Invariance Principles and Conservation Laws

Recall from last lecture:

Parity is the operation of changing all coordinates to their negative values. Parity of a state of orbital angular momentum:

P Ylm = (-1)lYlm
or P = (-1)l


Parity is a multiplicative quantity.

Parity is conserved in strong and electromagnetic interactions.

photon: JP = 1-

proton and neutron: JP = ½+

deuteron: J=1

pion: J=0

Pion Spin and Parity (cont'd)

Charged pion parity

The parity of the charged pion was originally deduced from the existence of the reaction:

π- d --> n n
There are a number of steps in the logic which I'll enumerate:
  1. The pion is captured by a deuteron nucleus. Since the pion is more massive than an electron by a factor of 250, the radius of a bound state is much smaller than that for an electron. Recall that the Bohr radius is 4πε0hbar2/mee2 for an electron; for a π- it is the same except to replace me by the reduced mass of the pion-deuteron system. Hence when a slow pion approaches a nucleus it can easily replace the electron, forming a mesic-deuterium.
  2. As the pion drops to lower energy bound states it emits X-rays. The emitted X-rays tell us that the π- settles into an S-state before interacting.
  3. The total angular momentum of the initial state must be J=1 since the pion is spin 0, the deuteron is spin 1, and the orbital angular momentum is 0. (More on addition of angular momenta later.
  4. The final state consists of two non-relativistic neutrons. Therefore we can write a wave function for their state, with space and spin separated (only possible in the non-relativistic limit):
    ψ = φ(space) α(spin) = φ(r) α(S, Sz)
  5. For a state consisting of two spin ½ neutrons, the possible values of combined spin are S=0 or 1.
  6. There are 3 substates for S=1 and 1 for S=0. Using the ket notation |S, Sz> (different from the text) and denoting the two neutrons as |Sz1, Sz2>, the four states are:
    |1,1> = |½,½>
    |1,0> = (|½,-½> + |-½,½>)/sqrt(2)
    |1,-1> = |-½,-½>
    |0,0> = (|½,-½> - |-½,½>)/sqrt(2)
    The S=1 states are symmetric upon exchange of particles, while the S=0 state is anti-symmetric. Thus the exchange symmetry of the spin function is (-1)S+1.
  7. The space function can be further decomposed into a radial part and an angular part -- the potential between the neutrons depends on their relative separation only. Therefore, the exchange symmetry of the spatial part of the wave function is (-1)L.
  8. The overall exchange symmetry of the final state wavefunction is (-1)L+S+1.
  9. Since the state consists of two fermions, the overall exchange symmetry must be negative. Therefore L+S must be even.
  10. Since the initial state is J=1, so must be the final state (conservation of angular momentum).
  11. Given that S=0 or 1, a state with J=1 can be formed by combining states with {L=0, S=1}, {L=1, S=0}, {L=1, S=1}, or {L=2, S=1}. Of these, only the combination of {L=1, S=1} also satisfies the requirement that L+S be even.
  12. Thus the two neutrons must be in a state with L=1 and S=1 which has parity (-1)L = -1. The parity of the neutrons is positive, thus the parity of the final state is -1.
  13. The parity of the initial state is the parity of the pion times the parity of the deuteron. The deuteron is a neutron and proton state with zero orbital angular momentum and symmetric spin wave function. The parity of the deuteron is positive. Therefore, the parity of the pion must be negative.

Parity of the neutral pion

The parity of the neutral pion was determined by looking at the so called double Dalitz decay π0-->e+e-e+e-. The determination is that the π0 also has negative parity.

At the time, it was not clear if the particles π± were somehow related, or distinct. The fact that they have the same spin and parity is an indication that they are related, and is why they are grouped as pions. That they are not identical particles is clear from the difference in their masses, almost 5MeV, and the tremendous difference in their lifetimes.

Copyright © Robert Harr 2003