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Section 2.5 (4.3) of the Qiskit textbook, see here, discusses the conjugation of $R_x(\theta)$ by $CNOT$. The following expression is given:

$$CX_{j,k}(R_x(\theta)\otimes 1) CX_{j,k}=\color{brown}{CX_{j,k}e^{i\frac{\theta}{2}(X\otimes1)}CX_{j,k}=e^{i\frac{\theta}{2}CX_{j,k}(X\otimes 1)CX_{j,k}}}=e^{i\frac{\theta}{2}X\otimes X}$$

I am confused by the part highlighted in yellow. What exponentiation rules allow this? Could someone show me the intermediate steps or the relevant identities/rules to take us from the LHS to RHS of the highlighted part?

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This is an application of the following identity

$$ Be^AB^{-1} = e^{BAB^{-1}}\tag1 $$

where $A$ is any $n\times n$ real or complex matrix and $B$ is any invertible $n\times n$ real or complex matrix.

Proof of $(1)$. First, recall that the matrix exponential of $A$ is defined as

$$ e^A = \sum_{k=0}^\infty \frac{1}{k!}A^k. $$

Next, note that for any integer $k$

$$ (BAB^{-1})^k = BAB^{-1}BAB^{-1}\dots BAB^{-1} = BAA\dots AB^{-1} = BA^kB^{-1}\tag2. $$

Finally, calculate

$$ e^{BAB^{-1}} = \sum_{k=0}^\infty \frac{1}{k!}(BAB^{-1})^k = \sum_{k=0}^\infty \frac{1}{k!}BA^kB^{-1} = B \left(\sum_{k=0}^\infty \frac{1}{k!}A^k\right) B^{-1} = Be^AB^{-1} $$

where we used $(2)$ in the second step. $\square$

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  • $\begingroup$ Thanks so much, Adam! It now makes sense. $\endgroup$ – Othman Jan 10 at 21:44
  • $\begingroup$ You're welcome! :-) $\endgroup$ – Adam Zalcman Jan 10 at 21:45

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