Lecture 16: Mar 15th 2012
Reading Griffiths chapter 7
Midterm Exam Due Today
HW: 7.4, 7.7, 7.8
1) Klein-Gordon equation
We want to do the same quantization for a relativistic theory starting with the energy,
momentum equation. We will neglect any contribution from a potential since we intend
to treat the effect of a potential perturbatively later on.
E2 = p2 + m2
or more compactly
pp - m2 = 0
Introducing a relativistic notation where there will be a sum over the indices  = 1 to 4
and a negative sign for the upper indices 2-4 in order to give the proper relativistic
invariant expression.
Quantizing
p  i ¶, ¶ = ¶/ ¶ t, ¶ / ¶ x …
results in the Klein-Gordon equation
-
2
¶  ¶   - m2  = 0
or
With a solution:
f (x,t) = Ne ip×x-iEt
Note that this is the one of the simplest solutions. A more general solution includes a
function of the four momentum u(p). Also there is an ambiguity, since there is a solution
with an overall negative sign in the exponential corresponding to the negative energy in
solutions to the energy momentum equation, E=(p2-m2)1/2.
This solution also doesn’t include all the features we want for relativistic particles such as
spin, which is why we used a scalar wave function .
However, it will let us investigate the normalization of the relativistic wave equation and
the concept of antiparticles.
Consider multiplying the Klein Gordon -i* and subtracting the complex conjugate of the
equation multiplied by -i. The mass terms don’t involve operators and cancel out.
Brining out a time and space derivative
One can define a continuity equation as the change in time of the probability density in a
given volume is equal to the current of probability out of the volume:
The first term above is probability density  and the second the probability current j out
of the volume. In four vector notation this is a complete probability current.
j m = i (f *¶ mf - f¶ mf * )
Calculating for  using the wave function.  = 2E|N|2
Note the factor of 2E that we identified as the density of states before. Also the factor of
2E makes the probability density element d3x Lorentz invariant. If you consider a
particle in a Lorenzt boosted from space contracts since d3x d3x(1-v2)1/2. However,
energy, the time like component dilates and the probability goes up making the
probability density element constant.
Now consider the wave functions with a negative sign in the exponential. These
solutions have negative probability density!
This problem can be resolved by noting two things.
First, fundamentally the two energy solutions to the wave equations are independent.
y(x,t) = Ne ip× x-iEt and y (x, t) = Ne-ip×x+iEt .
It makes sense to identify them with independent particles both with energy E. This
eliminates the problem with negative energies where the positive energy particles would
step down in energy and populate the lower available negative energy states.
Second for Electricity and magnetism the fundamental element that is conserved and
interacts with other particles via the photon is the charge. If you had one particle with
probability density  = 2E|N|2 , E positive, and one antiparticle with probability density
 = -2E|N|2 , then the total probability density of zero and you can annihilate them
without changing the probability density and the conserved element is total charge!
On normalization. The probability should be unit-less. Therefore N ~ root(E)
2) Dirac Equation
Now consider a more general solution. The Dirac equation was developed while trying
to simplify the Klein-Gordon equation by factoring it into two first order differential
equations.
Consider the original energy momentum equation.
pp - m2 = 0, and wave equation - ¶  ¶  - m2 = 0
To factorize this we need to solve the problem
(kpk + m)(lpl - m) = klpkpl +m(k - k) pk - m2 = 0
for the 8 constants k and l
For this to be true clearly kl = 1,-1 for k=l and 0 for some combination kl terms to get
back to pp, and k = k to get rid of the term linear in p.
There is no solution to this problem using real or complex numbers.
There is a solution to this problem if k = k are 4x4 matrices. If you anticipate that for
the simplest massive particle, the electron, you would like to have solutions for spin ½
and -½ electrons and for antiparticle positrons then this seems natural. The wave
equation y (x)solutions are four component vectors with four orthogonal eigenvector
solutions to the wave equation, which involves the four by four matrices:
é 0 s iù
éI 0 ù
i
g =ê
ú,
ú, g = ê i
0û
ë0 -Iû
ë-s
0
é0 1ù
ú,
ë1 0û
s1 = ê
é0 -iù
é1 0 ù
3
ú, s = ê
ú,
ëi 0 û
ë0 -1û
s2 =ê
with these definitions then
(0)2 = 1,
(i)2 = -1,
and  -  = 0 for   , or {,} = 2g,
which defines 2g,, which can be used to translate between lower and upper four
momentum vectors.
(kpk + m)(lpl - m) = 0
kpk is often written as p with a slash through it: p/
The four vector solutions can come from either term giving 0 when multiplied by the
wave function.
Quantizing
i  ¶ - m = 0 or i  ¶  + m = 0
The solutions to these equations are similar to Schrödinger equation or the Klein-Gordon
equation solutions (in four vector notation).
The solution
y(x) = ae
-(i / )x m p m
u(p) or y (x) = ae
(i/ ) x m pm
u(p)
Where u(p) is four vector known as a Dirac spinor.
Inserting this into the wave equation gives us the momentum space Dirac equations
(p - mc)u(p) = 0 or (p + mc)u(p) = 0
The first equation corresponds to normal particle wave functions with energy.
