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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 1, 2008, p. 1. MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Physics Department
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Corrections to Problem Set 9: Problem 2: Notational improvement. The two equations should read:
3 Λ 1 B3 (η ) ≡ e−iηK = e−iηS
2
3 Λ 1 R3 (θ ) ≡ e−iθJ = e−iθS
2 03 12 .
. i
Problem 4: Should read S µν = 4 [γ µ , γ ν ] in the preamble, and identity (vi)
should read
/q
p / = 2p · q − q / = p · q − 2iS µν pµ qν .
/p ���� ����
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������ ��� �� ���� –1– Alan Guth, 8.323 Lecture, May 1, 2008, p. 2. ������� �� ����� �� ���
Poincar´ symmetry implies
e
µ µ eiPµ x ψa (y ) e−iPµ x = ψa (x + y ) ,
U −1 (Λ)ψa (x)U (Λ) = Mab ψb (Λ−1 x) , (68) where M (Λ) is a representation of the Lorentz group. Following Dirac we have
chosen
µν
i
i
(69)
Mab ≡ Λ 1 ,ab = e− 2 ωµν S , with S µν = [γ µ , γ ν ] ,
2
4
and following P&S we have made a particular choice of the γ matrices. P 2
is a Casimir operator, which we take to equal m2 , the square of the particle
mass, and we assume that P 0 > 0.
���� ���� –2– �����
������� ��������� �� ��
������� ������ ��� �� ���� Creation operators transform as U (Λ) a† (p ) U −1 (Λ) =
s EΛp
Ep s� a†� (Λp ) Ds� s RW (Λ, p ) .
s (70) Taking the adjoint: U (Λ) as (p ) U −1 (Λ) = = EΛp
Ep
EΛp
Ep ∗
as� (Λp ) Ds� s RW (Λ, p )
s� s� (71)
−
Dss1 RW (Λ, p ) as� (Λp ) ,
� ���� ����
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�������
������ ��� �� ���� –3– Alan Guth, 8.323 Lecture, May 1, 2008, p. 3. Expand the ﬁeld as
d3 p
( 2π )3 ψa (x) = 1
2E p µ s µ s
as (p ) us (p ) e−ipµ x + b† (p ) va (p ) eipµ x
a
s , (72)
where p0 = + p 2 + m2 . The sign of p0 , which follows from our assumption that
the operator P 0 was positive, will be important. The above equations imply that
� −
us (Λp ) = Λ 1 ,ab us (p )Ds� 1 RW (Λ, p ) .
b
a
s
2 (73) Writing ψa (x) as ψ1 (x) ψ (x) ψa (x) = 2
=
ψ3 (x)
ψ4 (x) ψL (x)
ψR (x) , (74) ���� ����
�����
������� ��������� �� ��
������� –4– ������ ��� �� ���� we found that rotations in the rest frame implied that
us ,a (p = 0) =
L √ s
m δa , (75) up to an arbitrary normalization. This is often written as
us (p = 0) =
L √ m ξ s , where ξ + = 1
0 , ξ− = 0
1 , (76) where ξ + corresponds to spinup, and ξ − corresponds to spindown, along the
z axis. The constraints on uR are identical, so we similarly choose
us (p = 0) =
R √ m ξs . (77) In both cases the normalization is an arbitrary convention, and we can cho ose them
to have the same normalization without any loss of generality.
���� ����
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������ ��� �� ���� –5– Alan Guth, 8.323 Lecture, May 1, 2008, p. 4. To discuss nonzero momenta, we bo ost from the rest frame:
us ( q ) = √ ξs
ξs m Λ 1 (B q )
2 . (78) We will calculate this explicitly in the canonical basis:
ˆ
Λ 1 (Bp ) = e−iηp·K ,
2 (79) where η is the rapidity, with cosh η = E /m, sinh η = p /m, and tanh η = v .
(Earlier I called the rapidity ξ , but ξ is now being used for the basis spinors.)
In our basis
i σj
0
.
K j = J 0j = −
0 −σ j
2
Λ 1 (Bp ) = exp
2 1
η
2 −p · σ
ˆ
0 0
p·σ
ˆ . Note that (p · σ )2 = 1, so
ˆ
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������� ��������� �� ��
������� –6– ������ ��� �� ���� 1
η
2 −p · σ
ˆ
0 0
p·σ
ˆ =1+ η
2 −p · σ
ˆ
0 0
p·σ
ˆ + 1
3! 1
0 0
1 Λ 1 (Bp ) = exp
2 = η
2 3 cosh + 1
2! −p · σ
ˆ
0 0
p·σ
ˆ η
+
2 −p · σ
ˆ
0 η
2 2 1
0 0
1 + +...
