Tuesday updated 23rd March, 2010 06:10
5. spontaneous symmetry breaking and Goldstone bosons (a) the linear -model (b) the non-linear -model (c) SU (2) SU (2) and SU (3) SU (3) effective chiral theory (d) accidental symmetry and the Coleman-Weinberg potential (e) not sure whether to do CCWZ and matter fields - maybe later ????
the physics of spontaneously broken continuous symmetry -- a mechanical analogy transverse waves in a string described by wave function (x)
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the physics of spontaneously broken continuous symmetry -- a mechanical analogy transverse waves in a string described by wave function (x) system invariant under translations x x + c normal modes (x + c) (x) normal modes eikx
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the physics of spontaneously broken continuous symmetry -- a mechanical analogy transverse waves in a string described by wave function (x) system invariant under translations x x + c normal modes (x + c) (x) normal modes eikx
......... .... ......... ............. ...................................................... ...................................................... ............
..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... .....
the physics of spontaneously broken continuous symmetry -- a mechanical analogy transverse waves in a string described by wave function (x) system invariant under translations x x + c normal modes (x + c) (x) normal modes ei kx
......... .... ......... ............. ...................................................... ...................................................... ............
dispersion relation 2 (k 2 ) restoring force for the k mode from mechanical properties of the string (tension, stiffness, etc)
..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... .....
the physics of spontaneously broken continuous symmetry -- a mechanical analogy transverse waves in a string described by wave function (x) system invariant under translations x x + c normal modes (x + c) (x) normal modes eikx
......... .... ......... ............. ...................................................... ...................................................... ............
dispersion relation 2 (k 2 ) restoring force for the k mode from mechanical properties of the string (tension, stiffness, etc) system invariant under translations y y + d -- normal modes already fixed symmetry spontaneously broken tells you about the dispersion relation 2 0 and k 2 0
..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... .....
the physics of spontaneously broken continuous symmetry -- a mechanical analogy transverse waves in a string described by wave function (x) system invariant under translations x x + c normal modes (x + c) (x) normal modes eikx
......................................................... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ............................................................................................................................................................................................. ..... . . .......................................... . . . . . . . . . . . ......
dispersion relation 2 (k 2 ) restoring force for the k mode from mechanical properties of the string (tension, stiffness, etc) system invariant under translations y y + d -- normal modes already fixed symmetry spontaneously broken tells you about the dispersion relation 2 0 and k 2 0
the physics of spontaneously broken continuous symmetry -- a mechanical analogy transverse waves in a string described by wave function (x) system invariant under translations x x + c normal modes (x + c) (x) normal modes eikx
................................................................................... ...................................................... ......... .......
system invariant under translations y dispersion relation 2 (k 2 ) y + d -- normal modes already fixed restoring force for the symmetry spontaneously broken k mode from mechanical tells you about the dispersion properties of the string (tension, stiffness, etc) relation 2 0 and k 2 0 k 0 gets closer and closer to a translation of the whole string the spontaneously broken symmetry guarantees that there is no restoring force as k 0 -- but notice that the system never gets to k = 0 -- that would require motion of the whole infinite system
..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... .....
I like this example because I think it survives the onslaught of quantum mechanics! normally quantum mechanics destroys spontaneous symmetry breaking -- even if a classical system has degenerate ground states that break symmetry -- like the double well potential -- the quantum mechanical ground state is symmetric
(x)
V
x
the reason is superposition and the uncertainty principle -- the states spread out around the classical minima because of the uncertainty principle -- then the wave functions around the classical minima interfere constructively and the symmetric state has the lowest energy -- a very general phenomenon -- continuous symmetry also -- like a particle moving freely on a circle
I like this example because I think it survives the onslaught of quantum mechanics! normally quantum mechanics destroys spontaneous symmetry breaking -- even if a classical system has degenerate ground states that break symmetry -- like the double well potential -- the quantum mechanical ground state is symmetric
(x)
V
x
the reason is superposition and the uncertainty principle -- the states spread out around the classical minima because of the uncertainty principle -- then the wave functions around the classical minima interfere constructively and the symmetric state has the lowest energy -- a very general phenomenon -- continuous symmetry also -- like a particle moving freely on a circle
I like this example because I think it survives the onslaught of quantum mechanics! normally quantum mechanics destroys spontaneous symmetry breaking -- even if a classical system has degenerate ground states that break symmetry -- like the double well potential -- the quantum mechanical ground state is symmetric
(x)
V
x
the reason is superposition and the uncertainty principle -- the states spread out around the classical minima because of the uncertainty principle -- then the wave functions around the classical minima interfere constructively and the symmetric state has the lowest energy -- a very general phenomenon -- continuous symmetry also -- like a particle moving freely on a circle
the physics of spontaneously broken continuous symmetry -- a mechanical analogy transverse waves in a string described by wave function (x) system invariant under translations x x + c normal modes (x + c) (x) normal modes eikx
................................................................................... ...................................................... ......... .......
system invariant under translations y dispersion relation 2 (k 2 ) y + d -- normal modes already fixed restoring force for the symmetry spontaneously broken k mode from mechanical tells you about the dispersion properties of the string (tension, stiffness, etc) relation 2 0 and k 2 0 k 0 gets closer and closer to a translation of the whole string the spontaneously broken symmetry guarantees that there is no restoring force as k 0 -- but notice that the system never gets to k = 0 -- that would require motion of the whole infinite system
..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... .....
