Vector space of $\mathbb{C}^4$ and its basis, the Pauli matrices

Question

How do I write an arbitrary $2\times 2$ matrix as a linear combination of the three Pauli Matrices and the $2\times 2$ unit matrix?

Any example for the same might help ?

Answer 1

A slow construction would go... $$$$

$$ \begin{pmatrix}a&b\\c&d\end{pmatrix} = a\begin{pmatrix}1&0\\0&0\end{pmatrix} +b\begin{pmatrix}0&1\\0&0\end{pmatrix} +c\begin{pmatrix}0&0\\1&0\end{pmatrix} +d\begin{pmatrix}0&0\\0&1\end{pmatrix} $$

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$$ \begin{pmatrix}1&0\\0&0\end{pmatrix} =\frac{1}{2} \begin{pmatrix}1&0\\0&1\end{pmatrix} + \frac{1}{2} \begin{pmatrix}1&0\\0&-1\end{pmatrix} =\frac{1}{2}1_2+\frac{1}{2}\sigma_3 $$

$$ \begin{pmatrix}0&1\\0&0\end{pmatrix} =\ ... $$

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$$ \Longrightarrow \begin{pmatrix}a&b\\c&d\end{pmatrix} = \frac{a}{2}1_2+\frac{a}{2}\sigma_3+\ ...\ (\text{other combintations of the four matrices}) $$

Answer 2

I like to put it this way:
$$\left(\begin{array}{cc} w+z&x-iy\\ x+iy&w-z\end{array}\right)$$

So, for example:
$$\left(\begin{array}{cc} 1&5\\1&2\end{array}\right) = \left(\begin{array}{cc} (1.5)+(-0.5)&(3)-i(2i)\\ (3)+i(2i)&(1.5)-(-0.5)\end{array}\right)$$ So $w=1.5, x=3, y=2i, z=-0.5$ and $$\left(\begin{array}{cc} 1&5\\1&2\end{array}\right) = 1.5 + 3\sigma_x + 2i\sigma_y -0.5\sigma_z.$$

You can solve for $w,x,y,z$ from the entries in the array easily. I.e. $x$ is the average of the top right and bottom left entries, etc.

Answer 3

The matrices $\sigma_0\equiv \boldsymbol{1}_2$, $\sigma_x$, $\sigma_y$ and $\sigma_z$ form an orthonormal basis of your vector space w.r.t. the scalar product $$ (X,Y) \equiv \frac{1}{2}\operatorname{tr}(X\cdot Y), $$ where $X$ and $Y$ label any two complex $2\times 2$ matrices. The factor $1/2$ is just for convenience, you may as well normalise your Pauli matrices by dividing them by $2$.

All you want to do now is to decompose an arbitrary element $$ M = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $$ of your vector space into the above basis and figure out the coefficients. As usual, this is done by projecting onto that basis by means of the scalar product $$ M = (\sigma_0,M)\cdot\sigma_0 + (\sigma_x,M)\cdot\sigma_x + (\sigma_y,M)\cdot\sigma_y + (\sigma_z,M)\cdot\sigma_z\ . $$

This has essentially been said in the above comments, particularly in the link posted by Kostya.