2
$\begingroup$

As a study case, consider the following Lagrangian for a left-handed Weyl field $\chi \in \mathbb{C}^{2}$:

$$\mathcal{L} = \chi^{\dagger} \mathrm{i} \overline{\sigma}^{\rho} \partial_{\rho} \chi$$

where $\overline{\sigma}^{\rho} \equiv (1_{2},-\sigma^{i})$, with $\sigma^{i}$ the standard Pauli matrices. The corresponding Hamiltonian (density) is obtained, as customarily, by a Legendre transformation [with $a$ running, of course, over the two components of $\chi$]:

$$\mathcal{H} = \mathcal{L}\frac{\overleftarrow{\partial }}{\partial \dot{\chi^{a}}}\dot{\chi^{a}}-\mathcal{L} = \mathrm{i} \chi_{a} ^{\dagger }\dot{\chi}^{a}-\mathcal{L},$$

correctly yielding a Hamiltonian containing no time-derivatives: $\mathcal{H} = -\chi^{\dagger} \mathrm{i} \overline{\sigma}^{i} \partial_{i} \chi.$ Now, consider the above Lagrangian augmented by the four-divergence $-\frac{\mathrm{i}}{2}\partial _{\rho }\left( \chi ^{\dagger } \overline{\sigma }^{\rho }\chi \right)$ to bring it on the following explicitly hermitian form:

$$\mathcal{L}'=\frac{\mathrm{i}}{2}\left[ \chi ^{\dagger }\overline{\sigma } ^{\rho }\partial _{\rho }\chi -\left( \partial _{\rho }\chi \right) ^{\dagger }\overline{\sigma }^{\rho }\chi \right];$$

and consider the following ansatz for the associated Hamiltonian:

$$\mathcal{H}' = \mathcal{L}'\frac{\overleftarrow{\partial }}{\partial \dot{\chi }^{a}}\dot{\chi}^{a}+\mathcal{L}'\frac{\overleftarrow{\partial }}{\partial \left( \dot{\chi}^{a}\right) ^{\ast }}\left( \dot{\chi}^{a}\right) ^{\ast }- \mathcal{L}' = \frac{\mathrm{i}}{2}\chi ^{\dagger }\dot{\chi}^{a}-\frac{\mathrm{i}}{2} \chi _{a}\left( \dot{\chi}^{a}\right) ^{\ast }-\mathcal{L}'.$$

Note that the minus sign of the second term comes from passing the derivative through from the right, using anticommutativity. Unlike the former case, here $\mathcal{H}'$ will contain no time-derivatives only if for the second term the relation $\frac{\mathrm{i}}{2} \chi _{a}\left( \dot{\chi}^{a}\right) ^{\ast } = -\frac{\mathrm{i}}{2} \left( \dot{\chi}^{a}\right) ^{\ast } \chi _{a}$ is applied.

At the classical level where the components of $\chi$ are Grassmann-valued, $\chi$ being a spinor field, such a relation is of course fully valid, but I would be more comfortable if the time-derivatives would readily drop out. Is that possible? And if affirmative, then how should the above ansatz for the Legrendre transformation be modified?

$\endgroup$
  • $\begingroup$ The case of Dirac spinors is sketched in 198054/84967. It might be useful to give it a read. $\endgroup$ – AccidentalFourierTransform Mar 11 '17 at 9:39
  • $\begingroup$ One should stress that the Hamiltonian is not guessed: there is no need to postulate an ansatz. There is a perfectly well-defined method to find the Hamiltonian of constrained systems (irrespective of whether the variables are even or odd). Google "Dirac-Bergmann algorithm". $\endgroup$ – AccidentalFourierTransform Mar 11 '17 at 9:58
  • $\begingroup$ Happy to hear that. My use of "ansatz" above was also meant in the rhetoric sense, as it was certainly my hope, and in fact expectation, that there exists some "well-defined method to find the Hamiltonian", as you put it. I have very little experience on these matters of quantization, and certainly so when dealing with constrained systems. Quite frustratingly, I generally feel lost, having never been properly trained on these matters during my education. There are all these different concepts as well as methods floating around, and I often feel I need some simple to follow 'recipes'. $\endgroup$ – John Fredsted Mar 11 '17 at 13:19
2
$\begingroup$
  1. The main point is that since e.g. the Lagrangian density ${\cal L}$ should be real, the complex Grassmann-odd Weyl spinors $\chi$ and $\chi^{\dagger}$ are not independent variables. See also this Phys.SE post.

  2. This leads to constraints. When the singular Legendre transformation is performed correctly, the Hamiltonian density ${\cal H}$ does not depend on velocity variables. See also this, this this, this and this related Phys.SE posts.

  3. For a bosonic analogue, see e.g. this Phys.SE post.

$\endgroup$
  • $\begingroup$ Thanks for your answer. I will try to study your provided links. For the next couple of hours, though, I will not be at my computer. $\endgroup$ – John Fredsted Mar 11 '17 at 9:58
  • $\begingroup$ I have looked through your links, one of which refers to a question of my own (somewhat embarrasingly, I am afraid). It would seem that there is no way around getting better acquainted with the (super) Dirac bracket which, as far I as understand it (please correct me if I am wrong), reduces to the (super) Poisson bracket only if the coordinates and generalized momenta are all independent. $\endgroup$ – John Fredsted Mar 11 '17 at 15:08

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.