# Light-cone gauge quantisation of point particle

1. Apr 8, 2009

### ismaili

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,

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

Choose the light-cone gauge

$$X^+(\tau) = \tau$$

Then the action becomes,

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

Thus, the Hamiltonian is

$$H = p_-\dot{X}^- + p_i\dot{X}^i - L = \frac{p^ip^i+m^2}{2p^+}$$

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

Q1: How does he get this relation $$H=-p_+ = p^-$$?

Q2: Why $$H$$ is active and $$p_+$$ is passive? What does he mean by active and passive?

Thanks very much for any instructions!