Download Introduction to Spontaneous Symmetry Breaking

Introduction to Spontaneous Symmetry Breaking
Ling-Fong Li
Carnegie Mellon University
2011 BCVSPIN, 25, July 2011
Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking
2011 BCVSPIN, 25, July 2011
1 / 30
Outline
1
2
3
4
5
6
7
Introduction
Symmetry and Conservation Law
Explicit vs Spontaneous Symmetry Breaking
Goldstone Theorem
Spontaneous symmetry breaking in global symmetry
Spontaneous symmetry breaking in Local
symmetry-Higgs Phenomena
Standard Model of Electroweak Interaction
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2011 BCVSPIN, 25, July 2011
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Introduction
Fundamental Interactions in nature
1
2
3
Strong interactions–Quantum Chromodynamics(QCD) : gauge theory based on SU (3 )
symmetry
Electromagnetic interaction
Electroweak interaction–gauge theory based on
Weak interaction
SU (2 ) U (1 ) symmetry with spontaneous symmetry breaking
Gravitational interaction: gauge theory of local Lorentz symmetry
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Symmetry and Conservation Law
Noether’s Theorem: Any continuous transformation which leaves the action
S =
Z
d 4x L
invariant, will give a conserved current
∂µ J µ = 0
which gives conserved charge
dQ
= 0,
dt
Q =
Symmetry Transformation
time translation
t ! t +a
!
!
!
space translation x ! x + b
rotation
Z
d 3 x J0
Conserved Charge
Energy
Momentum
Angular momentum
Other conserved quantities: Electric charge, Baryon number,. . .
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Remarks
Conservation laws =) stability of particles
e.g. Baryon number conservation =) proton is stable, Charge conservation =) electron
is stable, Dark Matter ?
Conservation laws might change as we gain more knowledge
e.g. muon number is violated only when ν oscillations were observed
Parity violation was discovered only in late 50’s, CP violation ...
In quantum system, symmetries =) degenercies of energy levels
e.g. rotational symmetry =) (2l + 1 ) degeneracies
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2) Symmetry Breaking
Most of the symmetries in nature are approximate symmetries.
(a ) Explicit breaking–
H = H0 + H1 ,
H1
does not have the symmetry of H 0
!
Example: Hydrogen atom in external magnetic …eld B
H0 =
!2
p
2m
Ze 2 1
,
4πε0 r
H1 =
! !
µ B
!
H 0 is invariant under all rotations while H 1 is invariant only for rotation along direction of B
Add small non-symmetric terms to the Hamiltonian =) degeneracies reduced or removed
(b ) Spontaneous breaking:
Hamiltonian has the symmetry, [Q , H ] = 0 but ground state does not, Q j0 i 6= 0 (Nambu 1960,
Goldstone 1961)
Example: Ferromagnetism
T > T C (Curie Temp) all magnetic dipoles are randomly oriented–rotational symm
T < T C all magnetic dipoles are in the same direction-not invariant under rotation
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Ginsburg-Landau theory:
!
Write the free energy u as function of magnetization M in the form,
!
!
!
α2 > 0,
α1 = α (T
!
!
!
u (M ) = ( ∂t M )2 + α1 (T ) (M M ) + α2 (M M )2 ,
where
TC )
α>0
Here u has O (3 ) rotational symmetry. The ground state is at
!
!
!
M (α1 + 2α2 M M ) = 0
!
T > T C minimum at M = 0. r
!
T < T C minimum at M =
!
α1
6= 0 . If we choose M to be in some direction the
2α2
rotational symmetry is broken.
Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking
2011 BCVSPIN, 25, July 2011
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Goldstone Theorem
Noether theorem: continuous symmetry =) conserved current,
∂µ J µ = 0,
Q =
Z
d 3 xJ 0 (x ) ,
dQ
=0
dt
Suppose A (x ) and B (x ) are some local operators and transform into each other under the
symmetry charge Q ,
[Q , A (0 )] = B (0 )
Suppose
h0 j[Q , A (0 )]j 0 i = h0 jB (0 )j 0 i = v 6= 0,
symmetry breaking condition
This implies
Q j0 i 6= 0,
and Q is a broken charge. Then
E n = 0,
as
!
p n = 0,
for some state n
This means zero energy excitations.
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To show this, write
h0 j[Q , A (0 )]j 0 i =
Z
d 3 x 0 J 0 (x ) , A (0 ) 0
Inserting a complete set of states and integating over x
h0 j[Q , A (0 )]j 0 i
=
!
