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Hilbert Space Selection

  1. Jan 28, 2005 #1

    Kane O'Donnell

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    Hi everyone,

    This summer (it's summer in Australia) I have been studying quantum mechanics from a mathematician's perspective, and the physical interpretation has become a little more difficult as the theory has become more in-depth.

    Do we have a particular method for choosing which Hilbert space we use for a given system?

    There seem to be two answers to my question: (let's just neglect spin here)

    1) No. The underlying space is always chosen to be [tex]L^2(\mathbb{R}^{mn})[/tex], where n is the number of particles and m is the number of physical degrees of freedom each particle has. Alternatively we can use some closed subspace of [tex]\mathbb{R}[/tex] in the case of a finite-'volume' system, but that is sort of not an issue here.

    2) Yes - the Hilbert space we work in is the closed linear span of the set of eigenvectors.

    In the first case, my notes say that we consider that each [tex]\phi[/tex] satisfying [tex] ||\phi || = 1 [/tex] to be a pure state of the system, where pure states are unique only up to a numerical factor of modulus one.

    If we take this approach, there exist pure states that are not in the domain of the position operators. Indeed, since the position operators are the same for every physical problem, there will *always* be states (supposed to be physical) on which the position operators cannot be applied. This would seem to bring interpretational difficulties - can we exclude these states *just because*?

    Introducing the continuity/differentiability requirement of course solves this problem, but from a mathematical perspective this is a cop out, especially since the theory itself shows why wavefunctions need to be continuous and differentiable.

    The problem is, you can't avoid this difficulty by switching to interpretation (2) - if you do, you lose a lot of physical systems, for example the unbound states in a Coulomb potential. Also, to my mind you lose quite a bit of beauty in the theory - by using a large Hilbert space, you can decompose a system into it's components mathematically as a direct sum of closed subspaces. For example, you can write the larger Hilbert space as the direct sum of the spaces of bound and unbound states.

    Is there a rigourous explanation resolving the difficulties I've posed here, or has it been largely ignored? Interpretation (2) seems to be the one used by physicists when they are just considering specific parts of a system, but I'm more concerned about the whole range of states.

    Cheerio,


    Kane
     
  2. jcsd
  3. Jan 28, 2005 #2

    dextercioby

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    The fundamental observable of a quantum system is its Hamiltonian.This densly-defined self-adjoint linear operator determines not only the physical states (at a given moment of time),but also their dynamics...
    The fundamental equation is:
    [tex] H\psi =E\psi [/tex](1)

    Those "\psi" are not necessary vectors...That's why i avoided the "ket" notation...Generally,the "E"-s are spectral values of the operator...All we know about them is that they have to be real...Since the operator is self-adjoint,it doesn't have residual spectral values,which is a good thing.It means that the spectrum can be either continuous,discrete or both at the same time...For the case of generality,i'll consider both character of the spectrum.
    The first principle introduces the HSS (separable Hilbert space of states) and asks for NORMALIZABLE vectors in that space to describe physical quantum states...
    Then the II-nd principle,combined with the IV-th (Schroedinger picture,Dirac formulation) reveals the importance of equation (1).
    If H does not have a PURELY discrete spectrum,then it is not CONTINUOUS/BOUNDED on H (defined by the I-st principle).It means that it is continuous only on a subset [itex] M\subset H [/itex].Now we make use of the HILBERTIAN TRIADE.That is,we move the operator H into the Hilbert space of continuous linear functionals on M,call it [tex] \tilde{M} [/tex]
    and then in this (enlarged) space the spectral equation (1) admits solutions...

    Each time an operator is unbounded on H,we make use of the Hilbertian triade to solve the spectral problem...
    So in this enlarged space the problem is completely solved.

    As you may see,i made no reference of the H space...[tex]\mathbb{L}^{2}(\mathbb{R}^{3n}) [/tex] or whatever...

    Daniel.
     
  4. Jan 28, 2005 #3

    Kane O'Donnell

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    But you've just gone and done exactly what I was asking for you *not* to do. That is, solve the mathematical problem using physical axioms. In particular, you've implicitly taken the approach outlined in interpretation (2) in my first post, by using the closed solution space of the Schrodinger equation. What I'm saying is, if you do this, you're only looking at the 'friendly' part of the quantum mechanical system. I'll try to explain a case in point.

