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

  1. Jan 5, 2014 #1
    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!

    *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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  3. Jan 6, 2014 #2

    edguy99

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    I believe your calculations are correct and I am also struck by the implications. I have not seen much discussion on the issue so it is hard to evaluate its meaning. Certainly the "Dirac Electron" is one example of discussing this, where one equation for a massive particle can be rewritten as two equations for two interacting massless particles, where the coupling constant of the interaction is the mass of the electron.

    The visual image you have constructed matches a trapped current (in blue rotating horizontally, with the axis shown in blue), trapping an orthogonal magnetic field (in red rotating vertically).

    electron_and_anti.jpg
     
  4. Jan 10, 2014 #3

    tom.stoer

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    I guess this can be resolved by treating the hydrogen atom as a two-particle system.

    Unfortunately this is often ignored, but in non.-rel. QM there's a nice procedure via unitary transformations to map a system S with canonical variables r1, r2, p1, p2 and H[p1,p2,r1-r2] to a new system S' with, r,R,p,P and H'[p,P,r]. Note that H' does not depend on R, therefore [H',P] = 0. The system S' still lives in a two-particle Hilbert space, but usually (in many textbooks) this information gets lost. Therefore they only deal with one-particle states, i.e. |nlm>, neglecting the important information that there is another degree of freedom, namely the canonical pair R,P. The complete quantum state including the total momentum reads |nlm,K> and b/c [H',P] = 0 we have conservation of P which means plane waves exp(iKX).

    For the binding energy only |nlm> is relevant, but the total spectrum is Enlm + K2/2M.

    Now I would guess that something similar might by relevant for other operators. So it could very well be that there are additional contributions to certain operators sensitive to the two-particle nature of the problem. Therefore I would suggest to check what happens to the two-particle operators for spin, current etc. when one applies the above mentioned transformation.
     
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