Number operator - annihilation operator commutation

Question

Is there a rigorous way to prove that $$ (N+1)^{-1/2} a = a N^{-1/2} $$ where $a$ is a bosonic annihilation operator and $N=a^\dagger a$ is the corresponding number operator?

Answer 1

Consider the action of the two operators on an arbitrary number state $|n\rangle$:

• for the left-hand side, \begin{align} (N+1)^{-1/2}a|n\rangle & = (N+1)^{-1/2}\sqrt{n}|n-1\rangle = \sqrt{n}(N+1)^{-1/2}|n-1\rangle \\& = \sqrt{n}(n-1+1)^{-1/2}|n-1\rangle = \frac{\sqrt{n}}{\sqrt{n}}|n-1\rangle \\& = |n-1\rangle, \end{align}
• whereas on the right-hand side \begin{align} aN^{-1/2}|n\rangle & = a \frac{1}{\sqrt{n}}|n\rangle = \frac{1}{\sqrt{n}} a |n\rangle = \frac{1}{\sqrt{n}} \sqrt{n} |n-1\rangle \\& = |n-1\rangle. \end{align}

Since the two operators agree on a basis, they are equal as operators.