# Understanding $\beta$ Decay Transition Classifications

I'm currently trying to understand the allowed transitions of Beta Decay through conservation of momentum and parity.

I'm currently confused on how multiple types are allowed for the same decay in certain situations.

An example is the decay of just the neutron into a proton.

$N \rightarrow P + e^- + \bar{v}$

Making use of the momentum conservation equation:

$J_f = J_i + \bar{L} + \bar{S}$

We are dealing with fermions so we see a $\Delta J$ of $0$, as we are transferring from $\frac{1^+}{2}$ to $\frac{1^+}{2}$. This gives us a $\Delta \pi$ that is positive, meaning we must have a decay that must be $l = even$.

Now if we were to perform a Fermi decay and set $S = 0$, this would give us a super-allowed transition as the equation would be satisfied by $l = 0$. However, in my notes it is described as a mixed transition with a Gamow-Teller transition also possible. I am confused by this as I cannot see how to satisfy the equation with the Gamow-Teller condition of $S = 1$.

Please let me know about what silly mistake I am making here!

Cheers

• What do $L$ and $S$ represent in your notation, and what do the bars in $\bar{L}$ ad $\bar{S}$ represent?
– user4552
Apr 17, 2018 at 21:29
• Hi Ben. $L$ represents the angular momentum of the emission, with $S$ representing the spin. The bars are included to represent the vector expressions. Apr 18, 2018 at 12:11
• When you say "of the emission," do you mean of both the electron and the antineutrino?
– user4552
Apr 18, 2018 at 21:24

Don't forget that in the addition of angular momentum vectors in quantum mechanics,

$$\vec J_\text{tot} = \vec J_1 + \vec J_2,$$

the allowed total angular momentum takes on a range of values,

$$\left|J_1 - J_2\right| \leq J_\text{tot} \leq J_1 + J_2.$$ A semi-classical way to think of it is to imagine the maximum $J_\text{tot}$ as what happens when the individual $J_i$ are aligned, and the minimum as what happens when they are anti-aligned. It's an integer-arithmetic version of the triangle inequality.

Your concern seems to be satisfying $\vec J_f = \vec J_i + \vec L + \vec S$ with $J_f = J_i = \frac12$, $L=0$, and $S=1$. But that's totally allowed. Classically, think of $\vec J_i$ and $\vec S$ as being antiparallel.

• Hi Rob, thanks for your help. Am I right then in thinking, the possible values on the LHS would be the $\bar{J_{tot}}$ range discussed, in this case both $0$ and $1$? So the Fermi transition is satisfied when $\bar{J_{tot}} = 0$ & $L = 1, S = 0$. While the GT transition would just be when $\bar{J_{tot}} = 1$? Apr 19, 2018 at 13:11
• We must have $J_\text{total}=1/2$ before and after the decay, because that's what the free neutron has. My point is that you can do this with both $S=0$ and $S=1$, no need for $L\neq0$.
– rob
Apr 19, 2018 at 13:58