    For each observable represented by a self-adjoint operator [tex]\hat{T}[/tex] there exists a unique spectral measure [tex]E:\mathcal{B}\to\mathcal{P}[/tex] such that:

    [tex] \langle \psi | E_{B}^{T}\psi\rangle [/tex]​

    is the probability of a measurment of the (normalised) state [tex]\psi[/tex] with respect to the observable [tex]T[/tex] will give a result in the subset [tex]B[/tex] of [tex]\mathbb{R}[/tex]. Here, [tex]\mathcal{B}[/tex] is the set of Borel subsets of [tex]\mathbb{R}[/tex] and [tex]\mathcal{P}[/tex] is the set of projections. The preceding is a rigourous statement of the observables postulate. The point is, in the case where the operator *has* eigenvalues, then the unique spectral measure will typically have a form such that the the usual observables postulate involving eigenvalues and so forth follows immediately.

    On the other hand, there are *plenty* of normalised states [tex]\psi[/tex] which are *not* eigenfunctions of the Hamiltonian, for example, yet are physically real. The standard example is the unbound states of the Coulomb potential Hamiltonian - these *cannot* be decomposed as a sum of bound states, since all the bound states are orthogonal to them. On the other hand, if we can find a form for the spectral measure for positive energy, we can still talk about the probability of an unbound electron having energy in a certain range (although now, the probability of it having a definite energy is now zero). A form for this spectral measure can be found in, for example, Levitan & Sargsjan, Introduction to Spectral Theory: Self-Adjoint Ordinary Differential Operators.

    It's not pretty, but that's not the point.

    So the observables postulate can be stated in a way which encapsulates the traditional way in which it is stated, but it includes more.

    Problem: The spectral measure for the position operator in say a one dimensional problem has the form:

    [tex] \langle \psi | E_{B}^{x}\psi\rangle = \int_{B} |\psi |^2 dx[/tex]​

    which corresponds to the usual interpretation of the wavefunction being a probability distribution of position. The problem is, there are states for which the spectral measure makes sense but the expectation value for the observable is undefined. That is, we can find the probability of a state being in a particular range, but we can't find the average value. Take for example the state on [tex]L^2(\mathbb{R})[/tex] which is zero for x < 1 and equal to [tex]\frac{1}{x}[/tex] for x greater than or equal to 1. It is easy to see that this is a unit vector but that it is not in the domain of the position operator. The function is not continuous everywhere of course, but is this a good justification for excluding the state?

    Just as an aside, if you treat the Hamiltonian as the fundamental observable, then the time-dependent Schrodinger equation arises naturally out of the statement that [tex] e^{\frac{it\hat{T}}{\hbar}} [/tex] is unitary provided that [tex]\hat{T}[/tex] is self-adjoint (and vice versa via Stone's theorem).

    I hope you're not implying that boundedness is equivalent to having a complete set of eigenvectors with discrete eigenvalues - the Hamiltonian for the quantum harmonic oscillator in one dimension, for example, has a complete set of eigenvectors and the eigenvalues are all discrete, yet the operator H is certainly not bounded.

    Besides, this is irrelevant - I'm not lost on the mathematical details, what I'm asking is whether there are more precise QM postulates rather than the ones traditionally given to physicists.

    Let me digress again.

    I don't understand the reason why. Look, you've gone and made the mathematics a bit wobbly on this point. The Hamiltonian is only defined on a dense subset of [tex]L^2[/tex]. It's obviously not bounded, otherwise we wouldn't have the problem of finding a self-adjoint extension, come on, I know all this. The point is, if you restrict the Hilbert space in question to the closed solution space of the Schrodinger equation, then H *is* bounded, but you lose out on the other aspects of the system (for example, the unbounded states of Hydrogen).

    For calculation purposes obviously we only care about say, atomic states, or whatever, but I'm more interested in the technical details of how the QM postulates can be reduced as far as possible. I suspect all of my questions have been answered long ago - they must have been in order to have a useful scattering theory, for example, but I haven't seen a modern discussion on the subject.

    Cheerio,

    Kane
     
    Last edited: Jan 28, 2005
  5. Jan 28, 2005 #4

    dextercioby

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    My QM teacher said to me:"I didn't treat all mathematics behind QM's postulates" and invited me to read & understand the famous book
    J.Prugoveçki:"Quantum Mechanics in Hilbert Space"...

    I didn't have time for that,so i invite you to read it...

    And yes,scattering theory will be much easier after reading Prugoveçki...

    Daniel.
     
  6. Jan 28, 2005 #5

    Kane O'Donnell

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    Thanks :smile:

    Kane
     
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