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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 13, 2008, p. 1. MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Physics Department
������ ����������� ������� ����� ������ � ������
����
����� ��������
�������� ��� ��� ���� — ���� ���� ���� ����
�����
������� ��������� �� ��
�������
������ ��� ��� ���� ������ ������� λφ4 �������������� ������������ ������ � � = 1
1
λ
(∂µ φ)2 − m2 φ2 − φ4 .
2
2
4! Then
H = H0 + Hint ,
where
Hint = d3 x λ4
φ (x ) .
4! Goal: to perturbatively calculate matrix elements of the Heisenberg ﬁeld
φ(x , t) = eiH t φ(x , 0)e−iHt
in the state Ω , the true ground state of the interacting theory.
–1– Alan Guth, 8.323 Lecture, May 13, 2008, p. 2. ������
���� ��
����� Cho ose any reference time t0 , at which the interaction picture operators and
Heisenberg operators will coincide. Deﬁne
φI (x , t) ≡ eiH0 (t−t0 ) φ(x , t0 )e−iH0 (t−t0 ) .
At t = t0 , can expand Heisenberg φ and π in creation and annihilation operators:
φ(x , t0 ) =
˙
π (x , t0 ) = φ(x , t0 ) = d3 p
( 2π )2 1
ap eip ·x + a† e−ip ·x
p
2E p d3 p
( 2π )2 . 1
−iEp ap eip ·x + iEp a† e−ip ·x
p
2E p . †
ap creates state of momentum p , but not energy Ep — not single particle. But [φ(x , t0 ) , π (y , t0 )] = iδ 3 (x − y ) =⇒ †
ap , aq = (2π )3 δ (3) (p − q ) . –2– Can write φI (x , t) for all t:
d3 p
φI (x , t) =
( 2π )2 1
2E p ap e−ip·x + a† eip·x
p x0 =t−t0 . How to express φ(x , t):
φ(x , t) = eiH (t−t0 ) e−iH0 (t−t0 ) φI (x , t)eiH0 (t−t0 ) e−iH (t−t0 )
≡ U † (t, t0 )φI (x , t)U (t, t0) ,
where U (t, t0 ) = eiH0 (t−t0 ) e−iH (t−t0 ) .
Diﬀerential equation for U :
∂
i U (t, t0 ) = eiH0 (t−t0 ) (H − H0 )e−iH (t−t0 )
∂t
= eiH0 (t−t0 ) Hint e−iH (t−t0 )
= eiH0 (t−t0 ) Hint e−iH0 (t−t0 ) eiH0 (t−t0 ) e−iH (t−t0 )
= HI (t)U (t, t0 ) ,
where
HI (t) = eiH0 (t−t0 ) Hint e−iH0 (t−t0 ) = d3 x λ4
φ (x , t) .
4! I
–3– Alan Guth, 8.323 Lecture, May 13, 2008, p. 3. �������� �� ������������ ��������� ∂
U (t, t0 ) = HI (t)U (t, t0 )
∂t
implies the integral equation
i t U (t, t0 ) = I − i t0 U (t0 , t0 ) = I with dt HI (t )U (t , t0 ) . To ﬁrst order in HI , t U (t, t0 ) = I − i dt1 HI (t1 ) . t0 To second order in HI ,
t U (t, t0 ) = I − i t0 dt1 HI (t1 ) + (−i) t 2 t0 t1 dt1 t0 dt2 HI (t1 )HI (t2 ) . To third order,
U (t, t0 ) = . . . + (−i) t 3 t0 t1 dt1 t0 t2 dt2 t0 dt3 HI (t1 )HI (t2 )HI (t3 ) .
–4– Note that t1 ≥ t2 ≥ t3 . Can rewrite 3rd order term as
U (t, t0 ) = . . . + (−i)
= ...+ 3 t
t0 (−i)3
3! dt1 t
t0 t1
t0 dt1 t
t0 dt2
dt2 t2
t0
t
t0 dt3 HI (t1 )HI (t2 )HI (t3 ) dt3 T {HI (t1 )HI (t2 )HI (t3 )} , where T {} is timeordered pro duct (earliest time to right). Finally,
U (t, t0 ) = I + (−i)
≡T t
t0 exp −i (−i)2
dt1 HI (t1 ) +
2!
t
t0 dt HI (t ) t
t0 dt1 t
t0 dt2 T {HI (t1 )HI (t2 )} + . . . . –5– Alan Guth, 8.323 Lecture, May 13, 2008, p. 4.
