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Using the first of these, then a single equation for can be obtained. Since , the equation is only satisfied if the quantity in the brackets vanishes, which yields the eigenvalues.

We see that the eigenvalues can be positive or negative. A plot of the energy levels is shown below:. There is a continuum for turquoise and for periwinkle. There is also a gap between and. We will show that for , an appropriate solution is to take. Note that when , the third and fourth components of vanish. Authorization Required.

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Perhaps after quantizing? If so how do they differ? EDIT: To put what I'm saying in other words: Is the field theory of a Dirac spinor obeying the Dirac equation equivalent to the field theory of a left handed spinor obeying the KG equation? The Dirac equation is more restrictive than the Klein-Gordon equation.

L4.4 Dirac equation for the electron and hydrogen Hamiltonian

For every solution to the Dirac equation, its components will be a solution of the Klein-Gordon equation, but the converse isn't true: if you form a spinor whose components are solutions of the Klein-Gordon equation, it might not solve the Dirac equation. But, what happens if we plug these solutions in the original Dirac equation?

The Dirac Equation and Its Solutions pdf

If you consider only the Klein-Gordon equation, you're introducing extra "solutions" that don't really solve the Dirac equation. Why does this happen? You can regard the Klein-Gordon equation as the "squared" version of the Dirac equation. And when you square an equation, you always get this nasty false solutions: if you have e.

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The first equation implies the second, but the converse isn't true. Sign up to join this community. The best answers are voted up and rise to the top. Home Questions Tags Users Unanswered.

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Is the Dirac equation equivalent to the Klein-Gordon equation for its left handed component? Ask Question.

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