simplest field theory model with spontaneously broken continuous symmetry 1 L = 2 show a contradition invariant under +c
cannot be generated by a unitary transformation that leaves the vacuum invariant Uc Uc = + c and Uc |0 = |0
impossible for c = 0 this means that the unitary operator Uc must take the vacuum into some other states which means, as promised that although the Lagrangian is invariant, the vacuum is not and the symmetry is spontaneously broken 0||0 = 0|Uc Uc |0 = 0|( + c)|0 = 0||0 + c so what DOES this transformation do if it does not leave the vacuum invariant? -- what "transformed vacuum state" does it produce? -- and how does this keep QM from spoiling SSB? a more limited but more quantitative way to ask this quesion is to ask about the matrix element the conserved Noether 0|Uc |0 current is 0
simplest field theory model with spontaneously broken continuous symmetry 1 L = 2 show a contradition invariant under +c
cannot be generated by a unitary transformation that leaves the vacuum invariant Uc Uc = + c and Uc |0 = |0
impossible for c = 0 this means that the unitary operator Uc must take the vacuum into some other states which means, as promised that although the Lagrangian is invariant, the vacuum is not and the symmetry is spontaneously broken 0||0 = 0|Uc Uc |0 = 0|( + c)|0 = 0||0 + c so what DOES this transformation do if it does not leave the vacuum invariant? -- what "transformed vacuum state" does it produce? -- and how does this keep QM from spoiling SSB? a more limited but more quantitative way to ask this quesion is to ask about the matrix element the conserved Noether 0|Uc |0 current is 0
simplest field theory model with spontaneously broken continuous symmetry 1 L = 2 show a contradition invariant under +c
cannot be generated by a unitary transformation that leaves the vacuum invariant Uc Uc = + c and Uc |0 = |0
impossible for c = 0 this means that the unitary operator Uc must take the vacuum into some other states which means, as promised that although the Lagrangian is invariant, the vacuum is not and the symmetry is spontaneously broken 0||0 = 0|Uc Uc |0 = 0|( + c)|0 = 0||0 + c so what DOES this transformation do if it does not leave the vacuum invariant? -- what "transformed vacuum state" does it produce? -- and how does this keep QM from spoiling SSB? a more limited but more quantitative way to ask this quesion is to ask about the matrix element the conserved Noether 0|Uc |0 current is 0
the conserved Noether current is 0 so we should consider the operator Uc = e i c Q where Q= 0 (y, t) d3 y
the trouble is that this is not very well defined -- so consider a smeared version of Q 2 2 which goes to Q Q (t) e-|y| /2 0 (y, t) d3 y as goes to define ,c (x, t) eicQ (t) (x, t) e-icQ (t) so ,0 (x, t) = (x, t)
now to see why we chose Q compute =i =i e-|y| e-|y|
2 /22
,c (x, t) = i eicQ (t) [Q (t), (x, t)] e-icQ (t) c eicQ (t) [ 0 (y, t), (x, t)] e-icQ (t) d3 y
2 /22
2 /22
eicQ (t) -i 3 (x - y e-icQ (t) d3 y = e-|x|
2 2 using c = 0 (x, t) + c eicQ (t) (x, t) e-icQ (t) = (x, t) + c e-|x| /2 value as is the size of the region over which is translated
the conserved Noether current is 0 so we should consider the operator Uc = e i c Q where Q= 0 (y, t) d3 y
the trouble is that this is not very well defined -- so consider a smeared version of Q 2 2 which goes to Q Q (t) e-|y| /2 0 (y, t) d3 y as goes to define ,c (x, t) eicQ (t) (x, t) e-icQ (t) so ,0 (x, t) = (x, t)
now to see why we chose Q compute =i =i e-|y| e-|y|
2 /22
,c (x, t) = i eicQ (t) [Q (t), (x, t)] e-icQ (t) c eicQ (t) [ 0 (y, t), (x, t)] e-icQ (t) d3 y
2 /22
2 /22
eicQ (t) -i 3 (x - y e-icQ (t) d3 y = e-|x|
2 2 using c = 0 (x, t) + c eicQ (t) (x, t) e-icQ (t) = (x, t) + c e-|x| /2 value as is the size of the region over which is translated
the conserved Noether current is 0 so we should consider the operator Uc = e i c Q where Q= 0 (y, t) d3 y
the trouble is that this is not very well defined -- so consider a smeared version of Q 2 2 which goes to Q Q (t) e-|y| /2 0 (y, t) d3 y as goes to define ,c (x, t) eicQ (t) (x, t) e-icQ (t) so ,0 (x, t) = (x, t)
now to see why we chose Q compute =i =i e-|y| e-|y|
2 /22
,c (x, t) = i eicQ (t) [Q (t), (x, t)] e-icQ (t) c eicQ (t) [ 0 (y, t), (x, t)] e-icQ (t) d3 y
2 /22
2 /22
eicQ (t) -i 3 (x - y e-icQ (t) d3 y = e-|x|
2 2 using c = 0 (x, t) + c eicQ (t) (x, t) e-icQ (t) = (x, t) + c e-|x| /2 value as is the size of the region over which is translated
the conserved Noether current is 0 so we should consider the operator Uc = e i c Q where Q= 0 (y, t) d3 y
the trouble is that this is not very well defined -- so consider a smeared version of Q 2 2 which goes to Q Q (t) e-|y| /2 0 (y, t) d3 y as goes to define ,c (x, t) eicQ (t) (x, t) e-icQ (t) so ,0 (x, t) = (x, t)
now to see why we chose Q compute =i =i using c = 0 value e-|y| e-|y|
2 /22
,c (x, t) = i eicQ (t) [Q (t), (x, t)] e-icQ (t) c eicQ (t) [ 0 (y, t), (x, t)] e-icQ (t) d3 y
2 /22
2 /22
eicQ (t) -i 3 (x - y e-icQ (t) d3 y = e-|x|
2 /22
eicQ (t) (x, t) e-icQ (t) = (x, t) + c e-|x|
(x, t) + c as
is the size of the region over which is translated
the conserved Noether current is 0 so we should consider the operator Uc = e i c Q where Q= 0 (y, t) d3 y
the trouble is that this is not very well defined -- so consider a smeared version of Q 2 2 which goes to Q Q (t) e-|y| /2 0 (y, t) d3 y as goes to define ,c (x, t) eicQ (t) (x, t) e-icQ (t) so ,0 (x, t) = (x, t)
now to see why we chose Q compute =i =i using c = 0 value e-|y| e-|y|
2 /22
,c (x, t) = i eicQ (t) [Q (t), (x, t)] e-icQ (t) c eicQ (t) [ 0 (y, t), (x, t)] e-icQ (t) d3 y
2 /22
2 /22
eicQ (t) -i 3 (x - y e-icQ (t) d3 y = e-|x|
2 /22
eicQ (t) (x, t) e-icQ (t) = (x, t) + c e-|x|
(x, t) + c as
is the size of the region over which is translated
(x, t) = 0 (x, t) =
1 d3 k a eikx + ak e-ikx (2)3 2k k k 0 = k = |k | |k | d3 k (2)3/2
d3 k 1 i k 0 a eikx - ak e-ikx k (2)3 2k 3 (y, 0) d y = i 2
0 3
Q (0)
e
-|y|2 /22
e
-2 |k |2 /2
a k
- ak
3 i a - a 2 [a , a ] = 4 = e-
2 |k |2
where a and a are annihilation and creation operators for particles 2 2 with wave-fuction e-|x| /2
2 (|k |2 +|k |2 )/2
e-
[ak , a ] k
|k ||k |
d3 k d3 k (2)3 A = 22 a
|k |
d3 k 1 = 2 4 3 (2) 4
define Q (0) = i
A = 22 a A - A 4
[A , A ] = 1
(x, t) = 0 (x, t) =
1 d3 k a eikx + ak e-ikx (2)3 2k k k 0 = k = |k | |k | d3 k (2)3/2
d3 k 1 i k 0 a eikx - ak e-ikx k (2)3 2k 3 (y, 0) d y = i 2
0 3
Q (0)
e
-|y|2 /22
e
-2 |k |2 /2
a k
- ak
3 i a - a 2 [a , a ] = 4 = e-
2 |k |2
where a and a are annihilation and creation operators for particles 2 2 with wave-fuction e-|x| /2
2 (|k |2 +|k |2 )/2
e-
[ak , a ] k
|k ||k |
d3 k d3 k (2)3 A = 22 a
|k |
d3 k 1 = 2 4 3 (2) 4
define Q (0) = i
A = 22 a A - A 4
[A , A ] = 1
(x, t) = 0 (x, t) =
1 d3 k a eikx + ak e-ikx (2)3 2k k k 0 = k = |k | a k |k | d3 k (2)3/2
d3 k 1 i k 0 a eikx - ak e-ikx k (2)3 2k 3 (y, 0) d y = i 2
0 3
Q (0)
e
-|y|2 /22
e
-2 |k |2 /2
- ak
3 i a - a 2 [a , a ] = 4 = e
-2 |k |2
