Lagrangian of Schrodinger field

The usual Schrodinger Lagrangian is

$$\tag 1 i(\psi^{*}\partial_{t}\psi ) + \frac{1}{2m} \psi^{*}(\nabla^2)\psi,$$ which gives the correct equations of motion, with conjugate momentum for $\psi^{*}$ vanishing. This Lagrangian density is not real but differs from a real Lagrangian density $$\tag 2 \frac{i}{2}(\psi^{*}\partial_{t}\psi -\psi \partial_{t}\psi^{*} ) + \frac{1}{2m} \psi^{*}(\nabla^2)\psi$$ by a total derivative.

My trouble is that these two lagrangian densities lead to different conjugate momenta and hence when setting equal time commutation relations, I am getting different results (a factor of 2 is causing the problem). I can rescale the fields but then my Hamiltonian also changes. Then applying Heisenberg equation of motion, I don't get the operator Schrodinger equation.

Is it possible to work with the real Lagrangian density and somehow get the correct commutation relations? I would have expected two Lagrangians differing by total derivative terms to give identical commutation relations (since canonical transformations preserve them). But perhaps I am making some very simple error. Unless all conjugate momenta are equivalent for two Lagrangians differing by total derivatives, how does one choose the correct one?

I guess the same thing happens for other first order systems like Dirac Lagrangian also.