A Different Hamiltonian

String Theory without the Baggage, Episode 6.

Hey Friends!

Yes. I know. We’re still discussing the relativistic point particle. We’re going to go all the way through BRST quantization of it. That will make our discussion of strings so much easier. I promise.

Sean

The first snow of the year has arrived!

Let's revisit the Action for a relativistic point particle¹:

\(S = -mc \int_{\gamma} ds.\)

As we've seen, s is a dummy variable - a parametrization of the worldline, which we can expand as

\(S = -mc \int_{\gamma} ds\, \sqrt{\eta_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}},\)

where now

\(\dot{x}^{\mu} = \frac{dx^{\mu}}{ds}.\)

Let us consider a framework that will freak out some canonical² thinkers. Let us flout concerns about convexity³ and consider the naïve Legendre transform of the four-velocity:

\(p_{\mu} = -mc \frac{d}{d\dot{x}^{\mu}}\sqrt{\eta_{\alpha\beta}\dot{x}^{\alpha}\dot{x}^{\beta}} = -mc\frac{\eta_{\mu\nu}\dot{x}^{\nu}}{\sqrt{\eta_{\alpha\beta}\dot{x}^{\alpha}\dot{x}^{\beta}}}.\)

In this context then the Hamiltonian should be

\(H = p_{\mu}\dot{x}^{\mu} - L,\)

but notice that

\(p_{\mu}\dot{x}^{\mu} = -mc\frac{\eta_{\mu\nu}\dot{x^{\mu}}\dot{x}^{\nu}}{2\sqrt{\eta_{\alpha\beta}\dot{x}^{\alpha}\dot{x}^{\beta}}} = L.\)

Hence, from this perspective, H should vanish. Clearly something is wrong with this manifestly Lorentz covariant Legendre transform.

The problem involves the mass-shell constraint. To properly compute H, let us supplement L with the mass-shell constraint using a Lagrange multiplier 𝜆

\(\begin{equation}L = p_{\mu}\dot{x}^{\mu} - \frac{\lambda}{2} \left(\eta^{\mu\nu}p_{\mu}p_{\nu} +m^{2}c^{4}\right).\end{equation}\)

Variation of L with respect to 𝜆, having no kinetic terms, essentially enforce the constraint. In this framework, the Hamiltonian (which still vanishes on-shell) is given by

\(\begin{equation} H = \frac{\lambda}{2} \left(\eta^{\mu\nu}p_{\mu}p_{\nu} +m^{2}c^{4}\right).\end{equation}\)

Working instead with the so-called “first order” form of L, given in above, we can integrate out p_𝜇 (which has no s derivatives), so that the Euler-Lagrange equation for p_𝜇 reads:

\(\frac{\partial L}{\partial p_{\mu}} = \dot{x}^{\mu} - \lambda p_{\mu}= 0,\)

so that we can identify the canonical momentum with the constraints employed⁴ as

\(p_{\mu} = \frac{1}{\lambda}\eta_{\mu\nu}\dot{x}^{\nu},\)

which allows us to represent the Lagrangian L purely in terms of the x^𝜇
variables:

\(\begin{equation}L = \frac{1}{2\lambda} \eta_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu} - \frac{\lambda}{2}m^{2}c^{4}.\end{equation}\)

As pointed out in GSW, the nice thing about this “second order” form of the constrained Lagrangian is that it allows us to apply it to the case of massless particles as well⁵. As an exercise, you can compute Hamiltonian based on this second order form and verify that it returns H given above.

In summary, the price we pay for using the four-vector Legendre transform of dx/ds, rather than the explicit dx/dt variable is a Hamiltonian that vanishes on-shell and a new, auxiliary field 𝜆.

Note that these are different Hamiltonians because we are using different coordinates. For the canonical pendants out there, we saw last time that a full symplectic analysis of the actual dynamical coordinates x yields

\(H = \frac{mc^{2}}{\sqrt{1-|\dot{\mathbf{x}}/c|^{2}}}, \)

which is the nothing more than the 0^th component of the energy-momentum four-vector p_𝜇.

Next time we’ll explore the technical implications of - and the associated headaches with - switching to this different Hamiltonian.

1. Notice the issue with units here. One copy of c got absorbed implicitly into the definition of s.

2. For our mathematics readers, “canonical” here refers to symplectic, rather than the generic sort of “standard” typically implied in Mathematics. The thinkers I have in mind are the kind of folks who generate lists of QFT textbooks that “correctly” deal with canonical variables or whatever, and those that don’t. As if it matters.

3. Typically we consider the Legendre transform of specific variables. Doing so with respect to a four-vector implicitly suggests doing so component-wise, which is fine for the spatial components. In this interpretation, there is a specific convexity issue around the dt/ds component. Of course, it is the mass-shell constraint which keeps things consistent, however implicitly.

4. This sort of double Legendre transformation is the price we must pay for specifying the constraints in terms of the variable p_𝜇.

5. The massless free particle - something like a photon, for example - sits in a slightly different representation of the Lorentz / Poincaré group. These vectors are isotropic, meaning that they’re "norm” is zero, despite affording any positive value of energy.