where a and a are annihilation and creation operators for particles 2 2 with wave-fuction e-|x| /2
2 (|k |2 +|k |2 )/2
e-
[ak , a ] k
|k ||k |
d3 k d3 k (2)3 A = 22 a
d3 k 1 |k | = 2 4 3 (2) 4 [A , A ] = 1
define Q (0) = i
A = 22 a A - A 4
(x, t) = 0 (x, t) =
1 d3 k a eikx + ak e-ikx (2)3 2k k k 0 = k = |k | a k |k | d3 k (2)3/2
d3 k 1 i k 0 a eikx - ak e-ikx k (2)3 2k 3 (y, 0) d y = i 2
0 3
Q (0)
e
-|y|2 /22
e
-2 |k |2 /2
- ak
3 i a - a 2 [a , a ] = e-
2 |k |2
where a and a are annihilation and creation operators for particles 2 2 with wave-fuction e-|x| /2
-2 (|k |2 +|k |2 )/2
=4
e
[ak , a k
]
d3 k d3 k |k ||k | (2)3 A = 22 a
|k |
d3 k 1 = 2 4 3 (2) 4
define Q (0) = i
A = 22 a A - A 4
[A , A ] = 1
(x, t) = 0 (x, t) =
1 d3 k a eikx + ak e-ikx (2)3 2k k k 0 = k = |k | a k |k | d3 k (2)3/2
d3 k 1 i k 0 a eikx - ak e-ikx k (2)3 2k 3 (y, 0) d y = i 2
0 3
Q (0)
e
-|y|2 /22
e
-2 |k |2 /2
- ak
3 i a - a 2 [a , a ] = 4 = e-
2 |k |2
where a and a are annihilation and creation operators for particles 2 2 with wave-fuction e-|x| /2
2 (|k |2 +|k |2 )/2
e-
[ak , a ] k
|k ||k |
d3 k d3 k (2)3 A = 22 a
|k |
d3 k 1 = 2 4 (2)3 4
define Q (0) = i
A = 22 a A - A 4
[A , A ] = 1
[A , A ] = 1 now we want to calculate
Q (0) = i
0|eic Q (0) |0
A - A 4 the overlap between the transformed vacuum and the original vacuum
d 0|eic Q (0) |0 = i 0|Q (0) eic Q (0) |0 dc = - 0| (A - A )eic Q (0) |0 = 0|A eic Q (0) |0 4 4 =
=
0|[A , eic Q (0) ]|0 4
now because [A , Q (0)] commutes with Q (0)
2 0|[A , c A - A ]eic Q (0) |0 0|[A , ic Q (0)]eic Q (0) |0 = - 4 16 2
2 2 2 2 = -c 0|eic Q (0) |0 0|eic Q (0) |0 = e-c /32 16 2 the important point is that 0|eic Q (0) |0 0 as for c fixed
[A , A ] = 1 now we want to calculate
Q (0) = i
0|eic Q (0) |0
A - A 4 the overlap between the transformed vacuum and the original vacuum
d 0|eic Q (0) |0 = i 0|Q (0) eic Q (0) |0 dc = - 0| (A - A )eic Q (0) |0 = 0|A eic Q (0) |0 4 4 =
=
0|[A , eic Q (0) ]|0 4
now because [A , Q (0)] commutes with Q (0)
2 0|[A , c A - A ]eic Q (0) |0 0|[A , ic Q (0)]eic Q (0) |0 = - 4 16 2
2 2 2 2 = -c 0|eic Q (0) |0 0|eic Q (0) |0 = e-c /32 16 2 the important point is that 0|eic Q (0) |0 0 as for c fixed
[A , A ] = 1 now we want to calculate
Q (0) = i
0|eic Q (0) |0
A - A 4 the overlap between the transformed vacuum and the original vacuum
d 0|eic Q (0) |0 = i 0|Q (0) eic Q (0) |0 dc = - 0| (A - A )eic Q (0) |0 = 0|A eic Q (0) |0 4 4 =
=
0|[A , eic Q (0) ]|0 4
now because [A , Q (0)] commutes with Q (0)
2 0|[A , ic Q (0)]eic Q (0) |0 = - 0|[A , c A - A ]eic Q (0) |0 4 16 2
2 2 2 2 = -c 0|eic Q (0) |0 0|eic Q (0) |0 = e-c /32 16 2 the important point is that 0|eic Q (0) |0 0 as for c fixed
[A , A ] = 1 now we want to calculate
Q (0) = i
0|eic Q (0) |0
A - A 4 the overlap between the transformed vacuum and the original vacuum
d 0|eic Q (0) |0 = i 0|Q (0) eic Q (0) |0 dc = - 0| (A - A )eic Q (0) |0 = 0|A eic Q (0) |0 4 4 =
=
0|[A , eic Q (0) ]|0 4
now because [A , Q (0)] commutes with Q (0)
2 0|[A , ic Q (0)]eic Q (0) |0 = - 0|[A , c A - A ]eic Q (0) |0 4 16 2
2 2 2 2 = -c 0|eic Q (0) |0 0|eic Q (0) |0 = e-c /32 16 2 the important point is that 0|eic Q (0) |0 0 as for c fixed
[A , A ] = 1 now we want to calculate
Q (0) = i
0|eic Q (0) |0
A - A 4 the overlap between the transformed vacuum and the original vacuum
d 0|eic Q (0) |0 = i 0|Q (0) eic Q (0) |0 dc = - 0| (A - A )eic Q (0) |0 = 0|A eic Q (0) |0 4 4 =
=
0|[A , eic Q (0) ]|0 4
now because [A , Q (0)] commutes with Q (0)
2 0|[A , ic Q (0)]eic Q (0) |0 = - 0|[A , c A - A ]eic Q (0) |0 4 16 2
2 2 2 2 = -c 0|eic Q (0) |0 0|eic Q (0) |0 = e-c /32 16 2 the important point is that 0|eic Q (0) |0 0 as for c fixed
0|eic Q (0) |0 = e-c
2 2
/32 2
now consider the overlap between the transformed vacuum state and a one particle state using the fact that |k | d3 k 3 2 2 Q (0) = i e- |k | /2 a - ak k 2 (2)3/2
0|ak eic Q (0) |0 = 0|[ak , eic Q (0) ]|0 3 -2 |k |2 /2 ic Q (0) = 0|[ak , ic Q (0)] e |0 = i e |k | 0|eic Q (0) |0 2 3 2 2 2 2 2 = i e- |k | /2 |k | e-c /32 2
as this goes to zero for any non-zero k independent of c keeps happening for any finite number of particles with nonzero momentum, or localized in any finite region -- the transformed vacuum becomes orthogonal to all states built on the original vacuum by ordinary creation operators as and of course you can build a perfectly good Hilbert Space by acting on the transformed vacuum with transformed creation operators -- all these state are also orthogonal to all normal states continuously # of equivalent spaces labeled by c -- supersection rules
the key, clearly, is the particles which transform inhomogeneously under the symmetry -- they have to be massless as the is trivially in our toy theory -- but the problem is precisely the "global" nature of the symmetry -- this mess happened because we insisted on thinking about a symmetry that is spread out over all of infinite space -- the transformed vacuum is built out of "particles" whose wave function is similarly spread out, and this has negligible overlap -- this saves SSB from the ravages of QM -- and it means that what the symmetry tells us is quite different from what happens in Gell-Mann's SU (3) -- we are not interested in Q -- be we are interested in Q for finite because even though the limit is seriously diseased -- the approach to the limit provides very interesting information -- SSB relates states with different numbers of low momentum particles -- this is an absolute set-up for the kind of momentum expansion that we use in EFT -- and in fact, the clearest formulation of the effective field theory paradigm arose from the attempts of Steve Weinberg and others to make sense of spontaneously broken global symmetries -- the fields like in our stupid little example are massless in the symmetry limit and must ALWAYS be included in the EFT -- and the symmetry tells us interesting things about them because they ARE the transformations of the vacuum
the Goldstone theorem in perturbative QFT -- let's do something simple -- see how this works if all the couplings are small and all the fields have direct physical meanings -- so consider some real scalar fields 1 is some multiplet L() = - V () of spinless fields 2
T V () and L() invariant under where Ta = -Ta = -Ta = i a Ta symmetry generated by Ta because s are real these fields transform homogeneously -- SSB arises if these are not the right fields to use at low energies which is what happens if the minimum of V () is not at = 0 -- suppose there is a minimum of V () at = F = 0 VEV
for convenience in the analysis define F is an extremum if perturbing around F shifted fields =-F create particles
Vj1 jn () =
n V () j1 . . . jn
the condition that F be an extremum of V () as Vj (F) = 0 and F is a minimum if Vjk (F) 0
1 V () = V (F) + j Vj (F) + j k Vjk (F) + 2
perturbing around F shifted fields =-F create particles
..... .... 1 ........ V () = V (F) + j V......(F) + j k Vjk (F) + ....j ....... 2 ...
