# Proof that adjoint is equal to complex conjugate in a quantum information theory viewpoint

† 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: $$\left = \left(\left\right)^*=\left^*= \left$$ 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 $$\left = (\left)^*$$ and $$\left$$

• 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. Jun 21, 2023 at 14:37
• 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. Jun 21, 2023 at 14:40
• 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 $A^\dagger=\begin{pmatrix} -i& 0\\1&0\end{pmatrix}$ Jun 21, 2023 at 14:48
• In general (for matrices) $A^\dagger = A^{t*}$, where $A^t$ is the transpose. Jun 21, 2023 at 14:49

The answer was explicitly given by Luftbahnfahrer, whereby $$\left^*$$ and the correct equality is $$\left^\dagger$$. 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., $$\left< \psi \right| = \alpha^{*} \left< 0 \right| + \beta^{*} \left< 1 \right|$$
• Complex conjugation does distribute, if this is what you mean: $(A|v\rangle)^*=A^*|v\rangle^*$ Jun 21, 2023 at 14:43