∑ (2π )3 δ3 ( p n )
n
n
0 J 0 (0 ) n hn jA j 0 i e
iE n t
h0 jA j n i n J 0 (0 ) 0 e iEn t
RHS is time independent while LHS depends explicitly on time from e
satis…ed only if there exists an intermediate state jn i for which
E n = 0,
for
iE n t .
o
= υ 6= 0
This relation can be
!
pn = 0
For relativistic system, energy momentum relation yields
En =
q
!2
pn
+ m n2
)
m n = 0, Goldstone boson
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Remarks
1
State jn i with property
h0 jA (0 )j n i 6= 0,
is the massless Goldstone boson while local …eld B corresponds to some massive particle.
This state will also have the property
hn jQ j 0 i 6 = 0
2
This means jn i can connect to vacuum through broken charge Q .
Spectrum
Here Q is a broken symmetry charge. In general, each broken charge will have a Goldstone
boson,
# of Goldstone bosons= # of broken generators
In many cases, there are symmetry charges which remain unbroken,
Q i j0 i = 0,
i = 1, 2,
,l
These unbroken charges Q 1 , Q 2 ,
Q l form a group H , a subgroup of original symmetry
group G . As a consequence, the particles will form multiplets of symm group H . For
example, if symmetry SU (2 ) SU (2 ) is spontaneously broken to SU (2 ) , particles will
form SU (2 ) multiplets
Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking
2011 BCVSPIN, 25, July 2011
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3
In relativistic …eld theory, massless particle =) long range force. In real world , no
massless particls except photon.
In strong interaction, we have approximate spontaneous symmetry breaking. =) almost
Goldtone bosons
4
Goldstone bosons can either be elementary …elds or composite …elds
a) Elementary …elds
Here local operators A, B are …elds in the Lagrangian–this is the case in Standard
electroweak theory
b) Composite …elds
In this case A, B correspond to some bound states of elementary …elds.
For example in QCD the ‡avor chiral symmetry is broken by quark condensates. Here we
have
_
_
Q 5 , qγ5 q = qq
Q 5 some
chiral charge, q quark …eld
_
_
If 0 qq 0 6= 0 ( quark condensate), then qγ5 q will correspond to a Goldstone boson.
This is usually referred to as dynamical symmetry breaking.
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2011 BCVSPIN, 25, July 2011
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Global symmetry
We now discuss the spontaneou symmetry in some simple relativistic …eld theory. As an
example, consider
1
L = [(∂µ σ)2 + (∂µ π )2 ] V (σ2 + π 2 )
2
with
V ( σ2 + π 2 ) =
λ
µ2 2
( σ + π 2 ) + ( σ 2 + π 2 )2
2
4
(1)
This potential has O (2 ) symmetry
σ
π
!
σ0
π0
=
cos α
sin α
sin α
cos α
σ
π
where α is a constant, global symm
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To get the minimum we solve the equations
∂V
=
∂σ
µ2 + λ ( σ 2 + π 2 ) σ = 0
∂V
=
∂π
µ2 + λ ( σ 2 + π 2 ) π = 0
Solution
σ2 + π 2 =
µ2
= ν2
λ
circle in σ
π plane
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Choose σ = ν, π = 0 which are …elds con…gurations for classical vacuum. New quantum …elds
σ0 = σ
ν , π0 = π
correspond to oscillations around the minimum. Note that
hσi = v 6= 0
is the symmetry breaking condition in Goldstone’s theorem. The new Lagrangian is
L=
1
[(∂µ σ02 + (∂µ π )2 ]
2
µ2 σ 02
λνσ0 (σ02 + π 02 )
λ 02
( σ + π 02 )2
4
There is no π 02 term, ) π 0 massless Goldstone boson
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Local symmetry: gauge symmetry
Maxwell Equations:
! !
r E =
!
r
! !
ρ
,
ε0
!
r B =0
!
∂B
E+
= 0,
∂t
1 !
r
µ0
!
!
!
∂E
B = ε0
+J
∂t
!
Source free equations can be solved by introducing scalar and vector potentials, φ, A,
!
!
B =r
!
A,
!
E =
!
!
rφ
∂A
∂t
These solutions are not unique because of gauge invariance,
φ !φ
∂α
,
∂t
!
A
!
!
! A + rα
Or
Aµ
! Aµ
∂ µ α (x )
Classically, it is not so clear what the physical role is played by gauge invariance.
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In quantum mechanics, Schrodinger eq for charged particle is,
"
1
2m
h!
r
i
! 2
eA
#
eφ ψ = i h
∂ψ
∂t
This requires transformation of wave function,
ψ ! exp i
e
α (x ) ψ
h
to get same physics. Thus gauge invariance is now connected to symmetry (local)
transformation.
