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We can say that

$X (\cos \frac{\theta}{2} |0\rangle + e^{i \phi}\sin \frac{\theta}{2} |1\rangle) = \cos \frac{\pi-\theta}{2} |0\rangle + e^{-i \phi}\sin \frac{\pi-\theta}{2} |1\rangle$,

a fact that can be derived by multiplying the $X$ matrix and the state vector and using angle identities:

$X (\cos \frac{\theta}{2} |0\rangle + e^{i \phi}\sin \frac{\theta}{2} |1\rangle) = e^{i \phi}\sin \frac{\theta}{2} |0\rangle + \cos \frac{\theta}{2} |1\rangle = e^{i \phi}\cos \frac{\pi - \theta}{2} |0\rangle + \sin \frac{\pi - \theta}{2} |1\rangle = \cos \frac{\pi-\theta}{2} |0\rangle + e^{-i \phi}\sin \frac{\pi-\theta}{2} |1\rangle$

Is there a writeup somewhere or a standard way to derive this identity for other gates, i.e. $Y, Z, H, \sqrt{Y}$?

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  • $\begingroup$ Do you know about the Bloch sphere? Any unitary gate acting on a qubit is equivalent to a rotation of the Bloch vector (describing the state) about some axis by some angle (axis and angle determined by the unitary. $\endgroup$ May 11, 2021 at 18:17

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As Quantum Mechanic said in the comments, the Bloch vector is the way to go. Any one-qubit pure state $|\psi\rangle$ can be written in the form $$ |\psi\rangle\langle\psi|=\frac12(I+n_XX+n_YY+n_ZZ) $$ where $n_X^2+n_Y^2+n_Z^2=1$. Moreover, the vector $\vec{n}=(n_X,n_Y,n_Z)$ can also be written as $(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$. So, if we can tell what happens to the Bloch vector, you can easily map this through to the angles.

So, consider the action of the $X$ matrix of $|\psi\rangle$. $$ |\psi\rangle\langle\psi|\mapsto X|\psi\rangle\langle\psi|X=\frac12(I+n_XX-n_YY-n_ZZ). $$ You just need to use the commutation properties of the Pauli matrices. So, we see that $$ (n_X,n_Y,n_Z)\xrightarrow X (n_X,-n_Y,-n_Z) $$

By the same token, you very quickly get \begin{align*} (n_X,n_Y,n_Z)&\xrightarrow Z (-n_X,-n_Y,n_Z) \\ (n_X,n_Y,n_Z)&\xrightarrow H (n_Z,-n_Y,n_X) \end{align*} I've left the other couple for you to do.

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