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?

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

Thanks a lot in advance :)

  • $\begingroup$ Write them down interchanging 1s with 2s. Then find the similarity transformation connecting the two sets. $\endgroup$ – Cosmas Zachos Jan 9 at 3:13
  • $\begingroup$ Thanks for the advice. Appreciate if you could elaborate it a bit. $\endgroup$ – Toshi_H Jan 9 at 5:56
  • $\begingroup$ 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. $\endgroup$ – Frobenius Jan 9 at 6:04
  • $\begingroup$ Similarity transformation by $\sigma_x$. Do it in your question. $\endgroup$ – Cosmas Zachos Jan 9 at 11:06
  • $\begingroup$ 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. $\endgroup$ – Frobenius Jan 11 at 0:05

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}

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