Generalize to arbitrary t0 : U ( t2 , t1 ) ≡ T exp −i t2
t1 dt HI (t ) , where (for t1 < t0 < t2 )
U ( t 2 , t1 ) = T exp −i t2
t0 dt HI (t ) T exp −i t0
t1 dt HI (t ) = U (t2 , t0 )U −1 (t1 , t0 ) .
Given
have U (t, t0 ) = eiH0 (t−t0 ) e−iH (t−t0 ) ,
U (t2 , t1 ) = eiH0 (t2 −t0 ) e−iH (t2 −t0 ) eiH (t1 −t0 ) e−iH0 (t1 −t0 )
= eiH0 (t2 −t0 ) e−iH (t2 −t1 ) e−iH0 (t1 −t0 ) .
–6– U ( t2 , t1 ) ≡ T exp −i t2
t1 dt HI (t ) . Properties:
• U is unitary.
• U (t3 , t2 )U (t2 , t1 ) = U (t3 , t1 ) .
• U (t2 , t1 )−1 = U (t1 , t2 ) . –7– Alan Guth, 8.323 Lecture, May 13, 2008, p. 5. �������
���� �� ���� ������ ����� Ω � Assume that 0 has nonzero overlap with Ω :
e−iHT 0 = e−iEn T n n 0 . n If T had large negative imaginary part, all other states would be suppressed relative
to  Ω .
Ω =
= lim e−iH (T +t0 ) 0 lim e−iH (T +t0 ) eiH0 (T +t0 ) 0 T →∞(1−i ) T →∞(1−i ) Recall e−iE0 (T +t0 ) Ω 0 −1 −1 e−iE0 (T +t0 ) Ω 0 . U (t2 , t1 ) = eiH0 (t2 −t0 ) e−iH (t2 −t1 ) e−iH0 (t1 −t0 ) , so
Ω = lim T →∞(1−i ) U (t0 , −T ) 0 e−iE0 (T +t0 ) Ω 0 −1 .
–8– Similarly,
Ω =
Recall lim T →∞(1−i ) e−iE0 (T −t0 ) Ω 0 −1 0 U (T, t0 ) . φ(x , x0 ) = U † (x0 , t0 )φI (x , x0 )U (x0 , t0 ) . So, for x0 > y 0 ,
Ω φ(x)φ(y ) Ω = lim T →∞(1−i ) 0 U (T, t0 )U (t0 , x0 )φI (x , x0 )U (x0 , t0 ) × U (t0 , y 0 )φI (y , y 0 )U (y 0 , t0 )U (t0 , −T ) 0
× Normalization factor
= lim T →∞(1−i ) 0 U (T, x0 )φI (x , x0 )U (x0 , y 0 ) × φI (y , y 0 )U (y 0 , −T ) 0
× Normalization factor .
But
Normalization factor = Ω Ω −1 , –9– Alan Guth, 8.323 Lecture, May 13, 2008, p. 6.
so
0T
Ω φ(x)φ(y ) Ω = φI (x)φI (y ) exp −i lim T →∞(1−i ) 0T exp −i T
−T T
−T dt HI (t) 0
. dt HI (t) 0 ���������� HI (t) = d3 x � (x , t) , so
0T
Ω φ(x)φ(y ) Ω = φI (x)φI (y ) exp −i
0T � exp −i dz
4 d4 z � � I (z ) I (z ) 0
. 0 � If z1 and z2 are spacelikeseparated, their time ordering is framedependent. Need
I (z1 ) ,
I (z2 ) = 0 to get same answer in all frames.
–10– 0T
Ω φ(x)φ(y ) Ω = �������������� ������������ ������� φI (x)φI (y ) exp −i
0T exp −i dz
4 d4 z � � I (z ) I (z ) 0
. 0 –11– Alan Guth, 8.323 Lecture, May 13, 2008, p. 7. ��
��� ������� GianCarlo Wick
October 15, 1909 – April 20, 1992 For more information see
The National Academies Press
Biographical Memoir
http://www.nap.edu/readingroom.php?book=biomems&page=gwick.html –12– T {φ(x1 )φ(x2 ) . . . φ(xm )} = N {φ(x1 )φ(x2 ) . . . φ(xm )+ all possible contractions} .
Example: Corollary:
0 T {φ(x1 )φ(x2 ) . . . φ(xm )} 0 = all possible FULL contractions} .
Example:
0 T {φ(x1 )φ(x2 )φ(x3 )φ(x4 )} 0 = ∆F (x1 − x2 )∆F (x3 − x4 )
+ ∆F (x1 − x3 )∆F (x2 − x4 ) + ∆F (x1 − x4 )∆F (x2 − x3 ) .
