# Power of adjoint operators

If operator $$\hat{A^{\dagger}}$$ is the hermitian conjugate (adjoint) of $$\hat{A}$$, i.e. $$\left\langle \hat{A^{\dagger}}\psi \middle|\psi \right\rangle = \left\langle \psi \middle|\hat{A}\psi \right\rangle$$, is $$\left\langle \left( \hat{A^{\dagger}} \right)^n\psi \middle|\psi \right\rangle = \left\langle \psi \middle| \left(\hat{A}\right)^n\psi \right\rangle \hspace{1mm}?$$ In the context i'm working on, $$\hat{A}$$ is the ladder operator and $$n$$ is a real number.

## 1 Answer

I am assuming here that $$n$$ is an integer. It is true for $$n=1$$ . Now, put $$n=2$$ . Then, $$\left\langle{(\hat {A^\dagger})}^2\psi\middle |\psi\right\rangle=\left\langle\hat {A^\dagger}(\hat {A^\dagger}\psi)\middle |\psi\right\rangle=\left\langle\hat {A^\dagger}\psi\middle|\hat {A}\psi\right\rangle=\left\langle\hat {A^\dagger}\psi\middle|\phi\right\rangle=\left\langle\psi\middle|\hat A\phi\right\rangle=\left\langle\psi\middle|{\hat A}^2\psi\right\rangle$$

where $$\phi=\hat A|\psi\rangle$$ . At the third and fifth steps, we use the definition of the self adjoint vector $$\hat A$$ . So the given relation is true for $$n=2$$ . Similarly, you can show that if it is true for some $$n$$, then it is also true for $$n+1$$. In this way by mathematical induction, you can prove that your relation is true for any $$n$$.

However, in case of a harmonic oscillator, if $$\hat A$$ is a ladder operator, specifically the lowering operator and $$\hat {A^{\dagger}}$$ is the raising operator such that $$\hat {A^{\dagger}}\psi_{k}=\sqrt {k+1}\,\psi_{k+1}$$ and $$\hat {A}\psi_{k}=\sqrt {k}\,\psi_{k-1}$$ . Then, $$\left\langle{(\hat {A^\dagger})}^n\psi_k\middle |\psi_k\right\rangle= \sqrt {(k+1)(k+2)\cdot\cdot\cdot(k+n)}\left\langle\psi_{k+n+1}\middle |\psi_k\right\rangle\tag{1}$$

Again, $$\left\langle\psi_k\middle|{\hat A}^n\phi_k\right\rangle=\sqrt {(k(k-1)\cdot\cdot\cdot(k-n+1)}\left\langle\psi_k\middle |\psi_{k-n}\right\rangle\tag{2}$$

$$(1)$$ and $$(2)$$ are not equal from their forms but as $$\psi_k$$ and $$\psi_{k^{'}}$$ are orthogonal for all non-negative integers $$k\neq k^{'}$$, so, both of these terms equal to $$0$$ and hence are equal to each other.

This is what I get after calculation. Any correction to this post is welcome.

• Thank you very much! At the 6th step of your demonstration for $n=2$, wouldn't it be $\left\langle \psi\middle|{\hat{A}}^2\psi \right\rangle$? – Cerlosh Jun 8 '20 at 14:08
• oh yes.. sorry, my mistake. – Alice Jun 8 '20 at 14:08
• You are welcome. – Alice Jun 8 '20 at 14:25