# Conjugation of $R_x(\theta)$ with $CNOT$

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?

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. $$\square$$

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