127 - pr 03 - linear sigma model: The interactions of pions at low energy can be described by a phenomenological model called the “linear sigma” model. Essentially: Es sentially: this model consists of N real scalar sca lar i 4 fields coupled by a φ -interaction that is symmetric under rotations of the N fields. More specifically: let Φ ( x) , i = 1,..., N be a set of N fields governed by the Hamiltonian ,
∫
H = H ⋅ d 3 x =
∑ ∫( i = N i =1
1 2
(Π i ) 2 + 12 (∇Φ i )2 + V i (Φ 2 )) ⋅ d 3 x; (Φ i )2 = Φ i • Φ i ; V i (Φ 2 ) =
m2 (Φ i )2 + 14 λ ((Φ i )2 )2 ; (1.1)
1 2
Note V (Φ 2 ) is a function symmetric under rotations of Φ . For (classical) field-configurations of Φi ( x) that are constant in space and time, this term gives the only contribution to H = H int int ; hence, V is the field potential energy. Sidenote: What does this Hamiltonian have to do with the strong interactions? There are two types of light quarks: u, d . d . These quarks have identical strong interactions, but different masses. If these quarks are massless, the Hamiltonian of the strong interactions is invariant to unitary transformations of the 2-component object (u , d ) ; i.e., (u, d ) → (u , d )′ = e iα •σ /2 (u , d ) . This transformation is called an isospin rotation. If, in addition, the strong interactions are described by a vector “gluon” field, as is true in QCD, the strong interaction Hamiltonian is invariant to the isospin rotations done separately on the left-handed and righthanded components of the quark fields. Thus, the complete symmetry of QCD with two massless quarks is SU (2) × SU (2 ) . It happens that SO(4), the group of rotations in 4 dimensions, is isomorphic to SU (2) × SU (2 ) , so for N = 4, the linear sigma sigma model has has the same symmetry symmetry group group as the strong strong interactions interactions . (a) analyze the linear sigma model for m2 > 0 by noticing that, for λ = 0 , the Hamiltonian Hamiltonian given given above is exactly N copies of the Klein-Gordon Hamiltonian.
Putting λ = 0 into (1.1), we have that V i (Φ 2 ) = 12 m 2 (Φ i )2 , so we get, H =
∑ ∫ i=N i =1
i
i
i
( 12 (Π ) 2 + 12 (∇Φ )2 + 12 m2 (Φ ) 2 ) ⋅ d 3 x =
∑
i= N i =1
(
∫
i
H KG ⋅ d
3
x) =
∑
i= N
i
H KG
i =1
(1.2)
We can then calculate scattering amplitudes as perturbation series in the parameter λ . Show that the propagator propagator is,
Φ ( x )Φ ( x′) = δ DF ( x − x′); DF = [ Klein-Gordon propagator for mass m ] i
Using the reasoning
H
j
ij
(1.3)
= H0 + H int = [pro [propa paga gato tors rs]] + [int [inter erac acti tion ons] s] = [ext [exteernal rnal lines lines]] +[ver [verti tice ces] s] , and
constructing the λ = 0 diagrams, we have no vertices, which topologically-mandates lines only, as, i i j F ɶ i ( p) • − − − − •Φɶ j ( p′); Φ ( x ) • − − − − •Φ ( x′) = Dij ( x − x′) =ɺ =Φ 2 2 ( p − p′) − m + iε
(1.4)
In which we have,
F
F
i
i+
j
Dij = Dij ( x − x′) = Φ ( x)Φ ( x′) =
j−
)] Θ( x − x′)[φ ( x), φ ( x′)] i+
j−
+Θ ( x′ − x)[φ ( x′), ), φ ( x)]
i ; φ+≡
3 d pi ap e
∫ (2π )
3
µ
− i p µ x
2 E i
i+ †
= (φ ) ;
Zero interactions means zero decay, and thus zero particle-transmutations: However, since there is no interactions, interactions, we have, from [4.86] in the text, M = M decay = 0 → d Γ = 0 , and thus the propagator (1.4) is diagonal, DiFj = δ ij DiFj (summation not implied), and corresponds to mass m i,
(1.5)
i j ij F Φ ( x)Φ ( x′) = δ Dmi ( x − x′)
(1.6)
Show that there is one type of vertex given by , ij
= −2iλ (δ δ
iM ≡
kℓ
iℓ
+δ δ
jk
ik
jℓ
+δ δ )
(1.7)
Sticks: All possible vertices are those which can be constructed from two propagators (“sticks”). We could then draw the following diagrams:
(1.8) However, only the we first saw in (1.7) satisfies the constraint of “connected, amputated”. It can be shown that connected and amputated diagrams cancel out in the expectation of any operator (c.f., Peskin and Schroeder chapter 7.2; a proper proof is for now deferred). 1
2
That is: the vertex between two Φ ' s and two Φ ' s has the value iM ≡ has the value iM ≡ = − 6iλ .