Vjk (F) is the meson mass-squared matrix -- there are no tachyons in the free theory about which we are perturbing = i a Ta = i a Ta + i a Ta F Ta F is a set of vectors in field space ab F T Ta Tb F because F doesn't change
is a real, symmetric and positive assume we've diagonalized
two kinds of generators -- "unbroken" Sa | Sa F = 0 -- "broken" Xa | Xa F = 0 -- the shifted fields transform homogeneously under Sa and inhomogenously under Xa -- Sa form a subalgebra because Sa F = 0 and Sb F = 0 [Sa , Sb ]F = 0 so Sa generate the "unbroken subalgebra"
V ( + ) - V () = iVk () a [Ta ]kl l = 0 now differentiate and set = F Vjk (F) a [Ta ]kl Fl + Vk (F) a [Ta ]kj = 0 Vjk (F)[Xa ]kl Fl = 0 the nonzero vectors we get by acting on the VEV F with the broken generators Xa are eigenvectors of the mass matrix with eigenvalue 0 -- massless particles -- this is the Goldstone theorem -- F describes the "vacuum" -- Xa F is the change we make to try to "transform" the vacuum -- but instead of getting a new vacuum, we get massless fields F T Xa that look like transformed vacuum in some limited region of of space -- these are the Goldstone bosons
F T Xa
example -- the linear -model 2 2 matrix of spinless fields = 2 2 = + i transformation of under "chiral" SU (2)L SU (2)R where and are real fields and a are the Pauli matrices -- quaternions
L Ta
L R = i a Ta - i a Ta L R
where R = Ta = a /2
or we can exponentiate the infinitesimal transformations leaves invariant the quaternion "absolute value"
LR
L L = exp ila Ta R R = exp ira Ta
2 = = 2 + a
the magic of Pauli matrices
1 1 2 2 2 L = + a a - + a - f 2 2 2 4 equivalent 1 T formulation 1 L = T - - f2 with real 2 2 4 fields 3 SU (2) SU (2) algebra SO(4) algebra
2
with real fields we can apply the Goldstone analysis
1 T L = T - - f2 2 4
2
L R = i a Ta - i a Ta R L
where L R Ta = Ta = a /2 ( + i ) = - 2
L
use ( a) ( b) = a b + i (a b) + i 2 TL =
L
L = L ( + i ) = i L = -
L
L
- 2
L
2
2
L =
L
- 2
L
2
R
or i = 2 or i
R
L
0
L /2
- L /2 - L /2
R
R = R ( + i ) = -i( + i ) R =
R
+ i - 2
R
R
- 2
2
2 f 0
R = -
L
R
- 2
R
2 0 L /2
R T
=
0 R /2 - R /2 - R /2 0 - R /2
F=
TL F =
R
TR F =
with real fields we can apply the Goldstone analysis L = - R =
L
- 2
2 2
1 T L = T - - f2 2 4 or i or i
L
2
2
L = R = -
L
L
TL =
R
0 L /2 0
- L /2 - L /2
R /2
R
2 f 0
R
- 2
R
R T
=
- R /2 - R /2 0 - R /2
F=
L
TL F =
0 L /2
V
R
TR F =
S = TV TL + TR =
V
S =
V
(T L + T R ) F vector or "diagonal" symmetry is unbroken chiral symmetry is spontaneiously broken F T X
T L F +
V
T R F = i
0 0 + - V /2 V /2
A
=0
X = TA TL - TR the shifted fields are
XF = =-f =
0 f /2 A
==F
the Goldstone bosons are
with real fields we can apply the Goldstone analysis F= f 0
L
1 T L = T - - f2 2 4 0 L /2
V
2
TL F =
R
TR F =
0 - R /2
S = TV TL + TR =
V
S =
V
(T L + T R ) F vector or "diagonal" symmetry is unbroken chiral symmetry is spontaneiously broken F T X
T L F +
V
T R F = i
0 0 + - V /2 V /2
A
=0
X = TA TL - TR the shifted fields are
XF = =-f =
0 A f /2
==F
the Goldstone bosons are
1 1 2 L = + - ( + f )2 + a a - f 2 2 2 4 2 1 1 no 2 = + - + a a + 2f mass 2 2 4
-model deceptively simple because SU (2) SU (2) algebra SO(4) 3 3 matrix SU (3) SU (3) symmetry tr - LR
L L = exp ila Ta a R R = exp ir Ta
L=
~ ~ 1 2 tr( ) - + m2 2 2 1 2
ii1 i2 jj1 j2 i1 j1 i2 j2
+
(det + h.c.) 3
cofij =
det =
1 3
ij cofij
ij
1 det = tr T cof 3 L = tr - -
invariance of det
3f 2 4 2
cof L cofRT - 2 tr 2 -
f2 4 2
1 tr( ) - 2
3 = f I/2 tr cof - f cof - f T 2 2 2 = f I/2 breaks SU (3) SU (3) l = r L = R = u uu SU (3) Goldstone bosons l = -r L = R =
the physics of spontaneously broken continuous symmetry -- a mechanical analogy transverse waves in a string described by wave function (x) system invariant under translations x x + c normal modes (x + c) (x) normal modes eikx
................................................................................... ...................................................... ......... .......
system invariant under translations dispersion y relation 2 (k 2 ) y + d -- normal modes already fixed restoring force for the symmetry spontaneously broken k mode from mechanical tells you about the dispersion properties of the string (tension, stiffness, etc) relation 2 0 and k 2 0 k 0 gets closer and closer to a translation of the whole string the spontaneously broken symmetry guarantees that there is no restoring force as k 0 -- but notice that the system never gets to k = 0 -- that would require motion of the whole infinite system
..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... .....