–13– Alan Guth, 8.323 Lecture, May 13, 2008, p. 8. ������� �������� Example:
0 T {φ(x1 )φ(x2 )φ(x3 )φ(x4 )} 0 = ∆F (x1 − x2 )∆F (x3 − x4 )
+ ∆F (x1 − x3 )∆F (x2 − x4 ) + ∆F (x1 − x4 )∆F (x2 − x3 ) .
Feynman diagrams: [Note: the diagrams and some equations on this and the next 12 pages were taken from An
Introduction to Quantum Field Theory, by Michael Peskin and Daniel Schroeder.] –14– ���������� �������� Ω T {φ(x)φ(y )} Ω =
0T �
0T φ(x)φ(y ) −i =3· −iλ
4! + 12 · I (z ) = φ(x)φ(y ) + φ(x)φ(y ) −i
d4 z � DF (x − y )
−iλ
4! λ4
φ (z )
4! I (z ) d4 z � I (z ) +... 0 0 d4 zDF (z − z )DF (z − z ) d4 zDF (x − z )DF (y − z )DF (z − z ) = –15– . Alan Guth, 8.323 Lecture, May 13, 2008, p. 9. ���� ���������� �������� How many identical contractions are there? Overall factor: 1
3! × 13
4! × 3! × 4 · 3 × 4! × 4 · 3 × 1
2 = 1
8 ≡ 1
symmetry factor .
–16– �������� ��
����� –17– Alan Guth, 8.323 Lecture, May 13, 2008, p. 10. 0T φ(x)φ(y ) exp −i d4 z � I (z ) 0 λφ4 = ������ ������� ����� ��� sum of all possible diagrams
with two external points Rules: –18– DF (x − y ) = Vertex: �������� ���
� ������� ������ d4 p
i
e−ip·(x−y) .
4 p2 − m2 + i
(2π ) d4 z e−ip1 z e−ip2 z e−ip3 z e−ip4 z
= (2π )4 δ (4) (p1 + p2 + p3 + p4 ) . –19– . Alan Guth, 8.323 Lecture, May 13, 2008, p. 11. –20– ���
����
��� �������� Consider Momentum conservation at one vertex implies conservation at the other. Graph is
proportional to (2π )4 δ (4) (0) Use
(2π )4 δ (4) (0) = (volume of space) × 2T .
Disconnected diagrams: –21– Alan Guth, 8.323 Lecture, May 13, 2008, p. 12.
Give each disconnected piece a name: Then
Diagram = (value of connected piece) ·
i 1
(Vi )ni .
ni ! So, the sum of all diagrams is: Factoring, –22– Factoring even more: –23– Alan Guth, 8.323 Lecture, May 13, 2008, p. 13.
For our example, Now lo ok at denominator of matrix element: –24– Finally, Summarizing, Vacuum energy density: –25– Alan Guth, 8.323 Lecture, May 13, 2008, p. 14. Final sum for fourpoint function: –26– ������� �� ��������������
������������ ������ Example: The fourpoint function: –27– Alan Guth, 8.323 Lecture, May 13, 2008, p. 15. ���������� �� ����� ��
����� ����� ��
����� ��� ��
�� ����� where
ρA and ρB = number density of particles A = cross sectional area of beams A and B Then = lengths of particle packets σ≡ Number of scattering events of speciﬁed type
.
ρA A ρB B A Note that σ depends on frame.
Special case: 1 particle in each beam, so ρA AA = 1 and ρB BA = 1. Then Number of events = σ/A .
–28– Γ≡ ���������� �� ��
�� ����� Number of decays per unit time
.
Number of particles present Number of surviving particles at time t:
N (t) = N0 e−Γt .
Mean lifetime:
τ= 1
N0 Halflife:
e−Γt = ∞ dt 0 1
2 =⇒ − dN
dt t = 1/Γ . t1 / 2 = τ l n 2 . –29– Alan Guth, 8.323 Lecture, May 13, 2008, p. 16. �������� �����
��� �� �������
�� ���������� Unstable particles are not eigenstates of H ; they are resonances in scattering
experiments.
In nonrelativistic quantum mechanics, the BreitWigner formula
f (E ) ∝ 1 E − E0 + iΓ/2 =⇒ σ ∝ 1
.
(E − E0 )2 + Γ2 /4 The “full width at half max” of the resonance = Γ.
