Symmetry breaking: Pion as a Nambu-Goldstone boson

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1 Symmetry breaking: Pion as a Nambu-Goldstone boson Stefan Kölling Universität Bonn Seminar zur theoretischen Teilchenphysik,

2 Structure 1 General theory of sponatneous symmetry breaking What it is Spontaneous breaking of a global continuous symmetry (classical) Spontaneous breaking of a global continuous symmetry (QFT) Quantum fluctuations 2 Pion as a Nambu-Goldstone Boson Weak decays Pion as a Nambu-Goldstone boson 3 Summary

3 What it is A symmetry is said to be spontaneously broken, if the Lagrangian exhibits a symmetry but the ground state/vacuum does not have this symmetry.

4 What it is A symmetry is said to be spontaneously broken, if the Lagrangian exhibits a symmetry but the ground state/vacuum does not have this symmetry. Goldstone-theorem If the Lagrangian is invariant under a continuous global symmetry operation g G and the vacuum is invariant under a subgroup H G, then there exist n(g/h)=n(g)-n(h) massless spinless particles. With n(g) the number of generators of the group G.

5 Spontaneous breaking of a global continuous symmetry (classical) L(Φ i, Φ i ) = 1 2 ( µφ i ) 2 V (Φ i ) with V (Φ i )invariant under a global continous symmetry group G i.e. V (Φ i ) = V (ρ(g)φ i ) with g G Φ i Φ, i = ρ(g)φ i Φ i + δφ i = Φ i + iɛ a T a Φ i

6 Spontaneous breaking of a global continuous symmetry (classical) L(Φ i, Φ i ) = 1 2 ( µφ i ) 2 V (Φ i ) with V (Φ i )invariant under a global continous symmetry group G i.e. V (Φ i ) = V (ρ(g)φ i ) with g G Φ i Φ, i = ρ(g)φ i Φ i + δφ i = Φ i + iɛ a T a Φ i We demand that the vacuum has a minimal energy minimize V Φ i,min =< 0 Φ i 0 >= const. χ i := Φ i Φ i,min V (Φ i ) = V (Φ i,min ) + V (Φ i,min )χ i V χ i χ j +... Φ i 2 Φ i Φ j

7 Spontaneous breaking of a global continuous symmetry (classical) V (Φ i,min ) is a minimum, therefore V Φ i (Φ i,min ) = 0 2 V Φ i Φ j : = M 2 ij is semi-positive definite

8 Spontaneous breaking of a global continuous symmetry (classical) V (Φ i,min ) is a minimum, therefore V Φ i (Φ i,min ) = 0 2 V Φ i Φ j : = M 2 ij is semi-positive definite Now use the symmetry: V (Φ i,min ) = V (ρ(g)φ i,min ) = V (Φ i,min ) M2 ij(δφ i,min )(δφ j,min ) 0 = M 2 ijδφ j,min 0 = M 2 ijt a Φ j,min i i

9 Spontaneous breaking of a global continuous symmetry (classical) important result 0 = M 2 ij T a Φ j,min i

10 Spontaneous breaking of a global continuous symmetry (classical) important result 0 = M 2 ij T a Φ j,min i 1 T a H G with T a Φ j,min = 0 Every generator of H leaves the vacuum invariant.

11 Spontaneous breaking of a global continuous symmetry (classical) important result 0 = M 2 ij T a Φ j,min i 1 2 T a H G with T a Φ j,min = 0 Every generator of H leaves the vacuum invariant. T a / H T a Φ j,min 0 T a Φ j,min is an eigenstate of M 2 with 0 eigenvalue T a Φ j,min is a Nambu-Goldstone boson The number of Nambu-Goldstone bosons is n(g)-n(h)=n(g/h).

12 Pictures L = 1 2 ( µφ) µ2 φ 2 λ 4! φ4 with φ min = φ = φ min + σ 6 λ µ2 L = 1 2 ( µσ) (2µ2 )σ 2 λ 4! σ4 λ 6 µσ3

13 Pictures L = 1 2 ( µφ) µ2 φ 2 λ 4! φ4 with φ min = φ = φ min + σ 6 λ µ2 L = 1 2 ( µσ) (2µ2 )σ 2 λ 4! σ4 λ 6 µσ3 Massive modes correspond to fluctuations in radial direction. Massless modes correspond to fluctuations in angular direction.

