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?