0
p·σ
ˆ sinh η
2 . To simplify, look at 2 × 2 blo cks:
ΛL
0 Λ 1 (B p ) =
2 0
ΛR , where Λ L = cosh
R η
η
∓ p · σ sinh
ˆ
2
2 . Then remember that e±η = (E ± p )/m, so
ΛL = 1
2 E + p 
ˆ
[1 − p · σ ] +
m E − p 
ˆ
[1 + p · σ ]
m . ���� ����
�����
������� ��������� �� ��
�������
������ ��� �� ���� –7– Alan Guth, 8.323 Lecture, May 1, 2008, p. 5. ΛL = 1
2 E + p 
[1 − p · σ ] +
ˆ
m The eigenvalues of p · σ are ±1, so
ˆ ΛL = = E − p 
[1 + p · σ ]
ˆ
m E −p 
m if p · σ = 1
ˆ E +p 
m . if p · σ = −1
ˆ E−p ·σ
.
m If the eigenvalues of a matrix are positive, we deﬁne its square ro ot by diagonalizing
the matrix and taking the positive square ro ot of its eigenvalues. ���� ����
�����
������� ��������� �� ��
������� –8– ������ ��� �� ���� Similarly,
ΛR = E +p ·σ
.
m The notation can be simpliﬁed further by deﬁning 4vectors σ µ = (1, σ i ) and
σ µ = (1, −σ i ), so
¯
p·σ
p·σ
¯
, ΛR =
,
ΛL =
m
m
and ﬁnally
us ( p ) = √
p · σ ξs
√
p · σ ξs
¯ . (80) ���� ����
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������ ��� �� ���� –9– Alan Guth, 8.323 Lecture, May 1, 2008, p. 6. ������ ���������� ��� us(p ) (p · σ )(p · σ ) = p2 = m2 .
¯ (81) Normalization:
† u ( p ) u( p ) = ξ †√ p·σ ξ †√ p·σ
¯ √
p · σ ξs
√
p · σ ξs
¯ . = ξ † (p · σ + p · σ )ξ
¯ (82) = 2E p ξ † ξ .
¯
u† ( p ) γ 0 u( p ) ≡ u( p ) u( p )
√
√
√
√
¯
¯
= ξ † p · σ p · σξ + ξ † p · σ p · σξ (83) = 2m ξ † ξ . �����
������� ��������� �� ��
������� –10– ���� ���� ������ ��� �� ���� √
s √p · σ ξ u (p ) =
p · σ ξs
¯
s u s ( p ) us ( p ) = γ · p + m ,
¯ =⇒ (84) s ¯
where us (p ) ≡ us† (p )γ 0 . For the v s,
� −
us (Λp ) = Λ 1 ,ab us (p )Ds� 1 RW (Λ, p ) .
b
a
s
2 (73) is replaced by
� −
s
s
va (Λp ) = Λ 1 ,ab vb (p )Ds� 1 ∗ RW (Λ, p ) .
s
2 (85) ���� ����
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������ ��� �� ���� –11– Alan Guth, 8.323 Lecture, May 1, 2008, p. 7. From lo oking at the rest frame, we use
D ∗ (R ) = σ 2 D (R )σ 2 ,
and following the same logic that we used to show that ua s (p ) must be a
multiple of the identity, here we show that (vσ 2 )a s must be a multiple of the
identity. The normalization is arbitrary, so we can write
√
√
s
2
s
2
(86)
vL,a (p = 0) = −i m σas βL , vR,a (p = 0) = i m σas βR ,
where βL and βR are used here to indicate our freedom to cho ose the
normalization of these terms arbitrarily, as far as the Lorentz transformations
of the ﬁeld are concerned. We will so on ﬁnd, however, that these constants
are constrained.
Bo osting to an arbitrary momentum,
√ 2s
p √ · σ (−iσ ξ )βL , v (p ) = − p · σ (−iσ 2 ξ s )βR
¯
s (87) ���� ����
�����
������� ��������� �� ��
������� –12– ������ ��� �� ���� 1 and ξ − = 0 ξ=
0
1
describe spinup and spindown particles. We can also deﬁne
where as before + η s ≡ −iσ 2 ξ s , (88) and the two spinors η s , for s = + and s = −, will form a pair of orthonormal
vectors in the space of 2component spinors. Then one has
√ s
p √ · σ η βL , v (p ) = − p · σ η s βR
¯
s (89) The spin sum for the v ’s then becomes ∗
−mβL βR v (p ) v (p ) = ¯
p · σ βR 2
¯
s s s p · σ βL 2 ∗ .
