Suppose I have a cost function $C = \langle \psi \rvert U^\dagger O U \rvert \psi \rangle$ for a fixed observable $O$ and a fixed state $\rvert \psi \rangle$. I know that usually people take the derivative of the cost function wrt parameters, but here I want to take the derivative of $C$ wrt $U$: $\frac{\partial C}{\partial U} = \langle \psi \rvert \frac{\partial U^\dagger}{\partial U} O U \rvert \psi \rangle + \langle \psi \rvert U^\dagger O \rvert \psi \rangle$, and I am not sure how to deal with $\frac{\partial U^\dagger}{\partial U}$ part. Any help would be appreciated!

  • 2
    $\begingroup$ What is the origin of the unitary matrix and how can it change? (I hope you're not allowing an entirely arbitrary $n$-qubit unitary!) $\endgroup$
    – DaftWullie
    Commented Jul 3, 2023 at 6:13

1 Answer 1


If I take your questions literally, then the answer is fairly straightforward: $$ \frac{dU^\dagger}{dU}=-1. $$ You should think about it this way: we know $UU^\dagger=I$. If $U$ changes by $\delta U$ (small), $U^\dagger$ must change to compensate: $$ (U+\delta U)(U-\delta U)=I+O(\delta U^2) $$ Thus, to first order, $U^\dagger$ changes to $U^\dagger-\delta U$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.