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2011 BCVSPIN, 25, July 2011
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Spontaneous symmetry breaking in local symmetry
Local symmetry : symm parameter depends on space-time.
Consider a …eld theory with U (1 ) local symmetry,
L = Dµ φ
†
(D µ φ ) + µ2 φ † φ
λ φ† φ
2
1
F µν F µν
4
where
D µ φ = ∂µ
F µν = ∂µ A ν
igA µ φ,
∂ν A µ
This local U (1 ) phase transformation is of the form,
φ (x ) ! φ 0 (x ) = e
iα
φ (x ) ,
α = α (x )
The derivative transforms as
∂ µ φ (x ) ! ∂ µ φ 0 (x ) = e
iα
∂ µ φ (x )
i ∂µ α
which is not a phase U (1 ) transformation.
Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking
2011 BCVSPIN, 25, July 2011
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Introduce A µ to form covariant derivative,
D µ φ = ∂µ
igA µ φ
which will a simple U (1 ) transformation
Dµ φ
0
=e
iα
Dµ φ
if we require
A µ (x ) ! A 0µ (x ) = A µ (x )
∂ µ α (x )
which is the gauge transformation in Maxwell’s equations.
Important features of local symmetry: gauge coupling is universal. Also gauge …eld (photon) is
massless(long range force), because mass term A µ A µ is not gauge invariant.
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Symmetry Breaking
When µ2 > 0, minimum of potential
V (φ) =
is
φ† φ =
Write
µ2 φ † φ + λ φ † φ
v2
,
2
with
v2 =
2
µ2
λ
1
φ = p ( φ1 + i φ2 )
2
Quantum …elds are the oscillations around the minimum,
h0 j φ1 j 0 i = v ,
h0 j φ2 j 0 i = 0
As before φ2 is a Goldstone boson.
New feature : covariant derivative produces mass term for gauge boson,
Dµ φ
2
=
∂µ
igA µ φ
2
' ...
g 2v 2 µ
A Aµ + . . .
2
with mass
M = gv
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Write the scalar …eld in polar coordinates
1
φ = p [v + η (x )] exp (i ξ/v )
2
We can now use gauge transformation to transform away ξ. De…ne
1
φ" = exp ( i ξ/v ) φ = p [v + η (x )]
2
and
Bµ = Aµ
1
∂µ ξ
gv
ξ disappears. In fact ξ becomes the longitudinal component of B µ .
massless gauge boson + Goldstone boson=massive vector meson
all long range forces disappear. This was discovered in the 60 0 s by Higgs, Englert & Brout,
Guralnik, Hagen & Kibble independently and is usually called Higgs phenomena
The extension to non-abelian local symmetry(Yang-Mills …elds) is straightforward and no
new features emerge.
Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking
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Connection with superconductivity
Equation of motion for scalar …eld interacting with em …eld,
!
!
r
with
!
J = ie φ†
!
r
!
B =J
!
ie A
!
!
r + ie A φ† φ
φ
Spontaneous symmetry breaking ) φ = v and
!
!
J = e2v 2 A
This is London equation. Then
!
r
!
r
!
B
!
=r
This gives Meissner e¤ect.
!
!
J,
=)
!
!
r2 B = e 2 v 2 B
!
For the static case, ∂0 A = 0, A 0 = 0, we get E = 0 and from Ohm’s law
!
!
E = ρJ ,
ρ resistivity
we get ρ = 0, i.e. superconductivity.
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Standard Model of Electroweak Interactions
Important features of Weak Interactions :
Universality of coupling strength
e.g. β decay and µ decay can be described by same coupling G F '
theory,
G
Lwk = pF J µ J µ† + h.c .
2
h_
where J µ = νγµ (1
10 5
M p2
in 4-fermion
i
γ5 ) e +
Short range interaction
If weak interaction is mediated by a massive vector bosons W
Lwk =g J µ W µ + h.c . ,
g2
G
= pF
M w2
2
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Then W boson is massive
These features suggest a theory gauge interaction with massive gauge boson.
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Standard Model (Weinberg 1967, ’t Hooft 1971)
This is a gauge theory with spontaneous symmetry breaking.
!
Gauge group: SU (2 ) U (1 ) gauge bosons: A µ , B µ –includes both weak and em interaction
Spontaneous Symmetry Breaking :SU (2 ) U (1 ) ! U (1 )em
Simple choice for scalar …elds
φ+
φ=
,
φ0
The Lagrangian containing φ is
where
Dµ φ =
V (φ) =
∂µ
†
(D µ φ )
V (φ)
ig ! !
