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

Insights Mathematical Quantum Field Theory - Fields - Comments

  1. Dec 2, 2017 #76

    Urs Schreiber

    User Avatar
    Science Advisor
    Gold Member

    It's formal power series algebras equipped with star-algebra structure and positivity is defined in terms of the star algebra structure.
  2. Dec 2, 2017 #77


    User Avatar
    Science Advisor

    The issue is that I've heard various purist advocates of the algebraic approach suggesting that Hilbert space is not essential to quantum physics, but only an afterthought. I claim Hilbert space is essential, and no one has yet satisfactorily refuted this by deriving the quantum angular momentum spectrum without reliance on Hilbert space.

    Well, the ladder operators come later in a treatment that relies on nothing more than the algebra and abstract Hilbert space. Cf. Ballentine section 7.1. The extra baggage of a harmonic oscillator is unnecessary for deriving the spectrum.
  3. Dec 3, 2017 #78


    User Avatar
    Science Advisor

    In the Heisenberg picture, there is an initial state which does not evolve with time. The initial state can be any state in the Hilbert space. How can one do away with this arbitrary initial state?
  4. Dec 4, 2017 #79

    A. Neumaier

    User Avatar
    Science Advisor

    But as I had mentioned, the positivity obtained is not the physical one, as for formal power series in a variable ##x##, the rule ##\xi-x\ge 0## holds for no real ##\xi## while after picking the physical value of ##x## (in a nonperturbative theory) one has ##\xi-x\ge 0## for every real ##\xi## exceeding the physical value.
  5. Dec 4, 2017 #80


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    Well, I also think, the Hilbert-space structure is an essential element in teaching at least QM. Since relativistic QFT in (1+3) dimensions is not rigorously defined, I understand that mathematicians try a different approach to define states. Of course, in QT it is of utmost importance to distinguish between observables and states. It's the very point dinstinguishing QT from classical theories that observables and states are disinct entities of the theory.

    Concerning the treatment of angular momentum, I never understood, why one should bother students with the wave-mechanical derivation of the angular-momentum eigenvectors, i.e., an old-fashioned treatment of the spherical harmonics. It's so much more transparent to treat the algebra su(2) and its representations. The only cumbersome point is to show that the special case of orbital angular momentum has no half-integer representations, and for that you need the "harmonic-oscillator approach". See, e.g.,

    D. M. Kaplan, F. Y. Wu, On the Eigenvalues of Orbital Angular Momentum, Chin. Jour. Phys. 9, 31 (1971).

    Of course, with that analysis at hand, you can very easily derive all properties of the spherical harmonics by using the position representation (in spherical coordinates).
  6. Dec 22, 2017 #81


    User Avatar
    Science Advisor

    This is not correct. Notions such as completeness (by a norm) and continuity (i.e., boundedness) of any element of an operator algebra need to be defined with respect to some vector space topology. Hermitian adjoint can only be defined on a vector space with scalar product. Moreover, every (abstract) non-commutative [itex]C^{*}[/itex]-algebra can be realized as (i.e., isomorphic to) a norm-closed , *-closed subalgebra of [itex]\mathcal{L}(\mathcal{H})[/itex], the algebra of bounded operators on some Hilbert space [itex]\mathcal{H}[/itex]. Precisely speaking, for every abstract [itex]C^{*}[/itex]-algebra [itex]\mathcal{A}[/itex], there exists a Hilbert space [itex]\mathcal{H}[/itex] and injective *-homomorphism [itex]\rho : \mathcal{A} \to \mathcal{L}(\mathcal{H})[/itex]. That is [itex]\mathcal{A} \cong \rho (\mathcal{A}) \subset \mathcal{L}(\mathcal{H})[/itex], as every *-homomorphism is continuous (i.e., norm-decreasing).

    In general, one can say the following about quantization: Given a locally compact group [itex]G[/itex], its (projective) unitary representation on some Hilbert space [itex]\mbox{(p)Urep}_{\mathcal{H}}(G)[/itex] and the group (Banach) *-algebra [itex]\mathcal{A}(G)[/itex], then you have the following bijective correspondence [tex]\mbox{(p)URep}_{\mathcal{H}}(G) \leftrightarrow \mbox{Rep}_{\mathcal{H}}\left(\mathcal{A}(G)\right) \ , \ \ \ \ (1)[/tex] where [itex] \mbox{Rep}_{\mathcal{H}}\left(\mathcal{A}(G)\right)[/itex] is the representation of the (Banach) *-algebra [itex]\mathcal{A}(G)[/itex] on the same Hilbert space [itex]\mathcal{H}[/itex], i.e., *-homomorphism from [itex]\mathcal{A}(G)[/itex] into the algebra of bounded operators [itex]\mathcal{L}(\mathcal{H})[/itex] on [itex]\mathcal{H}[/itex]. Similar bijective correspondence exists when [itex]\mathcal{A}[/itex] is a C*-algebra. And both ends of the correspondence lead to quantization. When [itex]G = \mathbb{R}^{2n}[/itex] is the Abelian group of translations on the phase-space [itex]S = T^{*}\left(\mathbb{R}^{n}\right) \cong \mathbb{R}^{2n}[/itex] (or its central extension [itex]H^{(2n+1)}[/itex], the Weyl-Heisenberg group) then (a) the left-hand-side of the correspondence leads (via the Stone-von Neumann theorem) to the so-called Schrodinger representation on [itex]\mathcal{H} = L^{2}(\mathbb{R}^{n})[/itex] [Side remark: of course Weyl did all the work, but mathematicians decided (unjustly) to associate Heisenberg’s name with the group [itex]H^{2n+1}[/itex]], while (b) the right-hand-side of the correspondence leads to the Weyl quantization which one can interpret as deformation quantization (in effect, Weyl quantization induces a non-commutative product (star product) on the classical observable algebra, thus deforming the commutative associative algebra of functions [itex]C^{\infty}(\mathbb{R}^{2n})[/itex]).
  7. Dec 22, 2017 #82


    Staff: Mentor

    I have said it before and will say it again - I wish Samalkhaiat had the time to post more. His answers cut straight though.

    The c*Algebra approach is, as it must be, equivalent to the normal Hilbert-Space approach - but can be formulated in a way where its association with classical mechanics is clearer:

  8. Jan 17, 2018 #83

    A. Neumaier

    User Avatar
    Science Advisor

    Actually it is more general, as the same algebra may have states corresponding to different Hilbert spaces (more precisely, unitarily inequivalent representations).

    Thus it is able to account for superselection rules (restrictions of the superposition principle), which have no natural place in a pure Hilbert space approach.

    Also it accounts for quantum systems having no pure states (such as those required in interacting relativistic quantum field theory).
    Last edited: Jan 19, 2018
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted