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.

  • $\begingroup$ I think if you want to write down how the probability density changes in time, you basically get the continuity equation. As commented in the answer below, $\rho$ does not give all information of the quantum state. $\endgroup$
    – Zheng Liu
    Commented Dec 7, 2017 at 7:42
  • $\begingroup$ @Zheng Liu I'm not so worried not having all information in $\psi$ if you do not need it to find solutions for $\rho$. But even so, following AFT's response you can express the complex phase of $\psi$ in $\rho$ though it is a functional form and cumbersome. So all information in the quantum state can still be found if you want it. $\endgroup$
    – Jan Bos
    Commented Dec 9, 2017 at 2:19
  • $\begingroup$ Right. It can be recovered if you know the complex phase. $\endgroup$
    – Zheng Liu
    Commented Dec 14, 2017 at 3:42

3 Answers 3


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.

  • $\begingroup$ Shouldn't the answer then be along the lines of "Yes you can but the equation involves a complicated functional of $\rho$ and is not practical to use". In fact, the 2nd part of your answer seems to contradict the first part since you showed that the degrees of freedom are coupled albeit in a complicated way. It is interesting that the relative simple probability density of say a 1s electron in hydrogen is a solution of this very tedious equation. $\endgroup$
    – Jan Bos
    Commented Dec 7, 2017 at 12:55
  • $\begingroup$ Appreciate the link to the Quantum Hamilton-Jacobi Equation (the 2nd differential equation in the couples set). There seems to be some work done on that but it seems not straightforward. $\endgroup$
    – Jan Bos
    Commented Dec 7, 2017 at 14:22
  • 2
    $\begingroup$ any books on this? It's more mathematically cumbersome but I like this so much better conceptually. $\endgroup$
    – Mike Flynn
    Commented Feb 8, 2018 at 5:26
  • $\begingroup$ @MikeFlynn Bohm wrote several books himself, so you should definitely check them out. I've heard they are quite good (even by those that dislike Bohm's interpretation). $\endgroup$ Commented Feb 8, 2018 at 14:51
  • 1
    $\begingroup$ It seems there are missing $\hbar$ here? See this answer: physics.stackexchange.com/a/812408/127307 $\endgroup$
    – user56834
    Commented Apr 29 at 19:17

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.

  • $\begingroup$ You can write an equation for $\rho$ and $J$ (the probability current) though. $\endgroup$
    – Mauricio
    Commented Dec 7, 2017 at 9:59
  • $\begingroup$ Mauricio Yes, but $\mathbf{j}$ is $\psi$-dependent $\endgroup$
    – J.G.
    Commented Dec 7, 2017 at 10:01
  • $\begingroup$ Yeah, but if I remember correctly you can solve certain scattering introductory problems using continuity equation only and without using $\psi$. $\endgroup$
    – Mauricio
    Commented Dec 7, 2017 at 10:06
  • $\begingroup$ @Mauricio If you come up with an example, you should probably mention it in an answer here, even if it's only "a long comment". $\endgroup$
    – J.G.
    Commented Dec 7, 2017 at 12:44
  • 1
    $\begingroup$ Your answer does not prove that such an equation does not exist. It just gives an example of one that does not work. Per my comment under my question in response to Zheng Liu I also am not convinced on your statement that there is more information in $\psi$ than in $\rho$. You probably could say at most that the state at point $x = x_1$ contains more information than $\rho$ at the single point $x = x_1$. $\endgroup$
    – Jan Bos
    Commented Dec 9, 2017 at 2:27

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.

  • $\begingroup$ Thanks for your reference to the quantum analogue of the Liouville equation. It looks like something I was looking for. It must be related to the equation in the response of AccidentalFourierTransform. About your 1st paragraph I'm not so worried about it. On a higher level I was wondering if QM can be done without talking about states. Let's say you take all the probability densities $\rho_{nlm}$ of the hydrogen atom there must be some way to connect $m$ to the angular momentum without reference to states but that could be another question. $\endgroup$
    – Jan Bos
    Commented Dec 8, 2017 at 1:00

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