# How does the parity eigenvalue change in cobalt-60 beta decay parity violating process?

All the explanations for why the beta decay of cobalt 60 seem to revolve around 'in a mirror flipped world the process is different', and while I understand the intrinsic relationship between symmetries and conserved quantities I would like to see properly in the maths how the parity eigenvalue is not conserved in this process.

So if we assume that parity is conserved this is equivalent to assuming $$[H,P]=0$$ right? But how from this do we show that the cobalt-60 initial state must be a parity eigenstate in the first place? Why can't it be in a superposition of different parity states, and in this case, how could we ever know that parity has been violated?

I'm assuming we somehow know that the cobalt-60 must be in an energy eigenstate, and for some reason we know it is in a non-degenerate energy eigenstate or something so that we know for sure it is also a parity eigenstate. But how?

You’ve probably gone through the exercise of computing the orbital wavefunctions for electrons around a hydrogen-like atom, and the observation that the angular momentum eigenfunctions have definite parity $$(-1)^L$$. A transition between two orbitals emits a photon that carries away both energy and angular momentum, and the photon field also has an associated parity. There are rules for deciding whether a transition is “electric” or “magnetic,” and whether its angular momentum distribution is “dipole,” “quadrupole,” or some higher order, and those rules include the parity difference between the initial and final states. A transition $$2^- \to 1^+$$ is mostly “electric dipole,” or $$E1$$; a transition $$2^+ \to 1^+$$ is mostly “magnetic dipole,” $$M1$$, but winds up having strong contributions from “electric quadrupole,” $$E2$$.

You’ve probably also learned, studying the hydrogen atom, that the states with various quantum numbers are also energy eigenstates, whose energies are

$$E_n = \frac{\alpha m c^2}{n^2}$$

where $$\alpha$$ is the fine-structure constant, $$m$$ is the reduced mass of your atom (the electron mass, to four significant figures), and $$n$$ is the principal quantum number.

Unfortunately for your intro quantum notes (but fortunately for life), the energy eigenstate thing is a lie.

The energy-time version of the uncertainty principle,

$$\Delta E \Delta t \gtrsim \hbar$$

says that an energy eigenstate with $$\Delta E=0$$ has to live forever, never decaying, and never radiating. A state that participates in transitions must have a “width” $$\Gamma \sim \hbar/\Delta t$$ which is broader for short-lived states than for long-lived ones. Long-lived states, like a picosecond atomic excitation, might have a fractional width $$\Delta E/E$$ which is a challenge for a spectroscopist to measure, so it’s a useful approximation to call them energy eigenstates and to say that the energy is a “good quantum number.” But the physical reality is different from our model in a nuanced way (see comments below).

Likewise, it’s a lie that any particular quantum state is an exact eigenstate of parity. But, because the only parity-mixing interaction we’ve discovered is weak, assigning a definite parity to a particular state is a very good approximation.

Cobalt-60 is an odd-odd nucleus whose ground state has angular momentum $$5\hbar$$. This means it has a nonzero magnetic moment and can be oriented in a magnetic field. A clever experimenter can arrange a cobalt sample so that it’s substantially polarized, with a nonzero (pseudovector) spin $$\vec\sigma_\text{Co}$$.

When the decay happens in a polarized sample, the electrons are more likely to be emitted from the spinning cobalt’s “north pole” than from its “south pole.” (Or perhaps vice-versa; I don’t actually remember the sign of the asymmetry.) A position-sensitive detector tells you something about the direction of the (polar vector) momentum $$\vec p_\text{e}$$ of the electron.

So what you have in the Wu experiment is a measurement of a scalar quantity: the differential decay rate into various pieces of solid angle. However that decay rate is experimentally not a scalar, but is a mix of scalar and pseudoscalar terms,

$$\frac{d\Gamma}{d\Omega} \propto \left( 1 + A \hat\sigma_\text{Co} \cdot \hat p_\text{e} \right) \propto \left( 1 + A \cos\theta_{\sigma p} \right)$$

The mixing parameter $$A$$ is called the “asymmetry,” and Wu’s famous result was $$A \neq 0$$ with shockingly high confidence. This demonstrates clearly that beta decay is a parity-mixing interaction. But identifying where the parity mixing has happened is tough because there are a number of other parity-violating observables. For instance the neutrino and electron both have parity-violating helicities which get stronger as the particles become more relativistic. (That’s why $$\pi\to\mu\nu$$ decay occurs and is a source of polarized muons, while $$\pi\to \text{e}\nu$$ is forbidden.)

One of the things I did as a graduate student was to study parity-mixing electromagnetic transitions in nuclei. We would create polarized, excited compound nuclei by putting polarized cold neutrons on various targets, and they’d decay by emitting a messy cascade of gamma rays that we’d capture in a direction-sensitive calorimeter. The largest asymmetries came from nuclei which had a short-lived state and a long-lived state with approximately the same energy, but opposite parities. Since angular momentum is a better quantum number than parity, an excited nucleus that’s in, say, a $$2^-$$ state with a long life, has some probability to decay as if it were in a $$2^+$$ with a much shorter lifetime. The faster the other state, the wider its $$\Delta E$$ and the bigger the overlap. The asymmetry in the decay radiation (which is the physical observable) comes about from the quantum-mechanical interference between the two pathways. There’s not a super-large literature on the subject, because the experimental effects are very small and the theory is challenging.

• Thanks! So are you essentially saying that while its easy to say that the conservation law no longer holds since the symmetry is no longer valid (as measured by the experiment), it is very difficult to express which part of the wavefunction breaks the parity symmetry due to the how complicated the wavefunctions are in these cases? Jan 3 at 19:19
• At high energy, the parity violation in the charged current is really simple: the $W^\pm$ field interacts with right-handed antiparticles and left-handed particles but not vice-versa. But even at high energy, the neutral current (which is one mixture of $W^0$ and $B$ after symmetry breaking, the other mixture being the photon) has parity-mixing effects which require care to describe. Put it in a bottle with a few dozen nucleons, whose strong interactions we only know phenomenologically, and it becomes a real challenge to say things that are both succinct and correct. That is: yes. It's messy.
– rob
Jan 3 at 19:33
• It's worth mentioning that, for bound states of atomic electrons, the eigenstates truly are eigenstates of the atomic hamiltonian. The difference is that in real life the full hamiltonian includes the energy of interaction with the EM field, and that's where the non-eigenstate-ness comes in. Jan 4 at 0:18
• @EmilioPisanty The approximation that parity is a good quantum number is (even) better in atomic states than within nuclei, but the atomic hamiltonian still includes parity-mixing perturbations from electron-nucleus and electron-electron interactions via the neutral current. And the purely hadronic weak interaction can cause parity mixing in electronic transitions thanks to the hyperfine interaction. If the nuclear ground state is secretly a parity mixture, adding electrons and calling the assembly an atom doesn’t remove the parity mixing.
– rob
Jan 4 at 2:26
• @rob I fully agree with your last comment -- but it is out of step with the text of the answer. If nothing else, the claim that "energy is not a good quantum number" should be removed or better explained (e.g. by explaining that "energy" means different things and that we often approximate it to mean the restricted atomic-only (EM+strong-only) hamiltonian and that this approximate energy is a reasonable, but not perfect, quantum number). Jan 7 at 12:03