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We factored the Klein-Gordon equation into

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We factored the Klein-Gordon equation into

then found solutions for:

Free particle solution to Dirac’s equation

(x) = ue-ixp/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

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

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”

Now watch this:

= I

The transformation matrices

must be ORTHOGONAL!

So

must mean

So

must mean

=

(a-1)

(a-1)

Finally

chain rule (4 terms!)

or

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?

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

S-1

S-1

which is invariant provided

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!

†

†

( )

(S 0S† 0)

(S 0S† 0) = (S 0S† 0)

whereS, 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!

=† 0

Defining:

as the “adjoint” spinor

then notice:

†

†

†

†

=†0 0S†0S

S-1

(x)

(x)

=†0

=

(x)

(x)

(x)

(x)