Return to Answer

You appear to have two unrelated gaps, the final one being the rotation formula for the vector (triplet) representation, the celebrated Rodrigues rotation formula. $\vec \sigma$ rotates like a vector. Adjoint indices rotate like a vector in all representations!! This is standard textbook stuff, so I'll assume it.

Your question then is straightforward to answer via direct calculation, since the 2$\times$2 Pauli vector exponentials have an explicit and tractable form, $$ e^{i\theta n\cdot \sigma/2}= {\mathbb I} \cos(\theta/2) +in\cdot \sigma \sin(\theta/2) . $$ Thus, $$ e^{i(\theta/2)(\hat{n}\cdot\vec \sigma)} \vec \sigma e^{-i(\theta/2)(\hat{n}\cdot\vec \sigma)}\\ = \Bigl ( \cos(\theta/2) +i\hat n\cdot\vec \sigma \sin(\theta/2)\Bigr ) \sigma^k\Bigl ( \cos(\theta/2) -i\hat n\cdot\vec \sigma \sin(\theta/2)\Bigr )\\ =\cos\theta~\sigma^k + \sin\theta ~\epsilon _{kjm} n^j \sigma^m -(\cos\theta -1) n^k n^m \sigma^m \\ =\cos\theta~\vec \sigma +\sin\theta ~\hat n \times \vec \sigma + (1-\cos\theta )\hat n ~~\hat n \cdot \vec \sigma , $$ by elementary (tedious) Pauli matrix algebra.

But this is just the celebrated Euler-Rodrigues vector rotation formula linked and explained above, and written in terms of matrix exponentials later in the article. Note how the prefactor trigonometric function coefficients have doubled the angle and rearranged themselves in the respective expressions. Dozens of students miss the magic and get confused by this and attempt to "correct" formulas which are already correct!

The (antisymmetric) logarithm of a rotation matrix is is found here, so (Tr$R -1)/2=\cos\theta$, etc.

You appear to have two unrelated gaps, the final one being the rotation formula for the vector (triplet) representation, the celebrated Rodrigues rotation formula. $\vec \sigma$ rotates like a vector. Adjoint indices rotate like a vector in all representations!! This is standard textbook stuff, so I'll assume it.

Your question then is straightforward to answer via direct calculation, since the 2$\times$2 Pauli vector exponentials have an explicit and tractable form, $$ e^{i\theta n\cdot \sigma/2}= {\mathbb I} \cos(\theta/2) +in\cdot \sigma \sin(\theta/2) . $$ Thus, $$ e^{i(\theta/2)(\hat{n}\cdot\vec \sigma)} \vec \sigma e^{-i(\theta/2)(\hat{n}\cdot\vec \sigma)}\\ = \Bigl ( \cos(\theta/2) +i\hat n\cdot\vec \sigma \sin(\theta/2)\Bigr ) \sigma^k\Bigl ( \cos(\theta/2) -i\hat n\cdot\vec \sigma \sin(\theta/2)\Bigr )\\ =\cos\theta~\sigma^k + \sin\theta ~\epsilon _{kjm} n^j \sigma^m -(\cos\theta -1) n^k n^m \sigma^m \\ =\cos\theta~\vec \sigma +\sin\theta ~\hat n \times \vec \sigma + (1-\cos\theta )\hat n ~~\hat n \cdot \vec \sigma , $$ by elementary (tedious) Pauli matrix algebra.

But this is just the celebrated Euler-Rodrigues vector rotation formula linked and explained above, and written in terms of matrix exponentials later in the article. Note how the prefactor trigonometric function coefficients have doubled the angle and rearranged themselves in the respective expressions. Dozens of students miss the magic and get confused by this and attempt to "correct" formulas which are already correct!

The (antisymmetric) logarithm of a rotation matrix is is found here, so (Tr$R -1)/2=\cos\theta$, etc., so $R=e^{\theta \hat n\cdot \vec J}$.

You appear to have two unrelated gaps, the final one being the rotation formula for the vector (triplet) representation, the celebrated Rodrigues rotation formula. $\vec \sigma$ rotates like a vector. Adjoint indices rotate like a vector in all representations!! This is standard textbook stuff, so I'll assume it.

Your question then is straightforward to answer via direct calculation, since the 2$\times$2 Pauli vector exponentials have an explicit and tractable form, $$ e^{i\theta n\cdot \sigma/2}= {\mathbb I} \cos(\theta/2) +in\cdot \sigma \sin(\theta/2) . $$ Thus, $$ e^{i(\theta/2)(\hat{n}\cdot\vec \sigma)} \vec \sigma e^{-i(\theta/2)(\hat{n}\cdot\vec \sigma)}\\ = \Bigl ( \cos(\theta/2) +i\hat n\cdot\vec \sigma \sin(\theta/2)\Bigr ) \sigma^k\Bigl ( \cos(\theta/2) -i\hat n\cdot\vec \sigma \sin(\theta/2)\Bigr )\\ =\sigma^k + \sin\theta ~\epsilon _{kjm} n^j \sigma^m -(\cos\theta -1) n^k n^m \sigma^m \\ =\vec \sigma +\sin\theta ~\hat n \times \vec \sigma + (1-\cos\theta )\hat n ~~\hat n \cdot \vec \sigma , $$ by elementary (tedious) Pauli matrix algebra.

But this is just the celebrated Euler-Rodrigues vector rotation formula linked and explained above, and written in terms of matrix exponentials later in the article. Note how the prefactor trigonometric function coefficients have doubled the angle and rearranged themselves in the respective expressions. Dozens of students miss the magic and get confused by this and attempt to "correct" formulas which are already correct! The (antisymmetric) logarithm of a rotation matrix is is found here.