τ Aµ
2
ig 0
Bµ
2
Lφ = D µ φ
µ2 φ † φ + λ φ † φ
φ
2
Spontaneous symmetry breaking: SU (2 ) U (1 ) ! U (1 )em
Minimum
h
i
∂V
=
µ2 + 2 ( φ † φ ) φi = 0
∂φi
=)
µ2 + 2 ( φ † φ ) = 0
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Simple choice
1
h φ i0 = p
2
0
ν
ν=
r
µ2
λ
Expand the quanutm …elds around the minimum
φ = φ 0 + h φ i0 =
φ0+
v
φ +p
!
00
2
the quadratic terms in V are
V 2 (φ) =
µ2 φ 0 † φ 0 +
0
λv 2 0† 0
φ φ + 2 Re φ 0
2
2
0
v 2 = 2 Re φ 0
2
v2
0
Thus φ0+ and Im φ 0 are Goldstone bosons and the symmetry breaking is
SU (2 )
U (1 ) ! U (1 )em
From covariant derivative in Lφ
Lφ =
0
0 0
0
0 0
g Bµ
~τ ~A µ
v 2 † ~τ ~A µ
g Bµ
χ (g
+
)(g
+
)χ +
2
2
2
2
2
,
χ=
0
1
Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking
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we get the mass terms for the gauge bosons,
Lφ
=
=
v2 2
fg [(A 1µ )2 + (A 2µ )2 ] + (gA 3µ
8
1
2
MW
W +µ W µ + M Z2 Z µ Z µ +
2
0
g B µ )2 g +
where
1
W µ+ = p (A 1µ
2
The …eld
Zµ = p
1
g2 + g
02
iA 2µ ),
0
(g A 3
Aµ = p
gB µ ),
1
g2 + g
2
MW
=
02
M Z2 =
g 2v 2
4
g2 + g
4
02
v2
0
(g A 3µ + gB µ )
is massless photon.
We can write the scalar …elds in the form
! !
φ = exp i τ
ξ (x ) /v
0
v + η (x )
Ling-Fong Li (Carnegie Mellon University) Introduction to Spontaneous Symmetry Breaking
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!
where ξ (x ) are Goldstone bosons. We can make a " gauge transformation"
!
0
v + η (x )
φ0 = U ( ξ )φ =
! !0
τ Aµ
2
!
= U(ξ )
! !
τ Aµ
U
2
1
!
i
g
(ξ )
!
∂µ U ( ξ ) U
1
!
(ξ )
B µ0 = B µ
where
!
U ( ξ ) = exp
! !
iτ
ξ (x ) /v
!
Again the Goldstone bosons ξ (x ) will be eaten up by gauge bosons to become massive. The
left over …eld η (x ) is usually called Higgs Particle.
Massive gauge bosons:
Wµ =
p1
2
A 1µ
g2v2
4
g2
2
4 cos θ W
iA 2µ
2 =
MW
Z µ = cos θ W A 3µ sin θ W B µ
A µ = sin θ W A 3µ + cos θ W B µ
where
tan θ W =
M Z2 =
Mγ = 0
v2
g0
g
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Note that
M Z2
2
MW
=1
cos2 θ W
The mass for the Higgs particle is
m η2 = 2λv 2
From comparing Standard Model with 4-fermion theory, we get
1
v = qp
' 246 Gev
2G F
Since magnitude of Higgs self coupling λ is not known, we do not know the mass of Higgs
particle.
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Fermion mass and spontaneous symmetry breaking
Another role played by spontaneous symmetry breaking is to give masses to the fermions.
Fermions:
a) Leptons
Li =
νe
e
,
L
νµ
µ
,
L
ντ
τ
,
L
R i = e R , µR , τ R ,
b) Quarks (Glashow, Iliopoupos, and Maiani, Kobayashi and Maskawa)
q iL =
u0
d
,
L
c0
s
,
L
t0
b
,
U iR = u R , cR , tR , D iR = d R , sR , b R
L
All left-handed fermions are in SU (2 ) doublets and right-handed fermions are all singlets.
Yukawa coupling:
_
LY = fij L i R j φ + h.c . +
Fermions get their masses from spontaneous symmetry breaking through Yukawa couplings,
m ij = fij v
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2011 BCVSPIN, 25, July 2011
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This implies that the Yukawa couplings ∝ masses. This an example of spontaneous breaking of
global symmetry(chiral symmetry),
Li
! L i0 = exp ( i
! !
τ α
)L i ,
2
Ri
! R i0 = R i
Remark:
The scalar particles φ plays 2 important roles in Standard Model:
1
It breaks the gauge symmetry through the universal gauge coupling
2
It give mass to fermions through Yukawa couplings which are not universal
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