We factored the Klein-Gordon equation into

1. We factored the Klein-Gordon equation into then found solutions for:

2. Free particle solution to Dirac’s equation (x) = ue-ixp/h u(p) cpz E-mc2 c(px+ipy) E-mc2 c(px-ipy) E-mc2 -cpz E-mc2 1 0 0 1 c(px-ipy) E+mc2 -cpz E+mc2 cpz E+mc2 c(px+ipy) E+mc2 1 0 1 0

3. What if we tried to solve: We would find 4 nearly identical Dirac spinors with the uA, uB (matter/antimatter entries) interchanged: E+mc2 E-mc2

4. In general, anyROTATION or LORENTZ Transformation mixes vector components: space-time coordinates not the spinor components! amn= sin, cos, 1, 0 forR = , , 1, 0 for If we want to preserve “lengths” and “distances”          

5. Now watch this: = I The transformation matrices must be ORTHOGONAL! So must mean

6. So must mean   = (a-1) (a-1)  

7. Finally chain rule (4 terms!) or

8. In general we can expect that any DIRAC SPINOR (x) when transformed by a  or R matrix: (x) '(x') ( its spinor components each a function of the space-time 4-vector (ct;r) ) is STILL expressible as a linear combination of the components in the initial (un“rotated”) basis: 4 space-time coordinates '(x') = S(x) column vector of the 4 spinor components How does the DIRAC EQUATION transform? Is IT invariant?

9. Multiple (both sides) through, from the left with: S-1 S-1 which is invariant provided

10. or † Warning! S is not unitary: Taking hermitian conjugate: † † † † † † Recall:g 0g mg 0= g m† Multiply both sides by:  0[ ] 0 =  0[ ] 0 † † since:g 0g 0= I † † † † obviously inverses!

11. † † ( ) (S 0S† 0) (S 0S† 0)   =  (S 0S† 0) whereS, 0, do not commute S 0S† 0= I  0S† 0= S-1

12. What will be preserved under  transformations? What are the invariant quantities? † (' )†' = (S)†S = †S†S = † † cannot be the probability density!  =† 0 Defining: as the “adjoint” spinor then notice: † † † † =†0 0S†0S S-1 (x) (x) =†0 = (x) (x) (x) (x)