Lagrangian of Schrodinger field

Here we will for simplicity only consider the Schrödinger system. We will assume that

$$phi~=~(phi^1+iphi^2)/sqrt{2} tag{A}$$

is a bosonic complex field, and that

$$phi^*~=~(phi^1-iphi^2)/sqrt{2} tag{B} $$

is the complex conjugate, where $phi^a$ are the two real component fields, $a=1,2$. [Note the change in notation $psilongrightarrowphi$ as compared with the OP's question (v1).]

1. The Lagrangian density

$${cal L}~:=~ iphi^{*}dot{phi} + frac{1}{2m} phi^* nabla^2phi tag{C} $$

for the Schrödinger field $phi$ is already on the Hamiltonian form

$${cal L}~=~ pidot{phi} - {cal H}. tag{D} $$

Simply define complex momentum

$$pi~:=~i phi^{ast}, tag{E} $$

and Hamiltonian density

$${cal H}~:=~-frac{1}{2m} phi^{ast} nabla^2phi. tag{F} $$

More generally, this identification is a simple example of the Faddeev-Jackiw method.

2. 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

$$begin{align}{cal H} ~~longrightarrow~~ {cal H}^{prime}~:=~&frac{1}{2m}|nablaphi|^2cr ~=~&frac{1}{4m}(nablaphi^1)^2 +frac{1}{4m}(nablaphi^2)^2,end{align}tag{H} $$

and we can (via temporal integrations by part) choose an equivalent kinetic term

$$begin{align} iphi^*dot{phi}~=~ pidot{phi} ~~longrightarrow~&~ frac{1}{2}(pidot{phi}-phidot{pi})cr ~=~& frac{i}{2}(phi^*dot{phi}-phidot{phi}^*)cr ~=~&frac{1}{2}(phi^2dot{phi}^1-phi^1dot{phi}^2)cr ~~longrightarrow~&~phi^2dot{phi}^1. end{align}tag{I} $$

The last expression shows (in accordance with the Faddeev-Jackiw method) that

$$ text{The second component }phi^2 text{ is the momenta for the first component }phi^1. tag{J}$$

3. Alternatively, we can perform a Dirac-Bergmann analysis$^1$ directly. Consider for instance the Lagrangian density

$${cal L}^{prime}~=~ (alpha+frac{1}{2})phi^2dot{phi}^1+(alpha-frac{1}{2})phi^1dot{phi}^2 - {cal H}^{prime},tag{K} $$

where $alpha$ is an arbitrary real number. [The term $d(phi^1phi^2)/ dt$, 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 $alpha$.

4. In all cases, the canonical equal-time commutator relations for the corresponding operators become

$$begin{align} [hat{phi}^1({bf x},t), hat{phi}^2({bf y},t)] ~=~& ihbar {bf 1}~delta^3 ({bf x}-{bf y}), cr [hat{phi}({bf x},t), hat{phi}^{dagger}({bf y},t)] ~=~& hbar {bf 1}~delta^3 ({bf x}-{bf y}), cr [hat{phi}({bf x},t), hat{pi}({bf y},t)] ~=~& ihbar {bf 1}~delta^3 ({bf x}-{bf y}).end{align} tag{R}$$

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$^1$ See, e.g., M. Henneaux and C. Teitelboim, Quantization of Gauge Systems, 1992.