Lecture 6: Symmetries I

• Isospin • Parity • Parity Conservation and Assignment • Charge Conjugation

Useful Sections in Martin & Shaw: Section C.1 - C.2, Sections 5.2, 4.3, 4.4 Isospin

) θ n p

) θ n p proton & neutron appear to be swapping identities ⇒ ''exchange force" This means, to conserve charge, pions must come in 3 types: q = -1, 0, +1

Note also that mp = 938.3 MeV mn = 939.6 MeV

So there appears to an '' approximate " symmetry here Noether’s theorem says something must be conserved... Call this ''Isospin" in analogy with normal spin, so the neutron is just a ''flipped" version of the proton

I=1/2 System ⇒ p n (just 2 states) I3 : 1/2 -1/2 Impose isospin Some way pions can be produced conservation + + p + p → p + n + πππ I3(πππ ) = +1 000 000 → p + p + πππ I3(πππ ) = 0 + −−− − → p + p + π + πππ I3(πππ ) = -1

⇒ I=1 for the pions (similar arguments for other particle systems)

So we can think of these particle ''states" as the result of a (continuous) ''rotation" in isospin-space Example: What are the possible values of the isotopic spin and it’s z-component for the following systems of particles: a) π+ + p b) π− + p

+ p π a) p : I = 1/2, I = +1/2 π+ : I = 1, I = +1 π0 3 n 3 π−

total I3 = 1 + 1/2 = 3/2 thus, the only value of total Isospin we can have is also I = 3/2

π− : I = 1, I = −1 b) p : I = 1/2, I 3 = +1/2 3

total I3 = −1 + 1/2 = −1/2 thus, possible values of total Isospin are: I = 1 − 1/2 = 1/2 or I = 1 + 1/2 = 3/2 Parity P F(x) = F(-x) ⇒ discreet symmetry (no conserved ''currents")

consider the scattering y probability of the following: m1 m2 x

y P x = −x

m1 m2 x

P dx/dt = − dx/dt y

m1 m2 x

So, even though x and dx/dt are each odd under parity, the scattering probability, i.e. P PS , is even ( PS = PS) ⇒ parity is multiplicative, not additive Also note that parity does not reverse the direction of spin! +z −z −z +z

flip z flip ''velocity" −z +z direction +z −z parity (stand on your head )

But, for orbital angular momentum in a system of particles, it depends on the symmetry of the spatial wave function!! +z −z −z −z

flip z flip ''velocity" flip x & y −z +z direction +z positions +z parity +z

not the same ! −z (stand on your head) Intrinsic parity of the photon from ''first-principles":

∇ • E(x,t) = ρ(x,t)/ ε0

P ρ(x,t) = ρ(−x,t) P ∇ = −∇

thus, we must have P E(x,t) = −E(-x,t) for Poisson’s equation to remain invariant

But also E = −∇Φ − ∂A/∂t = − ∂A/∂t (in absenceof free charges)

and since ∂/∂t doesn’t change the parity ⇒ P A(x,t) = −A(−x,t)

But A basically corresponds to the photon wave function: A(x,t) = N εεε(k) exp[i( k•••x−ω t)]

intrinsic Thus, the parity of the photon is −1 (or ℘γ = −1 ) However, the effective parity depends on the angular momentum carried away by the photon from the system which produced it: l Pγ = ℘γ (−1)

(i.e. radiation could be s-wave, p-wave etc.)

but for an isolated photon, this cannot be disentangled!! The η(547) meson has spin 0 and is observed to decay via the Example: electromagnetic interaction through the channels: η → π0 + π0 + π0 and η → π+ + π− + π0 From this, deduce the intrinsic parity of the η and explain why the decays: η → π0 + π0 and η → π+ + π− are never seen P = P P P π η π π π π1 3 L L L 3 12 3 L 3 ℘η = ( ℘π ) (−1) ( −1) 12 π2 but final state must have zero total angular momentum since the initial state has spin 0

Ltot = L12 + L3 = 0 ⇒ L12 = L 3 L 2 3 3 = (℘ )3 = (−1 )3 = −1 ℘η = (℘π ) {( − 1) } π

However, for 2-pion final states we would have: 2 L P = ( ℘π ) (−1) 2 but we must have L=0 , so P = ( ℘π ) = 1 and is thus forbidden Evidence for Parity Conservation (in strong/electromagnetic interactions) 1) Polarized protons scattering off a nucleus show no obvious asymmetry towards spin-up vs spin-down directions

2) Ground state of deuteron (np) has total angular momentum J=1 and spin S=1 . Thus, the orbital angular momentum could take on values of lll = 0 (m=1), lll = 1 (m=0) or lll = 2 (m = −1) But the observed magnetic moment is consistent with a superposition of only S and D waves (lll=0, 2) . This can be reconciled if P (p+n) = P (d) l ⇒ so lll must be even ℘p ℘n = ℘p ℘n (−1)

By convention , ℘p = ℘n ≡ +1 & also ℘e- ≡ +1 ⇒ and the relative parities of the other particles then follow (anti-fermions have the opposite parity, anti-bosons have the same parity) Parity is a different animal from other symmetries in many respects...

It is often impossible to determine the absolute parity (assigned +1 or -1) of many particles or classes of particles. So we essentially just assume that basic physical processes are invariant with respect to parity and construct theories accordingly, making arbitrary assignments of parity when necessary until we run into trouble.

Weak interaction violates parity !!

(so, ''left" and ''right" really matter... weird!) C-Parity (charge conjugation)

Changes particle to anti-particle (without affecting linear or angular momentum)

Electromagnetism is obviously symmetric with respect to C-parity (flip the signs of all charges and who would know?). It turns out that the strong force is as well. But, again, not the weak force (otherwise there would be left-handed anti-neutrinos... there aren’t !)

For particles with distinct anti-particles (x = e −, p, π−, n, ...)

a state characterized C ∣x, ψ〉 = ∣x, ψ〉 where ∣x, ψ〉 ≡ by a particle x and a wave function ψ

For particles which no not have distinct anti-particles (y = γ, π0, ...)

where C is a ''phase factor" of ±±±1 C ∣y, ψ〉 = C ∣y, ψ〉 y y (like for parity) used to determine C-conservation in interactions For multi-particle systems

C ∣ = ∣ , = ∣ , x, ψ1; x, ψ 2〉 x ψ1; x, ψ 2〉 ± x ψ1; x, ψ 2〉

depending on whether the system is symmetric or antisymmetric under the operation

Example: consider a π+π− pair in a state of definite orbital angular momentum L

C ∣π+π−,L〉 = (-1) L ∣π+π−,L〉

since interchanging π+ and π− reverses their relative position vector in the spatial wave function Example: Experimentally, π0 → γ + γ but never π0 → γ + γ + γ How is this reconciled with the concept of C-parity ??

C ∣ 0 = ∣ 0 using a similar argument π 〉 Cπ° π 〉 as for parity, Cγ = −1 C = +1 π° C ∣ = ∣ C ∣γγγ = C C C ∣γγγ γγ 〉 CγCγ γγ 〉 〉 γ γ γ 〉 = 2 ∣ = C 2 C ∣γγγ Cγ γγ 〉 γ γ 〉 = C ∣γγγ = ∣γγ 〉 γ 〉 = −∣γγγ 〉

Ah! So we can never get π0 → γ + γ + γ if C-parity is conserved !!