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

Algebra and physics

  1. Jun 26, 2013 #1
    Can someone tell me why the quantities having the same algebra structure can be indentified as the same dynamical variables? Are the Possion brackets and the quantum commutators two different presentation of the same Lie algebra?
     
  2. jcsd
  3. Jun 27, 2013 #2

    tom.stoer

    User Avatar
    Science Advisor

    I am not sure whether I fully agree with your first sentence. Suppose there is a) a three-dim. harmonic oscillator, b) QCD with color, c) three-quark flavor symmetry. In all cases one can construct su(3) charges Qa with a=1.8 and [Qa,Qb] = i fabc Qc. But of course the fundamental dynamics is different, the algebra does not fix the Hamiltonian (in the flavor case the Hamiltonian was unknown when Gell-Mann and others discovered the quark model with its symmetries). And the algebra does not fix the allowes representations: in the flavor case we know quarks, mesons and baryons in different multiplets, but the question which multiplets do exist in nature are not determined by the algebra but by the dynamics; in QCD we construct everything from fundamental and adjoint fields of su(3) color, but we know that in the physical Hilbert space only color singulets (trivial rep. singulet) are allowed.

    Regarding the second sentence: yes, the matrices, the classical objects and the quantum mechanical operators are different 'representations' of the same algebra. Note that in math the term 'representation' is used to distinguish different algebraic properties (fundamental rep., adjoint rep., irreduzible rep., ...) whereas here we use the same algebraic structure but constructed from different objects like a) creation and annihilation operatos, b) quark and gluon fields, c) only quark fields and acting on different vectors spaces (Hilbert spaces). Looking at the pure algebraic properties all we need are generators, commutation relations and especially their structure constants fabc; this defines the algebra uniquely, regardless from which entities the generators have been constructed.
     
  4. Jun 29, 2013 #3
    Sorry for the obscure question.
    I just wonder why we can identify the generators of Lorentz group as the physical variables, such as placement for momentum, rotation for angular momentum...
    Is it because they have the same algebra relation as the classical Possion brackets?
     
  5. Jun 30, 2013 #4

    tom.stoer

    User Avatar
    Science Advisor

    You can do that for an even larger group, the Poincare group.

    You get generators for translation (4-momentum), rotations (angular momentum) and boosts.
     
    Last edited: Jun 30, 2013
  6. Jun 30, 2013 #5

    samalkhaiat

    User Avatar
    Science Advisor

    The invariance of your theory under the Poincare’ group, say, leads (through Noether Theorem) to a set of 10 conserved quantities (Constants of Motion/ Noether Charges). Then we proceed to show that these quantities have the following properties: (i) they transform EXACTLY like the generators do under the Poincare’ group. (ii) they act on the fields of the theory generating the CORRECT Poincare’ transformations on them. (iii) they satisfy (through Possion Brackets or Commutators) the same Lie algebra of the Poincare’ group. So, we can identify them with the generators of the Poincare’group. It is like the saying: “If it SMELLS like an apple, LOOKS LIKE an apple and TASTES like an apple, it is an apple”
    What Noether theorem does is simply giving FIELD REALIZATION to the generators.

    Sam
     
  7. Jun 30, 2013 #6

    tom.stoer

    User Avatar
    Science Advisor

    I don't think you have 10 conserved quantities.

    The commutation relations for rotations L, boosts K, 3-momentum P with the Hamiltonian H are

    [Li,H] = [Pi,H] = 0

    [Ki,H] = -i Pi

    That means that the boosts K do not commute with H and can therefore not be 'conserved charges'.
     
