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Poincare Algebra from Poisson Bracket with KG Action

  1. Sep 14, 2012 #1
    1. The problem statement, all variables and given/known data
    Consider the Klein-Gordan action. Show that the Noether charges of the Poincare Group generate the Poincare Algebra in the Poisson brackets. There will be 10 generators.


    2. Relevant equations
    [tex]
    \{ A,B \}=\frac{\delta A}{\delta \phi}\frac{\delta B}{\delta \pi}-\frac{\delta A}{\delta \pi}\frac{\delta B}{\delta \phi}
    [/tex]
    [tex]
    j_{a}^{\mu}=\frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)}A_{a}^{\nu}\partial_\nu \phi - A_{a}^{\mu}\mathcal{L}
    [/tex]
    [tex]
    \pi = \frac{\partial \mathcal{L}}{\partial (\partial_0 \phi)}
    [/tex]
    [tex]
    \Box \phi - m^2 \phi=0
    [/tex]
    [tex]
    Q_a = \int d^3 x j^{0}_{a}
    [/tex]


    3. The attempt at a solution
    Starting with the action
    [tex]
    \mathcal{L}=\frac{1}{2}(\partial_\mu \phi \partial^\mu \phi -m^2 \phi^2)
    [/tex]
    for the conserved currents for both translations and lorentz transformations I obtain:
    [tex]
    p^\mu = \int d^3 x (\partial^0 \phi \partial^\mu \phi - g^{0\mu}\mathcal{L})
    [/tex]
    [tex]
    m^{\alpha\beta}=\int d^3 x (\theta^{0\alpha}x^\beta - \theta^{0\beta}x^\alpha)
    [/tex]
    with
    [tex]
    \theta^{\alpha\beta}=\partial^\alpha \phi \partial^\beta \phi - g^{\alpha\beta}\mathcal{L}
    [/tex]
    The for [itex]\{ p^\mu ,p^\nu \}[/itex] I get
    [tex]
    \frac{\delta p^\mu}{\delta \phi}=g^{0\mu}m^2 \phi \quad\text{and}\quad \frac{\delta p^\mu}{\delta \pi} = \partial^\mu \phi - g^{0\mu}\partial^0 \phi
    [/tex]
    which gives
    [tex]
    \begin{align}
    \{ p^\mu ,p^\nu \} &= \int d^4 x (g^{0\mu}m^2 \phi)(\partial^\nu \phi - g^{0\nu}\partial^0 \phi)-(\partial^\mu \phi - g^{0\mu}\partial^0 \phi)(g^{0\nu}m^2 \phi) \\
    &= \int d^4 x m^2(g^{0\mu} \phi \partial^\nu \phi - \partial^\mu \phi g^{0\nu}\phi) \\
    &= \int d^4 x m^2 g^{0\mu}\frac{1}{2} \partial^\nu (\phi^2) - \int d^4 x m^2 g^{0\nu}\frac{1}{2} \partial^\mu (\phi^2) =0
    \end{align}
    [/tex]
    since the field vanishes on the boundary. Now if I have done this right... to compute [itex]\{ p^\mu , m^{\alpha\beta} \}[/itex] I can borrow the momentum stuff, and for the Ms I get
    [tex]
    \frac{\delta m^{\alpha\beta}}{\delta \pi}=(\partial^\alpha \phi - g^{0\alpha}\partial^0 \phi)x^\beta - (\partial^\beta \phi - g^{0\beta}\partial^0 \phi)x^\alpha
    [/tex]
    and
    [tex]
    \frac{\delta m^{\alpha}}{\delta \phi}=g^{0\alpha}m^2 \phi x^\beta - g^{0\beta}m^2 \phi x^\alpha
    [/tex]
    Now at this point I have tried to compute the poisson bracket of those above guys, however I cannot seem to retrieve the appropriate form... I know I need to get a superpostion of momenta but I can't seem to get the momentum to come back out. The things I have tried range from integration by parts of various terms, to substituting in the the EOM for the [itex]m^2[/itex] term and trying to integrate by parts from that. But no luck as of yet. Could someone give me a solid push in how to manipulate these guys into the correct end form, thanks.
     
    Last edited: Sep 14, 2012
  2. jcsd
  3. Sep 15, 2012 #2
    just looking at the term [itex]\{ p^0 , m^{0i} \} [/itex] I get the following
    [tex]
    \begin{align}
    &=\int d^3 x m^2 \phi (-\partial^i \phi x^0) -0 \\
    &= -\int d^3 x (m^2 \phi)\partial^i \phi x^0 = -\int d^3 x (\partial_\lambda \partial^\lambda \phi ) \partial^i \phi x^0 \\
    &= \int d^3 x \partial^\lambda \phi \partial_\lambda (\partial^i \phi x^0) \\
    &= \int d^3 x \partial^\lambda \phi (\partial_\lambda \partial^i \phi)x^0 + \int d^3 x \partial^0 \phi \partial^i \phi \\
    &= \text{something}+p^i
    \end{align}
    [/tex]
    which is the desired result except for the "something". Can someone point out why that first term is zero, which it should be. Thanks.
     
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