1
= − 2iλ , and that between four Φ ' s
Compute, to leading order in λ , the differential cross section dσ / d Ω in the center-of-mass frame for the 1
2
2
1
1
1
2
2
1
1
1
1
scattering processes Φ Φ → Φ Φ ; Φ Φ → Φ Φ ; Φ Φ → Φ Φ as functions of the center-of-mass energy .
The differential cross section for amplitude [4.84] → dσ =
pi
M ( pi′ ,
p j′ → pi , p j )
out
p i p j | k i ′ k j′
2
4 Ei′ E j′ v i′ − v j′ ⋅ (2π ) 2 4 ECM
d Ω =
in
is given by,
(2π ) 4 δ (4) (ki + kj − pi′ − p j′ ) 4 Ei′ E j′ v i′ − v j′
i =n
d 3 pi M (ki′ , ki′ → pi )
∏ (2π )3
2
2 E i
i =1
(1.9)
Diagram amplitude is the labour-intensive part: In which: the amplitude of the diagram (1.7) has a somewhat-labour-intensive evaluation, iM ≡
≡
out
pi p j | T (
∫
λ 4
2 2 4 [(Φ ℓ ) ] ( x′′) d x′′) | k i′k j ′
= in
out
∫
′
′
pi p j | T ( Φℓ Φℓ Φℓ Φℓ d 4 x′′) | k i ′k j ′
in
; ℓ,ℓ ′ = 1, 2; (1.10)
Wick’s theorem allows us to write (1.10) as a cumbersome but straightforward permanent (for bosons), D13 D23 D73 D83 ′ ′ 4 4 | T ( Φ ℓ Φ ℓΦℓ Φℓ d x′′) | k i ′k j ′ D14 D24 D74 D84 out p i p j in iM ≡ɺ = ≡ perm D ≡ ∏ ai,σ (i ) D D D D 1 5 2 5 7 5 8 5 σ ∈ S 4 i =1 = ɺ 12 | T [3456] | 78 D16 D26 D76 D86 +
∫
∑
(1.11)
1
We use Mathematica to compute the permanent , and put the output in monospaced font so that it is easy to read. Let Dij be the contracted propagator. Then, we get,
1
Using: Permanent[m_L ist] := With[{v = Array[x, http://mathworld.wolfram.com/Permanent.html.
Length[m]]},Coefficien t[Times @@ (m.v), Times @@ v]] , found
at
iM =
D16 D25 D74 D83 + D15 D26 D74 D83 + D14 D26 D75 D83 + D16 D24 D75 D83 + D15 D24 D76 D83 + D14 D25 D76 D83 +
(1.12)
D16 D25 D73 D84 + D15 D26 D73 D84 + D13 D26 D74 D85 + D16 D23 D74 D85 + D15 D23 D76 D84 + D13 D25 D76 D84 + D16 D24 D73 D85 + D14 D26 D73 D85 + D13 D26 D75 D84 + D16 D23 D75 D84 + D14 D23 D76 D85 + D13 D24 D76 D85 + D15 D24 D73 D86 + D14 D25 D73 D86 + D15 D23 D74 D86 + D13 D25 D74 D86 + D14 D23 D75 D86 + D13 D24 D75 D86
To make sense of the jungle of terms in (1.12), one considers the physical picture: there are four possible scatterings: (1) ℓℓ′ → ℓℓ′ , (2) ℓℓ → ℓ′ℓ′ , (3) ℓℓ′ → ℓ′ℓ , and (4) ℓℓ → ℓℓ ; diagrammatically, these appear as,
(1.13) Correspondingly, the terms that contribute to each, using set-notation to fix what 1 and 2 are coupled with and leaving 7 and 8 to freely vary, are,
( (13) ∩ (25) ) ∪ ( (14) ∩ (26) )
( (13) ∩ (26) ) ∪ ( (15) ∩ (24) )
( (16) ∩ (25) ) ∪ ( (14) ∩ (23) ) ALL
D13 D13 D14 D14
D13 D13 D15 D15
D26 D26 D24 D24
D74 D75 D73 D76
D85 D84 D86 D83
D16 D16 D14 D14
D25 D25 D23 D23
D74 D73 D76 D75
D83 D84 D85 D86
∪ ( (15) ∩ (23) ∪ ( (16) ∩ (24) )
TWENTY FOUR OF ∪ ∪ ( (13) ∩ (24) ) ∪ ( (15) ∩ (26) ) ( (14) ∩ (25) ) ∪ ( (16) ∩ (23) ) THEM...