14 Spontaneous breaking of a global continuous symmetry (QFT) In QFT every continuous symmetry implies the existence of a conserved current J µ (x): µ J µ (x) = 0 Q = d D xj 0 ( x, t) is conserved, i.e. [H, Q] = 0

15 Spontaneous breaking of a global continuous symmetry (QFT) In QFT every continuous symmetry implies the existence of a conserved current J µ (x): µ J µ (x) = 0 Q = d D xj 0 ( x, t) is conserved, i.e. [H, Q] = 0 Q 0 > =0, for every energy-eigenstate Φ >= Φ 0 > define Φ > := e iθq Φe iθq 0 > H Φ > = He iθq Φe iθq 0 >= He iθq Φ 0 > = e iθq HΦ 0 >= e iθq EΦ 0 >= E Φ > Energy spectrum is organized in multiplets of degenerate states.

16 Spontaneous breaking of a global continuous symmetry (QFT) Q 0 > 0 Problem, Q is not defined < 0 Q 2 (t) 0 > = d D x < 0 J 0 ( x, t)q(t) 0 > = d D x < 0 e ip x J 0 ( 0, 0)e ip x Q(t) 0 > = d D x < 0 J 0 ( 0, t)q(t) 0 >

17 Spontaneous breaking of a global continuous symmetry (QFT) Q 0 > 0 Problem, Q is not defined < 0 Q 2 (t) 0 > = d D x < 0 J 0 ( x, t)q(t) 0 > = d D x < 0 e ip x J 0 ( 0, 0)e ip x Q(t) 0 > = d D x < 0 J 0 ( 0, t)q(t) 0 > But the commutator with Q may be well defined. Current conservation implies: 0 = d D x[ µ J µ ( x, t), Φ(0)] = 0 d D [J 0 ( x, t), Φ(0)] + d S [ J( x, t), Φ(0)]

18 Spontaneous breaking of a global continuous symmetry (QFT) d [Q(t), Φ(0)] = 0 dt The symmetry is broken, if there is an operator Φ s.t.: < 0 [Q(t), Φ(0)] 0 >= η 0

19 Spontaneous breaking of a global continuous symmetry (QFT) d [Q(t), Φ(0)] = 0 dt The symmetry is broken, if there is an operator Φ s.t.: < 0 [Q(t), Φ(0)] 0 >= η 0 then: η = n d D x{< 0 J 0 (x) n > < n Φ(0) 0 > < 0 Φ(0) n >< n J 0 (x) 0 >} = n (2π) D δ D ( p n ){< 0 J 0 (0) n > < n Φ(0) 0 > e ient < 0 Φ(0) n >< n J 0 (0) 0 > e ient }

20 Since this is time independent, there has to be a state with: Nambu-Goldstone Boson E n = 0 for p n = 0 m n = 0 < n Φ(0) 0 > 0 < 0 J 0 (0) n > 0

21 Since this is time independent, there has to be a state with: Nambu-Goldstone Boson E n = 0 for p n = 0 m n = 0 < n Φ(0) 0 > 0 < 0 J 0 (0) n > 0 So the Nambu-Goldstone boson has to carry all the quantum numbers of the conserved current. In general there will be n(g) currents J a µ. Of these n(h) generate the symmetry group of vacuum. So again there are n(g/h)=n(g)-n(h) Nambu-Goldstone bosons.

22 Quantum fluctuations Can the massless fields wander away from their ground state? Calculate mean square fluctuation: < b(0) 2 > = 1 Dbe (is[b]) b(0)b(0) Z 1 = lim Dse is(b) b(x)b(0) x 0 Z d d k e ik x = lim x 0 (2π) d k 2 The UV devergencies can be regulated by a cutoff. But for d 2 there is an IR divergence.

23 Quantum fluctuations Can the massless fields wander away from their ground state? Calculate mean square fluctuation: < b(0) 2 > = 1 Dbe (is[b]) b(0)b(0) Z 1 = lim Dse is(b) b(x)b(0) x 0 Z d d k e ik x = lim x 0 (2π) d k 2 The UV devergencies can be regulated by a cutoff. But for d 2 there is an IR divergence. Coleman-Mermin-Wagner theorem Spontaneous symmetry breaking is impossible in d 2.

