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- Thread starter galvin452
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Can't one use all of the Dirac four spinor in relativistic calculation of, e.g. the quantum states of the electron in hydrogen? Penrose in `The Road to Reality' (pg 629) specifically uses the four spinor to describe the electron only.

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Since the four spinor can be associated with only the electron, then if one associate a "spin up" state with one of the spinors what does one associate with the other three spinors?Can't one use all of the Dirac four spinor in relativistic calculation of, e.g. the quantum states of the electron in hydrogen? Penrose in `The Road to Reality' (pg 629) specifically uses the four spinor to describe the electron only.

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Bill_K

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The Dirac Equation has both positive frequency and negative frequency solutions. The negative frequency ones are sometimes interpreted as negative energy ("hole") states. But in the normal way of writing a four-spinor, the four solutions do not one-to-one correspond to the four components of the spinor. Each solution involves all four components.Can't one use all of the Dirac four spinor in relativistic calculation of, e.g. the quantum states of the electron in hydrogen? Penrose in `The Road to Reality' (pg 629) specifically uses the four spinor to describe the electron only.

If you write the four-spinor as a pair of two-spinors, Ψ

(E + eφ)Ψ

(E + eφ + 2mc

where E is kinetic plus potential energy (relativistic energy minus mc

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Does this mean one can not associate a "spin up" with any one of the spinors?The Dirac Equation has both positive frequency and negative frequency solutions. The negative frequency ones are sometimes interpreted as negative energy ("hole") states. But in the normal way of writing a four-spinor, the four solutions do not one-to-one correspond to the four components of the spinor. Each solution involves all four components.

If you write the four-spinor as a pair of two-spinors, Ψ_{1}, Ψ_{2}and put this into the Dirac Equation, you find they are coupled together:

(E + eφ)Ψ_{1}= cσ·pΨ_{2}

(E + eφ + 2mc^{2})Ψ_{2}= cσ·pΨ_{1}

where E is kinetic plus potential energy (relativistic energy minus mc^{2}) and φ is the electrostatic potential. Putting φ = e/r, you can solve this pair of equations to find the bound states of the hydrogen atom. As Penrose said, all four components of the spinor are involved in the solution.

Is there any interpretaion of the two-spinor Ψ

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Bill_K

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Yes, in the nonrelativistic limit. In the second equation, (E + eφ + 2mcDoes this mean one can not associate a "spin up" with any one of the spinors? Is there any interpretaion of the two-spinor Ψ_{1}?

2mc

This shows that Ψ

E Ψ

This is a direct generalization of the Schrodinger Equation for a particle having spin 1/2, and is known as the Schrodinger-Pauli Equation.

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