† While trying to prove that $$(A+B)^†=A^†+B^†$$ I have stumbled accross a self proof that seems to validly suggest that $A^† = A^*$ This intuitively seems false but I cannot find where in my proof my mathematical reasoning has failed.

I would like an answer to the question of if I have gone wrong somewhere or if that the adjoint of an operator is indeed equivalent to its complex conjugate within a quantum information viewpoint.

The proof is as follows: $$\left<v\right|A^†\left|w\right> = \left(\left<w\right|A\left|v\right>\right)^*=\left<w\right|^* A^* \left|v\right>^*= \left<v\right| A^* \left|w\right>$$ Where $^*$ is to notify complex conjugate and $^†$ is to signify the adjoint. This clearly implies that $A^† = A^*$.
I have used the quantum informational identities that $\left<v\right|A^†\left|w\right> = (\left<w\right|A\left|v\right>)^*$ and $ \left<w\right|^* = \left|w\right>$

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    $\begingroup$ In your last line, the equality $\langle w|^* = |w\rangle$ is not correct, as one side is a row vector and the other is a column vector. $\endgroup$ Jun 21, 2023 at 14:37
  • $\begingroup$ In your "proof", the equality $\langle w|^*A^*|v\rangle^* =\langle v|A^*|w\rangle$ is simply incorrect. I suggest replacing the variables with numerical vectors to see what is going on. $\endgroup$ Jun 21, 2023 at 14:40
  • $\begingroup$ Finally, to see an example of where your claim fails, consider $A=\begin{pmatrix} i& 1\\0&0\end{pmatrix}$ which has $A^*=\begin{pmatrix} -i& 1\\0&0\end{pmatrix}$ but $A^\dagger=\begin{pmatrix} -i& 0\\1&0\end{pmatrix}$ $\endgroup$ Jun 21, 2023 at 14:48
  • $\begingroup$ In general (for matrices) $A^\dagger = A^{t*}$, where $A^t$ is the transpose. $\endgroup$ Jun 21, 2023 at 14:49

1 Answer 1


The answer was explicitly given by Luftbahnfahrer, whereby $\left<v\right| \neq \left|v\right>^*$ and the correct equality is $\left<v\right| = \left|v\right>^\dagger$. The confusion arose because within the course vector notation is rarely used and it is more common to express a qbit as $\left|\psi\right> = \alpha\left|0\right> + \beta \left|1\right>$, this makes the transpose implicit once all ket's are converted to bra's and therefore only a complex conjugate must be taken to obtain the equality from there; i.e., $ \left< \psi \right| = \alpha^{*} \left< 0 \right| + \beta^{*} \left< 1 \right| $

It seems to be a defining property that the associated bra to a ket is the hermitian adjoint of the ket, (i.e. the complex conjugate of the ket expressed as a row vector).
In any case in light of this new information the proof is false, as it should be. Thanks Luftbahnfahrer!

  • $\begingroup$ Complex conjugation does distribute, if this is what you mean: $(A|v\rangle)^*=A^*|v\rangle^*$ $\endgroup$ Jun 21, 2023 at 14:43

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