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I'm not sure what the correct notation for $\beta^{+}$ decay is when using the shorthand notation.

For the reaction $$a+X\rightarrow b+Y$$ it would be $ X(a,b)Y$. However, for $\beta^{+}$ decay, $$ X\rightarrow Y+e^{+}+\nu_{e} $$ there is no reacting "$a$" particle and two "$b$" daughter particles.

Would the shorthand notation be $$ X(a+b)Y?$$ If so, is it taken for granted from common knowledge that this is $\beta^{+}$ decay, and so these must be the products (as there is no comma to place them either as reacting or product particles)? Or is there no shorthand notation for such a reaction?

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The notation (incoming, outgoing) is for scattering processes, not for spontaneous decays.

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The scattering notation includes hooks for the detected final state, e.g., deep inelastic electron scattering off a target proton is:

$$ e(p,e')X$$

indicating and electron ($e$) scatters off a proton ($p$) and is detected ($e'$) with kinematics such that proton breaks up ($X$).

Likewise, photo disintegration of the deuteron (were the final state proton has more energy than the beam, ensuring there is only an undetected neutron) is written:

$$ d(\gamma,p)n$$

If this notation were applied to beta decay (it's not), it would be:

$$ (n,e^-)pX$$

where modern understanding allows us to fill in $X$:

$$ (n,e^-)p\bar{\nu}_e$$

The $n$ is inside the parenthesis because the notation is generally used in fixed target experiments:

$${beam}({target}, { detected}) { undetected}$$

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  • $\begingroup$ Typo: it's usually "target(beam, detected)undetected". I have never see any variation of this notation used to describe a spontaneous decay, even for a beam experiment; do you have a reference to a publication that does so? $\endgroup$
    – rob
    Commented Oct 22, 2021 at 18:05

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