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:
−ℏ22m∇2ψ(r)+V(r)ψ(r)=Eψ(r)
Any solution will yield a probability density p(r)=ψ∗(r)ψ(r) and the question if an equation can be found of which p is the solution if ψ 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 ψ 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
No, you can't.
The function ψ∈C has two real degrees of freedom; they are coupled and dynamical (non-gauge). On the other hand, the function ρ∈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 ψ=√ρeiS, with ρ,S a pair of real variables. You may write the Schrödinger equation directly in terms of ρ,S as (cf. Madelung or Bohm) ∂√ρ∂t=−12m(√ρ∇2S+2∇√ρ⋅∇S)∂S∂t=−(|∇S|22m+V−ℏ22m∇2√ρ√ρ)
As you can see, you cannot write an equation for ρ 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 ρ, and plug the result into the equation for ρ, thus obtaining an equation for ρ alone. This is impractical because it is not really possible to solve for S=S[ρ] in general terms, and even if we could, the functional would be highly non-local so the resulting equation for ρ would be impossible to work with. The Schrödinger equation, written in terms of ψ, even if complicated, is as simple as it gets. Any other reformulation is way more cumbersome to use.
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