−mβR βL (90) ���� ����
�����
������� ��������� �� ��
�������
������ ��� �� ���� –13– Alan Guth, 8.323 Lecture, May 1, 2008, p. 8. ���� ��� ��������
� A key result of quantum ﬁeld theory is the SpinStatistics Theorem, which
says that particles of integer spin must be bosons, while particles of halfinteger
spin must be fermions. (Note that there is no such theorem in nonrelativistic
quantum theory.)
A full pro of is beyond this course. See, for example, PCT, Spin & Statistics,
and All That, by R.F. Streater and Arthur S. Wightman, 1964. Despite the
informal title and mo dest length, this bo ok is not easy reading for most of us.
We will, however, demonstrate a limited version of this theorem. We will
consider the possibility that the Dirac particles are either bosons or fermions,
and show that only the second choice gives a consistent theory. �����
������� ��������� �� ��
������� –14– ���� ���� ������ ��� �� ���� ��� ����� �����������
���� ����� ����������� Consider ﬁrst the boson possibility. Bosons are like the scalar particles that we have
studied, with creation/annihilation operator commutators
as (p ) , ar† (q ) = (2π )3 δ rs δ (3) (p − q ) .
bs (p ) , br† (q ) = (2π )3 δ rs δ (3) (p − q ) , (91h) with other commutators ([a , a], [b , b], and their adjoints) vanishing. The “h”
in the equation number stands for “hypothesis”. It will turn out NOT to be
consistent. We will now compute the equaltime commutator
ψ (x ,0) , ψ † (y ,0)
which is clearly a spacelike separation. There are two reasons why this should
vanish.
���� ����
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������� ��������� �� ��
�������
������ ��� �� ���� –15– Alan Guth, 8.323 Lecture, May 1, 2008, p. 9. ��� ��� ���������� ������ ������
�� ���
����� ����������� �� The ﬁelds should represent measurable quantities, so a nonzero commutator
implies an uncertainty principle — a precise measurement of one causes the
other to become completely uncertain. But if this happens at spacelike
separations, then information can be transmitted faster than light. �� Later we will use these ﬁelds to describe terms in an interaction Hamiltonian,
and perturbation theory will allow us to express evolution (and hence scattering
cross sections) in terms of timeordered products. However, the timeordering
between 2 spacetime points is ambiguous if they are spacelike separated, so
a unique answer is found only if the operators commute. So, the Lorentz
invariance of interactions calculated in QFT would be violated if ﬁelds did not
commute at spacelike separations.
The requirement that commutators vanish at spacelike
separations is called ���������� ���� ����
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������� ��������� �� ��
������� –16– ������ ��� �� ���� ���
������� �� ��� ����������
From Eq. (72), ψa (x , 0) =
= d3 p
( 2π )3 1
2E p 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 s
as (p ) us (p ) + b† (−p ) va (−p ) eip ·x .
a
s s (92)
So ¯
ψ (x ,0) , ψ (y ,0) =
× ei(p ·x −q ·y ) d3 p d3 q
( 2π )6 1
( 2E p ) ( 2E q ) a r ( p ) , a s † ( q ) u r ( p ) us ( q )
¯ (93) r,s + br† (−p ) , bs (−q ) v r (−p ) v s (−q ) .
¯
–17– Alan Guth, 8.323 Lecture, May 1, 2008, p. 10. Using the creation/annihilation operator commutators (91h) and the spin sums (84)
and (90), this becomes ¯
ψ (x ,0) , ψ (y ,0) = Ep d3 p 1 ip ·(x −y )
e
×
( 2π )3 2E p ∗
m (1 + βL βR )
1 − βR 2 + p · σ 1 + βR 2 Ep 1 − βL 2 − p · σ 1 + βL 2 .
∗
m (1 + βR βL )
(94) The expression has particle contributions, with no β ’s, and antiparticle contribu
tions, with 2 β ’s. The killer terms are the ones in p · σ , since the co eﬃcients
of particles and antiparticles add. Lo oking back to Eq. (93), one sees that the
antiparticle contribution has 2 sign changes relative to the particle contribution:
the signs of p and q are reversed, and the commutator has the creation operator
ﬁrst.