In the relativistic theory, the BreitWigner formula is replaced by a mo diﬁed
(Lorentzinvariant) propagator:
p2 1
1
≈
,
2 + imΓ
0 − E + i(m/E )Γ/2)
−m
2E p ( p
p
p which can be seen using
2
(p0 )2 − p 2 − m2 = (p0 )2 − Ep = p0 + Ep p0 − Ep ≈ 2Ep p0 − Ep . –30– ������� ��� ����� ������� ��� ��� ����������� ����� ��
����� ��� ��� �������� Recall our discussion of particle creation by an external source,
( + m2 )φ(x) = j (x) , where j was assumed to be nonzero only during a ﬁnite interval t1 < t < t2 .
• In that case, the Fock space of the free theory for t < t1 deﬁned the instates,
the Fock space of the free theory for t > t2 deﬁned the outstates, and we
could calculate exactly the relationship between the two.
• We started in the invacuum and stayed there. The amplitude p 1 p 2 . . . p N , out 0, in was then interpreted as the amplitude for pro ducing a set of ﬁnal particles with
momenta p 1 . . . p N .
–31– Alan Guth, 8.323 Lecture, May 13, 2008, p. 17. For interacting QFT’s, it is more complicated. The interactions do not turn oﬀ, and
aﬀect even the 1particle states. It is still possible to deﬁne in and outstates
p 1 . . . p N , in and p 1 . . . p N , out with the following properties:
• They are exact eigenstates of the full Hamiltonian.
• At asymptotically early times, wavepackets constructed from p 1 . . . p N , in
evolve as free wavepackets. (The pieces of this ket that describe the scattering
vanish in stationary phase approximation at early times.) These states are used
to describe the initial state of the scattering.
• At asymptotically late times, wavepackets constructed from p 1 . . . p N , out
evolve as free wavepackets. These states are used to describe the ﬁnal state. –32– ������
��� ������� Oneparticle incoming wave packet:
φ = d3 k
( 2π )3 1
φ(k ) k , in
2E k , φ φ = 1 =⇒ d3 k
φ(k )
( 2π )3 =1. where 2 Twoparticle initial state:
φA φB , b , in = d3 kA d3 kB φA (k A ) φB (k B )e−ib ·k B
k A k B , in
(2π )3 (2π )3
(2EA )(2EB ) , where b is a vector which translates particle B orthogonal to the beam, so that
we can construct collisions with diﬀerent impact parameters.
Multiparticle ﬁnal state: φ1 . . . φn , out = n f =1 d pf φf (p f ) p 1 . . . p n , out .
( 2π )3 2E f
3 –33– Alan Guth, 8.323 Lecture, May 13, 2008, p. 18. ��� ��������� Deﬁnition: S Ψ, out = Ψ, in . Therefore
Ψ, out Ψ, in = Ψ, out S  Ψ, out .
But S maps a complete set of orthonormal states onto a complete set of
orthonormal states, so S is unitary. Therefore
Ψ, out S  Ψ, out = Ψ, out S † S S Ψ, out
= Ψ, in S  Ψ, in ,
so P&S often do not label the states as in or out. –34– No scattering � S , T , and M =⇒ ﬁnal = initial, so separate this part of S : S ≡ 1 + iT . But T must contain a momentumconserving δ function, so deﬁne p 1 . . . p n iT  kA kB ≡ (2π )4 δ (4) kA + kB − pf · iM(kA kB → {pf }) . –35– Alan Guth, 8.323 Lecture, May 13, 2008, p. 19. ���������� �� ��� ����� ��
����� The probability of scattering into the speciﬁed ﬁnal states is just the square of the
S matrix element, summed over the ﬁnal states: � AB, b → p1 . . . pn = n d pf 1 ( 2π )3 2E f
3 f =1 p 1 . . . p n S  φA φB , b 2 . To relate to the cross section, think of a single particle B scattering oﬀ of a particle
A, with impact parameter vector b : Remembering that the cross section can be viewed as the cross sectional area
blo cked oﬀ by the target particle,
dσ = d2 b � . AB, b → p1 . . . pn –36– Substituting the expression for � and writing out the wavepacket integrals
describing the initial state, n
3
d pf 1 d3 kB φA (k A ) φB (k B )
d3 kA
dσ = d2 b ( 2π )3 2E f
( 2π )3
( 2π )3
(2EA )(2EB )
f =1 ¯
d3 k A
( 2π )3 × ¯
¯
¯
d3 k B φ∗ (kA ) φ∗ (k B ) ib ·(kB −k B )
¯
A
B
e
3
¯
¯
( 2π )
(2E A )(2E B ) × p 1 . . . p n S  k A k B ¯¯
p 1 . . . p n S  k A k B ∗ . This can be simpliﬁed by using
¯
¯
d2 b eib ·(kB −k B ) = (2π )2 δ (2) k ⊥ − k ⊥
B
B , p 1 . . . p n S  k A k B = iM k A k B → {p f } (2π )4 δ (4) kA + kB −
¯¯
p 1 . . . p n S  k A k B ∗ ¯
¯
¯¯
= −iM∗ k A k B → {p f } (2π )4 δ (4) k A + k B − , pf
pf –37– . Alan Guth, 8.323 Lecture, May 13, 2008, p. 20.
