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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?

Sergio Prats
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  • 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
  • Related, and Madelung background reading. – 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
  • What I see is that, while the current is always real, the momentum obtained for a wave function that has same phase everywhere is purely imaginary since it is -i(W)(dW/dx) where W is the wavefunction and W its conjugated. So maybe only the "real" momentum implies current and hence movement. – Sergio Prats Nov 18 '18 at 23:06
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    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

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I think that the apparent puzzle originates, at least in part, from our conceptual difficulty to abandon the macroscopic-world identification between momentum and velocity.

In classical mechanics, velocity is a derived concept, justified and rooted on the existence of a continuous trajectory, something which is well defined at the macroscopic scale. In the quantum description there is nothing allowing to justify the idea of a smooth trajectory. Thus, no velocity, at least according to the classical mechanics definition. Instead, in QM, one can still introduce a measurable quantity, the momentum of a particle, for example through a quantitative analysis of scattering experiments.

Taking into account the previous remark, let's analyze the situation of a hydrogen atom with its electron in the 1s state. On the one hand, no doubt that the local probability current must be zero everywhere, since the 1s state can always be described by a real function. In physical terms this is related to the absence of an electric current in the 1s state. On the other hand, a measurement of momentum may give any value according to a distribution probability provided by the squared modulus of the 3D Fourier transform of the 1s wave-function.

The two facts are not in contradiction: the vanishing of the local current is telling that in average, taking an ensemble of H atoms all in their 1s state, there is no electron displacement or circulation in a well defined direction. However, in a single measurement, nothing forbids the possibility of finding any value for the momentum.

GiorgioP-DoomsdayClockIsAt-90
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  • There is a point that looks unclear to me: the average of any quantity implies integration through the space while the current is zero everywhere meaning there is no internal movement of mass. In other hydrogen atom states there are circular currents that average zero movement although it exists a current). While mathematically I can understand that the momentum should not be zero everywhere as it is the 3D Fourier transform of the wave in the space, I see unclear the physical meaning of this momentum that does not mean movement of energy/mass. – Sergio Prats Nov 18 '18 at 15:31
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    @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
  • @SergioPrats ; the momentum is not the 3D Fourier transform of the wave in the space. This Fourier tranform, provides the density probabiliy of momentum. A value of momentum where this probability density is not zero can appear in a single experiment, even if the average of the vector momentum will turn to be zero. – GiorgioP-DoomsdayClockIsAt-90 Nov 18 '18 at 17:35