Quantum mechanics: Probability current density in terms of velocity vs. in terms of continuity equation

Question

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.

Answer 1

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.

Answer 2

Consider an arbitrary quantum system whose state evolves under the Schrödinger equation $$ i\frac{d}{dt}|\psi\rangle=\hat H|\psi\rangle.\tag{1}\label{eq1} $$ Let $\hat{\mathbf X}\triangleq\lbrace\hat X_1,\dots, \hat X_n \rbrace$ be any complete set of compatible observables. Let $\lbrace\mathbf x\rbrace$ be the corresponding spectrum and $\lbrace|\mathbf x\rangle\rbrace$ be the corresponding basis of eigenstates. In general, we can define the following probability density and current over the quantum numbers $\mathbf x$ [1]: \begin{align} \rho(\mathbf x,t)&\triangleq |\langle\mathbf x|\psi(t)\rangle|^2,\tag{2a}\label{eq2a}\\ j(\mathbf x,\mathbf x',t)&\triangleq 2\Im\left\lbrace\langle\psi(t)|\mathbf x\rangle\langle \mathbf x|\hat H|\mathbf x'\rangle\langle\mathbf x'|\psi(t)\rangle\right\rbrace.\tag{2b}\label{eq2b} \end{align} These can be shown to satisfy the following continuity equation as a consequence of eq. \eqref{eq1}: $$ \frac{\partial\rho(\mathbf x,t)}{\partial t}=\int d^n\mathbf x'\,j(\mathbf x, \mathbf x', t).\tag{3}\label{eq3} $$ The interpretation is that $\rho(\mathbf x,t)$ is the probability that the outcome $\mathbf x$ is obtained if a measurement of $\hat{\mathbf X}$ is performed at time $t$, while $j(\mathbf x, \mathbf x', t)=-j(\mathbf x', \mathbf x, t)$ is the rate at which that probability flows from $\mathbf x'$ to $\mathbf x$ if the system is left to evolve.

OP's question involves the special case in which the Hamiltonian is of the form $$ \hat H=\frac{1}{2}\hat{\mathbf P}\cdot\hat{\mathbf P}+V(\hat{\mathbf X}),\tag{4}\label{eq4} $$ where $\hat{\mathbf P}\triangleq\lbrace \hat P_1,\dots,\hat P_n\rbrace$, is the conjugate of $\hat{\mathbf X}$ (i.e. $[\hat P_i,\hat P_j]=0, [\hat X_i,\hat P_j]=i\delta_{ij}$). In this special case the integral in \eqref{eq3} reduces to $$ \int d^n\mathbf x'\,j(\mathbf x, \mathbf x', t)=-\frac{\partial}{\partial \mathbf x}\cdot\Im\left\lbrace\psi(\mathbf x,t)^*\frac{\partial}{\partial \mathbf x}\psi(\mathbf x,t)\right\rbrace,\tag{5}\label{eq5} $$ where $\psi(\mathbf x,t)\triangleq\langle\mathbf x|\psi(t)\rangle$. In this case the continuity equation reduces to the local form that OP seems to be expecting: $$ \frac{\partial\rho}{\partial t}+\frac{\partial}{\partial\mathbf x}\cdot\mathbf j=0,\tag{6}\label{eq6} $$ where $$ \mathbf j(\mathbf x,t)\triangleq \Im\left\lbrace\psi(\mathbf x,t)^*\frac{\partial}{\partial \mathbf x}\psi(\mathbf x,t)\right\rbrace.\tag{7}\label{eq7} $$ This expression for the current is equivalent to OP's eq. (1). Note that it is only valid for special types of Hamiltonians. The general definition is \eqref{eq2b}.

Side note: the definitions for the currents in eqs. \eqref{eq2b}, \eqref{eq7} are not the only possibilities. In the case of eq. \eqref{eq7}, $\mathbf j$ can be regarded as an $(n-1)$-differential form, and any closed $(n-1)$-form can be added to it without invalidating the continuity equation.

[1] There may be a more standard reference out there, but I learned \eqref{eq2b} by reading about Pilot Wave theory from J. Bell, Beables for quantum field theory, in Speakable and unspeakable in quantum mechanics, Cambridge University Press, 1987.