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http://ocw.mit.edu 8.323 Relativistic Quantum Field Theory I Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Alan Guth, 8.323 Lecture, May 6, 2008, p. 1. MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Physics Department
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�� �� ���� ���� �� �������� We learned that spin 1 particles cannot be bosons, but can be fermions.
2
From Eq. (100), βL 2 = βR 2 = 1, we learned that antiparticles are manda
tory. In the equaltime commutator, the antiparticle contribution canceled the
particle contribution, but only if the antiparticle creation/annihilation operators
are included in the ﬁeld with the same magnitude as the particle operators.
We found that the phases of βL and βR had to be equal to each other, but
were undetermined. This freedom to rotate both phases together should have
been expected: it corresponds to changing the phase of all antiparticle states.
We never deﬁned those phases in the ﬁrst place, so it should make no diﬀerence
if they are changed. (All antiparticle states must have their phases changed in
the same way, however, or else the representation of the Poincar´ group would
e
have to be changed.)
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������ ��� �� ���� –1– Alan Guth, 8.323 Lecture, May 6, 2008, p. 2. ���������� P0 > 0 ��� �� If we did not insist that P 0 > 0, then we could have replaced the b† (p ) in the
s
expression for the ﬁeld,
d3 p
( 2π )3 ψa (x , 0) = 1
2E p µ µ s
as (p ) us (p ) e−ipµ x + b† (p ) va (p ) eipµ x
a
s s by bs (p ). That is, we could use an operator that, instead of creating an
antiparticle with positive energy, would destroy a particle with negative energy.
This interchange of bs and b† would reverse the sign of b† (−p ) , bs (−q ) that
s
s
appeared in Eq. (93), allowing the bosonic commutator to vanish for spacelike
separations. Such negative energy particles, however, apparently do not exist. �����
������� ��������� �� ��
������� –2– ��� ����� ��������� ���� ���� ������ ��� �� ���� In this discussion, the Dirac equation is a byproduct. Now that we have constructed
the Dirac ﬁeld as a quantum operator, we can notice that it satisﬁes the Dirac
equation.
Recall from Eqs. (76) and (77) that
us (q = 0) = us (q = 0) =
L
R
and that √ m ξs , 0 1 . γ =
10
0 (103) (104) Thus γ 0 exchanges L and R, and for q = 0 the u s are equal. Thus
γ 0 us (q = 0) = us (q = 0) .
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������ ��� �� ���� –3– , Alan Guth, 8.323 Lecture, May 6, 2008, p. 3. γ 0 us (q = 0) = us (q = 0) .
Since q µ = (m, 0, 0, 0), this can be rewritten as
qµ γ µ us (q = 0) = mus (q = 0) .
We can show that it holds in all frames by applying Λ 1 (Bp ) to both sides, using
2
Λ 1 (Bp )us (q = 0) = us (p )
2
and Λ 1 γ µ Λ−1 = Λ−1
1
2 µ 2 ν γν , (105) (106) which follows from the commutation relations between γ µ and the Lorentz
generators, which show that γ µ transforms as a Lorentz 4vector. Remember that
Bp q = p.
