Matrix Examples

Any $2\times2$ complex matrix, $M$, can be written in the form \begin{equation} M = \begin{pmatrix} t+z& x-iy\\ x+iy& t-z\\ \end{pmatrix} \end{equation} with $x,y,z,t\in\RR$. Looking at the matrix coefficients of these variables, we can write \begin{equation} M = t \,I + x \,\sigma_x + y \,\sigma_y + z \,\sigma_z \end{equation} thus defining the three matrices \begin{align} \sigma_x &= \begin{pmatrix} 0& 1\\ 1& 0\\ \end{pmatrix} ,\\ \sigma_y &= \begin{pmatrix} 0& -i\\ i& 0\\ \end{pmatrix} ,\\ \sigma_z &= \begin{pmatrix} 1& 0\\ 0& -1\\ \end{pmatrix} , \end{align} which are known as the Pauli matrices. The Pauli matrices $\{\sigma_m\}$ have several interesting properties. First of all, each Pauli matrix squares to the identity matrix, is tracefree, and of course is Hermitian: \begin{align} \sigma_m^2 &= I ,\\ \tr(\sigma_m) &= 0 ,\\ \sigma_m^\dagger &= \sigma_m . \end{align} Also of interest is that these matrices anticommute, and that the product of any two is the third, for instance: \begin{equation} \sigma_x \sigma_y = i\,\sigma_z = -\sigma_y \sigma_x . \end{equation}

Thus, any $2\times2$ Hermitian matrix is a linear combination of the Pauli matrices and the identity matrix. What about $2\times2$ unitary matrices?

Now we must have $MM^\dagger=I$, that is, \begin{equation} \begin{pmatrix} a& b\\ c& d\\ \end{pmatrix} \begin{pmatrix} a^*& c^*\\ b^*& d^*\\ \end{pmatrix} = \begin{pmatrix} |a|^2+|b|^2& ac^*+bd^*\\ ca^*+db^*& |c|^2+|d|^2\\ \end{pmatrix} = I . \end{equation} The normalization conditions \begin{equation} |a|^2+|b|^2 = 1 = |c|^2+|d|^2 \end{equation} allow us to write \begin{align} a &= e^{i\alpha}\cos\theta ,\\ b &= e^{i\beta}\sin\theta ,\\ c &= e^{i\gamma}\sin\phi ,\\ d &= e^{i\delta}\cos\phi , \end{align} where we can assume that $\theta,\phi\in[0,\frac{\pi}{2}]$. The remaining condition becomes \begin{equation} 0 = ac^*+bd^* = e^{i(\alpha-\gamma)}\cos\theta\sin\phi + e^{i(\beta-\delta)}\sin\theta\cos\phi \end{equation} which forces in general \begin{align} \tan\theta &= \tan\phi ,\\ e^{i(\alpha-\gamma)} &= -e^{i(\beta-\delta)} \end{align} (with the second condition unnecessary if $\theta=0$ or $\theta=\frac{\pi}{2}$), so that \begin{align} \theta &= \phi ,\\ \alpha + \delta &= \beta + \gamma - \pi . \end{align} Some special cases are $\alpha=0=\delta$, $\beta=\frac{\pi}{2}=\gamma$, corresponding to \begin{equation} U_x = \begin{pmatrix} \cos\theta& i\sin\theta\\ i\sin\theta& \cos\theta\\ \end{pmatrix} , \end{equation} $\alpha=0=\delta$, $\beta=0$, $\gamma=\pi$, corresponding to \begin{equation} U_y = \begin{pmatrix} \cos\theta& \sin\theta\\ -\sin\theta& \cos\theta\\ \end{pmatrix} , \end{equation} and $\theta=0$, $\delta=\alpha$, corresponding to \begin{equation} U_z = \begin{pmatrix} e^{i\theta}& 0\\ 0& e^{-i\theta}\\ \end{pmatrix} , \end{equation} where we have replaced $\alpha$ by $\theta$ for consistency.

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