1 Dirac’s hole theory
Shown are (qualitatively) the discrete single particle levels (E, ~p, s), for example for an electron in a box. Here E =±Ep is the energy, ~pis the momentum, and s =±1/2 is the spin direction.
Vacuum
0 Energy
0 Energy
0 Energy One-particle state One-hole state
s=1/2 s=-1/2 s=1/2 s=-1/2 s=1/2 s=-1/2
(a)
(b)
1. Vacuum: All negative energy levels are filled, according to the Pauli principle. All positive energy levels are empty.
This “vacuum state” (reference state) has:
• energy E =−∞, but by “renormalization” it gets zero energy:
E =−∞renormalization
−→ E = 0;
• zero momentum (because for each occupied momentum p, also~ −~p is occupied):
P~ =~0;
• zero spin (because s = 1/2 and s=−1/2 are occupied):
S = 0;
• charge Q=−∞, but by “renormalization” is gets zero charge:
Q=−∞renormalization
−→ Q= 0.
2. One-particle state: All negative energy levels are filled. In the positive energy level (a), which has (E = +Ep, ~p, s), one electron is added.
This “one-particle state” has:
• energy E =−∞+Ep
renormalization
−→ E = +Ep;
• momentum P~ =~0 +~p=~p;
• spin S = 0 +s =s;
• charge Q=−∞+e renormalization
−→ Q=e <0.
3. Single-hole state: All positive energy levels are empty. In the negative energy level (b), which has (E =−Ep,−~p,−s), one electron is missing.
This “one-hole state” has:
• energy E =−∞ −(−Ep)renormalization
−→ E = +Ep;
• momentum P~ =~0−(−~p) = ~p;
• spin S = 0−(−s) =s;
• charge Q=−∞ −erenormalization
−→ Q=−e >0.
We see: The “one-particle state” and “one-hole state” have the same energy, momentum and spin, but opposite charge. ⇒One can interpret the “one-hole state” as the “antiparticle” (positron) state.
The free Dirac wave functions are (see No. 2):
1. For electron (“one-particle state”) with energy Ep >0, momentum ~p, spins, charge e <0:
ψp,s(+)~ (~x, t) = 1
√V s
mc2
Ep u(~p, s)e−i(Ept−~p·~x)/~
2. For positron (“one-hole state”) with energy Ep >0, momentum ~p, spin s, charge −e >0:
ψ−~(−)p,−s(~x, t) = 1
√V s
mc2
Ep v(~p,−s)ei(Ept−~p·~x)/~
However, there is a problem with time evolution:
time
t
0t
1
When a positron evolves from time t0 to a later time t1 > t0:
• At timet=t0: Creation of positive energy positron≡Annihilation of negative energy electron.
• At timet=t1: Annihilation of positive energy positron≡Creation of negative energy electron.
Result (Feynman, St¨uckelberg): If we describe the positron as a missing negative energy electron, then the negative energy electrons must move backward in time! (They first “die” at time t0, and later “are born” at timet1.)
2 Green function (propagator) for the Dirac equation
The Green functionS(x−x0) satisfies the wave equation1
(i6∂−m) S(x−x0) =δ(4)(x−x0) (2.1) Make a Fourier transformation to momentum space:
S(x−x0) =
Z d4p
(2π)4 S(p)e−ip·(x−x0) (2.2)
δ(4)(x−x0) =
Z d4p
(2π)4 1e−ip·(x−x0) (2.3)
Then (2.1) becomes simply
(6p−m)S(p) = 1⇒S(p) = 1
6p−m = 6p+m
p2−m2 (2.4)
We will show later: In order thatS(x−x0) propagates positive energy solutions forward in time, and negative energy solutions backward in time, we need to addi(where = 0+) in the denominator to get the “Feynman propagator”SF:
SF(p) = 1
6p−m+i = 6p+m p2−m2+i
= 6p+m
p20−Ep2+i = 6p+m
(p0−Ep+i) (p0+Ep −i) (2.5) Fourier transform back to coordinate space: From (2.2) and (2.5)
SF(x−x0) =
Z d4p (2π)4
p0γ0−~p·~γ+m
(p0 −Ep+i) (p0+Ep−i)e−ip0(t−t0)ei~p·(~x−~x0) (2.6) In order to perform the integral overp0, use the theorem of residues: For a closed contourC we have
1From here, we will use “natural system of units”, where~=c= 1.
if (t-t’)<0
if (t-t’)>0
X
X p = -E + i ε
0 p
p = E - i ε 0 p p0- plane
I
C
dp0F(p0) = 2πi X
z
Res [F(p0), p0 =z] (2.7)
Here the sum is over the poles (z) of F(p0) inside the contour C, and Res [F(p0), p0 =z] is the residue of F(p0) at the pole z. [Note: This formula holds for “positive” orientation of C. For negative orientation, there is a minus sign.]