3 3 matrix SU (3) SU (3) symmetry L = tr - -
LR
3f 2 4 2
L L = exp ila Ta R R = exp ira Ta
1 tr( ) - 2
-
2 tr 2
-
f2 4
2
3 tr cof - f cof - f T = f I/2 2 2 2 = f I/2 breaks SU (3) SU (3) l = r L = R = u uu SU (3) Goldstone bosons l = -r L = R = describe the GBs by a chiral transformation! = ei/f (h + f I/2) where h = h LR = ei /f (h + f I/2)
n
how do , h and transform? = because det L = det R = 1
tr is invariant so the eigenvalues of h don't change h = uhu (where u(L, R, )) L(h + f I/2)R = u(h + f I/2)u nonlinear transformation = Lu = uR 2 L 2 R homogeneous on h uhu crazy on Lu = uR
3 3 matrix SU (3) SU (3) symmetry L = tr - -
LR
3f 2 4 2
L L = exp ila Ta R R = exp ira Ta
1 tr( ) - 2
-
2 tr 2
-
f2 4
2
3 tr cof - f cof - f T = f I/2 2 2 2 = f I/2 breaks SU (3) SU (3) l = r L = R = u uu SU (3) Goldstone bosons l = -r L = R = describe the GBs by a chiral transformation! = ei/f (h + f I/2) tr( ) -
3f 2 4 2
where
f2 4 2
h = h
no dependence in the potential cof - f T 2
tr
-
tr cof - f 2
mass to trh SO(18) at low energies only GBs survive
mass to h chiral U (1) L=
mass to , h just SU (3) SU (3)
f2 tr U U where U = 2 4 very low energy physics completely determined by symmetry and f
1 L = (i D q - q Mq q) - G Ga - (v + a 5 ) q q q 4 a quark field q vector D = + igTa G igTa G = [D , D ] a a in 3D color space is "covariant G is "gluon a + 3D flavor space derivative" fieldstrength" Ta 3 3 traceless Hermitian matrices in color space 1 tr(Ta Tb ) = ab 2 Mq diagonal in flavor space u, d, s
i color Ta G U Ta G U - g U U a a gauge G U G U symmetry q U q Ta 's are "color" charges like EM charge in QED binds quarks and antiquarks into colorneutral combinations like photon exchange binds charged particles into electrically neutral atoms -- colorneutral combinations are
qq mesons
and
baryons
jkl qj qk ql
v = v t and a = a t sources for the "vector" and "axial vector" currents
where t are 3 3 hermitian flavor generators
1 L = i D q - G Ga - q (v + a 5 )q - q (s + ip5 )q q 4 a 1 = i D q - G Ga - qL qL - qR r qR - qR (s + ip)qL - qL (s - ip)qR q 4 a r , , s and p are classical Hermitian 3 3 matrix fields (tr = trr = 0) -- just assuming (at this point) no chiral U (1) symmetry (det term)??? r = v - a
= v + a
s Mq
r and fields are classical gauge fields for the SU (3)R and SU (3)L symmetries, and also the sources for the corresponding Noether currents we want to build a low energy theory with the same symmetries -- qR R q R qL L q L r R r R - iR R L L - iL L s + ip R (s + ip) L GB field U U L U R r R r R - iR R L L - iL L s + ip R (s + ip) L
we want to build a low energy theory with the same symmetries -- qR R q R qL L q L r R r R - iR R L L - iL L s + ip R (s + ip) L GB field U U L U R r R r R - iR R L L - iL L s + ip R (s + ip) L
low energy theory effective L depends on U and classical sources D U = U + i U - iU r D U = U + ir U - iU D U L D U R D U R D U L QCD also invariant under parity r -r U U r ss p -p
most general effective Lagrangian consistent with the symmetries -- only useful because we can expand in powers of derivatives sources and truncate the expansion after a small number of terms leading term for s = p = 0 L(U, r, , s, p) = f2 tr D U D U 4 like theory
most general effective Lagrangian consistent with the symmetries -- only useful because we can expand in powers of derivatives sources and truncate the expansion after a small number of terms leading term for s = p = 0 L(U, r, , s, p) = U = exp[2i/f ] 1 = a Ta = 2
1 0 2
f2 tr D U D U 4 expand in powers of
1 6
like theory
+ -
1 6
+ 1 - 2 0 + K0
K-
K+ K0 2 - 6
K- = K+ K0 = K0
expanding derivative term in =
f2 1 tr D U D U = a a 4 2
1 0 0 + + + - + K + K - + K 0 K 0 2 properly normalized KE for meson fields
U = exp[2i/f ] 1 = a Ta = 2
1 0 2
expand in powers of
1 6
+ -
1 6
+ 1 - 2 0 + K0
K-
K+ K0 2 - 6
K- = K+ K0 = K0
pure vector SU (3) transformation L = R = u U U = uU u = uu pure chiral transformation L = R = c U = eic U eic take c to be infinitesimal and write as a power series in c and (1 + 2i /f + ) = (1 + ic + )(1 + 2i/f + )(1 + ic + ). = + f c + or a = a + f c a + terms are odd in and c because of parity -- c and change signs under L R inhomogeneous term a signal of spontaneous symmetry breakdown mesons transform like SU (3) octet
1 tr U U with U = exp[2i/f ] 4 explore nonlinear terms -- for any matrix M the derivative of eM eM =
0 1
e(1-s)M M esM ds M eM =
0
1
e-sM M esM ds
1 1 1 tr U U = tr U U U U = - tr U U U U 4 4 4 1 2i U U = e-2is/f e2is/f ds f 0 2s2 2i 1 2is [, ] - 2 [, [, ]] + ds - f 0 f f because GBs are rotations of the vacuum, the only reason they interact at all in leading order is the non-Abelian nature of the symmetry -- introduces nonlinearity in the scattering of classical GB waves = try to visualize this with a model of SU (2) SU (2) SU (2) SU (2) SO(3) specify vacuum by orientation in isospin space -- gb-scattering3.nb
what happens when a Goldstone boson wave passes through?