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������� ��������� �� ��
������� –4– ������ ��� �� ���� Thus Λ 1 (Bp )qµ γ µ us (q = 0) = mus (p ) ,
2 where the LHS can be rewritten as
LHS = qµ Λ 1 (Bp ) γ µ Λ−1 (Bp ) Λ 1 (Bp ) us (q = 0)
1
2
2
2 −
= qµ Bp 1 µ
ν γ ν us ( p ) −
= q · (Bp 1 γ ) us (p ) = p · γ us (p ) where in the last line we used the fact that a dot product is Lorentz invariant,
so we can apply Bp to each factor. Finally,
(γ · p) us (p ) = m us (p ) . (107) ���� ����
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������ ��� �� ���� –5– Alan Guth, 8.323 Lecture, May 6, 2008, p. 4. For the v ’s, we had from Eq. (86),
s
s
vL (q = 0) = −vR (q = 0) . (108) When we wrote Eq. (86) we allowed for an arbitrary multiplicative factor between
s
s
vL (q = 0) and vR (q = 0), in the form of the β ’s, but later we found that causality
required βL = βR , so Eq. (108) is mandatory. The calculation for the v ’s is
otherwise identical, leading to
(γ · p) v s (p ) = −m v s (p ) . (109) ���� ����
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������� ��������� �� ��
������� –6– ������ ��� �� ���� Summarizing, we have
ψa (x) = d3 p
( 2π )3 1
2E p µ s µ s
as (p ) us (p ) e−ipµ x + b† (p ) va (p ) eipµ x
a
s ,
(72) (γ · p) us (p ) = m us (p ) , (107) (γ · p) v s (p ) = −m v s (p ) . (109) and The application of ∂µ to ψ (x) brings down a factor of −ipµ for the ﬁrst term and
ipµ for the second term, so the Dirac ﬁeld satisﬁes (iγ µ ∂µ − m) ψ (x) = 0 . (110) ���� ����
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������ ��� �� ���� –7– Alan Guth, 8.323 Lecture, May 6, 2008, p. 5. 2×2 ���
�� ��� ���� �������� �� It is sometimes useful to use our representation of the γ matrices, and the deﬁnitions
¯
σ µ = (1, σ i ) and σ µ = (1, −σ i ) to write the Dirac equation as: iσ · ∂ ψL =0. −m
ψR −m iσ · ∂
¯ (111) In this form we see that the spatial derivative mixes the upper and lower (L and
R) components, as we commented earlier that we expected, on the grounds
1
that (0, 1 ) and ( 1 , 1 ) can only produce ( 1 , 0) or ( 2 , 1).
2
22
2 �����
������� ��������� �� ��
������� –8– ��� ���� ���������� ���� ���� ������ ��� �� ���� The Dirac equation is (iγ µ ∂µ − m) ψ (x) = 0 . (110) This is a 4component equation, and the Lagrangian must be a real scalar.
The natural guess is therefore to contract these indices with ψ † (x), where ψ1 (x) ψ2 (x) ψ (x) ≡ ψ3 (x) , ψ4 (x) ψ † (x) ψ † (x) ψ † (x) ψ † (x) .
ψ (x) =
1
2
3
4
† (112)
Then we can try � maybe = ψ † (iγ µ ∂µ − m)ψ . (113) ���� ����
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������ ��� �� ���� –9– Alan Guth, 8.323 Lecture, May 6, 2008, p. 6. In principle we should vary the real and imaginary (e.g., hermitian and anti
hermitian) parts of ψ (x), but we can equivalently vary ψ and ψ † independently
by deﬁning
1
¯
∂ψ a ≡
2 ∂
∂ Re ψa 1
¯†
∂ψ a ≡
2 ∂
∂ Re ψa If we vary � maybe −i ∂
∂ Im ψ a +i ∂
∂ Im ψ a (114) = ψ † (iγ µ ∂µ − m)ψ with respect to ψ † , one gets the Dirac equation. Nonetheless, Peskin &
Schroeder reject
maybe b ecause it is not Lorentzinvariant. � � This reason for rejecting
maybe seems weak to me, since it did generate the
correct, Lorentzinvariant Dirac equation.
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������� –10– ������ ��� �� ���� ����
���� � maybe � A more serious problem with maybe is that it gives inconsistent equations. Varying
maybe with respect to ψ , one integrates the action by parts and ﬁnds � � maybe = ψ † (iγ µ ∂µ − m)ψ =⇒
←
−
ψ † (−iγ µ ∂ µ − m) = 0 . (115) To compare with the usual Dirac equation, we have to take the complex conjugate
†
(or adjoint) of this equation. To do this we need to compute (γ µ ) . For our
conventions σi 0
i
0 1
0 .
