For simplicity, consider a one-electron system. Some sources tell that the probability current density can be written in terms of the velocity operator $\mathbf{v} = -i[\mathbf{r}, H]$ as $$ \mathbf{j} = \frac{1}{2}(\psi^\ast \mathbf{v} \psi + (\mathbf{v}\psi)^\ast \psi). \quad\quad(1)$$ I try to reconcile this with the continuity equation. Using the Schrödinger equation, one can obtain $$ \frac{\partial \rho}{\partial t} = \frac{\partial}{\partial t}(\psi^\ast \psi) = -i[\psi^\ast H\psi -(H\psi)^\ast \psi] \stackrel{!}{=}-\nabla \cdot \mathbf{j}\quad\quad(2)$$ Combining these two equations would mean that $$ \frac{1}{2} \nabla \cdot (\psi^\ast[\mathbf{r}, H]\psi + ([\mathbf{r},H]\psi)^\ast \psi)= -[\psi^\ast H \psi - (H\psi)^\ast \psi].\quad\quad(3)$$

I have tried a lot to show explicitly that both sides of this equation are identical but have failed so far. I would be happy if someone knows how to do this, or can hint me to some additional assumption that is needed for this equation to be true.

Note that for specific forms of the Hamiltonian, e.g. a nonrelativistic electron in an electrostatic external potential (where only the momentum operator contributes to the velocity) or also including a static magnetic field (where also the vector potential contributes to the velocity), I am able to show the equality of both sides of Eq. (3). But I am interested in a general proof.

  • $\begingroup$ If you write $H=p^2/2m + V(\vec{r})$ you can find the form of the velocity operator explicitly. Position commutes with the potential. Then it should be doable. $\endgroup$ – doublefelix Mar 18 at 20:47
  • $\begingroup$ As I said, I am able to do the derivation for specific forms of the Hamiltonian, like the nonrelativistic one you wrote down. However, I am looking for a general proof. The reason is that the derivation of the current density from the continuity equation is quite cumbersome, while the commutator of the Hamiltonian and position is easily calculated. If I knew that both lead to the same results, I would always follow the latter approach. $\endgroup$ – LLang Mar 19 at 6:08
  • $\begingroup$ Sorry, didn't realize you meant even more general than that. It is an interesting question whether this is possible to show. Maybe it could be shown for a hamiltonian which is an arbitrary function of the position and momentum operators (written as a taylor series in powers of those variables). I think that's as general as it gets, until fields are quantized. $\endgroup$ – doublefelix Mar 19 at 15:21
  • $\begingroup$ That is an interesting idea. I assume that it is also possible to add operators that depend on spin, since it commutes with position. Having a look at operators from the Breit-Pauli Hamiltonian (en.wikipedia.org/wiki/Breit_equation), which are possible candidates to be added to the nonrelativistic Hamiltonian, it seems they can all be written in terms of these basic ingredients (position, momentum, spin). $\endgroup$ – LLang Mar 20 at 7:44

I normally approach this from a different angle, but I hope that it may give some insight. It seems like you want to show that the Schrödinger equation leads to a meaningful probability current, but I would start with probability and show that it requires unitary time evolution satisfying the conditions of Stone's theorem, and from there derive the Schrödinger equation. Then it must be clear that the Schrödinger equation is a statement of probability and must give a conserved probability current.

For a normalised state $|f\rangle$, the probability density $\rho(x) $at time $t$ is $$ \rho(x)=\langle f| x \rangle \langle x| f \rangle = |f(x)|^2 , $$ where $$\int d^3x \rho(x) = \int d^3x\langle f| x \rangle \langle x| f \rangle = \int d^3x\langle f| f \rangle = 1.$$ The evolution operator is constrained by unitarity to ensure that probability is conserved. Consequently, it obeys a conservation law with the form $$ \frac{\partial}{\partial t}\rho + \nabla . \mathbf{j}=0 $$ where $\mathbf{j}$ is the probability current. The expected flow of probability must correspond to a flow of mass, that is it relates to momentum, $$\mathbf{P}= - \int d^3x |x\rangle i\nabla \langle x|.$$ The Hermitian operator with the required property is the current density operator, $$\mathbf{J}(x)= - \frac{i}{2m} |x\rangle (\nabla - \overleftarrow \nabla) \langle x|$$ where the arrow means that differentiation acts to the left. I have used this notation to emphasise that $\mathbf J$ is a density at a point, which can get lost otherwise. This form is also used in some sources, and I think it should be clear that for the particular Hamiltonians you have considered it is equivalent to the form which you gave. I am not clear whether it is identically the same. If it is not, that may be the source of the problem. The form you have appears to have been taken from Ehrenfest's theorem, but as we are not dealing with an observable I am not sure if it is correctly applied.

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  • $\begingroup$ Dear Charles, if I am not mistaken, you also assume a specific form for the Hamiltonian. Namely, that the only operator that does not commute with position is the nonrelativistic kinetic energy. Otherwise, it is not the canonical momentum operator but the "velocity operator" that is relevant. My question is essentially: How can I show that the velocity operator defined as $\mathbf{v}= -[\mathbf{r},H]$ leads to a current that fulfills the continuity equation. Note that I am not sure if this is generally true. $\endgroup$ – LLang Mar 20 at 7:32
  • $\begingroup$ @Lang, it is true that my assumptions are different, because my work has been in mathematical foundations of relativistic quantum mechanics. This does place (relativistic) constraints on the Hamiltonian. It is implicit that non-relativistic Hamiltonians are derivable using approximation and semi-classical approximation (a potential is an expectation). It is not obvious what that means in all cases, as I only consider general derivations like the interacting Dirac equation. $\endgroup$ – Charles Francis Mar 20 at 7:52
  • $\begingroup$ That said, the current density I have given is clearly correct in qed, as its expectation is the classical current. Although related to momentum, it is actually a momentum density divided by m, which does correspond to a velocity (it is the form Bohmians use to define a velocity, although I do not accept their argument). As I suggested, the velocity operator you use seems very uncomfortable. I have previously not seen it used, as there is no meaningful velocity observable in qm. Everything depends on conjugacy of momentum and position. I really don't know if it can work generally. $\endgroup$ – Charles Francis Mar 20 at 8:03

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