D15 D15 D16 D16
D13 D13 D15 D15
D24 D24 D26 D26
D76 D75 D73 D74
D85 D86 D84 D83
D14 D14 D16 D16
D25 D25 D23 D23
D76 D73 D74 D75
D83 D86 D85 D84
D25 D25 D26 D26
D23 D23 D24 D24
D76 D74 D73 D75
D76 D74 D75 D73
D84 D86 D85 D83
D84 D86 D83 D85
Moreover: we notice that each of these propagators are diagonal, as we proved in (1.6), which means, iM ≡
(1) → −i 14 λ ⋅ (8 ⋅ δ ℓℓδ ℓ′ℓ′ ) (1) → − 2iλ = (2) → −i 14 λ ⋅ (8 ⋅ δ ℓℓδ ℓ′ℓ′ ) (2) → − 2iλ (1.14) ij kℓ iℓ jk ik j ℓ = = − + + i 2 λ ( δ δ δ δ δ δ ) ℓℓ ℓ′ℓ′ 1 (3) → −i 4 λ ⋅ (8 ⋅ δ δ ) (3) → −2iλ (4) → −i 1 λ ⋅ (24 ⋅ δ ℓℓδ ℓℓ ) (4) → − 6iλ 4
=
Thus, the differential cross section (1.9), using (1.14) is,
12 × λ E λ E 16π 16π 2 12 × λ λ ECM M ( pk , pi → pℓ , p j ) 16π E 16π E = = = 2 4 ECM ECM ⋅1⋅ (2π ) 4 ECM 12 × λ λ 16π E 16π E 2 3 × 16π λ E 9 16π λ E 2
2
2
2
2
2
CM
d σ dΩ
=
pi
M ( pi′ ,
p j ′ → pi , p j )
2
4 Ei′ E j′ v i′ − v j′ ⋅ (2π ) 2 4 E CM
CM
2
2
2
2
2
2
CM
CM
2
(1.15)
2
2
2
CM
2
2
CM
2
2
2
2
CM
2
CM
2
i
2
(b) Now consider the case m < 0 , In this case, V has a local maximum, rather than a minimum, at Φ = 0 . i
Since V is a potential energy: this implies that the ground state of the theory is not near Φ = 0 but rather is obtained by shifting Φ i toward the minimum of V. By rotational invariance, we can consider this shift to be in th the N direction. Write, then, i i N (1.16) Φ ( x) = π ( x); i = 1,..., N − 1; Φ ( x) = v + σ ( x);
where v is a constant chosen to minimize V. (The notation suggests a pion field and should not be confused with a canonical momentum.) Show that, in these new coordinates (and substituting for v its expression in terms of λ and µ ) we have a theory of a massive field and N massless pion π i fields interacting through cubic and quartic potential energy terms which all become small as λ → 0 .