¯
¯
¯
We ﬁrst integrate over k A and k B using δ (2) k ⊥ − k
⊥ and B
B
δ (4) ¯
¯
kA + kB − pf =δ ¯ ¯
k ⊥ + k ⊥ − A
B (2) p⊥ f ¯
¯
× δ EA + EB − ¯
¯z
δ k + k z − A
B
Ef pz f , ⊥
¯ ¯ where the beam is taken along the z axis. After integrating over k and k ⊥ , we A
B
are left with z z
¯z
¯ ¯z
¯ z ¯z
¯
¯ pz δ E A + E B − Ef = dk δ F (
¯ ) , k dk dk δ k + k − A
B
A
B
f
A
A ¯z
where the ﬁrst δ function was used to integrate k , and B ¯⊥
k A z
F (
¯ ) = k A Then 2 ¯z
+ k A 2 + m2 + z ¯z
k dk δ F (
¯ ) = A
A 1 dF dk ¯z ¯ 2 + k⊥ B ¯z
pz − k A
f 2 + m2 − Ef . , evaluated where F (
¯z ) = 0 . kA A –38– Rewriting
z
F (
¯ ) = k A one ﬁnds ¯⊥
k A 2 ¯z
+ k A 2 + m2 + ¯ 2 + k⊥ B z dF k ¯ = ¯A − dk E ¯z A
A ¯z
pz − k A
f z ¯ pf − k z A
¯
E
B 2 + m2 − Ef , z
¯ ¯z
Remembering the δ function constraint δ k + k − pz from the previous A
B
f
slide, one has ¯z
¯ dF
k kz = ¯A − ¯B = v z − v z  . ¯A ¯B dk E ¯z
E
B A
A
¯z
What values of k satisfy the constraint F (
¯z ) = 0? There are two solutions, since kA
A
z
¯ ) = 0 can be manipulated into a simple quadratic equation. (To see this, move F (k A
¯
one square ro ot to the RHS of the equation and square both sides. The (k ) term z 2
A
on each side cancels, leaving only linear terms and a square root on the LHS. Isolate
the square ro ot and square both sides again, obtaining a quadratic equation.) One
¯
solution gives k A = k A , and the other corresponds to A and B approaching each
other from opposite directions. Assume that the initial wavepacket is too narrow
to overlap the 2nd solution.
–39– Alan Guth, 8.323 Lecture, May 13, 2008, p. 21.
Then dσ = n d pf 1 ( 2π )3 2E f
3 f =1 d3 kA
( 2π )3 × M k A k B → {p f } 2 d3 kB φA (k A )2 φB (k B )2
z
z
(2π )3 (2EA )(2EB ) vA − vB 
(2π )4 δ (4) kA + kB − pf . Deﬁne the relativistically invariant nbo dy phase space measure dΠn (P ) ≡ n f =1 d pf 1 (2π )4 δ (4) P −
3 2E
( 2π )
f
3 pf , 2 z
z
and assume that EA (k A ), EB (k B ), vA − vB , M k A k B → {p f } , and
dΠn (kA + kB ) are all suﬃciently slowly varying that they can be evaluated at
the central momenta of the two intial wavepackets, k A = p A and k B = p B . Then
the normalization of the wavepackets implies that d3 kA
( 2π )3 d3 kB
φA (k A )2 φB (k B )2 = 1 ,
3
( 2π )
–40– so ﬁnally
2 M (p A p B → {p f })
dσ =
z
z dΠn (pA + pB ) .
