# Can the reaction ${\rm n}+{\rm p} \rightarrow {\rm d} + \pi^0$ give a deuteron in excited state?

I'm trying to find the isospin quantum numbers of the deuteron on the right hand side of this equation. I know that $|\rm p\rangle = |{1\over2},{1\over2}\rangle$, $|\rm n\rangle = |{1\over2},-{1\over2}\rangle$ and $|\pi^0\rangle = |1,0\rangle$. My problem is regarding the strong isospin $I$. I can say that on the left hand side we have $I=0$ or $1$, and on the right hand side I have $I=1 \pm I({\rm d})$ where $I({\rm d})$ is the isospin of the deuteron. Now if I choose $I({\rm d})=0$ then the right hand side is necessarily $1$, which means that the left hand side is also $1$. If I choose $I({\rm d})=1$, the right hand side can be $0$ or $2$. Only $0$ is compatible with the left hand side.

My question is, are both choices acceptable? Is there a reason why $I({\rm d})=1$ would not be okay?

Thanks a lot in advance.