# Time dependent Pauli matrix notation in quantum optics

I am reading a quantum optics paper on superradiance and subradiance. In building their Hamiltonian, the authors define an operator $$J_\pm(t)=\sum^N_{n=1}\sigma^{(n)}_\pm \exp{\left(\pm i \delta \omega_n t \right)} \tag{1},$$ and say that $$\sigma^{(n)}_\pm = \sigma^{(n)}_x \pm i\sigma^{(n)}_y \tag{2},$$ with "$$\sigma^{(n)}_{x,y,z}$$ representing the Pauli matrices for individual atoms". Later they define a squared atomic polarization term which includes the term $$I(t) = \sum^N_{n=1}\langle\sigma^{(n)}_{+}\sigma^{(n)}_{-}\rangle (t)\tag{3}.$$ Long ago I was taught that the Pauli matrices are $$2\times 2$$ complex matrices which form a nice basis set. However, this doesn't seem to work with how these author's are using the term Pauli matrices.

Here are my questions:

1. Why do the authors index their Pauli matrices atom-by-atom?
2. How do the authors sneak in time dependence in the expectation value of constant matrices multiplied together?

Pauli matrices are of course just a well-known set of 2-by-2 matrices, defined in basic quantum mechanics. One often has to deal with spin, polarization or other quantities, whose operators can be represented in terms of Pauli matrices, and this seems to be the case here. The operators associated with polarization/spin of different atoms are designated by different indices, and, considered in Heisenberg picture, these operators are time-dependent. The only rationale for calling them Pauli matrices in this context is that they do not contain $$\hbar/2$$ and possibly other proportionality factors that would clutter the calculation.
It could be an easy and clarifying exercise to calculate explicitly the time evolution of spin operators in a constant magnetic field: $$\hat{s}_x = \frac{\hbar}{2}\sigma_x, \hat{s}_y = \frac{\hbar}{2}\sigma_y, \hat{s}_z = \frac{\hbar}{2}\sigma_z,\\ \hat{H}=-g\mu_B B \hat{s_z}$$ Now we can calculate the time evolution of the operators in the Heisenberg picture as $$\hat{s}_i(t) = e^{\frac{i\hat{H}t}{\hbar}}\hat{s}_i(0)e^{-\frac{i\hat{H}t}{\hbar}}$$
1. It is possible to construct operators which only operate on a particular particle. This is done using the tensor product. Tensor products work as follows $$(\hat A\otimes\hat B)\cdot(\hat C\otimes\hat D)=(\hat A\cdot \hat C)\otimes(\hat B\cdot \hat D)$$ $$(\hat A\otimes\hat B)\cdot(|u\rangle\otimes|v\rangle)=(\,\hat A|u\rangle\,)\otimes(\,\hat B|v\rangle\,)$$ This behaviour can be roughly summarised as the left and the right positions acting separately from another. Often $$|u\rangle\otimes|v\rangle$$ is abbreviated as $$|u\rangle|v\rangle$$. If you have a state of $$n$$ particles you can easily contruct an operator that only acts on one particle using the tensor product. Take for example $$\sigma^{(1)}_x$$: $$\sigma^{(1)}_x|\chi\rangle=(\underbrace{\sigma_x\otimes I\otimes\dots\otimes I}_\text{n times})\,|\chi_1\rangle\dots|\chi_n\rangle$$ where $$I$$ is the identity operator. Similarly you can make $$\sigma^{(i)}$$ by replacing the $$i$$-th identity operator by $$\sigma$$