The definition of the adjoint operator of an operator $\hat A$ is

\begin{equation} (\vec x|\hat A \vec y) = (\hat A \vec x| \vec y) \quad \forall x, y \in \mathcal{H} \end{equation}

where $(\cdot|\cdot)$ is the inner product of a Hilbert space.

So before this definition came into play, I innocently tried to do the expected values of $\hat a$ and $\hat a^\dagger$ for the coherent state $\left|\alpha\right>$, getting:

\begin{equation} \left<\alpha\right|\hat a \left|\alpha\right> = \alpha \left<\alpha|\alpha\right> = \alpha \end{equation}


\begin{equation} \left<\alpha\right|\hat a^\dagger \left|\alpha\right> = \alpha^* \left<\alpha|\alpha\right> = \alpha^* \end{equation}

which I think are correct.

But then, I thought that the expected value of $\hat a$ and $\hat a^\dagger$ should be the same, since one is the transposed complex-conjugated operator of the other, and that usually means that one is the adjoint of the other, but with the definition of adjoint operator and getting different values for the brakets I did before, it's clear that they're not their adjoint.

So my question is:

  • Did I suppose or do anything wrong? Is this definition I gave for adjoint operator not applicable to this case?

  • If it's correct, do $\hat a$ and $\hat a^\dagger$ have any operator corresponding to its adjoint, or if the transposed complex-conjugated operator of $\hat A$ is not the adjoint of $\hat A$ then $$\hat A$ doesn't have an adjoint operator?

  • If they do have an adjoint, which are they?


2 Answers 2


There’s nothing wrong with what you did except with your incorrect notion that the expectation values and its adjoint should be the same.

Indeed you have just shown that, for $\hat a$ and $\hat a^\dagger$, one is the complex conjugate of the other. One intuitive way to see this is to remember that, up to constants $$ \hat a\sim \hat x+i\hat p\, , \qquad \hat a^\dagger \sim \hat x -i \hat p \tag{1} $$ with $\hat x$ and $\hat p$ self-adjoint, so their expectations values are real. Using this together with Eq.(1) clearly shows how one can anticipate that the expectation values of $\hat a$ and $\hat a^\dagger$ will be complex conjugate in pairs.


The creation and annihilation operators are true adjoint operators, yes. In the case of the coherent state, we have


I think you may be forgetting that the complex inner product satisfies $(c\vec{v}|\vec{w})=c^{*}(\vec{v}|\vec{w})$ and not $(c\vec{v}|\vec{w})=c(\vec{v}|\vec{w})$.

If this isn't where your confusion lies, comment below or edit your question to more clearly explain why you think that $a$ and $a^{\dagger}$ should have the same expectation value.

I hope this helps!

  • $\begingroup$ Ok, then after seeing this answer, I think that maybe my problem is with the inner product. When doing $(a\alpha|\alpha)$ you're treating the first $\alpha$ as a ket? In my definition of adjoint, I thought that $(\hat A \vec x|\vec y)=<\vec x|\hat A|\vec y>$, but in your answer it doesn't seem like it... I don't quite get it $\endgroup$
    – Mr. Nobody
    Mar 11, 2018 at 22:59
  • 1
    $\begingroup$ @Mr.Nobody - $({\hat A} \vec{x} | \vec{y} ) = \langle\vec{x} | {\hat A}^\dagger | \vec{y} \rangle$. $\endgroup$
    – Prahar
    Mar 11, 2018 at 23:10
  • $\begingroup$ So my mistake was trying to translate the definition so I could understand it into $\left<x\right|A\left|y\right>=\left<y\right|A^\dagger\left|x\right>$ which seems that isn't always true am I right? $\endgroup$
    – Mr. Nobody
    Mar 12, 2018 at 0:31

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