γ =
γ =
(116) 10
−σ i 0
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������ ��� �� ���� –11– Alan Guth, 8.323 Lecture, May 6, 2008, p. 7. One sees that
γ0 † = γ0 γi † = −γ i . (117) These signs are actually dictated by the original Dirac anticommutation relations
{γ µ , γ ν } = 2g µν . These imply that (γ 0 )2 = 1, and (γ i )2 = −1, so they must
be hermitian and antihermitian respectively. The anticommutation relations also
imply that γ µ and γ ν anticommute when µ = ν . Thus we can write for all µ that:
† γ 0 (γ µ ) γ 0 = γ µ . (118) So, taking the adjoint of Eq. (115),
←
−
0 = ψ † (−iγ µ ∂ µ − m)
This is ��� † † = i (γ µ ) ∂µ − m ψ = (iγ 0 γ µ γ 0 − m)ψ .
(119)
the Dirac equation. Note that this is a counterexample to the widely believed falseho od that equations
derived from a Lagrangian are necessarily consistent.
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������� –12– ������ ��� �� ���� ������� �� ���� �������� � maybe � maybe is not real. To understand, think about minimizing a function of a complex vector (z1 , . . . , zN ):
∗
L = za M ab zb , where M ab is a matrix. We can vary with respect to z or z ∗ by deﬁning
1
¯
∂za ≡
2 ∂
∂
−i
∂ Re za
∂ Im za 1
¯∗
∂za ≡
2 ∂
∂
+i
∂ Re za
∂ Im za . ���� ����
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������� ��������� �� ��
�������
������ ��� �� ���� –13– Alan Guth, 8.323 Lecture, May 6, 2008, p. 8. Setting
∗
∂za L = 0 =⇒ M ab zb = 0 , ∂za L = 0 =⇒ ∗
za M ab = 0 . and setting These two equations are consistent if and only if M ab is hermitian, M ab ∗ =
M ba . This is also the condition that L be real.
Trying to make a complex L stationary is like trying to make two Lagrangians —
i.e., the real and imaginary parts — stationary at the same time. It will usually
be inconsistent. ���� ����
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������� ��������� �� ��
������� –14– ������ ��� �� ���� ������ ��� ���� ���������� To cure the problem, introduce a factor of γ 0 to cancel the γ 0 ’s we found in
Eq. (119):
†
←
−
ψ † (−iγ µ ∂ µ − m) = (iγ 0 γ µ γ 0 − m)ψ .
Let us try � Dirac ¯
= ψ (iγ µ ∂µ − m) ψ , (120) where
¯
ψ (x) = ψ † (x) γ 0 †
0
¯
ψa (x) = ψb (x) γba . (121) ���� ����
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������ ��� �� ��� –15– Alan Guth, 8.323 Lecture, May 6, 2008, p. 9. Then we can show that the action S is real:
S† = † ¯
d4 x ψ (iγ µ ∂µ − m) ψ d4 x ψ † γ 0 (iγ µ ∂µ − m) ψ = † . Integrating by parts,
S† =
= ←
−
d4 x ψ † γ 0 −iγ µ ∂ µ − m ψ † d4 x ψ † iγ µ† ∂µ − m γ 0 ψ
(122)
4 † 0 0 iγ γ µ† 0 = d xψ γ γ ∂µ − m ψ = ¯
d4 x ψ (iγ µ ∂µ − m) ψ = S . ���� ����
�����
������� ��������� �� ��
������� –16– ������ ��� �� ���� † Note that in the last line we used Eq. (118), γ 0 (γ µ ) γ 0 = γ µ .
Similarly we can show that the equations of motion are now consistent, since the
equation obtained by varying ψ is
←
−
¯
ψ −iγ µ ∂ µ − m = 0 . (122) The adjoint of this equation is
←
−
¯
0 = ψ −iγ µ ∂ µ − m † = iγ µ† ∂µ − m γ 0† ψ
0 =γ γ 0 µ† 0 (123) iγ ∂µ − m γ ψ = γ 0 (iγ µ ∂µ − m) ψ .
This is exactly the Dirac equation, multiplied by the (invertible) matrix γ 0 .
���� ����
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�������
������ ��� �� ���� –17– Alan Guth, 8.323 Lecture, May 6, 2008, p. 10. ���� ������� We already learned how to write the Dirac equation in 2 × 2 blo cks: −m iσ · ∂
¯ iσ · ∂ ψL =0.