gb-b0.flv gb-a2.flv gb-b1.flv gb-b2.flv gb-a3.flv gb-a4.flv
we want to build a low energy theory with the same symmetries -- qR R q R qL L qL r R r R - iR R L L - iL L s + ip M R M L GB field U U L U R r R r R - iR R L L - iL L M RML
low energy theory effective L depends on U and classical sources D U = U + i U - iU r D U = U + ir U - iU D U L D U R D U R D U L QCD also invariant under parity r -r U U r M M
most general effective Lagrangian consistent with the symmetries -- only useful because we can expand in powers of derivatives sources and truncate the expansion after a small number of terms leading term for M = 0 L(U, r, , s, p) = f2 tr D U D U 4
including M -- parameter (mass) 1 tr D U D U + tr (U M + h.c.) + 4 1 0 1 + + K+ 1 2 - 6 K- = K+ 1 1 - 2 0 + 6 K0 = a Ta = K0 = K0 2 2 K- K0 - 6 L = f2 U = exp(2i/f ) and D U = U + i U - iU r (gives the currents) and symmetry breaking by M = M qL M qR + h.c. gives "GB" masses M= mu 0 0 0 md 0 0 0 ms
for and M real, the linear term cancels and the quadratic term is -2tr(M 2 ), which corresponds to a mass term (like baryons but only one term) 4tr(M 2 ) for the -- evaluate these masses in the limit of isospin invariance, ignoring weak and electromagnetic interactions and setting mu = md = m
1
4
m2 = 4tr M 0 0 m2 K = 4tr M 0 0
1 12 1 4
0 0 0 0 0 0
1 12 1 4
0 0 = 2m 0 0 0
1 4
= (m + ms ) 0 2 0 = (m + 2ms ) 3 1
3
m2 = 4tr M 0 0
0
The m2 determined in this way satisfy the Gell-Mann Okubo relation 3m2 + m2 = 4m2 K but here we are using the momentum expansion rather than expanding in powers of the symmetry breaking -- this depends on the SU (3) symmetric part of M and makes sense even for m 0 -- and explains why GMO for mesons works much better for m2 than for m 2 2 notice that M gives a potential - f2 tr U M - f2 tr (U M ) that is minimized for U = I -- until we turned on M all U were equally good vacua
if we don't assume mu = md
1
4
m2 0 = m2 = 4tr M 0 0 m2 = 4tr M K 0 0
1 4
0 0
1 4
0 0 = (mu + md ) 0 = (mu + ms )
0 0 0 0 0 0
1 12 1 4
0 0
1 4
m2 0 K
0 = 4tr M 0 0 1
12
0 0 = (md + ms )
1 4
m2 = 4tr M 0 0
0
0 0 = (mu + md + 4ms ) 3 1
3
m2 = 4tr M 0 - 0
1 2 3
0
1 - 23 0
0 2 0 = (mu - md ) 3 0
the small - 0 mixing term is higher order in isospin breaking and very small 1 0 0 4 m2 0 = m2 = 4tr M 0 1 0 = (mu + md ) 4 0 0 0 m2 K = 4tr M 0 0
1 4
0 0 0 0 0 0
1 12 1 4
0 0
1 4
= (mu + ms )
m2 0 K
0 M 0 = 4tr 0 1
12
0 0 = (md + ms )
1 4
m2 = 4tr M 0 0
0
0 0 = (mu + md + 4ms ) 3 1
3
quark masses do not split m from m0 to this order - both proportional to mu + md (the splitting is a I = 2 effect) but EM interactions do -- Q has m2 = m2 K tr(U QU Q) L and R part -- photon loops others 0 mq ratios
including M and Q 1 f2 tr D U D U + tr (U M + h.c.) + tr(U QU Q) + 4 2 1 0 1 + + K+ 2 6 1 K- = K+ 1 1 - - 2 0 + 6 K0 = a Ta = K0 = K0 2 2 - 0 K K - 6 L = f2 U = exp(2i/f ) and D U = U + i U - iU r (gives the currents) and symmetry breaking by s = M qL M qR + h.c. gives "GB" masses
M=
mu 0 0
0 md 0
0 0 ms
is the source the LH octet currents 1 1 if 2 tr U U - ( U ) U = -if 2 tr 4 2
1 U
U ds
=f
0
tr
- -
2is 2s2 [, ] + 2 [, [, ]] + f f
the general case - coset construction - and how G generated by Ta elements g = eia T broken to H generated by Sa
a
SU (3) SU (3)
L Ta elements L = ei a T a R Ta elements R = eira T
a
Ta
unbroken broken L R L R S Ta = Ta = Ta Xa Ta = -Ta = Ta generators a generators Goldstone bosons are the group elements associated with Xa
a GBs = eia X
L = = eia Ta and R = = e-ia Ta
general transformation can be uniquely decomposed into a product of a broken transformation (GB) and an unbroken transformation eixa X eisa S
a a
L = eixa T eisa T
a
a
and R = e-ixa T eisa T L = u(L, R, ) R = u(L, R, )
a
a
T S + X determines how the GBs transform eia Ta = eixa X eisa Sa u(, ) Goldstone bosons in the Coset space G/H
a
SU (3) SU (3) SU (3)
back to the heavy baryon effective theory using matrix B to describe the low-energy baryon -- matrix structure gets the flavor right -- spin is just ordinary angular momentum because we have broken Lorentz invariance by going to the frame in which the baryon is at rest -- so B is a Pauli spinor and a flavor matrix L = tr BD0 (B) + i f Bj
jk k [Ta va , B] +
d Bj
jk
k {Ta va , B} +
where
D (O) = O + iv [T , O]
how do the baryons transform under chiral SU (3) SU (3)? the answer is that they don't! baryons and all the other QCD bound states live in the vacuum defined by SU (3) SU (3) SU (3) -- we know that it doesn't make sense to make a global rotation of this vacuum and thus it only makes sense to ask how the baryons transform under the unbroken Gell-Mann's SU (3) -- and this we know -- they are an octet -- but this is enough to allow us to construct an SU (3) SU (3) invariant theory using the Goldstone Boson fields which transform under a general L and R like = Lu(L, R, ) = u(L, R, )R
= Lu(L, R, ) = u(L, R, )R u(L, R, ) is the vacuum preserving part of the transformation -- all the rest goes into transforming the s which are the physically sensible local rotations of the physical vacuum -- thus under a general L and R, the baryon transformation law is nonlinear and involves B B = u(L, R, ) B u(L, R, ) to build an invariant L, we have to deal with derivatives -- a little crazy because we now have 3 "local" symmetries -- L and R because we demand classical gauge invariance with the sources as gauge field -- and u which depends on space and time through its dependence on L, R and -- build two octets out of the GB fields with simple transformation laws under u
= Lu(L, R, ) = u(L, R, )R i ( + i ) + ( + ir ) 2 transforms like a gauge field for the local u transformation vector field V =- V u(L, R, )V u(L, R, ) - i u(L, R, ) u(L, R, ) i ( + i ) - ( + ir ) 2 transforms like an SU (3) octet field (note trV = trA = 0) axial vector field A = - A u(L, R, )A u(L, R, ) with V we can make an improved covariant derivative with chiral symmetry D (B) = B + i[V , B] u(L, R, )D (B) u(L, R, ) the leading derivative terms in L are L = tr B D0 (B) + d B {A, B} + if B [A, B] +
L = tr B D0 (B) + d B {A, B} + if B [A, B] + D (B) = B + i[V , B] V =- = ei/f
i + r ( + i ) + ( + ir ) = + 2 2 i - r A = - ( + i ) - ( + ir ) = + /f + 2 2 0 the D term is what we saw before -- it determines the couplings of the vector current
0 = 3
j = i1 j
5 = -i 0 1 2 3 = 2
0 j 5 = - j
the d and f terms determine both the baryon matrix element of the axial vector current (some of which can be measured in the semileptonic weak interactions) and the GB couplings to the baryons -- for the chiral SU (2) currents between proton and neutron states only the f term contributes and the matrix element of the axial vector current in decay is related to the pion coupling to nucleons -- Goldberger-Treiman relation
0 decay -- another EM effect -- could be produced by D U = U + ieA [Q, U ]. but this confused people (me included) for many years -- this simple substitution and related quantum effects are not very efficient in producing the decay angular momentum and parity in l = 1, j = 0 state associated with the pseudoscalar operator F . F an effective interaction of the following form would produce the decay 0
F F
but the quark charge matrix of QCD commutes with T3 because both are diagonal -- Q doesn't break the chiral T3 symmetry -- so (the old slightly wrong argument called the Sutherland theorem goes) all interactions produced by Q must have the chiral symmetry and so in our beautiful GB formalism, they must be derivative interaction from Lagrangian terms invariant under the chiral symmetry -- 0 field transforms inhomogeneously
we could write invariant interaction terms with derivatives, such as i
F F tr Q2 U U + h.c.
or terms involving the explicit chiral symmetry breaking, such as i
F F tr Q2 M U + h.c.
