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QFT and unitary Lorent representation

  1. Jul 16, 2006 #1
    Hey guys! A question:

    My QFT Lagrangian, fournishes through Noether's thm plus relativistic invariance a supposedly unitary and a supposed representation of the Lorentz group. These are the operators meant to act on my Hilbert space of possible states.

    What guarantees that this actually happens, ie: 1. The commutation relations of the lorentz algebra are actually satisfied, 2. The operators are hermitian (guarantiing a unitary representation). 3. They actually do what they are advertized to do ie: for example applying the z-direction angular momentum operator to a single particle state created by your creation operator does give what its suposed to. ie: the dirac field creates s=1/2, the scalar field s=0, etc..

    i find it amazing and non trivial if conditions 1-3 are generally satisfied just because of lorentz invariance plus the right commutation relations for the fields! (if true for those two reasons alone then it might motivate the supposedly put in by hand commutations relations!)

    maybe theres a good reference i could look at?


  2. jcsd
  3. Jul 18, 2006 #2


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    Last edited: Jul 18, 2006
  4. Jul 18, 2006 #3
    thanks! And i,ll take a look at the book.
  5. Jul 18, 2006 #4
    A very beautiful result..

  6. Jul 30, 2006 #5


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    Last edited: Jul 30, 2006
  7. Aug 2, 2006 #6
    Hey, I found a proof that the Noether charges do generate the symmetries of the theory whilst I was flicking through a book of mine (it's in a footnote Chp. 26 pg. 92 of Weinberg "Quantum Theory of Fields" Vol. III).

    Consider a Lagrangian (not density) [itex]L=L(q_n, \dot{q}_n)[/itex]. If the action has some symmetry [itex]q\rightarrow q+\delta q[/itex] then the Lagrangian will remain unchanged up to some time derivative of some suitable functional F (as when you integrate the perturbed Lagrangian over time you'll end up with [itex]F(t_1)-F(t_0)[/itex], and one can of course choose an F so that this vanishes, i.e. the action remains invariant, just as when one would usually choose the Lagrangian density to change by a four-divergence in relativistic field theory). Thus

    [tex]\delta L=\sum_n\frac{\partial L}{\partial\dot{q}^n}\delta\dot{q}^n+\sum_n\frac{\partial L}{\partial q^n}\delta q^n=\frac{d}{dt}F[/tex] (1)

    Then the charge Q ([itex]=\int d^3x j^0[/itex])associated with the conserved Noether current is

    [tex]Q=-\sum_n\frac{\partial L}{\partial\dot{q}^n}\delta q^n+F[/tex]

    Using the usual commutation relations

    [tex]\left[\frac{\partial L}{\partial\dot{q}^n}, q^m\right] =-i\delta^m_n[/tex]

    [tex]\left[q^n, q^m\right] =0[/tex]

    You get the commutator

    [tex]\left[Q, q^m\right]=-i\delta q^m-\sum_{nl}\frac{\partial L}{\partial\dot{q}^l}\frac{\partial\delta q^l}{\partial\dot{q}^n}\left[\dot{q}^n, q^m\right]+\sum_n\frac{\partial F}{\partial\dot{q}^n}\left[\dot{q}^n, q^m\right][/tex]

    From (1) we get (expand dF/dt and [itex]\delta\dot{q}[/itex] and equate coefficients of the second time derivatives of the qs)

    [tex]\sum_l\frac{\partial L}{\partial\dot{q}^l}\frac{\partial\delta q^l}{\partial\dot{q}^n}=\frac{\partial F}{\partial\dot{q}^n}[/tex]

    And thus

    [tex]\left[Q, q^m\right]=-i\delta q^m[/tex]


    [tex]\left[Q, \dot{q}^m\right]=-i\delta \dot{q}^m[/tex]

    Thus the Qs generate the symmetries of the theory. The above obviously generalises to field theory where index n of each configuration variable would refer to spin, species, spatial coordinates etc.
    Last edited: Aug 2, 2006
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