# Physical meaning of momentum without current in Quantum Mechanics

In Quantum Mechanics it is possible to have momentum at some point without having any current, that can be seen for example in the Hydrogen atom n=1 level using the Schrödinger equation.

My concern is, what is the physical meaning of this momentum which is not related to any movement of mass?

• What do you mean by “current”? The momentum is indeed the meaning you know which is the movement of mass etc. A particle can‘t ever have zero momentum, i.e. It can never be at rest, if it can be at rest, then it will be against the uncertainty principle. – Gradient137 Nov 18 '18 at 10:11
• Perhaps you are looking at the wrong state: m =0 excludes any "rotating waves", the origin of a magnetic moments. The probability density just sits there, and so is the phase, the quantity really controlled by the current density. – Cosmas Zachos Nov 18 '18 at 15:59
• – Cosmas Zachos Nov 18 '18 at 16:08
• Cosmas Zachos, I choosed n=1 to avoid any "rotating wave", because this status has the same phase on the space for a given time, why you say the phase is controlled by the current density? – Sergio Prats Nov 18 '18 at 23:00
• Indeed, only the real part of the momentum density indicates group motion. For the ground state of the Hydrogen atom, $\psi(x)\sim \exp (-r)$, the momentum density is a pure imaginary radial function, $\psi^* \vec P \psi \propto i \hat r \rho$, leading to zero current. For excited states, you might get a nontrivial phase S, and so a (barely) nontrivial current. Madelung rules! – Cosmas Zachos Nov 19 '18 at 15:22

• @SergioPrats ; The current over the probability density is basically the local group velocity density of the state. For a radially symmetric stationary state, that group velocity vanishes, but for an $m\neq 0$ there is constant angular phase rotation. – Cosmas Zachos Nov 18 '18 at 15:43