Does the quantum evolution equation has to be a complex first order equation?

Question

I was reading the Schrodinger's famous 1926 paper, and he introduces his Schrodinger's equation as a real second-order in time equation (I modified it a bit) $$ \partial_t^2 \psi = - \hat{H}^2 \psi,\tag{1} $$ where $\hat{H}$ is the usual Hamiltonian $\hat{H}=\frac{\hat{p}^2}{2 m}+V(r)$. The complex first-order version of this equation is rather a trick, used to find the real solution (notice, solution to $i\partial \psi=\hat{H}\psi$ is also a solution to Eq. 1).

When modeling quantum systems, we always use the first order complex equation, because it is simpler, and the Born rule looks nice there. But can be that the actual quantum equation is the second order real equation and all the complex stuff is just a mathematical trick? Notice, we can always make a second order real equation out of the first order complex Shroedinger equation, but the reverse does not seem to be necessarily true. Are we missing anything by sticking with the first order complex equation, similar to how one would lack understanding of open quantum systems, if avoided dealing with density matrices and worked only with the wave functions?

Answer 1

The problem of equation (1) is the difficulty of introducing probability. In regular quantum mechanics one of the ways to get probability is from the symmetry of the effective Lagrangian $$ L=i\psi^{\dagger}\partial_t \psi-\psi^{\dagger}\hat{H}\psi. $$ Here $\psi=(\psi_1,\psi_2,....)$ and $\psi^{\dagger}=(\psi^*_1,\psi^*_2,....)$, where $\psi^*_1, \psi^*_2,... $ are new auxiliary variables, independent of $\psi_1,\psi_2,...$. Under the symmetry $\psi\rightarrow\psi e^{i\alpha}$, $\psi^{\dagger}\rightarrow\psi^{\dagger} e^{-i\alpha}$ such Lagrangian generates a conserved quantity

$$Q=\psi^\dagger\psi,$$ which we associate with probability.

Doing a similar trick with the Eq. (1) yields a very different result. The Lagrangian in this case is $$ L=\partial_t\psi^{\dagger}\partial_t \psi-\psi^{\dagger}\hat{H}^2\psi. $$ The corresponding conserved quantity due to the phase symmetry is $$ Q=i(\psi^{\dagger}\partial_t \psi-\partial_t \psi^{\dagger} \psi), \tag{2} $$ which can be positive or negative. What is worse, for all real $\psi$ it is zero. In other words, $Q$ cannot be probability!

In the comments to the OP there was a suggestion that since the equation of motion (1) corresponds to a collection of coupled oscillators, may be the "energy" of such system (kinetic+potential), $$ Q=\partial_t \psi^{\dagger}\partial_t\psi+ \psi^{\dagger}\hat{H}^2\psi, $$ could be associated with probability. The problem with this attempt is that if Hamiltonian $\hat{H}$ depends on time, the "energy" (and therefore the corresponding probability) will not be conserved, which, in turn, would make non-local evolution of probability possible.