Here we will for simplicity only consider the Schrödinger system. We will assume that
$$\phi~=~(\phi^1+i\phi^2)/\sqrt{2}$$
is a bosonic complex field, and that
$$\phi^*~=~(\phi^1-i\phi^2)/\sqrt{2}$$
is the complex conjugate, where $\phi^a$ are the two real component fields, $a=1,2$. [Note the change in notation $\psi\longrightarrow\phi$ as compared with the OP's question(v1).]
1) The Lagrangian density
$${\cal L}~:=~ i\phi^{*}\dot{\phi} + \frac{1}{2m} \phi^* \nabla^2\phi$$
for the Schrödinger field $\phi$ is already on the Hamiltonian form
$${\cal L}~=~ \pi\dot{\phi} - {\cal H}. $$
Simply define complex momentum
$$\pi~:=~i \phi^{*}, $$
and Hamiltonian density
$${\cal H}~:=~-\frac{1}{2m} \phi^* \nabla^2\phi.$$
More generally, this identification is a simple example of the Faddeev-Jackiw method.
2) Recall that the classical equations of motion do not change by adding a $4$-divergence $d_{\mu}\Lambda^{\mu}$ to the Lagrangian density
$${\cal L} ~~\longrightarrow~~ {\cal L}'~:=~{\cal L} + d_{\mu}\Lambda^{\mu}.$$
[We use the symbol $d_{\mu}$ (rather than $\partial_{\mu}$) to stress the fact that the derivative $d_{\mu}$ is a total derivative, which involves both implicit differentiation through the field variables $\phi^a(x)$, and explicit differentiation wrt. $x^{\mu}$.]
Therefore, we can (via spatial integration by part) choose an equivalent Hamiltonian density
$${\cal H} ~~\longrightarrow~~ {\cal H}'~:=~\frac{1}{2m}|\nabla\phi|^2 ~=~\frac{1}{4m}(\nabla\phi^1)^2 +\frac{1}{4m}(\nabla\phi^2)^2,$$
and we can (via temporal integrations by part) choose an equivalent kinetic term
$$i\phi^*\dot{\phi}~=~ \pi\dot{\phi} ~~\longrightarrow~~$$
$$\frac{1}{2}(\pi\dot{\phi}-\phi\dot{\pi}) ~=~ \frac{i}{2}(\phi^*\dot{\phi}-\phi\dot{\phi}^*)~=~\frac{1}{2}(\phi^2\dot{\phi}^1-\phi^1\dot{\phi}^2)~~\longrightarrow~~\phi^2\dot{\phi}^1. $$
The last expression shows (in accordance with the Faddeev-Jackiw method) that
The second component $\phi^2$ is the momenta for the first component $\phi^1$.
$\qquad$(1)
3) Alternatively, we can perform a Dirac-Bergmann analysis$^1$ directly. Consider for instance the Lagrangian density
$${\cal L}'~=~ (\alpha+\frac{1}{2})\phi^2\dot{\phi}^1+(\alpha-\frac{1}{2})\phi^1\dot{\phi}^2 - {\cal H}', $$
where $\alpha$ is an arbitrary real number. [The term $d(\phi^1\phi^2)/ dt$, which is multiplied by $\alpha$ in ${\cal L}'$, is a total time derivative.] Let us check that the quantization procedure does not depend on this parameter $\alpha$. We introduce canonical Poisson brackets
$$ \{\phi^a({\bf x},t),\phi^b({\bf y},t)\}_{PB} ~=~0, $$
$$ \{\phi^a({\bf x},t),\pi_b({\bf y},t)\}_{PB}
~=~\delta^a_b ~ \delta^3 ({\bf x}-{\bf y}), $$
$$ \{\pi_a({\bf x},t),\pi_b({\bf y},t)\}_{PB} ~=~0, $$
in the standard way. The canonical momenta $\pi_a$ are defined as
$$ \pi_1~:=~\frac{\partial {\cal L}'}{\partial \dot{\phi}_1}~=~(\alpha+\frac{1}{2})\phi^2,$$
$$ \pi_2~:=~\frac{\partial {\cal L}'}{\partial \dot{\phi}_2}~=~(\alpha-\frac{1}{2})\phi^1.$$
These two definitions produce two primary constraints
$$\chi_1~:=~\pi_1-(\alpha+\frac{1}{2})\phi^2~\approx~0,$$
$$\chi_2~:=~\pi_2-(\alpha-\frac{1}{2})\phi^1~\approx~0,$$
where the $\approx$ sign means equal modulo constraints. The two constraints are of
second-class, because
$$ \{\chi_2({\bf x},t),\chi_1({\bf y},t)\}_{PB}~=~\delta^3 ({\bf x}-{\bf y})~\neq~0. \qquad(2) $$
Thus the Poisson bracket should be replaced by the Dirac bracket. [There are no secondary constraints, because
$$ \dot{\chi}_a({\bf x},t) ~=~\{\chi_a({\bf x},t), H'(t)\}_{DB} ~=~ 0,
\qquad H'(t)~:=~ \int d^3y \ {\cal H}'({\bf y},t), $$
are automatically satisfied.] The Dirac bracket between the two $\phi^a$'s is
$$\{\phi^1({\bf x},t),\phi^2({\bf y},t)\}_{DB}~=~\delta^3 ({\bf x}-{\bf y}), \qquad(3)$$
leading to the same conclusion (1) as the Faddeev-Jackiw method.
Note that the eqs. (2) and (3) are independent of the parameter $\alpha$.
4) In all cases, the canonical equal-time commutator relations for the corresponding operators become
$$ [\hat{\phi}^1({\bf x},t), \hat{\phi}^2({\bf y},t)] ~=~ i\delta^3 ({\bf x}-{\bf y}), $$
$$ [\hat{\phi}({\bf x},t), \hat{\phi}^*({\bf y},t)] ~=~ \delta^3 ({\bf x}-{\bf y}), $$
$$ [\hat{\phi}({\bf x},t), \hat{\pi}({\bf y},t)] ~=~ i\delta^3 ({\bf x}-{\bf y}). $$
--
$^1$: See, e.g., M. Henneaux and C. Teitelboim, Quantization of Gauge Systems, 1992.
