# I Forbidden decays and conservation laws

#### mPlummers

Problem Statement: Hello! I'm trying to learn how to know if a particular interaction is allowed or forbidden. I found 3 decays which i can't understand.
Relevant Equations: The decays are:
1) $\eta \rightarrow \pi ^{0}+\gamma$
2) $\phi \rightarrow \rho^{0}+\gamma$
3) $\eta \rightarrow \pi ^{0}+\pi ^{0}$
4) $\eta^{0} \rightarrow \gamma+\gamma+\gamma$

1) $\eta$ and $\pi$ have zero spin, while the photon has spin equal to 1. It should be forbidden because angular momentum is not conserved.
2) Here the angular momentum is conserved (they all have spin 1, so the total angular momentum for the final state can be 0, 1, 2, in this case has to be 1). But what about the charge conjugation? I know that they all have C = -1, so it should not be conserved because the total C for the final state is -1 x -1 = 1.
3) The angular momentum is conserved, so i would say that the decay is allowed, but i can't find it in the Particle Data Group, so i'm trying to understand if the parity is conserved. All the particles have P=-1. I know that the parity for the final state is $P_{tot} = P_{1}P_{2}(-1)^{L}$. The problem is that i don't know the value for L. Should i know that or can i set it equal to zero? If i can put L=0, the the parity is not conserved, so the decay is forbidden... but the PDG says it's possible.
4) I think this is the same case of point 3. Angular momentum is conserved, but i don't know how to add parity quantum numbers (how to choose the total L).

[Moderator's note: Moved from homework to a technical forum.]

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#### mPlummers

Up!
If the answer for this question is too long, i can isolate the most relevant problem:

If i have two particles, each one with a certain parity, how can i calculate the total parity? The formula $P_{tot} = P_{1}P_{2}(-1)^{L}$ includes the L value. How do i know its value? I usually put L=0, but as i can see that's not the case.

My guess is: is it possible that the orbital angular momentum changes between the intial (L=0) and final state (L=1)?

Staff Emeritus
η\eta and π\pi have zero spin, while the photon has spin equal to 1. It should be forbidden because angular momentum is not conserved.
That's completely wrong. It could decay via P-wave, i.e. there would be one unit of orbital angular momentum in the decay. If it's forbidden, it's forbidden for some other reason.

#### mPlummers

That's completely wrong. It could decay via P-wave, i.e. there would be one unit of orbital angular momentum in the decay. If it's forbidden, it's forbidden for some other reason.
Thank you for your answer. I think i found my doubt. I have never considered before the change in L. Indeed, i think this is the only explanation also for parity conservation in these decays. Is it always possible to change the orbital angular momentum value between initial and initial state? Does it depend on the interaction?

Staff Emeritus
Did you cover angular momentum in quantum mechanics? What did it say about orbital angular momentum?

#### mPlummers

Yes i did, in a previous course. I know that the total orbital angular momentum $L_{tot}$ can have values $|L_{1}-L_{2}|\leq L_{tot}\leq L_{1}+L_{2}$, and that for each value of $L_{tot}$ there are $2L_{tot}+1$ of $m_{L}$. Same rules for total angular momentum $J_{tot}$, which comes from the sum of $L$ and $S$. This means that a 2 particles state can have several values of $L_{tot}$, but i can't understand how to choose the right value.

#### mfb

Mentor
In general all orbital angular momentum values are possible, but large values are typically unlikely. You can calculate how far apart e.g. two 100 MeV pions have to be to have L=1 in classical physics. While quantum mechanics is not classical physics you can transfer the right idea here: It is unlikely.

Tantalum-180m is a famous example from nuclear physics. As the name says it is a nuclear isomer - but so long-living that it occurs in nature and no one has observed a decay yet. It has a very high spin while the ground state or the results of a beta decay have a much lower spin.

#### mPlummers

In general all orbital angular momentum values are possible, but large values are typically unlikely. You can calculate how far apart e.g. two 100 MeV pions have to be to have L=1 in classical physics. While quantum mechanics is not classical physics you can transfer the right idea here: It is unlikely.
Thank you for your answer. Does it mean that, while computing the total angular momentum in the final state, i can choose different values of L? If yes, are the following steps correct to check conservation of total angular momentum?

1 - Given an interaction (decay or scattering) find the spin values of particle in the initial state and compute all the possible values of $J_{tot,i}$, putting L = 0 (collision). In the case of a decay, there's only one value of $J_{tot,i}$, depending on the spin and with L = 0.

2 - Find the spin values of the particles in the final state and compute all the possible values of $J_{tot,f}$ with $L_{tot,f}=0$. If it's not equal to $J_{tot,i}$, add a unit of L and check again. The higher the L value, the less likely the interaction.

If it's correct, is it always possible to apply these rules?

#### mfb

Mentor
The higher the L value, the less likely the interaction.
That is a good general rule in particle physics.

#### mPlummers

What i'm trying to do is to understand if an interaction is possible in terms of conservation of total angular momentum. For example:

$\eta \rightarrow \pi ^{0}+\gamma$

- $\eta$ has spin 0
- $\pi ^{0}$ has spin 0
- $\gamma$ has spin 1

Initial state: $J _{tot,i} = 0$
To find $J _{tot,f}$ for he final state: $|L_{tot,f}-S_{tot,f}|\leq J_{tot,f} \leq L_{tot,f}+S_{tot,f}$, where $S_{tot,f} = 1$. Now, if $L_{tot,f}=0$, the total angular momentum is not conserved (it can only be 1), BUT if i put $L_{tot,f}=1$ then i can obtain $J _{tot,f}=0$. Is this reasoning correct?

#### mfb

Mentor
Yes, see above.

"Forbidden decays and conservation laws"

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