1. The Euler-Lagrange Equation
Finding a Klein-Gordon Lagrangian or
The Klein-Gordon Equation L
Provided we can identify the appropriate
this should be derivable by
The Euler-Lagrange Equation L L

2. I claim the expression L
serves this purpose

3. L L L L L L L L

4. LDIRAC(r,t) L(r,t) L = LK-G + LDIRAC + LINT
You can show (and will for homework!) show
the Dirac Equation can be derived from:
LDIRAC(r,t)
We might expect a realistic Lagrangian that involves systems of particles
L(r,t)
= LK-G LDIRAC
but each term
describes free
non-interacting
particles
describes
photons
describes
e+e- objects L
+ LINT
But what does terms look like?
How do we introduce the interactions the experience?

5. We’ll follow (Jackson) E&M’s lead:
A charge interacts with a field through:
current-field interactions
the fermion
(electron)
the boson
(photon) field
from the Dirac
expression for J
particle
state
antiparticle
(hermitian conjugate)
state
Recall the “state functions”
Have coefficients that must
satisfy anticommutation relations.
They must involve operators!
What does such a PRODUCT of states mean?

6. or We introduce operators (p) and †(p) satisfying either of 2 cases:
Along with a representation of the (empty) vacuum state: | 0 
such that:

7. † † The Creation Operator It's alive! ALIVE! X (p) X (p)

8. | 0  “creates” a free particle of 4-momentum p
The complex conjugate of this equation reads:
The above expression also tells us:
| 0 
which we interpret as:

9. THE
ANNIHILATOR
The Annihilation Operator X
(p) X
(p)
| p >
= 0

10. the orthonormal states we’ve assumed:
If we demand, in general,
the orthonormal states we’ve assumed: 1
an operation that
makes no contribution
to any calculation

11. | 0  †(p)|0 = [ (p-q)  †(p)(q) ] |0 = (p-q)|0
Then
†(p)|0 =
[ (p-q)  †(p)(q) ]
|0
zero
= (p-q)|0
This is how the annihilation operator works:
If a state contains a particle with momentum
q, it destroys it. The term simply vanishes
(makes no contribution to any calculation)
if p  q.
|p1 p2 p3 = †(p1)†(p2)†(p3)
| 0 

12. †(p)†(k)|0 = †(k)†(p)|0 †(p)†(k)|0 = -†(p)†(k)|0or
in contrast
†(p)†(k)|0 = †(k)†(p)|0
†(p)†(k)|0 = -†(p)†(k)|0
|pk = |kp
|pk = -|kp
|pp = |pp
and if p=k this gives
and if p=k this must give
This is perfectly OK!
These must be symmetric states
These are anti-symmetric states
BOSONS
FERMONS

13. The most general state

14.   dk3 g h a b† a b k s s s {(r,t), †(r´,t)} =d 3(r – r)
Recall the most general DIRAC solution: 
dk3  k s s g h s
that these Dirac
particles are fermions
If we insist:
{(r,t), †(r´,t)} =d 3(r – r)
we can identify (your homework) g as an annihilation
operator a(p,s) and h as a creation operator b†(-p,-s) a b† † a b

15. d d [A(r,t), A†(r´,t)] =d 3(r – r)
Similarly for the photon field (vector potential)
[A(r,t), A†(r´,t)] =d 3(r – r)
If we insist:
Bosons! † -s -k -s -k d d
Remember here there is no separate anti-particle
(but 1 particle with 2 helicities).
Still, both solutions are needed for mathematical completeness.

16. a†b†d† a†ad† a†ad† a†ad bb†d† bb†d bad† bad Now, since
interactions between Dirac particles (like electrons)
and photons appear in the Lagrangian as
It means these interactions involve operator products of
(a† b ) (a b† ) (d† d )
creates an
electron
annihilates
an electron
creates a
photon
annihilates
a photon
annihilates
a positron
creates a
positron
giving terms with all these possible combinations:
a†b†d†
a†ad†
a†ad†
a†ad
bb†d†
bb†d
bad†
bad

17. a†b†d† a†ad† a†ad† a†ad bb†d† bb†d bad† bad 0|daa†|0 a†b†d† a†ad†
What do these mean?
a†b†d†
a†ad†
a†ad†
a†ad
bb†d†
bb†d
bad†
bad
In all computations/calculations we’re interested in,
we look for amplitudes/matrix elements like:
0|daa†|0
Dressed up by the full integrals to
calculate the probability coefficients
creates a
positron e+ e- e-  
a†b†d†
a†ad† e-
creates an
electron
creates a
photon
annihilates
an electron

18. e- 
a†ad†
time e- e-
a†ad
time e- 

19. In Quantum Electrodynamics (QED)
All physically are ultimately reducible to this elementary 3-branched process.
We can describe/explain ALL electromagnetic processes
by patching together copies of this “primitive vertex” p3 p4
…two final state electrons exit. e- e- 
…a  is exchanged (one emits/one absorbs)…
Our general solution
allows waves traveling
in BOTH directions e- e- p1 p2
Calculations will include both
and not distinguish the
contributions from either case.
Two electrons (in momentum states p1 and p2) enter…
Coulomb repulsion (or “Møller scattering”)
Mediated by an
exchanged photon!

20. bad†  What does this describe? time e- e+
These diagrams can be twisted/turned as long as we preserve the topology
(all vertex connections) and describe an equally valid (real, physical) process 
bad†
What does this describe?
time e- e+

21. Bhaba Scattering