Question
The first Pauli matrix is
calculate
For real , show that U1 is unitary and has determinant
Answer :
Word Count : 511
The first Pauli matrix is [ \sigma_1 = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}. ] We are asked to calculate [ U_1(\theta) = \exp(i \theta \sigma_1) = 1 + i \theta \sigma_1 - \frac{\theta^2}{2}\sigma_1^2 - i \frac{\theta^3}{6}\sigma_1^3 + \frac{\theta^4}{24}\sigma_1^4 + \dots ] and show that it is unitary and find its determinant. First, note the powers of (\sigma_1): [ \sigma_1^2 = \begin{bmatrix}0 & 1 \ 1 & 0\end{bmatrix} \begin{bmatrix}0 & 1 \ 1 & 0\end{bmatrix} = \begin{bmatrix}1 & 0 \ 0 & 1\end{bmatrix} = I ] [ \sigma_1^3 = \sigma_1^2 \sigma_1 = I \sigma_1 = \sigma_1 ] [ \sigma_1^4 = \sigma_1^2 \sigma_1^2 = I I = I ] Thus, powers of (\sigma_1) cycle: (\sigma_1^0 = I, \sigma_1^1 = \sigma_1, \sigma_1^2 = I, \sigma_1^3 = __________ _______ ____ ________ _________ _________ _________.
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The first Pauli matrix is [ \sigma_1 = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}. ] We are asked to calculate [ U_1(\theta) = \exp(i \theta \sigma_1) = 1 + i \theta \sigma_1 - \frac{\theta^2}{2}\sigma_1^2 - i \frac{\theta^3}{6}\sigma_1^3 + \frac{\theta^4}{24}\sigma_1^4 + \dots ] and show that it is unitary and find its determinant. First, note the powers of (\sigma_1): [ \sigma_1^2 = \begin{bmatrix}0 & 1 \ 1 & 0\end{bmatrix} \begin{bmatrix}0 & 1 \ 1 & 0\end{bmatrix} = \begin{bmatrix}1 & 0 \ 0 & 1\end{bmatrix} = I ] [ \sigma_1^3 = \sigma_1^2 \sigma_1 = I \sigma_1 = \sigma_1 ] [ \sigma_1^4 = \sigma_1^2 \sigma_1^2 = I I = I ] Thus, powers of (\sigma_1) cycle: (\sigma_1^0 = I, \sigma_1^1 = \sigma_1, \sigma_1^2 = I, \sigma_1^3 = __________ _______ ____ ________ _________ _________ _________.
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