but the contribution of these terms to the 0 decay amplitude is suppressed by a power of m2 over the dimensional constant that governs higher order terms in the derivative expansion -- which is expected to be of the order of 1 GeV -- these terms are just too small to explain the observed decay
this is particularly embarrassing because it seems to contradict a trivial calculation that gives the right answer -- take our perturbative model of an octet of GBs -- with the 3 3 field, and couple it to three colors of quarks with an SU (3) SU (3) invariant interaction g(L qR + h.c.) which after symmetry q breaking gives mq = gf -- in the effective theory below the mass of the quarks there a coupling between the GB 0 and generated by matching from the quark triangle diagram (I'll call this the Steinberger calculation) q 0 q we did this calculation in the first lecture and the result here is 0 F F 8f notice that the coupling to the quarks has gone away because the amplitude had a g in the numerator from the ig 0 3 q 5 q coupling and mq = gf in the denominator -- there is something general going on q
what is going on here? is there something wrong with our fancy treatment of GBs with only derivative interactions? look at three flavors of free massless fermions coupled to flavor currents
v q a q v a = ( - r )/2 q
v = (
+ r )/2
as usual just using the classical sources as tools to calculate n-point functions clasically there is an SU (3) SU (3) symmetry so we would expect the axial current to be conserved -- but the graph has IR (as well as UV) divergences calculate straightforwardly but use Lorentz invariance to pull out a piece proportional to several factors of momentum so the loop integration is finite (done in Zee's book) or get it as a limit of massive fermions
v q a q q = m5 q q q
v
v v but without the m, the right hand side is our favorite calculation again -- it has a 1/m that cancels the m in the numerator -- thus this contribution persists even as m 0, leading to an "anomalous divergence" of the axial vector current at m = 0 -- the anomaly is the fact that the effect persists to m = 0 -- it's an IR effect -- so do we need some kind of extra term in our theory to account for this? for free massless fermions, obviously not -- the triangle graph is there in the theory to produce this effect in our theory -- still not -- we added massless fermions, which produce the anomalous divergence -- we coupled them to and the VEV generated a mass -- but as we have seen, the mass and the details of the coupling just cancels in the triangle diagram which still produces the "anomalous" divergence but if we integrate the fermions out of the theory - or if we do our chiral rotation so the fermions transform only under the u SU (3) then we need a new term
this resolution of the puzzle is called the anomaly -- loops of chiral fermions coupled to gauge fields break chiral symmetries -- integrating out chiral fermions, either massless fermions or fermions whose mass is generated by spontaneous breaking of the chiral symmetry, introduces a new term in the low energy L that is not invariant under chiral SU (3) SU (3) symmetry - the chiral symmetry is broken by quantum effects -- a part of this extra term in L involves the photon fields, and under a T3 chiral transformation changes by 3e2 tr(T3 Q2 ) F F + 2 16 (the factor of 3 comes from the the three colors of the quarks) -- but the effects of chiral symmetry have not completely disappeared -- under a GLOBAL T3 chiral transformation 0 0 + c, the change in this term is a total derivative -- (like the F F term) so the action is invariant if we can neglect the fields at infinity -- the anomaly is an IR effect that only breaks the global chiral symmetry at long distances various ways of seeing what this extra term is -- call it W (U, v , a ) -- easiest to find its change under infinitesimal chiral transformation = eic = 1 + ic + for general sources v and a , Wess and Zumino determined its form from the SU (3) SU (3) symmetry and Bardeen calculated it directly by looking at all the relevant loop graphs -- both imposing conservation of the vector currents -- the result for the change is -c3
dW = F (c, v , a ) = where f =-
f (c(x), v (x), a (x)) d4 x ,
1 + a a tr c(3v v 16 2 -8i(a a v + a v a + v a a ) -32a a a a ) ,
with
v = ( + r )/2, a = ( - r )/2, v = v - v + i[v , v ] + i[a , a ], a = a - a + i[v , a ] + i[a , v ]. there are other forms like the so-called Fujikawa form that do not satisfy ordinary SU (3) gauge invariance, but they can be obtained from this with the addition of polynomials in v and a to the action -- these ambiguities are UV effects associated with the different schemes for defining the currents as composite operators like s2 counterterm that one needs in a scalar theory for the source s of the composite operator 2 -- the anomaly itself is an IR effect and is unambigous -- you can move its effects around with different UV counterterms, but you can't get rid of it
contact term and anomaly for free massless fermions in 1+1 dimensions with only one flavor source s (i - s ) clasically source s
j5 =
j
0 j5 = -j 1
1 j5 = -j 0
j = j5 = 0
(i - s ) = 1 (i - - s- )1 + 2 (i + - s+ )2
a = a0 a1
clasically 1 ei1 1 2 ei2 2
+j - = -j + = 0 covariant derivative covariant derivative
s- s- - - 1 s+ s+ - + 2
D- = - + is- D+ = + + is+
s s - (1 + 2 )/2 -
(1 - 2 )/2
s0 s0 - 0 (1 + 2 )/2 + 1 (1 - 2 )/2 s1 s1 - 1 (1 + 2 )/2 + 0 (1 - 2 )/2
1 propagator
1 S0 (x)
-i
0|T 1 (x) 1 (0)|0
=
d2 p -ipx p+ 1 x+ e =- (2)2 p2 + i 2 x2 - i
- S 1 (x) = 2+ S 1 (x) = -i(x0 ) 0|[1 (x), 1 (0)]+ |0 = -i 2 (x)
-
1 x+ i 2+ 2 = -2i(x+ )(x- ) = -i 2 (x) 2 x -i (x2 - i )2 - = 2 (x) - i )2
funny representation of the 2D -function lim
0 (x2
imaginary part and everything away from the origin for real part just averages to zero over smooth functions.
in terms of the currents j + = 21 1 and j - = 22 2 , L is
source s
(i - s ) = 1 i - 1 + 2 i + 2 - s- j - - s+ j +
look at the two-point function of currents -- ignoring potential issues when x = 0 -- for example we are assuming that (x0 )(-x0 ) = 0 in replacing the T -product of the product operators with the product of the T -products.
0|T j + (x)j + (0)|0 = 4 0|T 1 (x)1 (x) 1 (0)1 (0)|0 = 4 0|T 1 (x) 1 (0)|0 0|T 1 (x)1 (0)|0 = -4 0|T 1 (x) 1 (0)|0 0|T 1 (-x)1 (0)|0 = -
1 x+ x2 - i 2 2
2
- 0|T j + (x)j + (0)|0 = 2+ 0|T j + (x)j + (0)|0 = 4 2 x- (x+ )2 x+ - 2 (x2 - i )2 (x - i )3 = 4 2 -i x+ (x2 - i )3 = - (x2 i - i )2
i - + 2 (x)
for the j - matrix element 1 2 and + - -- can interpret this by saying that j = 0 and there is a "Schwinger term" in the equal time commutator, i 0|T j (x)j (0)|0 = (x0 ) 0|[j (x), j (0)]|0 = - 2 (x) 0|T j (x)j (0)|0 = 0 in the language of sources, we can also describe this as an anomaly in the current "conservation equation" so that j = - 0| j |0 1 s 2 0|0 s 1 s 2
s=0 s
0 as s0 1 2
j = -
= 2i
j =
s
s s=0
0|T j (x)j (0)|0 = 2i s (0) -
0| j |0 s (0)
s s=0
1 s (x) 0|0 2
i = - 2 (x)
"Fujikawa form" in which both the vector and axial vector currents are anomalous but in which the two chiralities don't talk to one another
notice that in this case the "anomalies" in the conservation equations j = - 1 s 2
j =
1 2
s
are only there while the sources are turned on -- once we have used s to find the 2-point functions, we can set s to zero -- then the currents are conserved j (x) = 0 i 0|T j (x)j (0)|0 = (x0 ) 0|[j (x), j (0)]|0 = - 2 (x) 0|T j (x)j (0)|0 = 0 or (x0 ) 0|[j 0 (x), j 0 (0)]|0 = (x0 ) 0|[j 1 (x), j 1 (0)]|0 = - (x0 ) 0|[j 0 (x), j 1 (0)]|0 = (x0 ) 0|[j 1 (x), j 0 (0)]|0 = - i 0 2 (x) 2 i 1 2 (x) 2 0|T j (x)j (0)|0 = (x0 ) 0|[j (x), j (0)]|0 = 0
the "anomalies" in the conservation equations j = - 1 s 2
j =
1 2
s
can be moved around by redefining the composite currents with source terms -- for example if ~ j + 1 s then j 2 1 ~ = 0 j ~ = s j or if ^ j - 1 s = ~ - 1 s then j j 2 1 ^ = - s j ^ = 0 j these different definitions correspond to different values of the s s term in L
Dear Dr.