���� ����
�����
������� ��������� �� ��
������� –18– ������ ��� �� ���� The term proportional to p · σ does not vanish when integrated, as it is proportional
to
−iσ · ∇ d3 p 1 ip ·(x −y )
e
,
( 2π )3 2E p (95) where the integral does not vanish. ���� ����
�����
������� ��������� �� ��
�������
������ ��� �� ���� –19– Alan Guth, 8.323 Lecture, May 1, 2008, p. 11. ��� ������� �����������
���� ����� �����
���������� This time let us assume that the particles are fermions, which means that only one
particle can have a speciﬁed momentum and spin, and that the multiparticle
basis state
p 1 s1 , . . . , p N sN
is antisymmetric in its particle labels. For example,
p 1 s1 , p 2 s2 = − p 2 s2 , p 1 s1 . (96) For fermions the creation and annihilation operators satisfy anticommutation
relations
as (p ) , ar† (q ) = (2π )3 δ rs δ (3) (p − q ) .
(97)
bs (p ) , br† (q ) = (2π )3 δ rs δ (3) (p − q ) ,
���� ����
�����
������� ��������� �� ��
�������
������ ��� �� ���� –20– Note that the antisymmetry of states, as in Eq. (96), implies that
as1 † (p 1 ) as2 † (p 2 ) = −as2 † (p 2 ) as1 † (p 1 ) , (98) since the creation operators create multiparticle states with the particle
labels in the same order as the operators. Thus there is no possibility that
fermion creation/annihilation operators could obey commutation rather than
anticommutation relations.
Since the ﬁelds are linear in creation and annihilation operators, if the creation
and annihilation operators satisfy anticommutation relations, then so must the
ﬁelds.
If the ﬁelds anticommute at spacelike separations, that is okay. It means that
any measurable quantity — any quantity that can appear in the Hamiltonian
— must be bilinear in the fermi ﬁelds, and then they will commute. We might
have expected this anyway for a fermi ﬁeld, since under a 360◦ rotation it turns
into minus itself! If such a ﬁeld were measurable, what would it look like?
���� ����
�����
������� ��������� �� ��
�������
������ ��� �� ���� –21– Alan Guth, 8.323 Lecture, May 1, 2008, p. 12. ���
������� �� ��� ����
���������
The calculation lo oks much like the previous one, with one key diﬀerence. In the
boson calculation we found that the antiparticle contribution was proportional
to bs† (−p ) , bs (−q ) , which produced a minus sign because the creation and
annihilation operators were not in their usual order. This time, however, we
ﬁnd an anticommutator instead, so there is no change in sign. Our ﬁnal
answer will be identical to the previous one, except that the entire antiparticle
contribution will change sign:
¯
ψ (x ,0) , ψ (y ,0) = Ep d3 p 1 ip ·(x −y )
e
×
( 2π )3 2E p ∗
m (1 − βL βR )
1 + βR 2 + p · σ 1 − βR 2 Ep 1 + βL 2 − p · σ 1 − βL 2 .
∗
m (1 − βR βL )
(99) The change in sign makes all the diﬀerence!
���� ����
�����
������� ��������� �� ��
������� –22– ������ ��� �� ���� ���������
�� �� ��� ���
�������
¯
ψ (x ,0) , ψ (y ,0) = Ep d3 p 1 ip ·(x −y )
e
×
( 2π )3 2E p ∗
m (1 − βL βR )
1 + βR 2 + p · σ 1 − βR 2 Ep 1 + βL 2 − p · σ 1 − βL 2 .
∗
m (1 − βR βL )
(99) The previously fatal p · σ terms will cancel if
βL 2 = βR 2 = 1 . (100) The terms in m will cancel as long as βL and βR have the same phase. Thus,
we require
βL = βR .
(101)
From now on we will adopt the standard phase convention,
(102)
βL = βR = 1 .
���� ����
�����
������� ��������� �� ��
�������
������ ��� �� ���� –23– Alan Guth, 8.323 Lecture, May 1, 2008, p. 13. d3 p 1 ip ·(x −y )
e
×
( 2π )3 2E p ¯
ψ (x ,0) , ψ (y ,0) = Ep ∗
m (1 − βL βR )
1 + βR 2 + p · σ 1 − βR 2 Ep 1 + βL 2 − p · σ 1 − βL 2 .
∗
m (1 − βR βL )
(99) The terms in Ep do not cancel, but notice that the Ep cancels the same factor
in the denominator, pro ducing a Dirac δ function which vanishes when x = y . �����
������� ��������� �� ��
������� ���� ���� ������ ��� �� ���� –24– ...
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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
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 Mass, Quantum Field Theory

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