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 1


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.

  • $\begingroup$ Thank you very much! At the 6th step of your demonstration for $n=2$, wouldn't it be $\left\langle \psi\middle|{\hat{A}}^2\psi \right\rangle$? $\endgroup$
    – Cerlosh
    Jun 8, 2020 at 14:08
  • $\begingroup$ oh yes.. sorry, my mistake. $\endgroup$
    – Alice
    Jun 8, 2020 at 14:08
  • $\begingroup$ You are welcome. $\endgroup$
    – Alice
    Jun 8, 2020 at 14:25

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