I have confounded myself with the following observation. Take the standard expression for the lowest energy Dirac 4-spinor solution of the Dirac equation with a Coulomb potential (the H atom ground state). Plug this into the standard expression for the 4-vector current and use values for the gamma matrices in spherical polar coordinates. In the r and θ directions the current is zero. No surprise there. But the [itex]\phi[/itex] component of current is non-zero (unless I've screwed up). Moreover, this current results in a magnetic dipole moment which can be calculated by multiplying by r.sin([itex]\theta[/itex]) and integrating over all space. The result is a magnetic moment of exactly one Bohr magneton. The reason why this result confuses me is that the ground state has (roughly speaking) an orbital angular momentum quantum number of zero* and so one would not expect there to be any orbital current or a magnetic moment associated with an orbital current. However, one Bohr magneton is the correct magnetic moment for the ground state - but it arises entirely from the electron spin. So I have tentatively concluded that the [itex]\phi[/itex] current is a manifestation of the spin distribution. Is this right? And is this current 'real'? Can a current truly arise from spin only when the electron 'cloud' is stationary? The total current is of prodigious magnitude, ~1 milliamp for a single electron!(adsbygoogle = window.adsbygoogle || []).push({});

*To be more precise, the ground state does not have a well defined value for the operator L^2 since it does not commute with the Dirac Hamiltonian. The upper 2-spinor has L = 0 and whilst the lower 2-spinor has L = 1 the lower2-spinor is very small in magnitude so the expectation value of the orbital angular momentum is tiny, about 10^-5 in h-bar units. This would align with a similarly tiny magnetic moment, which also confirms that the above magnetic moment is not due to orbital motion.

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# Does the Dirac ground state of the H atom have an electron current?

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