The Lagrangian (“theory”) of a massive field and N massless pion fields having cubic and quartic interactions would appear as, N
L =
∑(
1 2
i i i i i i (∂ µ Φ ) 2 − 12 m 2 (Φ ) 2 + αλΦ + αλ (Φ )3 + βλ (Φ ) 4 ); αλΦ = [will eventually be linear-sigma term] (1.17)
i =1
Coordinate change: in order to effect (1.16), we need to Legendre transform into Φ , ∂ µ Φ -coordinates, so from
the Hamiltonian density (1.1) we get the Lagrangian density, N
L
=
∑
ɺ i − H = π iΦ
i =1 N −1
=
N
N
∑
( ∂ µ Φi ) 2 −
i =1
1
∑ [ 2 (∂
µ Φ
i 2
) −
i =1
1 2
∑(
1 2
( ∂µ Φi ) 2 + 12 m2 ( Φ i ) 2 + 14 λ (( Φi ) 2 ) 2 ) =
1 2
i =1
m2 ( Φi ) 2 −
1 4
1
1
1
2
2
4
( ∂µ Φi ) 2 −
1
1 i i m2 ( Φ ) 2 − λ (( Φ ) 2 ) 2 2 4 (1.18)
λ ((Φi ) 2 ) 2 ] + ( ∂ µ Φ N ) 2 − m2 ( Φ N )2 − λ (( Φ N ) 2 ) 2
Then, using the change of coordinates (1.16), we see this (1.18) becomes, N −1 1 1 1 1 1 1 i i i L = [ (∂ µ π ) 2 − m 2 (π ) 2 − λ ((π ) 2 ) 2 ] + (∂ µ ( v + σ ( x))) 2 − m2 ( v + σ ( x)) 2 − λ(( v + σ ( x )) 2 ) 2 2 4 2 2 4 i =1 2
∑
N −1
=
∑ [L
free
i =1
i
i
i
(π , ∂ µπ ) − L int,φ 4 (π )] +
N −1
L
=
∑ i =1
L 4 (π φ
i
i
, ∂ µ π ) +
1 2
2 (∂ µ σ ) −
1 2
1 2 2
(0 + ∂ µ σ ) 2 − 2
1 2
2
2
m (v + σ ) −
2
m (σ + v + 2vσ ) −
1 4
1 4
λ ( v + σ ) 4
(1.19)
λ ( v 4 + σ 4 + 4σ v3 + 4σ 3v + 6σ 2 v 2 ) ;
Minimizing the potential energy/equilibrium: meanwhile, the coordinate-change (1.16) gains meaning when i i we write v as a parameter the potential-functionale is extremized with respect to. Putting the vector Φ eˆ (summation is implied) into the functionale V = V [(Φ i eˆi ) 2 ] , we get, i i 2
V [(Φ eˆ ) ] = − 12 m →
∂V ∂Φ j
=− m 1 2
2
∑
i
2
∑ ΦΦ i
i
i
(δ ij Φ + Φ δ ij ) + i
1 2
∑ ΦΦ ) λ (∑ Φ Φ ) ∑ (δ Φ i
i
+ 14 λ ( i
i
i 2
i
i
i
ij
i
i
+ Φ δ ij )
On setting this derivative (1.20) to zero, we find the minimum of V with respect to the jth field, j j i i i i 2 2 2 ∂V = 0 → m Φ = 2λΦ ( i Φ Φ ) ↔ Φ Φ = m / λ = v ∂Φ j i
∑
∑
(1.20)
(1.21)
Putting (1.21) into (1.19) as an expression of equilibrium, and noting − 12 m 2 v 2 and − 14 λ v 4 are constant shifts that contain no physics, we get,
N −1
L′ = L
−
∑ L
i
4
φ
i =1
i
(π , ∂ µπ ) − [shift] =
2
2
2
= 12 (∂ µ σ ) − 2m σ − (2 N −1
L
=
∑L i =1
φ4
m3
λ
1 2
2 (∂ µ σ ) −
1 2
2
2
m (σ + 2
σ)− λ
m
1 4
λ (σ 4 + 4σ
3
m
3/2
λ
+ 4σ
3 m
+ 6σ λ
)
2 m2
λ
σ + λ mσ 3 + 14 λσ 4 )
(1.22)
12 (∂ µ σ ) 2 + 2 µ 2σ 2 12 (∂ µ σ )2 − 2m 2σ 2 N −1 i i = ∑ L φ (π , ∂ µ π ) + µ (π , ∂ µ π ) + 3 4 3 4 m 1 1 −(2 σ + λ mσ + 4 λσ ) i =1 +2i σ − λ iµσ − 4 λσ λ λ i
i
4
3
3
