^\alpha f(x)=\frac{\partial^{|\alpha|}}{\partial x_1^{\alpha_1}\cdots\partial x_n^{\alpha_n}}f(x)[/tex]

    The members of D are called test functions. Now we define a distribution as a linear function [itex]T:D\rightarrow\mathbb C[/itex], which is continuous in the following sense:


    I think I would prefer to do it a bit differently (if the following is in fact equivalent to the above, but it might not be). We define an inner product and the associated norm by

    [tex]\langle f,g\rangle=\int_D f g\ d\mu[/tex]

    where [itex]\mu[/itex] is the Lebesgue measure on [itex]\mathbb R^n[/itex]. Now we define the space of distributions to be the dual space of D. Is this definition equivalent to the first?
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