−m
ψR (111) For the special case of m = 0, the two pieces decouple, giving the Weyl
equations:
iσ · ∂ ψL = 0 ,
¯ iσ · ∂ ψR = 0 . (124) ψL (x) and ψR (x) are called Weyl ﬁelds. Since σ 2 σ µ∗ σ 2 = σ µ , one can deﬁne
¯
†
χR = σ 2 ψR iσ · ∂ χR = 0 .
¯ (125) �����
������� ��������� �� ��
������� –18– ������ ��� �� ���� ���� ���� =⇒ The Weyl ﬁelds can be extracted from the Dirac ﬁeld by deﬁning
γ5 ≡ i γ0 γ1 γ2 γ3
=− i
4! µνλσ
µνλσ γ γ γ γ , (126) where µνλσ is the fully antisymmetric LeviCivita tensor, with the sign convention
(following P&S)
0123 = −1 . (127) Note that this corresponds to 0123 = 1. γ 5 is Lorentzinvariant, which one can see
by using the fact that µνλσ is Lorentzinvariant, or by noting that
S µν , γ 5 = 0 . (128) where S µν are the generators of Lorentz transformations deﬁned in Eq. (45).
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������ ��� �� ���� –19– Alan Guth, 8.323 Lecture, May 6, 2008, p. 11. In our conventions
γ5 = −1
0 0
1 , (129) which clearly separates the upper and lower halves of the Dirac ﬁeld:
1
(1 + γ 5 )ψ =
2 00
01 ψL
ψR 1
(1 − γ 5 )ψ =
2 10
00 ψL
ψR = 0
ψR = ψL
0 (130)
. For all choices of γ matrices, γ 5 is deﬁned by Eq. (126); it is Lorentzinvariant,
and can be used to project L and R components of ψ (which are deﬁned by the
( 1 , 0) + (0, 1 ) decomposition, but need not be the upper and lower pieces of ψ ).
2
2
���� ����
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������� ��������� �� ��
������� –20– ������ ��� �� ���� Recall that √
s √p · σ ξ , u (p ) = p · σ ξs
¯
s �������� �� ���� ���������� �������� √
2s p √ · σ (−iσ ξ ) . v (p ) = − p · σ (−iσ 2 ξ s )
¯
s For m = 0 we can consider the case pµ = (E, 0, 0, E ), and we recall that
¯
σ µ = (1, σ i ) and σ µ = (1, −σ i ). Then
p · σ = p0 − p3 σ z = E (1 − σ z ) = 2EP− ,
where P− = 1 (1 − σ z ) is the projector onto σ z = −1 states, or equivalently
2
the projector onto negative helicity states. Similarly, p · σ = 2EP+ , where P+
¯
projects onto σ z = 1, or positive helicity states. Since σ 2 anticommutes with
σ 3 , one has P+ σ 2 = σ 2 P− , and P− σ 2 = σ 2 P+ . Putting all this together,
s u (p ) = √ s P− ξ , 2E P+ ξ s 2
s
√ σ P+ ξ . v (p ) = −i 2E −σ 2 P− ξ s
s (131)
���� ����
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������� ��������� �� ��
�������
������ ��� �� ���� –21– Alan Guth, 8.323 Lecture, May 6, 2008, p. 12. s u (p ) = √ s P− ξ , 2E
P+ ξ s 2
s
√ σ P+ ξ . v (p ) = −i 2E
−σ 2 P− ξ s
s (131) Recall that
ψa (x) = d3 p
( 2π )3 1
2E p µ µ s
as (p ) us (p ) e−ipµ x + b† (p ) va (p ) eipµ x
a
s s , so one sees that for massless particles the upper components, ψL , destroy negative
helicity particles (as destroys particles), and creates positive helicity antiparticles.
For ψR it is the reverse.
For m = 0, one can have a theory with just negative helicity particles have
and positive helicity antiparticles, or vice versa. One could have particles
of both helicities, but there is no need for both to build a Lorentzinvariant
ﬁeld theory.
