In the number state basis $|n\rangle$ of a 1D quantum oscillator, the action of lowering operator $a$ is:
(A)
a|n\rangle = \sqrt{n} |n-1\rangle
(B)
a|n\rangle = \sqrt{n+1} |n-1\rangle
(C)
a|n\rangle = n |n-1\rangle
(D)
a|n\rangle = \sqrt{n} |n+1\rangle