24 How do pions enter the game? We want to calculate dacay rates of semileptonic decays like: Use an effective Lagrangian n p + e + ν π π 0 + e + ν L = G[ēγ µ (1 γ 5 )ν](j µ J µ5 ) Where J µ and J µ5 are hadronic currents which include strong interaction effects. We have to calculate < p J µ (x) J µ5 (x) n > or < 0 J µ (0) J µ5 (0) π >

25 Nucleon β-decay < p(k ) J µ5 (x) n(k) > =< k e ip x J µ5 (0)e ip x k > =< k J µ5 (0) k > e i(k k) x Use Lorentz invariance to simplify < k J µ5 (0) k > with q := k k < k J µ5 (0) k > = ū p (k )[ iγ µ γ 5 F (q 2 ) + q µ γ 5 G(q 2 ) + (k µ + k µ )γ 5 F 2(q 2 ) + i[γ µ, γ ν ]γ 5 q ν F 3(q 2 )]u n (k)

26 Nucleon β-decay < p(k ) J µ5 (x) n(k) > =< k e ip x J µ5 (0)e ip x k > =< k J µ5 (0) k > e i(k k) x Use Lorentz invariance to simplify < k J µ5 (0) k > with q := k k < k J µ5 (0) k > = ū p (k )[ iγ µ γ 5 F (q 2 ) + q µ γ 5 G(q 2 ) + (k µ + k µ )γ 5 F 2(q 2 ) + i[γ µ, γ ν ]γ 5 q ν F 3(q 2 )]u n (k) And use the Gordan-Identity: ū p (k )[i 1 2 [γ µ, γ ν ]γ 5 q ν ]u n (k) = ū p (k )[i(k µ + k µ )γ 5 + (m p m n )γ µ γ 5 ]ū n (k)

27 Nucleon β-decay So if we define: F 3 (q 2 ) = F 3(q 2 ) + 1 2i F 2(q 2 ) F(q 2 ) = F (q 2 ) (m p m n )F 2(q 2 )

28 Nucleon β-decay So if we define: F 3 (q 2 ) = F 3(q 2 ) + 1 2i F 2(q 2 ) F(q 2 ) = F (q 2 ) (m p m n )F 2(q 2 ) we obtain: Decomposition of < k J µ5 (0) k > into Form factors < k J µ5 (0) k >= u p (k )[ iγ µ γ 5 F(q 2 ) + q µ γ 5 G(q 2 ) + i[γ µ, γ ν ]γ 5 q ν F 3 (q 2 )]u n (k) And similarly we obtain: < 0 J µ 5 (0) π(k) >= if πk µ

29 Problems with strongly interacting particles In general a pertubation theory is not well defined.

30 Breaking of chiral symmetry M π 139MeV << 938Mev m N M π 0 Is it possible that the pion is a Nambu-Goldstone Boson?

31 Breaking of chiral symmetry M π 139MeV << 938Mev m N M π 0 Is it possible that the pion is a Nambu-Goldstone Boson? Suppose µ J µ 5 (x)=0 i.e. chiral symmetry holds then: < 0 J µ 5 (0) π(k) >= if πk µ < 0 J 0 (0) n > 0 < π π(0) 0 >= 1 < n Φ(0) 0 > 0

32 Breaking of chiral symmetry M π 139MeV << 938Mev m N M π 0 Is it possible that the pion is a Nambu-Goldstone Boson? Suppose µ J µ 5 (x)=0 i.e. chiral symmetry holds then: < 0 J µ 5 (0) π(k) >= if πk µ < 0 J 0 (0) n > 0 < π π(0) 0 >= 1 < n Φ(0) 0 > 0 Pion as a Nambu-Goldstone Boson The Pion is a Nambu-Goldstone Boson in a world in which chiral symmetry holds.

33 Is the Pion massless? If chiral symmetry holds then: ik µ < 0 J µ 5 (0) π(k) > exp( ik x) = µ < 0 J µ 5 (0) π(k) > exp( ik x) =< 0 µ J µ 5 (x) π(k) >= 0

34 Is the Pion massless? If chiral symmetry holds then: ik µ < 0 J µ 5 (0) π(k) > exp( ik x) = µ < 0 J µ 5 (0) π(k) > exp( ik x) =< 0 µ J µ 5 (x) π(k) >= 0 k µ < 0 J µ 5 (0) π(k) >= if πk µ k µ = if π M 2 π Thus µ J µ 5 (x)=0 implies that M2 π=0

35 Goldberger-Treiman relation q µ < k J µ5 (0) k > = i < k µ J µ5 (x) k > exp( iq x) = 0 q µ < k J µ5 (0) k > = q µ ū p (k )[ iγ µ γ 5 F(q 2 ) + q µ γ 5 G(q 2 ) + i[γ µ, γ ν ]γ 5 q ν F 3 (q 2 )]u n (k) = ū N (k )[(k µ ( i)γ µ γ 5 F(q 2 ) k µ ( i)γ µ γ 5 F(q 2 ) + q 2 γ 5 G(q 2 ) + i[γ µ, γ ν ]γ 5 q ν q µ F 3 (q 2 )]u N (k) = γ 5 [2m N F(q 2 ) + q 2 G(q 2 )] = 0 by using the Dirac equation: ū N (k )(k µγ µ im N ) = (k µ γ µ im N )u N (k) = 0 But if q 0 this implies m N = 0!?!