(2EA )(2EB ) vA − vB 
This formula holds whether the ﬁnal state particles are distinguishable or not. In
calculating a total cross section, however, one must not doublecount ﬁnal states. If
the ﬁnal state contains n identical particles, one must either restrict the integration
or divide the answer by n!. –41– Alan Guth, 8.323 Lecture, May 13, 2008, p. 22. ���
��� ����� ���������
�� ����� ������ In the center of mass (CM) frame, p A = −p B and Ecm = EA + EB , so dΠ2 (pA + pB ) = 2 f =1 d pf 1 (2π )4 δ (4) pA + pB −
(2π )3 2Ef
3 pf = 1
d3 p1 d3 p2
(2π )4 δ (4) (pA + pB − p1 − p2 )
(2π )3 (2π )3 (2E1 )(2E2 ) = 1 d3 p1
(2π )δ (Ecm − E1 − E2 )
3 (2E )(2E )
(2π )
1
2 = dΩ 1
p2 dp1
1
(2π )δ Ecm −
3 (2E )(2E )
( 2π )
1
2 p1
1
p1
p2
+
= dΩ 1 2
(2π ) (2E1 )(2E2 ) E1
E2
p1
= dΩ
. 16π 2 Ecm p2 + m2 −
1
1 p2 + m2
1
2 −1 –42– The twoparticle ﬁnal state, centerofmass cross section is then
dσ
dΩ 2 =
cm p A  M (p A p B → p 1 p 2 )
z
z.
64π 2 EA EB (EA + EB ) vA − vB  If all four masses are equal, then
EA = EB =
and
z
z
vA − vB  = 1
Ecm
2 2 p A 
4 p A 
=
,
EA
Ecm so
dσ
dΩ =
cm M2
2
64π 2 Ecm (all masses equal) . –43– Alan Guth, 8.323 Lecture, May 13, 2008, p. 23. ���������� �� ��� ��
�� ����� The formula for decay rates is more diﬃcult to justify, since decaying particles have
to be viewed as resonances in a scattering experiment. For now we just state the
result. By analogy with the formula for cross sections,
2 M (p A p B → {p f })
dσ =
z
z dΠn (pA + pB ) ,
(2EA )(2EB ) vA − vB 
we write
2 M (p A → {p f })
dΓ =
dΠn (pA ) .
2E A
Here M cannot be deﬁned in terms of an S matrix, since decaying particles cannot
be described by wavepackets constructed in the asymptotic past. M can be
calculated, however, by the Feynman rules that Peskin & Schroeder describe in
Section 4.6. If some or all of the ﬁnal state particles are identical, then the same
comments that were made about cross sections apply here.
–44– ������� �� ��� ������� �� ��� Timedependent p erturbation theory:
Ω T {φ(x1 ) . . . φ(xn )} Ω
= 0T φI (x1 ) . . . φI (xn ) exp −i d4 z � I (z ) 0
connected = Sum of all connected diagrams with external points x1 . . . xn .
Status: derivation was more or less rigorous, except for ignoring problems connected
with renormalization: evaluation of integrals in this expression will lead to
divergences. These questions will be dealt with next term. If the theory is
regulated, for example by deﬁning it on a lattice of ﬁnite size, the formula
above would be exactly true for the regulated theory. One ﬁnds, however,
that the limit as the lattice spacing goes to zero cannot be taken unless the
parameters m, λ, etc., are allowed to vary as the limit is taken, and in addition
the ﬁeld operators must be rescaled.
–45– Alan Guth, 8.323 Lecture, May 13, 2008, p. 24.
Cross sections from S matrix elements:
S = 1 + iT ,
where
p 1 . . . p n iT  kA kB ≡ (2π )4 δ (4) kA + kB − pf · iM(kA kB → {pf }) . The relativistically invariant nbo dy phase space measure is n
3
d pf 1 (2π )4 δ (4) P −
dΠn (P ) ≡ 3 2E
(2π )
f pf , f =1 and the diﬀerential cross section is
dσ = 2 M (p A p B → {p f })
z
z dΠn (pA + pB ) .
(2EA )(2EB ) vA − vB  Status: this derivation was more or less rigorous, making mild assumptions about
in and outstates. These assumptions, and the formula above, will need to be
mo diﬁed slightly when massless particles are present, since the resulting longrange forces mo dify particle trajectories even in the asymptotic past. These
mo diﬁcations arise only in higher order perturbation theory, and are part of the
renormalization issue.
–46– Special case— twoparticle ﬁnal states, in the centerofmass frame:
dσ
dΩ 2 cm p A  M (p A p B → p 1 p 2 )
=
z
z
64π 2 EA EB (EA + EB ) vA − vB 
= M2
2
64π 2 Ecm (if all masses are equal) . Decay rate from S matrix elements:
2 dΓ = M (p A → {p f })
dΠn (pA ) .
