# Basis-independent form of Pauli matrices

For a system with two possible states $$|e\rangle$$ and $$|g\rangle$$, some sources refer to the Pauli matrices as,

$$\sigma_z = |e\rangle\langle e| - |g\rangle\langle g|\\ \sigma_x = |e\rangle\langle g| + |g\rangle\langle e|\\ \sigma_y = -i|e\rangle\langle g| + i|g\rangle\langle e|$$

Some sources present the Pauli matrices (without explicitly specifying the basis set) as,

$$\sigma_z = \left(\begin{array}{cc}1&0\\0&-1\end{array}\right)\\ \sigma_x = \left(\begin{array}{cc}0&1\\1&0\end{array}\right)\\ \sigma_y = \left(\begin{array}{cc}0&-i\\i&0\end{array}\right)$$

I observed that the two equation sets above tally for the basis set:

$$|e\rangle = \left(\begin{array}{cc}1\\0\end{array}\right)\\ |g\rangle = \left(\begin{array}{cc}0\\1\end{array}\right)$$

I would highly appreciate it if you could shed some light on the generic/basis independent form of the Pauli matrices. What form of Pauli matrices should I use for the following basis set?

$$|e\rangle = \left(\begin{array}{cc}0\\1\end{array}\right)\\ |g\rangle = \left(\begin{array}{cc}1\\0\end{array}\right)$$

• Write them down interchanging 1s with 2s. Then find the similarity transformation connecting the two sets. Commented Jan 9, 2020 at 3:13
• Thanks for the advice. Appreciate if you could elaborate it a bit. Commented Jan 9, 2020 at 5:56
• By the way, for an other point of view of Pauli matrices, that of the basis of the linear space of $2\times 2$ hermitian traces matrices, see equations (001)-(003) in my answer here : Why is there this relationship between quaternions and Pauli matrices?. Also SECTION B in my answer here : Construction of Pauli Matrices. Commented Jan 9, 2020 at 6:04
• Similarity transformation by $\sigma_x$. Do it in your question. Commented Jan 9, 2020 at 11:06
• Matrices are representations of operators on the Hilbert of states with respect to a basis in this space. If you change the basis, the matrix changes accordingly, not the operator. So,"...basis independent form of the Pauli matrices..." has no sense. Commented Jan 11, 2020 at 0:05

If the above distinction between operators and matrices seems two abstract, let us think about wave functions and the basis vectors. In one basis we may have wave function $$|\psi\rangle=\begin{pmatrix}1\\0\end{pmatrix}$$ In another basis the same wave function will be written as $$|\psi\rangle=\begin{pmatrix}\alpha\\\beta\end{pmatrix} = \alpha \begin{pmatrix}1\\0\end{pmatrix} + \beta \begin{pmatrix}0\\1\end{pmatrix}$$ is $$\begin{pmatrix}1\\0\end{pmatrix}$$ in the first equation different from $$\begin{pmatrix}1\\0\end{pmatrix}$$? Yes, these are unit vectors in different representations, but these are the same 2-by-1 matrix.
Set \begin{align} |e'\rangle & =|g\rangle=\left(\begin{array}{cc}0\\1\end{array}\right) \tag{01a}\label{01a}\\ |g'\rangle & =|e\rangle=\left(\begin{array}{cc}1\\0\end{array}\right) \tag{01b}\label{01b} \end{align} then \begin{align} \sigma'_z &= |e'\rangle\langle e'| - |g'\rangle\langle g'|= |g\rangle\langle g| - |e\rangle\langle e|=-\sigma_z \tag{02a}\label{02a} \\ \sigma'_x &= |e'\rangle\langle g'| + |g'\rangle\langle e'|= |g\rangle\langle e| + |e\rangle\langle g|=+\sigma_x \tag{02b}\label{02b} \\ \sigma'_y & = -i|e'\rangle\langle g'| + i|g'\rangle\langle e'|= -i|g\rangle\langle e| + i|e\rangle\langle g|=-\sigma_y \tag{02c}\label{02c} \end{align}