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Light-cone gauge quantisation of point particle

  1. Apr 8, 2009 #1
    Dear all,

    I'm reading Polchinski's text of string theory. In section 1.3, he demonstrates how to quantize the free point particle in the light-cone gauge. I'm confused with a step in the follows.
    Begin with the action,

    [tex]S = \frac{1}{2}\int d\tau\left(\eta^{-1}\dot{X}^\mu\dot{X}_\mu - \eta m^2\right)[/tex]

    Choose the light-cone gauge

    [tex] X^+(\tau) = \tau [/tex]

    Then the action becomes,

    [tex] S' = \frac{1}{2} \int d\tau \left(-2\eta^{-1}\dot{X}^- + \eta^{-1}\dot{X}^i\dot{X}^i - \eta m^2 \right)[/tex]

    Thus, the Hamiltonian is

    [tex] H = p_-\dot{X}^- + p_i\dot{X}^i - L
    = \frac{p^ip^i+m^2}{2p^+} [/tex]

    I can follow these till now, but he says later which I don't understand how he does that:
    "The remaining momentum component [tex]p_+[/tex] is determined in terms of the others as follows. The gauge choice relates [tex]\tau[/tex] and [tex]X^+[/tex] translations, so [tex]H=-p_+ = p^-[/tex]. The relative sign between [tex]H[/tex] and [tex]p_+[/tex] arises because the former is active, and the later passive."

    Q1: How does he get this relation [tex]H=-p_+ = p^-[/tex]?

    Q2: Why [tex]H[/tex] is active and [tex]p_+[/tex] is passive? What does he mean by active and passive?

    Thanks very much for any instructions!
  2. jcsd
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