2E A Status: completely nonrigorous at this point. Unstable particles should be treated
as resonances, an issue which is discussed in Peskin & Schroeder in Chapter 7. –47– Alan Guth, 8.323 Lecture, May 13, 2008, p. 25. ���� ������� �������� ��������� ��� �������� Complication:
p 1 . . . p n S  p A p B ≡ p 1 . . . p n , out p A p B , in ,
but the in and outstates are hard to construct: even singleparticle states are
mo diﬁed by interactions. The solution will make use of the fact that
Ω φ(x) p = Ω eiP ·x φ(0)e−iP ·x p = e−ip·x Ω φ(0) p
is an exact expression for the interacting ﬁelds, with the full operator P µ and
the exact eigenstate p . By generalizing this to in and outstates, it will be
possible to manipulate the correlation functions Ω T (φ1 . . . φn ) Ω by inserting
complete sets of in and outstates at various places. When the correlation function
is Fouriertransformed in its variables x1 . . . xn to produce a function of p1 . . . pn ,
one can show that it contains poles when any pi is on its mass shell, p2 = m2 , and
i
i
that the residue when all the pi are on mass shell is the S matrix element.
–48– A derivation will be given in Chapter 7, but for now we accept the intuitive notion
that U (t2 , t1 ) describes time evolution in the interaction picture, and that the
S matrix describes time evolution from minus inﬁnity to inﬁnity. So we write
p 1 . . . p n S  p A p B
= I p1 . . . pn T exp −i d4 x � I (x) p Ap B I connected,
amputated , where “connected” means that the disconnected diagrams will cancel out as before,
and the meaning of “amputated” will be discussed below. It will be shown in
Chapter 7 that this formula is valid, up to an overall multiplicative factor that
arises only in higherorder perturbation theory, and is asso ciated with the rescaling
of ﬁeld operators required by renormalization. –49– Alan Guth, 8.323 Lecture, May 13, 2008, p. 26.
2 → 2 �
��������� ���
������� �� Normalization conventions:
2Ep a† (p ) 0 , p =
To zeroth order in � a(q ) , a† p = (2π )3 δ (3) (q − p ) . I, p 1 p 2 S  p A p B = I p 1 p 2 p A p B
= I (2E1 )(2E2 )(2EA)(2EB ) 0 a1 a2 a† a† 0
AB = (2EA )(2EB )(2π )6 δ (3) (p A − p 1 )δ (3) (p B − p 2 )
+ δ (3) (p A − p 2 )δ (3) (p B − p 1 ) .
Graphically, Contributes only to “1” of S = 1 + iT . To ﬁrst order in I � p 1p 2 T
= I –50– I: −i λ
4! p 1p 2 N d4 x φ4 (x)
I
−i λ
4! where
contractions = −i p Ap B I d4 x φ4 (x) + contractions
I
λ
6
4! φ(x)φ(x) + 3 p Ap B I , . Uncontracted ﬁelds can destroy particles in initial state or create them in the ﬁnal
state:
φI (x) = φ+ (x) + φ− (x) =
I
I d3 k
( 2π )3 so
φ+ (x) p
I I 1
aI (k )e−ik·x + a† (k )eik·x
I
2E k = e−ip·x 0 I , , where φ+ and φ− refer to the parts of φI (x) containing annihilation and creation
I
I
operators, respectively.
–51– Alan Guth, 8.323 Lecture, May 13, 2008, p. 27. This leads to a new type of contraction: We show this kind of contraction in a Feynman diagram as an external line.
Lo oking at the contracted terms from the Wick expansion, the fully contracted
term pro duces a multiple of the identity matrix element,
−i λ
4! d4 xI p 1p 2 p Ap B I = −i λ
4! d4 x × I p 1 p 2 p A p B I , so this term also contributes only to the uninteresting 1 part of S = 1 + iT . –52– The singly contracted term
−i 6λ
4! d4 xI p 1p 2 φ(x)φ(x) p A p B I contains terms where one φ(x) contracts with an incoming particle and the other
contracts with an outgoing particle, giving the Feynman diagrams The integration over x gives an energymomentum conserving δ function, and
the uncontracted inner product produces another, so these diagrams are again a
contribution to the 1 of S = 1 + iT .
The contributions to T come from fully connected diagrams, where all external
lines are connected to each other.
–53– Alan Guth, 8.323 Lecture, May 13, 2008, p. 28. Nontrivial contribution to T: There are 4! ways of contracting the 4 ﬁelds with the 4 external lines, so the
contribution is
λ
d4 x e−i(pA +pB −p1 −p2 )·x
4! · −i
4!
= −iλ(2π )4 δ (4) (pA + pB − p1 − p2 )
≡ iM(2π )4 δ (4) (pA + pB − p1 − p2 ) ,
so
M = −λ . –54– Repeating,
M = −λ .
By our previous rules, this implies
dσ
dΩ =
cm λ2
.
