Why is a wave function $\psi$ needed for QM? Is it possible to make a differential equation involving just the p.d.f. $|\psi|^2$ of a particle?

Question

Why do you need a wave function $\psi$ for quantum mechanics? Can't you just make a differential equation involving just the p.d.f. $|\psi|^2$ of a particle? Since basically with quantum mechanics the point is, given $\psi(x,0)$, to find $\psi(x,t)$ with the help of the Schrödinger equation

$$i\hbar\frac{\partial}{\partial t} \Psi(x,t) = \left [ - \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x,t)\right ] \Psi(x,t).$$

Is it possible to write a differential equation involving just:

$$ \rho = |\Psi|^2$$

Answer 1

Consider a wavepacket wavefunction describing a particle moving in free-space: $$ \psi(\mathbf{x})=Ae^{-|\mathbf{x}-\mathbf{x_0}|^2/2\sigma^2}e^{-i\mathbf{k}\cdot \mathbf{x}} $$ It describes a particle centered at $\mathbf{x}_0$ with a momentum expectation value of $\hbar\mathbf{k}$. Over time, this wavepacket will move with velocity $\hbar\mathbf{k}/m$ (and it will spread out). The probability density, however, is not dependent on $\mathbf{k}$ in any way: $$ |\psi|^2(\mathbf{x})=A^2e^{-|\mathbf{x}-\mathbf{x_0}|^2/\sigma^2} $$ So there can be no differential equation which depends on only $|\psi|^2$ and describes the motion of this wavepacket, because $|\psi|^2$ is not aware of the particle's momentum. In other words, $|\psi|^2$ does not encode the 2d information (real and imaginary part) necessary to fully describe the state of the particle.

I'm not all that well versed in the history and underlying justification of the Schrodinger equation, but I believe there is some aspect that people were aware that everyday objects have a position and a momentum. And equations of motion are typically equations for the acceleration of an object in terms of its velocity and position. Therefore, in order to have two independent aspects to the state of the wavefunction, the wavefunction would need two outputs (real and imaginary part), and the Schrodinger equation would need to be a wave equation unlike, for example, the heat equation, which describes the isotropic spread of only one quantity.

Answer 2

Let me first note that QM can be formulated without wave functions at all - e.g., in matrix form or in terms of path integrals.

Probability amplitudes are used to incorporate interference - without them we would be dealing with a *classical * probabilistic model, described by something similar to Fokker-Planck equation. This would mean that the uncertainty is simply due to our ignorance, impossibility of observing all the specimen in the population, rather then due to some inherent different between classical and quantum laws of physics, exemplified by experimental facts cited in the introductory chapters of any QM textbook.

To return to my first sentence: we could formulate a version of QM without wave function, in terms of some probability function... then question is then how we would incorporate physics in it - I am not saying it is impossible, but it is a question of form, not the essence.