$a^{\dagger}a(a^{\dagger})^{n-1}+(a^{\dagger})^{n-1}=n(a^{\dagger})^{n-1}$

Question

Let $a$ and $a^{\dagger}$ be annihilation and creation operators.

I can't prove the following:

$$a^{\dagger}a(a^{\dagger})^{n-1}+(a^{\dagger})^{n-1}=n(a^{\dagger})^{n-1}$$

Maybe it's silly, in that case I'm sorry.

Answer 1

We have for the commutator $[a,a^\dagger] = 1$. It will be useful to prove that $a(a^\dagger)^m = (a^\dagger)^ma + m(a^\dagger)^{m-1}$ for $m\geq 1$. We do this by induction. Notice that for $m=1$, we have $aa^\dagger = a^\dagger a + 1$ because $[a,a^\dagger] = 1$. So suppose the claim holds true for $m = k-1$ and examine the case $m=k$. We have: \begin{align} a(a^\dagger)^k = aa^\dagger(a^\dagger)^{k-1} &= (a^\dagger a + 1)(a^\dagger)^{k-1}\\ &=a^\dagger a(a^\dagger)^{k-1}+(a^\dagger)^{k-1}\\ &= a^\dagger \left((a^\dagger)^{k-1}a + (k-1)(a^\dagger)^{k-2}\right) + (a^\dagger)^{k-1}\qquad \text{(induction hypothesis)} \\ &=(a^\dagger)^ka + (k-1)(a^\dagger)^{k-1}+(a^\dagger)^{k-1} \\ &= (a^\dagger)^ka+k(a^\dagger)^{k-1} \end{align} So the claim holds. Now we apply it to your equation: \begin{align} a^\dagger a (a^\dagger)^{n-1} + (a^\dagger)^{n-1} &= a^\dagger\left((a^\dagger)^{n-1}a + (n-1)(a^\dagger)^{n-2}\right) + (a^\dagger)^{n-1} \\ &= (a^\dagger)^{n-1}a^\dagger a+(n-1)(a^\dagger)^{n-1} + (a^\dagger)^{n-1}\\ &= (a^\dagger)^{n-1}a^\dagger a + n(a^\dagger)^{n-1} \end{align} It might be easier to see how this operator affects a wavefunction $\Psi$ in this form. If we apply $a^\dagger a$ to the eigenstate $\Psi_\ell$, we get $\ell\Psi_\ell$. So for the $\ell^{th}$ energy eigenstate, we can write: $$a^\dagger a (a^\dagger)^{n-1} + (a^\dagger)^{n-1}= (\ell + n)(a^\dagger)^{n-1}$$ Maybe this is what you meant?