You appear to have two unrelated gaps, the final one being the rotation formula for the vector (triplet) representation, the celebrated Rodrigues rotation formula. $\vec \sigma$ rotates like a vector. Adjoint indices rotate like a vector in all representations!! This is standard textbook stuff, so I'll assume it.

Your question then is straightforward to answer via direct calculation, since the 2$\times$2 Pauli vector exponentials have an explicit and tractable form, $$ e^{i\theta n\cdot \sigma/2}= {\mathbb I} \cos(\theta/2) +in\cdot \sigma \sin(\theta/2) . $$ Thus, $$ e^{i(\theta/2)(\hat{n}\cdot\vec \sigma)} \vec \sigma e^{-i(\theta/2)(\hat{n}\cdot\vec \sigma)}\\ = \Bigl ( \cos(\theta/2) +i\hat n\cdot\vec \sigma \sin(\theta/2)\Bigr ) \sigma^k\Bigl ( \cos(\theta/2) -i\hat n\cdot\vec \sigma \sin(\theta/2)\Bigr )\\ =\cos\theta~\sigma^k + \sin\theta ~\epsilon _{kjm} n^j \sigma^m -(\cos\theta -1) n^k n^m \sigma^m \\ =\cos\theta~\vec \sigma +\sin\theta ~\hat n \times \vec \sigma + (1-\cos\theta )\hat n ~~\hat n \cdot \vec \sigma , $$ by elementary (tedious) Pauli matrix algebra.

But this is just the celebrated Euler-Rodrigues vector rotation formula linked and explained above, and written in terms of matrix exponentials later in the article. Note how the prefactor trigonometric function coefficients have doubled the angle and rearranged themselves in the respective expressions. Dozens of students miss the magic and get confused by this and attempt to "correct" formulas which are already correct! 

The (antisymmetric) logarithm of a rotation matrix is is found here, so (Tr$R -1)/2=\cos\theta$, etc.

You appear to have two unrelated gaps, the final one being the rotation formula for the vector (triplet) representation, the celebrated Rodrigues rotation formula. $\vec \sigma$ rotates like a vector. Adjoint indices rotate like a vector in all representations!! This is standard textbook stuff, so I'll assume it.

Your question then is straightforward to answer via direct calculation, since the 2$\times$2 Pauli vector exponentials have an explicit and tractable form, $$ e^{i\theta n\cdot \sigma/2}= {\mathbb I} \cos(\theta/2) +in\cdot \sigma \sin(\theta/2) . $$ Thus, $$ e^{i(\theta/2)(\hat{n}\cdot\vec \sigma)} \vec \sigma e^{-i(\theta/2)(\hat{n}\cdot\vec \sigma)}\\ = \Bigl ( \cos(\theta/2) +i\hat n\cdot\vec \sigma \sin(\theta/2)\Bigr ) \sigma^k\Bigl ( \cos(\theta/2) -i\hat n\cdot\vec \sigma \sin(\theta/2)\Bigr )\\ =\sigma^k + \sin\theta ~\epsilon _{kjm} n^j \sigma^m -(\cos\theta -1) n^k n^m \sigma^m \\ =\vec \sigma +\sin\theta ~\hat n \times \vec \sigma + (1-\cos\theta )\hat n ~~\hat n \cdot \vec \sigma , $$ by elementary (tedious) Pauli matrix algebra.

But this is just the celebrated Euler-Rodrigues vector rotation formula linked and explained above, and written in terms of matrix exponentials later in the article. Note how the prefactor trigonometric function coefficients have doubled the angle and rearranged themselves in the respective expressions. Dozens of students miss the magic and get confused by this and attempt to "correct" formulas which are already correct!

You appear to have two unrelated gaps, the final one being the rotation formula for the vector (triplet) representation, the celebrated Rodrigues rotation formula. $\vec \sigma$ rotates like a vector. Adjoint indices rotate like a vector in all representations!! This is standard textbook stuff, so I'll assume it.

Your question then is straightforward to answer via direct calculation, since the 2$\times$2 Pauli vector exponentials have an explicit and tractable form, $$ e^{i\theta n\cdot \sigma/2}= {\mathbb I} \cos(\theta/2) +in\cdot \sigma \sin(\theta/2) . $$ Thus, $$ e^{i(\theta/2)(\hat{n}\cdot\vec \sigma)} \vec \sigma e^{-i(\theta/2)(\hat{n}\cdot\vec \sigma)}\\ = \Bigl ( \cos(\theta/2) +i\hat n\cdot\vec \sigma \sin(\theta/2)\Bigr ) \sigma^k\Bigl ( \cos(\theta/2) -i\hat n\cdot\vec \sigma \sin(\theta/2)\Bigr )\\ =\sigma^k + \sin\theta ~\epsilon _{kjm} n^j \sigma^m -(\cos\theta -1) n^k n^m \sigma^m \\ =\vec \sigma +\sin\theta ~\hat n \times \vec \sigma + (1-\cos\theta )\hat n ~~\hat n \cdot \vec \sigma , $$ by elementary (tedious) Pauli matrix algebra.

But this is just the celebrated Euler-Rodrigues vector rotation formula linked and explained above, and written in terms of matrix exponentials later in the article. Note how the prefactor trigonometric function coefficients have doubled the angle and rearranged themselves in the respective expressions. Dozens of students miss the magic and get confused by this and attempt to "correct" formulas which are already correct! The (antisymmetric) logarithm of a rotation matrix is is found here.