Electric current in quantum mechanics? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-08-17T13:11:42Z https://physics.stackexchange.com/feeds/question/95826 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/95826 6 Electric current in quantum mechanics? SuperCiocia https://physics.stackexchange.com/users/37677 2014-01-29T16:48:17Z 2019-05-01T20:50:46Z <p>Quick question:</p> <p>I want to find an expression for the (electric) current density of an electron, in quantum mechanics. Either a single electron or a general charge distribution $\rho$.</p> <p>Classically <strong>j</strong>=$\rho$ <strong>v</strong>.</p> <p>What should I use here?</p> <p>Maybe the electric charge multiplied by the probability current?</p> <p>Thanks.</p> https://physics.stackexchange.com/questions/95826/-/95834#95834 9 Answer by Luboš Motl for Electric current in quantum mechanics? Luboš Motl https://physics.stackexchange.com/users/1236 2014-01-29T17:18:08Z 2014-01-29T17:23:28Z <p>Yes, $\vec \jmath(x,y,z)$ should be defined as $e$ times the Schrodinger probability current. \begin{equation*} \vec \jmath = \frac{e\hbar}{2mi}\left(\Psi^* \frac{\partial \Psi }{\partial x}- \left(\frac{\partial \Psi^* }{\partial x}\right)\Psi \right) , \quad e\lt 0. \end{equation*} That's possible to explicitly see in the formalism of quantum field theory. The definition $\vec v(x,y,z)/\rho$ would be no good because "the velocity of the electron at a particular point $(x,y,z)$" isn't too well-defined due to the uncertainty principle (if the position is given, the velocity is not).</p> <p>One may be puzzled because the expression for $\vec\jmath$ above isn't an operator – it is quadratic in the wave function. But in quantum field theory, it <em>is</em> an operator – an observable – because it is a function of the field operators $\Psi$.</p> <p>If we consider non-relativistic quantum mechanics with fixed coordinates of particles and we still want to define $\vec\jmath(x,y,z)$ as a linear operator, an observable, we must appreciate that this operator is only nonzero is the particle is located in the infinitesimal vicinity of the point $(x,y,z)$. So we have $$\rho (x_0,y_0,z_0) = e \delta^{(3)}(\hat{\vec r}-\vec r_0)$$ and $$\vec\jmath (x_0,y_0,z_0) = \frac e2\{\delta^{(3)}(\hat{\vec r}-\vec r_0),\frac{\hat{\vec p}}{m} \}$$ I had to write one-half of the commutator with the velocity operator because functions of positions and velocities don't commute but we still need a Hermitian operator.</p> <p>If there are $N$ charged particles, the operators $\hat{\vec r}$ and $\hat{\vec p}$ acquire an extra index from $1$ to $N$ and $\rho(x_0,y_0,z_0)$ and $\vec\jmath(x_0,y_0,z_0)$ must be written as a sum of the expressions over this index.</p> <p>One may verify that e.g. for wave packets, the integrals over $\vec r_0$ (some regions) give us what we would expect.</p>