Do both selection rules need to be satisfied for a transition to be allowed?

Question

The selection rules for electron transition are $\Delta l=\pm 1$ and $\Delta m = 0,\pm1$. Do both conditions need to be satisfied for a transition to be allowed? For example, is the transition from the state ($n=3$ $l=1$ $m_l=0$ $m_s=+1/2$) to the state ($n=3$ $l=1$ $m_l=0$ $m_s=-1/2$) allowed where $\Delta m = 0,\pm1$ is satisfied but $\Delta l=\pm 1$ is not satisfied?

Answer 1

Both selection rules need to be satisfied for the transition to be allowed.

As an aside, in practice, different transitions can be allowed, singly-forbidden (violating one selection rule), doubly-forbidden (violating two selection rules), etc. For example, in an atom, an electron would decay via a transition that is singly-forbidden more often than one that is doubly-forbidden, but would decay most often via a transition that is allowed.

Answer 2

These are the selection rules for "electric dipole" transitions, in which the electromagnetic field (or equivalently, the emitted photon) carries away one unit of angular momentum. The dipole field has negative parity, so the orbital angular momentum $\ell$ must change by an odd number for the parity of the entire system to remain constant — but the photon only has one unit of angular momentum, so $|\Delta \ell|>1$ isn't allowed. Likewise $|\Delta m|>1$ is forbidden because there's not enough angular momentum in the dipole field. However $\Delta m=0$ is allowed because the projection quantum numbers $m$ are orientation-dependent. A transition with $\Delta m=0$ in your coordinate system might be a (coherent superposition of) transition(s) with $\Delta m=1$ in some rotated coordinate system.

Your proposed transition, where only the projection of the spin changes, is allowed as a "magnetic dipole (M1)" transition; the coupling for an M1 transition is generally weaker than the coupling for an E1 transition with the same energy.