���� ����
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������� ��������� �� ��
������� –22– ������ ��� �� ���� ��� ���� ����������� The Dirac Lagrangian was given as Eq. (120), � Dirac ¯
= ψ (i γ µ ∂µ − m) ψ . Following the canonical procedure, � � . i γ µ ∂µ = i γ 0 ∂0 + γ i ∂i = i γ 0 ∂0 + γ · ∇ . π= ∂
= iψ † , and H =
˙
∂ψ ˙
d3 x π ψ − (132) Being careful with the signs, Then
H= ¯
d3 x ψ −iγ · ∇ + m ψ . (133) ���� ����
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������� ��������� �� ��
�������
������ ��� �� ���� –23– Alan Guth, 8.323 Lecture, May 6, 2008, p. 13. ¯
d3 x ψ −iγ · ∇ + m ψ . H= (133) Expanding the ﬁeld, as in Eq. (92),
µ
µ
1
d3 p
s
ψa ( x , 0) =
as (p ) us (p ) e−ipµ x + b† (p ) va (p ) eipµ x
a
s
( 2π )3 2E p s
d3 p
( 2π )3 = 1
2E p s t=0 s
as (p ) us (p ) + b† (−p ) va (−p ) eip ·x .
a
s Inserting into Eq. (133),
d3 x H= d3 p
(2π )3 1
2E p d3 q
( 2π )3 1
×
2E q a† (p ) us (p ) + bs (−p ) v s (−p ) e−ip ·x
¯
¯
s ×
rs × −iγ · ∇ + m ar (q )ur (q ) + b† (−q )v r (−q ) eiq ·x .
r ���� ����
�����
������� ��������� �� ��
������� –24– ������ ��� �� ���� d3 x H=
× rs d3 p
( 2π )3 1
2E p d3 q
( 2π )3 1
×
2E q a† (p ) us (p ) + bs (−p ) v s (−p ) e−ip ·x
¯
¯
s × −iγ · ∇ + m ar (q )ur (q ) + b† (−q )v r (−q ) eiq ·x .
r One can replace −iγ · ∇ by γ · q , and then integrate over x , obtaining a δ (p − q )
which can be used to integrate over q :
H= d3 p 1
( 2π )3 2E p rs a† (p ) us (p ) + bs (−p ) v s (−p )
¯
¯
s × [γ · p + m] ar (p )ur (p ) + b† (−p )v r (−p )
r . ���� ����
�����
������� ��������� �� ��
�������
������ ��� �� ���� –25– Alan Guth, 8.323 Lecture, May 6, 2008, p. 14. H= d3 p 1
( 2π )3 2E p a† (p ) us (p ) + bs (−p ) v s (−p )
¯
¯
s rs × [γ · p + m] ar (p )ur (p ) + b† (−p )v r (−p )
r . Now use
(γ · p) us (p ) = m us (p ) , (γ · p) v s (p ) = −m v s (p ) , and
u† (p ) us (p ) = 2Ep δrs
r †
vr (p ) vs (p ) = 2Ep δrs u† (p ) vs (−p ) = 0
r †
vr (p ) us (−p ) = 0 to obtain
H= d3 p
Ep
( 2π )3 s a† (p ) as (p ) − bs (−p ) b† (−p )
s
s . ���� ����
�����
������� ��������� �� ��
������� –26– ������ ��� �� ���� H= d3 p
Ep
( 2π )3 s a† (p ) as (p ) − bs (−p ) b† (−p )
s
s . Now use p → −p in 2nd term, and reverse the order of the b and b† , using Eq. (97):
bs (p ) , b† (q ) = (2π )3 δrs δ (3) (p − q ) .
r
H= d3 p
Ep
( 2π )3 s a† (p ) as (p ) + b† (p ) bs (p ) + Evac ,
s
s (134) where
Evac = −2 d3 p Ep δ (3) (0) = −2 where I am using
δ (3) (p ) = d3 p
Ep × Volume of space ,
( 2π )3 (135) d3 x ip ·x
e
.
( 2π )3 ���� ����
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������� ��������� �� ��
�������
������ ��� �� ���� –27– Alan Guth, 8.323 Lecture, May 6, 2008, p. 15. d3 p
Ep
(2π )3 H= a† (p ) as (p ) + b† (p ) bs (p ) + Evac ,
s
s s (134) where
d3 p Ep δ (3) (0) = −2 Evac = −2 d3 p
Ep × Volume of space .
