Form of Schrödinger equation for the probability density

Question

Is it possible to formulate the Schrödinger equation (SE) in terms of a differential equation involving only the probability density instead of the wave function? If not, why not?

We can take the time independent SE as an example:

$$-\frac{\hbar ^{2}}{2m}\nabla ^{2}\psi (\mathbf {r} )+V(\mathbf {r} )\psi (\mathbf {r} )=E\psi (\mathbf {r} )$$

Any solution will yield a probability density $p(\mathbf {r}) = \psi^*(\mathbf {r})\psi(\mathbf {r})$ and the question if an equation can be found of which $p$ is the solution if $\psi$ is a solution of the SE.

I assume not since it would have been widely known but I have not seen the arguments why this would be impossible. I understand the wave function contains more information than the probability density (e.g. the phase of $\psi$ which is relevant in QM drops out of $p$) but I do not see that as sufficient reason against the existence of such an equation.

Answer 1

No, you can't.

The function $\psi\in\mathbb C$ has two real degrees of freedom; they are coupled and dynamical (non-gauge). On the other hand, the function $\rho\in\mathbb R$ has one real degree of freedom. It is impossible to reduce the dynamics of the system from two variables to one variable without losing information in the process.

(But, in a formal sense: Yes, you can)

Let $\psi=\sqrt{\rho}\mathrm e^{iS}$, with $\rho,S$ a pair of real variables. You may write the Schrödinger equation directly in terms of $\rho,S$ as (cf. Madelung or Bohm) \begin{equation} \begin{aligned} \frac{\partial\sqrt{\rho}}{\partial t}&=-\frac{1}{2m}\left(\sqrt{\rho}\nabla^2S+2\nabla\sqrt{\rho}\cdot\nabla S\right)\\ \frac{\partial S}{\partial t}&=-\left(\frac{|\nabla S|^2}{2m}+V-\frac{\hbar^2}{2m}\frac{\nabla^2\sqrt{\rho}}{\sqrt{\rho}}\right) \end{aligned} \end{equation}

As you can see, you cannot write an equation for $\rho$ alone, because its equation is coupled to a second unknown, $S$. Two real degrees of freedom, not one. Formally speaking, you may solve the equation for $S$ as a functional of $\rho$, and plug the result into the equation for $\rho$, thus obtaining an equation for $\rho$ alone. This is impractical because it is not really possible to solve for $S=S[\rho]$ in general terms, and even if we could, the functional would be highly non-local so the resulting equation for $\rho$ would be impossible to work with. The Schrödinger equation, written in terms of $\psi$, even if complicated, is as simple as it gets. Any other reformulation is way more cumbersome to use.

Answer 2

We have $\psi^\ast\nabla^2\psi=\dfrac{2m}{\hbar^2}(V-E)\rho$ so by complex conjugation $\psi\nabla^2\psi^\ast=\dfrac{2m}{\hbar^2}(V-E)\rho$. Hence $$\nabla^2 \rho=\psi\nabla^2\psi^\ast+\psi^\ast\nabla^2\psi+2\boldsymbol{\nabla}\psi^\ast\cdot\boldsymbol{\nabla}\psi=\dfrac{4m}{\hbar^2}(V-E)\rho+2\boldsymbol{\nabla}\psi^\ast\cdot\boldsymbol{\nabla}\psi.$$It's that last term that gets in the way. There's more quantum-mechanical information in $\psi$ than in $\rho$, so we can't in general rewrite everything in terms of $\rho$ alone.

Answer 3

The probability density isn't a great point of comparison, because it has absolutely no information about the momentum properties of the state.

This goes a bit further in that the correct classical point of comparison for any quantum-mechanical formalism isn't really a single-trajectory Newtonian perspective; instead, it is the Liouville mechanics of the phase-space density $\rho(x,p)$ of a particle which obeys classical hamiltonian mechanics but whose state is only known down to a probability distribution on phase space, and whose density then obeys the Liouville equation $$ \frac{\partial\rho}{\partial t}=-\{\,\rho,H\,\}. $$

Once you do that, then there is a quantum analogue of the Liouville equation, given in this answer by Qmechanic, where you need to change the standard function multiplications for a $\hbar$-dependent Moyal product; the dynamical equation then reads $$ \frac{d\rho}{dt} = \frac{1}{i\hbar} [\rho\stackrel{\star}{,}H]. $$ I've never seen this used in anger, but that might just be because I've never looked at the places that do use it.