36 Goldberger-Treiman relation this diagram gives a contribution if π q µ i q 2 G πnn ū N (k )γ 5 u N (k) Note that there are still infinitely many diagrams which have a pole at q 2 = 0 G(q 2 ) F π 1 q 2 G πnn for q 0 Goldberger-Treiman relation G πnn = 2m Ng A F π with g A = F (0)

37 Experimental test of Goldberger-Treiman relation m N = (m p + m n ) 2 g A = 1, 257 F π = 93MeV G GTR πnn 25.4 G πnn = 27.0 = 939.9MeV In good agreement with the experiment.

38 Experimental test of Goldberger-Treiman relation m N = (m p + m n ) 2 g A = 1, 257 F π = 93MeV G GTR πnn 25.4 G πnn = 27.0 = 939.9MeV In good agreement with the experiment. Please note that this result is quite general. No assumptions about the broken symmetry have been made.

39 More about chiral symmetry The Lagrangian of strong interaction: L = ūγ µ D µ u dγ µ D µ d + m u (ū R u L + ū L u R ) + m d ( d L d R + d R d L ) +... has a SU(2) V SU(2) A symmetry in the case m u = m d = 0. q q = exp(iθ V τ + iγ 5ΘA τ)q with τ the isospin (Pauli) matrices.

40 More about chiral symmetry The Lagrangian of strong interaction: L = ūγ µ D µ u dγ µ D µ d + m u (ū R u L + ū L u R ) + m d ( d L d R + d R d L ) +... has a SU(2) V SU(2) A symmetry in the case m u = m d = 0. q q = exp(i Θ V τ + iγ 5 ΘA τ)q with τ the isospin (Pauli) matrices. And the conserved currents: with charges: J µ = i qγ µ τq and J µ 5 = i qγ µγ 5 τq Q V = d 3 x J 0 ( x, t) Q A = d 3 x J5 0 ( x, t)

41 More about chiral symmetry They act in the following way on q: [ Q V, q] = τq [ Q A, q] = γ 5 τq So if the symmetry is unbroken Q A transforms a state h> into a state of Q A h > of opposite parity. No such parity doubling is observed in the hadron spectrum. Conclude that SU(2) V SU(2) A is broken to SU(2) V isospin group. The three Pions are the three Nambu-Goldstone Bosons of the three currents J µ 5.

42 Summary/Remarks Breaking of a continuous global symmetry leads to Nambu-Goldstone Bosons. No information about how the breaking occurs is needed. Natural explanation for smallness of Pion mass. Found a technique to relate infinitly many diagramms.

43 Summary/Remarks Breaking of a continuous global symmetry leads to Nambu-Goldstone Bosons. No information about how the breaking occurs is needed. Natural explanation for smallness of Pion mass. Found a technique to relate infinitly many diagramms. From the assumption that Pions and even Kaons and the Eta are Goldstone bosons one can construct effective field theories like non-linear sigma model or chiral pertubation theory. The QCD Lagrangian for massless Quarks has two other global symmetries. One, U(1) V implies Baryon-number conservation. The other one, U(1) A will be treated next week.

44 Summary/Remarks Breaking of a continuous global symmetry leads to Nambu-Goldstone Bosons. No information about how the breaking occurs is needed. Natural explanation for smallness of Pion mass. Found a technique to relate infinitly many diagramms. From the assumption that Pions and even Kaons and the Eta are Goldstone bosons one can construct effective field theories like non-linear sigma model or chiral pertubation theory. The QCD Lagrangian for massless Quarks has two other global symmetries. One, U(1) V implies Baryon-number conservation. The other one, U(1) A will be treated next week. Thank you

45 Sources A. Zee, Quantum field theory in a nutshell, Princeton University press 2001 S. Weinberg, The quantum theory of fields II, Camebridge University Press 1996 Ta-Pei Cheng and Ling-Fong Li, Gauge theory of elementary particles, Oxford University Press 1984 Michael Peskin and Daniel Schroeder, An introduction to quantum field theory, Perseus Books, 1995

Group Structure of Spontaneously Broken Gauge Theories

Group Structure of SpontaneouslyBroken Gauge Theories p. 1/25 Group Structure of Spontaneously Broken Gauge Theories Laura Daniel ldaniel@ucsc.edu Physics Department University of California, Santa Cruz