The following is taken from pages 83-84 of Tom Hartman's lectures on Quantum Gravity and Black Holes.

See related questions here and here.

Consider a one-dimensional system with

  1. a single degree of freedom $q(t)$ and
  2. a conjugate momentum $\frac{\partial L}{\partial \dot{q}}$.

The action for the system is \begin{align}I = \int dt\ L = \int dt\ \left[ \frac{\partial L}{\partial \dot{q}} \dot{q}-H(p,q)\right].\end{align}

We can introduce a fake time-reparameterization symmetry by

  1. introducing a fake degree of freedom $t(\tau)$ and
  2. imposing the constraint $H(p,q) = -\frac{\partial L}{\partial t'},$

where $'$ denotes differentiation with respect to $\tau$.

Then, we find that the action becomes

$$I = \int d\tau \left[ \frac{\partial L}{\partial q'} q' + \frac{\partial L}{\partial t'} t' - N\left(H(p,q)+\frac{\partial L}{\partial t'}\right)\right],$$

where $N$ is a Lagrangian multiplier.

Then, the new Hamiltonian of the system is

$$H' = N \left( H(p,q)+\frac{\partial L}{\partial t'} \right).$$

The Hamiltonian $H'$, of course, vanishes on-shell, but I do not see why this is because $H'$ generates $\tau$-translations.

  • $\begingroup$ The question formulation (v1) seems to mix Lagrangian and Hamiltonian formalisms. Tom Hartman's lectures p. 83-84 only discuss the Hamiltonian formalism. $\endgroup$
    – Qmechanic
    Jun 25, 2017 at 11:48

1 Answer 1

  1. It is non-standard to call the constraints "the Hamiltonian". Generically, one would rather write a generic constrained Hamiltonian action as $$ S_\text{Ham} = \int (p_i \dot{q}^i - \lambda^i \chi_i(p,q) - H(p,q))\mathrm{d}t,\tag{1}$$ where the $\lambda^i$ are Lagrange multipliers enforcing the constraints $\chi_i \approx 0$ and $H$ is the what we actually call "the Hamiltonian". $H' = \lambda^i \chi_i - H$ is the extended Hamiltonian.

  2. Your $H'$ consists entirely of constraints. It's therefore trivial that it vanishes on-shell since the very definition of "on-shell" includes the constraints being satisfied. That one may view the constraints as the generators of gauge transformations, and that the $\tau$-transformations are such gauge transformations generated by $H'$ because it's just a constraint, are secondary concerns that are not really necessary to see why it vanishes on-shell.

  3. The non-trivial part here is not why the constrained part of the extended Hamiltonian $H'$ vanishes, it's why there is no $H$ part in it! This is because for the action (1) to be well-defined, $H(p,q)$ has to transform as a scalar density under $t$-transformations. But if $p,q$ transform as scalars (i.e. are invariant) under such transformation, as it is the case here and always in the case where one just artificially made the "actual" time variable a phase space variable, then every function of them is a scalar, not a scalar density, too. Therefore $H=0$ in time-reparametrization invariant systems.

Altogether, the reasoning "$H'\approx 0$ because it generates $\tau$-transformations" is circular - it generates $\tau$-transformations because it consists purely of constraints, which by definition vanish on-shell, but the fact you actually have to explain is why there's only constraints in it.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.