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Why can't the Klein-Gordon equation with a Couloumb potential describe the hydrogen atom?

Why can the first order Dirac equation explain it? What are the failures?

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  • $\begingroup$ ...because it fails to take into account the electron's spin, the equation predicts the hydrogen atom's fine structure incorrectly, including overestimating the ... : Klein–Gordon equation $\endgroup$ – Frobenius Dec 25 '16 at 20:14
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Electrons are spin-1/2 particles (this qualifies them as fermions, but is not immediately relevant for this discussion).

KG equation is for spin-0 particles, whereas Dirac equation is for spin-1/2 particles. Therefore the proper equation to describe an electron is the latter.

Note that, however, the low-energy limit of both of them gives back Schrödinger equation, therefore in this approximation you can put the spin "by hand" and long as you need it, and this works both for fermions and for bosons.

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  • $\begingroup$ Also the field operator for electrons also obeys the Klein Gordon equation. It will give some informations on it but not all. $\endgroup$ – Slereah Dec 27 '16 at 9:03
  • $\begingroup$ It obeys KG equation (or better, it satisfies the mass-shell condition $P^2\psi=m^2\psi$) as you can see by taking the "square" of Dirac's differential operator, but the latter has a much richer structure due to the fact that it carries spin. $\endgroup$ – user139175 Dec 27 '16 at 9:06
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The reason is electron spin. The energy of the Dirac and Klein-Gordon solution differ precisely by the substitution l<->j. The Klein-Gordon equation needs to be extended with a term describing the interaction of electron spin with the nuclear potential. The resulting equation is known as the squared Dirac equation. See Itzykson and Zuber, Ch. 2.3, for a solution of the Dirac equation for the single electron case via the squared Dirac equation.

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The application of the Klein-Gordon equation for the hydrogen atom leads to logical contradictions. Allowance for relativism in the Klein-Gordon equation should lead to more accurate solutions for the hydrogen atom compared with the Schrodinger equation. However, on the contrary, solutions became increasingly worse with an increase in the nuclear charge. Finally, for the element Z = 69 the solution of the Klein-Gordon equation terminates. The Klein-Gordon equation has other drawbacks. Therefore, we can say that the Klein-Gordon equation is erroneous.

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  • $\begingroup$ Note that the naive Dirac equation also has no solution for the $S$ state when $Z$ becomes large enough. $\endgroup$ – WAH Feb 9 '18 at 15:14
  • $\begingroup$ The Dirac equation has solutions up to Z = 137. $\endgroup$ – Mareta Arakelyan Feb 9 '18 at 17:46
  • $\begingroup$ Are you sure? All the hydrogen derivations I've seen for the Dirac equation have imaginary energies for $S$ states once $Z\alpha=1/2$. See, for example, quantummechanics.ucsd.edu/ph130a/130_notes/node501.html or physics.sharif.edu/~qmech/puppel.pdf. $\endgroup$ – WAH Feb 10 '18 at 6:41
  • $\begingroup$ I'm not sure which formula you mean. I ask you to give a formula for calculating the energy of the ground state. And an example of calculating where the imaginary energy is obtained. $\endgroup$ – Mareta Arakelyan Feb 10 '18 at 15:24

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