0))((

mcimci

We factored the Klein-Gordon equation into

then found solutions for:

0)( mci

Free particle solution to Dirac’s equation

(x) = ue-ixp/h

u(p)

1

0

cpz

E+mc2

c(px+ipy)E+mc2

0

1

c(pxipy)E+mc2

cpz

Emc2

1

0

cpz

Emc2

c(px+ipy)

Emc2

1

0

c(pxipy)Emc2

cpz

Emc2

0)( mci

What if we tried to solve:

We would find 4 nearly identical Dirac spinors with the uA, uB (matter/antimatter entries) interchanged:

E+mc2 Emc2

0))((

mcimci

xax In general, any ROTATION or LORENTZ Transformation mixes vector components:

33221100 xaxaxaxaxa

space-time coordinates

not the spinor components!

a = sin, cos, 1, 0 for R

= , , 1, 0 for

If we want to preserve “lengths” and “distances”

33221100 xxxxxxxxxxxx

xxaaxaxa ))(( aa

aa

Now watch this:

)( aa

aa

aa

t

t

1aat The transformation matrices must be ORTHOGONAL!

axx ' 'xax tSo must mean

= I

axx ' 'xax tSo must mean

xaaxa 1'

xxa '

xax ' (a1)(a1)

xxaa

xxa '

xax'

a

xx

x

xx

''chain rule (4 terms!)

xa

x

' x

ax

'

or

Finally

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:

'(x') = S(x)

4 space-time coordinates

column vector of the4 spinor components

How does the DIRAC EQUATION transform? Is IT invariant?

)()(02

xmc

xx

i

)()(02

xmc

xx

i

0)()(02

xS

mcxS

xai

0)()(2

xSmc

xSx

ai

S

-1 S -1

Multiple (both sides) through, from the left with:

0)()(2

1 x

mcx

xSaSi

which is invariant provided

SaS 1

SaS 1 aSS 1 or

Warning! S is not unitary: ISS

aSS 1

Taking hermitian conjugate:

†

aSS 1† † † †

Recall: = † 00100

aSS † †

Multiply both sides by: 0[ ] 0 = 0[ ] 0

aSS 01000 since: =

SSSS 101000

101000 SSSS

† †

† †

† †obviously inverses!

( ) 101000 SSSS (S 0S† 0)

(S 0S† 0) = (S 0S† 0)

where S, 0, do not commute

S 0S† 0= I

0S† 0= S-1

† †

What will be preserved under transformations?What are the invariant quantities?

† (' )†' = (S)†S = †S†S = †

† cannot be the probability density!

Defining: =† 0 as the “adjoint” spinor

then notice:

)()( xx)(

0)()()( xxxx

)(0

)()(0

)( )( xxxx SSSS

†

† † †

† 0 0S† 0S (x) (x)

† 0

S-1

(x)(x) (x)(x)