Here we will for simplicity only consider the Schrödinger system. We will assume that
$$\phi~=~(\phi^1+i\phi^2)/\sqrt{2} \tag{A}$$
is a bosonic complex field, and that
$$\phi^*~=~(\phi^1-i\phi^2)/\sqrt{2} \tag{B} $$
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).]
- The Lagrangian density
$${\cal L}~:=~ i\phi^{*}\dot{\phi} + \frac{1}{2m} \phi^* \nabla^2\phi \tag{C} $$
for the Schrödinger field $\phi$ is already on the Hamiltonian form
$${\cal L}~=~ \pi\dot{\phi} - {\cal H}. \tag{D} $$
Simply define complex momentum
$$\pi~:=~i \phi^{\ast}, \tag{E} $$
and Hamiltonian density
$${\cal H}~:=~-\frac{1}{2m} \phi^{\ast} \nabla^2\phi. \tag{F} $$
More generally, this identification is a simple example of the Faddeev-Jackiw method.
- Recall that the Euler-Lagrange equations do not change by adding a $4$-divergence $d_{\mu}\Lambda^{\mu}$ to the Lagrangian density
$${\cal L} ~~\longrightarrow~~ {\cal L}^{\prime}~:=~{\cal L} + d_{\mu}\Lambda^{\mu},\tag{G}$$
cf. e.g. this Phys.SE post.
[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 parts) choose an equivalent Hamiltonian density
$$\begin{align}{\cal H} ~~\longrightarrow~~ {\cal H}^{\prime}~:=~&\frac{1}{2m}|\nabla\phi|^2\cr ~=~&\frac{1}{4m}(\nabla\phi^1)^2 +\frac{1}{4m}(\nabla\phi^2)^2,\end{align}\tag{H} $$
and we can (via temporal integrations by part) choose an equivalent kinetic term
$$\begin{align} i\phi^*\dot{\phi}~=~ \pi\dot{\phi} ~~\longrightarrow~&~ \frac{1}{2}(\pi\dot{\phi}-\phi\dot{\pi})\cr
~=~& \frac{i}{2}(\phi^*\dot{\phi}-\phi\dot{\phi}^*)\cr
~=~&\frac{1}{2}(\phi^2\dot{\phi}^1-\phi^1\dot{\phi}^2)\cr
~~\longrightarrow~&~\phi^2\dot{\phi}^1. \end{align}\tag{I} $$
The last expression shows (in accordance with the Faddeev-Jackiw method) that
$$ \text{The second component }\phi^2 \\ \text{ is the momenta for the first component }\phi^1.
\tag{J}$$
- Alternatively, we can perform a Dirac-Bergmann analysis$^1$ directly. Consider for instance the Lagrangian density
$${\cal L}^{\prime}~=~ (\alpha+\frac{1}{2})\phi^2\dot{\phi}^1+(\alpha-\frac{1}{2})\phi^1\dot{\phi}^2 - {\cal H}^{\prime},\tag{K} $$
where $\alpha$ is an arbitrary real number. [The term $d(\phi^1\phi^2)/ dt$, which is multiplied by $\alpha$ in ${\cal L}^{\prime}$, is a total time derivative.] Let us check that the quantization procedure does not depend on this parameter $\alpha$. We introduce canonical Poisson brackets
$$\begin{align} \{\phi^a({\bf x},t),\phi^b({\bf y},t)\}_{PB} ~=~&0, \cr
\{\phi^a({\bf x},t),\pi_b({\bf y},t)\}_{PB}
~=~&\delta^a_b ~ \delta^3 ({\bf x}-{\bf y}), \cr
\{\pi_a({\bf x},t),\pi_b({\bf y},t)\}_{PB} ~=~&0,\end{align} \tag{L}$$
in the standard way. The canonical momenta $\pi_a$ are defined as
$$\begin{align} \pi_1~:=~&\frac{\partial {\cal L}^{\prime}}{\partial \dot{\phi}^1}
~=~(\alpha+\frac{1}{2})\phi^2,\cr \pi_2~:=~&\frac{\partial {\cal L}^{\prime}}{\partial \dot{\phi}^2}
~=~(\alpha-\frac{1}{2})\phi^1.\end{align}\tag{M}$$
These two definitions produce two primary constraints
$$\begin{align}\chi_1~:=~&\pi_1-(\alpha+\frac{1}{2})\phi^2~\approx~0,\cr
\chi_2~:=~&\pi_2-(\alpha-\frac{1}{2})\phi^1~\approx~0,\end{align}\tag{N}$$
where the $\approx$ sign means equal modulo constraints. The two constraints are of second-class, because
$$ \{\chi_2({\bf x},t),\chi_1({\bf y},t)\}_{PB}~=~\delta^3 ({\bf x}-{\bf y})~\neq~0. \tag{O} $$
Thus the Poisson bracket should be replaced by the Dirac bracket. [There are no secondary constraints, because
$$\begin{align} \dot{\chi}_a({\bf x},t) ~=~&\{\chi_a({\bf x},t), H^{\prime}(t)\}_{DB} ~=~ 0, \cr
H^{\prime}(t)~:=~& \int d^3y \ {\cal H}^{\prime}({\bf y},t),\end{align} \tag{P} $$
are automatically satisfied.] The Dirac bracket between the two $\phi^a$'s is
$$\{\phi^1({\bf x},t),\phi^2({\bf y},t)\}_{DB}~=~\delta^3 ({\bf x}-{\bf y}), \tag{Q}$$
leading to the same conclusion (J) as the Faddeev-Jackiw method.
Note that the eqs. (O) and (Q) are independent of the parameter $\alpha$.
- In all cases, the canonical equal-time commutator relations for the corresponding operators become
$$\begin{align} [\hat{\phi}^1({\bf x},t), \hat{\phi}^2({\bf y},t)] ~=~& i\hbar {\bf 1}~\delta^3 ({\bf x}-{\bf y}), \cr [\hat{\phi}({\bf x},t), \hat{\phi}^{\dagger}({\bf y},t)] ~=~& \hbar {\bf 1}~\delta^3 ({\bf x}-{\bf y}), \cr
[\hat{\phi}({\bf x},t), \hat{\pi}({\bf y},t)] ~=~& i\hbar {\bf 1}~\delta^3 ({\bf x}-{\bf y}).\end{align} \tag{R}$$
--
$^1$ See, e.g., M. Henneaux and C. Teitelboim, Quantization of Gauge Systems, 1992.