There are many ways to think of the second equation. I prefer to think of it as an
independent solution with Energy E and momentum p that when examined
is found to give a negative contribution to an overall probability density allowing for
interactions such as pair production and or annihilation without violating conservation of
overall probability and thus charge.
Alternatively as indicated in Feynman diagrams you can think of this as a normal particle
moving backwards in space and time with normal energy and momentum.
For the first equation.
æ E - m - p × s öæ uA ö
(p - mc)u(p) = (0E – m - .p)u(p) = ç
÷ç ÷ = 0
è p × s -E - møè uB ø
Then (E-m)uA = p× s uB
and (E+m)uB = p× s uA
Then linearly independent solutions for the second equation are:
æu ö
u( p)1 = N ç A ÷,
è uB ø
N=(|E|+m)
æ 1ö
uA = ç ÷,
è 0ø
uB =
æ1ö
1
1 æ pz ö
( p × s )ç ÷ =
ç
÷,
E+m
è0ø E + m è px + ipy ø
æu ö
u( p) 2 = N ç A ÷,
è uB ø
æ 0ö
uA = ç ÷,
è 1ø
uB =
æ0ö
1
1 æ px - ipy ö
( p × s )ç ÷ =
ç
÷
E+m
è1ø E + m è - pz ø
and for (p + mc)u(p) = 0
æ E + m - p × s öæ uA ö
(p + mc)u(p) = (0E + m - p)u(p) = ç
÷ç ÷ = 0
è p × s -E + møè uB ø
Then (E+m)uA = p× s uB
and (E-m)uB = p× s uA
Then linearly independent solutions (also linearly independent of the to the first
equations) for the 1st equation are:
æu ö
u( p) = N ç A ÷,
è uB ø
3
æ 1ö
uB = ç ÷,
è 0ø
æ
æ 1 ö
pz
1
1 ç
uA =
(p × s )ç
÷=
E+m
è 0 ø E + m çè px + ipy
æ 0ö
uB = ç ÷,
è 1ø
uB =
ö
÷,
÷
ø
N=(|E|+m)
æu ö
u( p) = -Nç A ÷,
è uB ø
4
æ
æ 0 ö
1
1 ç px - ipy
(p × s ) ç
=
÷
E+m
è 1 ø E + m çè -pz
ö
÷
÷
ø
Interpreting the other two solutions which have “negative” energy and momentum (-p)
as positive energy antiparticles.
æu ö
u ( p) 2 = Nç A ÷,
è uB ø
N=(|E|+m)
æu ö
u ( p)1 = -Nç A ÷,
è uB ø
æ 1ö
uB = ç ÷,
è 0ø
uA =
æ1ö
1
1 æ pz ö
( p × s )ç ÷ =
ç
÷,
E+m
è0ø E + m è px + ipy ø
æ 0ö
uB = ç ÷,
è 1ø
uA =
æ0ö
1
1 æ px - ipy ö
( p × s )ç ÷ =
ç
÷
E+m
è1ø E + m è - pz ø
ui+ui= 2|E|, the density of states constant we used in the toy theory and from the KleinGordon equation. Also for different solutions ui+uj = 0, they are orthogonal.
We typically rewrite (kpk + m)(lpl - m)u(p) = 0, as H u(p) = 0 and H(p) = 0
(kpk - m) u(p) = 0, (lpl + m)(p) = 0
where H = (kpk - m) and H = (lpl + m)
for the particles, electrons u(p) and the antiparticles, positrons (p)
with complete solutions y(x) = ae
-(i / )x m p m
u(p), y(x) = ae
(i / )x m p m
u ( p)
Initially we were seeking solutions that intrinsically include the concepts of particles and
anti particles (done), spin, total angular momentum, parity and charge conjugation.
3) Consider spin: If we pick p such that the particle is moving in z then these wave
equations are eigenvectors of
æs z 0 ö
Sz = ç
÷
2è 0 szø
take
æu ö
æ 1ö
u( p)1 = N ç A ÷, uA = ç ÷,
è uB ø
è 0ø
æu ö
æ 0ö
u( p) 2 = N ç A ÷,= uA = ç ÷,
è uB ø
è 1ø
with eigenvalues:
2
,-
2
In general the wave functions are not eigenvectors of spin but as expected they are a
linear combination of eigenvectors of spin. However, they can be shown to be
eigenvectors of the helicity operator. i.e. when their spin vector is pointed along the
momentums vector spin is a value. Homework 7.7
æs × pˆ
0 ö
Sz × pˆ = ç
÷
s × pˆ ø
2è 0
and of J=L+S using quantized L= rxp
where Sz × pˆ , S, L and H can all be written as four by four matrices. An easy way to
demonstrate that there are wave functions that are simultaneous eigenvectors of these
operators is to the check the commutation relationship [H,J] = 0. Homework 7.8.
æs × pˆ
0 ö
Sz × pˆ = ç
÷
s × pˆ ø
2è 0
and of J=L+S using quantized L= rxp
where Sz × pˆ , S, L and H can all be written as four vectors or four by four matrices. An
easy way to demonstrate that there are wave functions that are simultaneous eigenvectors
of these operators is to the check the commutation relationship [H,J] = 0. Homework 7.8.