(2π )3 (135) Note that Fermi statistics caused the antiparticle energy to be positive (go o d!),
and the vacuum energy to be negative (surprising?). The negative vacuum energy,
although illdeﬁned, is still welcome: allows at least the hope that one might
get the positive (bosonic) contributions to cancel against the negative (fermionic)
contributions, giving an answer that is ﬁnite and hopefully small. Note that if we
had 4 free scalar ﬁelds with the same mass, the cancelation would be exact: this
is what happens in EXACTLY supersymmetric mo dels, but it is spoiled as so on as
the supersymmetry is broken. �����
������� ��������� �� ��
������� –28– ���� ���� ������ ��� �� ���� ���� ���� ������ Unoccupied
states E Electron +mc2 +mc2
0 Radiation 0 Electron Radiation mc2 mc2
Occupied
states E Hole Hole Electronpositron pair creation Electronpositron pair annihilation Figure by MIT OpenCourseWare. Adapted from Bjorken & Drell, vol. 1, p. 65. In the 1particle quantum mechanics formulation, positrons show up as negative
energy states. Dirac proposed that in the vacuum, the negative energy “sea”
was ﬁlled. Physical positrons, in this view, are holes in the Dirac sea. In QFT,
on the other hand, particles and antiparticles are on equal fo oting. Nonetheless,
the Dirac sea allows an intuitive way to understand the negative vacuum energy.
From Bjorken & Drell, vol. 1, p. 65
–29– Alan Guth, 8.323 Lecture, May 6, 2008, p. 16. ��� ���� ���������� This is straightforward, so I will only summarize the results. ¯
0 ψa (x) ψb (y ) 0 = d3 p 1
( 2π )3 2E p = (i ∂ x + m)ab s us (p ) us (p ) e−ip·(x−y)
¯b
a d3 p 1 −ip·(x−y)
e
( 2π )3 2E p
(136) = (i ∂ x + m)ab D(x − y )
¯
0 ψb (x) ψa (y ) 0 = d3 p 1
( 2π )3 2E p s s
va (p ) vb (p ) e−ip·(y−x)
¯s = −(i ∂ x + m)ab D(y − x) ,
where ∂ = γ µ ∂µ and D(x) is the scalar 2point function 0 φ(x)φ(0) 0 .
���� ����
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������� ��������� �� ��
������� –30– ������ ��� �� ���� ab
¯
SR (x − y ) ≡ θ (x0 − y 0 ) 0 ψa (x) ψb (y ) 0 = (i ∂ x + m) DR (x − y ) , ��� �������� ���� ���������� (137) where DR (x − y ) is the scalar retarded propagator. One can show
(i ∂ x − m)SR (x − y ) = iδ (4) (x − y ) · 14×4 . (138) The Fourier expansion is
SR (x) = i( p + m)
d4 p −ip·x ˜
˜
SR (p) , where SR (p) = 0
e
.
4
(2π )
(p + i )2 − p 2 − m2
(139) ���� ����
�����
������� ��������� �� ��
�������
������ ��� �� ���� –31– Alan Guth, 8.323 Lecture, May 6, 2008, p. 17. SF (x − y ) ≡ ¯
0 ψ (x) ψ(y ) 0
¯
− 0 ψ ( y ) ψ ( x) 0 ¯
≡ 0 T ψ (x) ψ(y ) ��� ������� ���������� for x0 > y 0
for y 0 > x0 (140) 0. The Feynman propagator also satisﬁes Eq. (138). The Fourier expansion is
d4 p −ip·x ˜
i( p + m)
˜
SF (p) , where SF (p) = 2
.
e
(2π )4
p − m2 + i SF (x) = �����
������� ��������� �� ��
������� ������ ��� �� ���� ���� ���� (141) –32– ...
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 Spring '08
 staff
 Mass, Quantum Field Theory, Positron, Paul Dirac, Dirac equation, Alan Guth, Dirac sea

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