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The Klein-Gordon Equation

We noted difficulties trying to work with:. The Klein-Gordon Equation. or if you prefer:. stemming from the fact that its 2 nd order in t. In 1-dim , might try factoring p 2 - m 2 c 2 = 0 ( p + mc )( p - mc ) = 0.  two equations , 1 st order in t. but can’t add a

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The Klein-Gordon Equation

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  1. We noted difficulties trying to work with: The Klein-Gordon Equation or if you prefer: stemming from the fact that its 2nd order in t

  2. In 1-dim, might try factoringp2-m2c2= 0 (p+ mc)(p - mc) = 0 two equations, 1storderin t but can’t add a vector and scalar! If eitherbkpk+mc = 0orgl pl-mc = 0, theKlein Gordon Equation is satisfied, and the relativistic energy-momentum relation automatically holds!

  3. are all equal!

  4. p1, p2, p3 operators do commute! (It’s angular momentum operators which don’t!) For this factoring exercise to work, need (g 0)2 = 1 (g 1)2= -1 (g 0 g 1 + g 1 g 0 ) = 0

  5. For this factoring exercise to work, need (g 0)2 = 1 (g 1)2= -1 (g 0 g 1 + g 1 g 0 ) = 0 but if you naively chooseg 0 = 1 andg 1 = ithen (g 0 g 1 + g 1 g 0 ) = i More generally these g quantities must satisfy (g 0)2 =I(g i)2=-I {g m,g n } = (g mg n + g ng m) = 2gmn(anti-commute!) they are all UNITARY: g m†gm = g mgm†= I but not HERMITIAN: g 0g mg0= gm†

  6. Some g properties: (g 0)2 =I(g i)2=-Ifor i=1,2,3 {g m,g n } = (g mg n + g ng m) = 2gmn(anti-commute!) they are all UNITARY: g m†gm = g mgm†= I but not HERMITIAN: g 0g mg0= gm† since: g 0g mg0gm = g 0(2gm 0-g 0gm)gm =2gm 0 g 0 gm- g 0g 0gmgm =2(1)(g 0 )2- (g 0 )2(g 0 )2 = 2 -1 ? for m = 0 1 =2(-1)g 0 gm- (g 0 )2(g m)2 = 0--1 ? for m = 0 g 0g mg0=(g 0)-1 (g0gmg0)gm =1 which means:

  7. Diracdiscovered this could only be resolved withmatrices of at least order 4×4 0 si -si 0 I 0 0-I g0 = gi = 1 0 0 0 0 1 0 0 0 0 -1 0 00 0 -1 0 0 0 1 0 0 1 0 0 -1 0 0 -1 0 0 0 0 0 0 -i 0 0 +i 0 0 +i 0 0 -i 0 0 0 0 0 1 0 0 0 0 -1 -10 0 0 0 1 0 0 g0 = g1 = g2 = g3 =

  8. With  now a 4-element column matrix, for example with the basis: 1 2 3 4 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 1 +2 +3 +4  = = and Dirac’s equation is now a matrix equation a set of 4 simultaneous equations Warning: Although the columns above carry four components they are NOT 4-vectors! They are “Dirac spinors” or “bi-spinors.” does not act on them directly. 

  9. 1 0 0 0 0 1 0 0 0 0 -1 0 00 0 -1 0 si -si 0 I 0 0-I g0 = gi = g0 = The block diagonal form suggests it may sometimes be simpler to work with the “reduced” notation of 0 0 0 1 0 0 1 0 0 -1 0 0 -1 0 0 0 g1 = A 0 0 0 -i 0 0 +i 0 0 +i 0 0 -i 0 0 0  = g2 = B 0 0 1 0 0 0 0 -1 -10 0 0 0 1 0 0 1 3 g3 = A B = = where 2 4

  10. i ħ E c i ħ i ħ i ħ Recall that a “free particle” has, in general, a PLANE WAVE solution: e-Ete-p·r evolution of the time-dependent part solution to the space-dependent part e-[(ct)-pjrj ] = e-pmxm

  11. Assume a form (free particle of4-momentum pm) u a 4-component “Dirac spinor” carrying any needed normalization factors If we can find u’s that satisfy this, then the  above will be a solution to the Dirac equation

  12. Now, note that So our equation looks like: a two component vector (of 2 component vectors)

  13. and

  14. So returning to: we must have and ? which, notice together give:

  15. and We can start picking uA’s and solve for uB’s and/or uB’s and solve for uA’s We need 4 linearly independent solutions, right? An obvious starting point: What’s wrong with the simpler basis:

  16. What do these components mean? Let’s look at them in the limit where p  0

  17. in the limit where p  0 for Emc2 for E-mc2

  18. These ARE eigenvectors of with “spin” +1/2 -1/2 +1/2 -1/2

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