Georgi
I would be very grateful if you could send me a copy of your paper: Anomalies of the Axial - V ector Current which appeared in my notes in 1968 I suspect Professor Lowell Brown, Physics Dept Univsersity of Washington, Seattle, WA 98195
Dear Dr.
Georgi
I would be very grateful if you could send me a copy of your paper: Anomalies of the Axial - V ector Current which appeared in my notes in 1968 I suspect Professor Lowell Brown, Physics Dept Univsersity of Washington, Seattle, WA 98195
Dear Dr.
Georgi
I would be very grateful if you could send me a copy of your paper: Anomalies of the Axial - V ector Current which appeared in my notes in 1968 I suspect Professor Lowell Brown, Physics Dept Univsersity of Washington, Seattle, WA 98195
now back to 3+1 with 3 flavors dW = F (c, v , a ) = f =- f (c(x), v (x), a (x)) d4 x where
1 tr c(3v v + a a 2 16 -8i(a a v + a v a + v a a ) -32a a a a )
v = ( + r )/2 a = ( - r )/2 v = v - v + i[v , v ] + i[a , a ] a = a - a + i[v , a ] + i[a v ] there are other forms like the so-called Fujikawa form that do not satisfy ordinary SU (3) gauge invariance, but they can be obtained from this with the addition of polynomials in v and a to the action -- these ambiguities are UV effects associated with the different schemes for defining the currents as composite operators like s2 counterterm that one needs in a scalar theory for the source s of the composite operator 2 -- the anomaly itself is an IR effect and is unambigous -- you can move its effects around with different UV counterterms, but you can't get rid of it
with
now we can put the GB fields -- define U (s) = e2is/f (s) = eis/f (s) = (1 - s) (1 - s) - i (1 - s) (1 - s) r (s) = (1 - s)r (1 - s) - i(1 - s) (1 - s) v (s) = (s) + r (s) /2 a (s) = (s) - r (s) /2.
different values of s are related by chiral gauge transformations, which depend on the Goldstone boson field -- in particular d W U (s), v (s), a (s) ds ds = ds/f W = ds F /f, v (s), a (s) . we can integrate this! W U, v , a = W U (1), v (1), a (1)
1
=
0
F /f, v (s), a (s) ds - W U (0), v (0), a (0)
W U, v , a = W U (1), v (1), a (1)
1
=
0
F /f, v (s), a (s) ds - W U (0), v (0), a (0)
U (s) = e2is/f (s) = eis/f (s) = (1 - s) (1 - s) - i (1 - s) (1 - s) r (s) = (1 - s)r (1 - s) - i(1 - s) (1 - s) v (s) = (s) + r (s) /2 a (s) = (s) - r (s) /2.
U (0) = 1 (0) = - i r (0) = r - i v (0) and am (0) are the V and A we discussed earlier which transform only under the u SU (3) i V = - ( + i ) + ( + ir ) = v (0) 2 V u(L, R, )V u(L, R, ) - i u(L, R, ) u(L, R, ) i A = - ( + i ) - ( + ir ) = a (0) 2 A u(L, R, )A u(L, R, ) W (1, v (0), a (0)) is invariant under ordinary SU (3) gauge transformations!
thus W (1, (0), v (0)) can be absorbed into the ordinary terms in the action that are invariant under SU (3)L SU (3)R and we can write the extra term that incorporates the effect of the anomaly as W U, v , a =
0 1
F /f, v (s), a (s) ds
U (s) = e2is/f (s) = eis/f (s) = (1 - s) (1 - s) - i (1 - s) (1 - s) r (s) = (1 - s)r (1 - s) - i(1 - s) (1 - s) v (s) = (s) + r (s) /2 a (s) = (s) - r (s) /2
F (c, v , a ) = where f =- 1 16 2
f (c(x), v (x), a (x)) d4 x
tr c (3v v + a a
-8i(a a v + a v a + v a a )-32a a a a ) ~ ~ F F 3 F 5
thus W (1, (0), v (0)) can be absorbed into the ordinary terms in the action that are invariant under SU (3)L SU (3)R and we can write the extra term that incorporates the effect of the anomaly as W U, v , a =
0 1
F /f, v (s), a (s) ds
U (s) = e2is/f (s) = eis/f (s) = (1 - s) (1 - s) - i (1 - s) (1 - s) r (s) = (1 - s)r (1 - s) - i(1 - s) (1 - s) v (s) = (s) + r (s) /2 a (s) = (s) - r (s) /2
F (c, v , a ) = where f =- 1 16 2
f (c(x), v (x), a (x)) d4 x
tr c (3v v + a a
-8i(a a v + a v a + v a a )-32a a a a ) ~ ~ F F 3 F 5
thus W (1, (0), v (0)) can be absorbed into the ordinary terms in the action that are invariant under SU (3)L SU (3)R and we can write the extra term that incorporates the effect of the anomaly as W U, v , a =
0 1
F /f, v (s), a (s) ds
U (s) = e2is/f (s) = eis/f (s) = (1 - s) (1 - s) - i (1 - s) (1 - s) r (s) = (1 - s)r (1 - s) - i(1 - s) (1 - s) v (s) = (s) + r (s) /2 a (s) = (s) - r (s) /2
F (c, v , a ) = where f =- 1 16 2
f (c(x), v (x), a (x)) d4 x
tr c (3v v + a a
-8i(a a v + a v a + v a a )-32a a a a ) ~ ~ F F 3 F 5
thus W (1, (0), v (0)) can be absorbed into the ordinary terms in the action that are invariant under SU (3)L SU (3)R and we can write the extra term that incorporates the effect of the anomaly as W U, v , a =
0 1
F /f, v (s), a (s) ds
U (s) = e2is/f (s) = eis/f (s) = (1 - s) (1 - s) - i (1 - s) (1 - s) r (s) = (1 - s)r (1 - s) - i(1 - s) (1 - s) v (s) = (s) + r (s) /2 a (s) = (s) - r (s) /2
F (c, v , a ) = where f =- 1 16 2
f (c(x), v (x), a (x)) d4 x
tr c (3v v + a a
-8i(a a v + a v a + v a a )-32a a a a ) ~ ~ F F 3 F 5