• For (t−t0)<0, we close the contour in the upper plane. The residue at the polep0 =−Ep+i is calculated as
Res
p0γ0−~p·~γ+m
(p0−Ep+i) (p0+Ep−i)e−ip0(t−t0), p0 =−Ep+i
= −Epγ0−~p·~γ+m
−2Ep eiEp(t−t0)
• For (t−t0)>0, we close the contour in the lower plane. The residue at the pole p0 =Ep−i is calculated as
Res
p0γ0−~p·~γ+m
(p0−Ep+i) (p0+Ep −i)e−ip0(t−t0), p0 =Ep−i
= Epγ0−~p·~γ+m 2Ep
e−iEp(t−t0)
We then obtain for the propagator (2.6):
SF(x−x0) =−i
Z d3p
(2π)3 ei~p·(~x−~x0)
θ(t−t0)Epγ0−~p·~γ+m
2Ep e−iEp(t−t0)+θ(t0−t)−Epγ0−~p·~γ+m
2Ep eiEp(t−t0)
(2.8)
We set~p→ −~p in the second term, and use the result for the energy projection operators (No. 6) Epγ0−~p·~γ+m
2m = Λ+(~p) = X
s
u(~p, s)u(~p, s)
−Epγ0+~p·~γ+m
2m = Λ−(~p) = −X
s
v(~p, s)v(~p, s) Then we get
SF(x−x0) = −i
Z d3p (2π)3
m Ep
"
θ(t−t0) X
s
u(~p, s)u(~p, s)e−iEp(t−t0)ei~p·(~x−~x0)
− θ(t0−t) X
s
v(~p, s)v(~p, s)eiEp(t−t0)e−i~p·(~x−~x0)
#
= −iθ(t−t0) Z
d3p X
s
ψ(+)~ps (~x, t)ψ(+)~ps (~x0, t0) + iθ(t0−t)
Z
d3p X
s
ψ−~(−)p−s(~x, t)ψ(−)−~p−s(~x0, t0) (2.9)
where we denoted the positive and negative energy solutions of the Dirac equation by ψ~ps(+)(~x, t) =
rm Ep
1
(2π)3/2 u(~p, s)e−i(Ept−~p·~x) ψ−~(−)p−s(~x, t) =
rm Ep
1
(2π)3/2 v(~p,−s)ei(Ept−~p·~x) The normalization of these wave functions is as follows:
Z
d3x ψ~p(a)†0s0(~x, t)ψ~ps(b)†(~x, t) =δabδ(3)(~p0−~p)δs0s
We then see from (2.9) that the Feynman propagator acts as a “time evolution operator” in the following sense:
i Z
d3x0 SF(x−x0)γ0 ψps(+)~ (~x0, t0) = θ(t−t0)ψ~ps(+)(~x, t) (2.10) i
Z
d3x0 SF(x−x0)γ0 ψ(−)−~p−s(~x0, t0) = −θ(t0−t)ψ−~(−)p−s(~x, t) (2.11) These relations imply that “SF(x−x0) propagates the positive energy solutions forward in time, and the negative energy solutions backward in time”.
Note: One can confirm directly that the propagator (2.9) satisfies the equation for the Green function, Eq.(2.1). This follows from:
• The wave equations
(i6∂−m)ψ(~x, t) = 0
• The completeness relation (for fixed time t) Z
d3p X
s
ψ~ps(+)(~x, t)ψ(+)†~ps (~x0, t) +ψps(−)~ (~x, t)ψ~ps(−)†(~x0, t)
=δ(3)(~x−~x0)
(2.12)
• The identities
iγ0 ∂
∂tθ(t−t0) = iγ0δ(t−t0) iγ0 ∂
∂tθ(t0−t) = −iγ0δ(t−t0)
It follows that the expression (2.9), and also (2.10), (2.11), hold more generally in the presence of a time-independent external field.