# Spin up and four spinor

Is there a connection between the Dirac four spinor and "spin up", i.e one of the four spinor states is spin up or are these two separate unconected things.

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Yes there is a connection - depending on how you write the spinor, one of the four components can be interpreted as the up-component of the electron, one as the up-component of the positron (the other two are the down components).

Yes there is a connection - depending on how you write the spinor, one of the four components can be interpreted as the up-component of the electron, one as the up-component of the positron (the other two are the down 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.

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.
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?

Bill_K
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.
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, Ψ1, Ψ2 and put this into the Dirac Equation, you find they are coupled together:

(E + eφ)Ψ1 = c σ·p Ψ2
(E + eφ + 2mc22 = c σ·p Ψ1
where E is kinetic plus potential energy (relativistic energy minus mc2) 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.

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, Ψ1, Ψ2 and put this into the Dirac Equation, you find they are coupled together:

(E + eφ)Ψ1 = c σ·p Ψ2
(E + eφ + 2mc22 = c σ·p Ψ1
where E is kinetic plus potential energy (relativistic energy minus mc2) 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.
Does this mean one can not associate a "spin up" with any one of the spinors?
Is there any interpretaion of the two-spinor Ψ1?

Bill_K
Does this mean one can not associate a "spin up" with any one of the spinors? Is there any interpretaion of the two-spinor Ψ1?
Yes, in the nonrelativistic limit. In the second equation, (E + eφ + 2mc22 = c σ·p Ψ1, the rest energy mc2 is the largest energy, so we can approximate

2mc2Ψ2 = c σ·p Ψ1 or Ψ2 = (σ·p/2mc) Ψ1

This shows that Ψ2 << Ψ1 in this limit. Consequently Ψ1 is called the "large" component and Ψ2 the "small" component. Furthermore, we can eliminate Ψ2 in the first equation, getting an (approximate) second-order equation:

E Ψ1 = (σ·p σ·p/2m + eφ) Ψ1

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