2
64π 2 Ecm For σtotal one uses the fact that the two ﬁnal particles are identical. If we integrate
over all ﬁnal angles we have doublecounted, so we divide the answer by 2!. σtotal = λ2
1
=
× 4π ×
2E 2
64π cm
2! λ2
.
2
32πEcm –55– Alan Guth, 8.323 Lecture, May 13, 2008, p. 29.
����������� Consider the 2nd order diagram Contribution is
1
2 d4 p
i
4 p 2 − m2
( 2π ) i
d4 k
4 k 2 − m2
( 2π ) × (−iλ)(2π )4 δ (4) (pA + pB − p1 − p2 ) × (−iλ)(2π )4 δ (4) (pB − p ) .
Note that δ (4) (pB − p ) =⇒ p 2 = m2 , so
1
1
=,
2
p −m
0
2 which is inﬁnite. Any diagram in which all the momentum from one external line
is channeled through a single internal line will produce an inﬁnite propagator.
–56– Note that δ (4) (pB − p ) =⇒ p 2 = m2 , so
1
1
=,
2
p −m
0
2 which is inﬁnite. Any diagram in which all the momentum from one external line
is channeled through a single internal line will pro duce an inﬁnite propagator.
Amputation: Eliminate all diagrams for which cutting a single line results
in separating a single leg from the rest of the diagram.
For example, –57– Alan Guth, 8.323 Lecture, May 13, 2008, p. 30. ������� ����� ��� λφ4 �� �������� ���
�� iM · (2π )4 δ (4) (pA + pB − pf ) = (sum of all connected, amputated diagrams) ,
where the diagrams are constructed by the following rules: –58– ������� ����� ��� λφ4 �� �������� ���
�� iM = (sum of all connected, amputated diagrams) ,
where the diagrams are constructed by the following rules: –59– Alan Guth, 8.323 Lecture, May 13, 2008, p. 31. ������� ����� ��� �������� � Timedependent p erturbation theory:
Generalizes easily, since
relations:
¯
Ω T φ...ψ ...ψ ...
= I 0T I is bilinear in Fermi ﬁelds, so it obeys commutation Ω ¯
φI . . . ψI . . . ψ I . . . exp −i d4 z � I (z ) 0 I ,connected = Sum of all connected diagrams with speciﬁed external points .
But, to use Wick’s theorem, we must deﬁne timeordering and normal ordering
for fermion operators. –60– ������������ �����
�� �� ����� ������� Suppose x and y are spacelike separated, with y 0 > x0 . Then
ψ (x)ψ (y ) = −ψ (y )ψ (x) .
The RHS is already timeordered, so T {−ψ (y )ψ (x)} = −ψ (y )ψ (x). If T is to
act consistently on both sides, then
T {ψ (x)ψ (y )} = −ψ (y )ψ (x) .
Generalizing,
T {ψ1 ψ2 . . . ψn } = (product of ψ ’s ordered by time, earliest to right) × (−1)N , where N is the number of interchanges necessary to bring the ordering on the
LHS to the ordering on the RHS. (Here ψ represents a general Fermi ﬁeld, ψ
¯
or ψ .)
–61– Alan Guth, 8.323 Lecture, May 13, 2008, p. 32. ��� ������� ����������� By this deﬁnition
¯
T {ψ (x)ψ(y )} ≡ ¯
ψ (x)ψ(y )
for x0 > y 0
¯
−ψ (y )ψ (x) for y 0 > x0 . In free ﬁeld theory we have already learned that ¯
0 T { ψ ( x) ψ ( y ) } 0 = d4 p i( p + m) −ip·(x−y)
e
≡ SF (x − y ) .
(2π )4 p2 − m2 + i –62– For p = q , �������������� �����
�� �� ����� ���������� as (q )as † (p ) = −as † (p )as (q ) . The RHS is normalordered, so one presumably deﬁnes N {−as † (p )as (q )} = −as † (p )as (q ). If N is to act consistently on both sides, then
N {as (q )as † (p )} = −as † (p )as (q ) .
Generalizing,
N {product of a’s and a† ’s} = (pro duct with all a’s to the right) × (−1)N , where N is the number of interchanges necessary to bring the ordering on the
LHS to the ordering on the RHS.
–63– Alan Guth, 8.323 Lecture, May 13, 2008, p. 33. ��
��� �������� ¯
¯
T {ψ1 ψ 2 ψ3 . . .} = N {ψ1 ψ 2 ψ3 . . . + (all possible contractions)} .
A sample contraction would be –64– ...
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This note was uploaded on 11/08/2011 for the course PHY 8.323 taught by Professor Staff during the Spring '08 term at MIT.
 Spring '08
 staff
 Mass, Quantum Field Theory

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