This (1.22) is of the form of (1.17) if we identify the mass of the σ -field under interaction to be i 2 2 2 2 2 2 2 µ σ ≡ 12 mσ σ ↔ mσ = 2 µ . We also notice that 0 ≡ 12 mπ (π ) ↔ mπ = 0 . That is, we have neglected the pion-mass for simplicity (note that mπ = 139.6 MeV ; the pion is the lightest meson). 2 c Construct the Feynman rules by assigning values to the propagators and vertices,
σσ =
F
≡ Dm = mσ (x − x′ )
===== ⊳ =====
=
(1.23)
π iπ j =
F
− − − −− ⊳ − − − − − Dmπ =0
=
;
;
(1.24)
Using (1.14), and Mathematica to compute all the permanents, we have interactions from the π 2σ and πσ 2 dimensional terms in (1.22),
∫
π 2σ +πσ 2
= 0 p f | T ( Vint
(12)
4
i
∫ ∑ π π σ )d
j
d x) | p1p1 = 0 p f | T ( −i λµ (
(52)
i′
i′
i′
4
i
j
x) | p1p1 =ɺ −i λµ 1| 234 | 56
(62)
(1.25)
∫
ii ′
i ′j
4
i 1
j 2
4
= − i λ µ (13)
(53)
(63) = − i λ µ δ δ ⋅ δ ( p f − p − p ) d x = −2i λ µ ⋅ δ
(14)
(54)
(64) +
ij
= contributions from all = −i λ µ ⋅ 6 = − i λ µ ⋅ 3! = − i λ µ ⋅ S 3 ; S 3 = [permutation group] (1.26)
For the vertices in (1.24), we have bosons, and we get, ℓ
∫
ℓ
4
∫
4
ℓ
4
k i j k i j k i j = 0 p1 p1 | T ( Vint d x) | p1p1 = 0 p1 p1 | T ( ( − 14 λσ ) d x) | p1p1 = −i 14 λ p1 p1 | T (σ σ σ σ ) | p1p1
= ɺ − iλ 12 | T [3456] | 78 = − iλ 1 4
1 4
D13
D 23
D73
D83
D14
D 24
D74
D84
D15
D2 5
D75
D85
D16
D 26
D76
D86 +
(1.27) ij
= −2iλ (δ δ
kℓ
iℓ
+δ δ
jk
ik
jℓ
+ δ δ )
= using (1.25) with a multiplicity of 8 = −2iλ ⋅ δ ij
= contributions from all = −i 14 λ ⋅ 24 = − i 14 λ ⋅ 4! = − i 14 λ ⋅ S 4 ; S 4 = [permutation group]
(1.28)
(1.29)
appendix I – position and momentum space Feynman rules (respectively)
(1) for each propagator, x − − − − − − x′ ≡ DF ( x − x′) (2) for each vertex
= (− iλ )
particles at a vertex, and
∫(
∫(
) d 4 z . Corollary:
−iλ is the amplitude for emission and/or absorption of
) d 4 z is the sum of all points where the process can occur.
(3) for each external point: x − −− = 1 (4) divide by the symmetry factor.
(1) for each propagator, x − − − − − − x′ ≡ DF ( x − x′) = i ( p 2 − m 2 + iε )−1 (2) for each vertex
= (− iλ )
∫(
) d 4 z = − iλ
absorption of particles at a vertex, and
V V
= − iλ . Corollary: −iλ is the amplitude for emission and/or
2 4 is the sum of all points where the process can occur . d z ( ) ∫ µ − i p x
(3) for each external point: x − −− = 1 → F[1] = e µ . (4) impose momentum-conservation at each vertex (5) integrate over each undetermined momentum:
∫ (2π )
−4
d4 p.
(6) divide by the symmetry factor.
2
This is just the superposition-principle of quantum mechanics: when a process can happen in alternative ways, we add the amplitudes for each possible way. To compute each individual amplitude: the Feynman rules tell us t o multiply the amplitudes, propagators, and vertex factors for each independent part of the process.