  8. Jul 1, 2013 #7

    samalkhaiat

    User Avatar
    Science Advisor

    The invariance under the Poincare’ group implies that the energy-momentum 4-vector
    [tex]
    P_{ a } = \int d^{ 3 } x \ T_{ 0 a } = \int d^{ 3 } x \left( \frac{ \partial \mathcal{ L }}{ \partial ( \partial_{ 0 } \phi ) } \ \partial_{ a } \phi - \eta_{ 0 a } \mathcal{ L } \right) , \ \ (1)
    [/tex]
    and the angular momentum tensor
    [tex]
    M_{ ab } = \int d^{ 3 } x \ \left( T_{ 0 b } x_{ a } - T_{ 0 a } x_{ b } + \frac{ \partial \mathcal{ L } }{ \partial ( \partial_{ 0 } \phi ) } \ \Sigma_{ a b } \phi \right) , \ \ (2)
    [/tex]
    are CONSTANTS OF MOTION. These are the (4+6=10) conserved Noether CHARGES. It is very easy to show that
    [tex]\frac{d}{dx^{ 0 }} P_{ a } = \frac{d}{dx^{ 0 }} M_{a b} = 0 . \ \ (3)[/tex]


    This is very common misunderstanding. The components [itex]M_{ i 0 }[/itex] has an EXPLICIT time dependence which has to be accounted for when writing Heisenberg (Poisson) equation of motion. So, you need to write
    [tex]
    \frac{ d }{ dx^{ 0 } } M_{ i 0 } = \partial_{ 0 } M_{ i 0 } + [ i P_{ 0 } , M_{ i 0 } ] .
    [/tex]
    The conservation of [itex]M_{ i 0 }[/itex], [Eq(3)], therefore implies
    [tex]
    [ i P_{ 0 } , M_{ i 0 } ] = - \partial_{ 0 } M_{ i 0 } = - \partial_{ 0 } \int d^{ 3 } x \ \left( - \pi \partial_{ i } \phi \right) \ x^{ 0 } = \int d^{ 3 } x \ \pi \partial_{ i } \phi = P_{ i } .
    [/tex]
    So, the non-vanishing commutator [itex][ i H , M_{ i 0 } ][/itex] DOES NOT mean that [itex]M_{ i 0 }[/itex] is NOT CONSERVED.

    Sam
     
  9. Jul 1, 2013 #8

    tom.stoer

    User Avatar
    Science Advisor

    I missed the t-dependency; thx for clarification
     
    Last edited: Jul 2, 2013
  10. Jul 8, 2013 #9
    thanks, Sam.
    It seems that the most important property of physical quantities is that they are conserved, is it?
    And why the conserved quantities just happen to have those properties, is there some mechanism guaranteeing this?
    By the way, which algebra dose the commutator [x,p] or the classical Possion brackets belong to? Are they generators of some group?
     
  11. Jul 9, 2013 #10

    samalkhaiat

    User Avatar
    Science Advisor

    Yes, it is important, I suppose. Current conservation, [itex]\partial_{ a } J^{ a } ( x ) = 0[/itex], has a remarkable consequence for any matrix element of [itex]J_{ a }[/itex]. For any arbitrary states [itex]| I \rangle[/itex] and [itex]| F \rangle[/itex], we take the matrix of the divergence of Noether current and use Heisenberg equation, we find
    [tex]
    0 = \langle F | \partial_{ a } J^{ a } ( x ) | I \rangle = i \langle F | [ P_{ a } , J^{ a } ( x ) ] | I \rangle ,
    [/tex]
    or, if we define the momentum transfere 4-vector by [itex]q_{ a } = p_{ a }( F ) - p_{ a }( I )[/itex], we find
    [tex]q_{ a } \langle F | J^{ a } (0) | I \rangle = 0 .[/tex]
    This equation is an example of a “Ward-Takahashi” identity, a relation that must be satisfied by the matrix element of any operator that possesses some conservation property. Relations of this type play a vital role in proving the renormalizability of a theory.

    The only Mechanism I know of is Mathematics. We can SHOW that Noether charge has those properties.

    Yes. [itex]( x_{ i } , p_{ i } )[/itex] form 2n-dimensional Lie algebra called Heisenberg/ Poisson algebra.

    Sam
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Algebra and physics
  1. On current algebras (Replies: 1)

  2. Hamiltonian algebras (Replies: 2)

Loading...