What does it mean to generate the full algebra of $SU(2)$ by commutation?

Question

I am reading the paper 'Coherent Quantum Feedback ' by Seth Lloyd, which unfortunately is paywalled. In Section (IV. A) he states that, for a quantum system consisting of a single spin in an NMR setup, the two Hamiltonians $B \sigma_z$ and $B_x \sigma_x\text{sin}(\omega t)$ 'can be easily seen to generate the full algebra of $SU(2)$ by commutation'.

Could someone explain what this means? I know that $SU(2)$ is the group of $2 \times2$ unitary matrices with determinant 1, but have a very poor understanding of group theory and abstract algebra. What does it mean to generate an algebra?

Answer 1

Edit: I made a mistake in interpreting your question. Since you're only interested in knowing how to generate the full algebra this is how it is done: To get a matrix which rotates a spinor about an axis $\hat{n}$ by angle $\phi$ we just have to calculate $$\exp\Big(i(\sigma_x \hat{n}_x+\sigma_y \hat{n}_y+\sigma_z \hat{n}_z)\frac{\phi}{2}\Big)$$

All this belongs to the representation theory of groups. I take my words back Wikipedia (though I have added the link to the necessary sections) won't be good for a beginner since it's filled with math jargon and you're probably new to this representation business.

Old: Everything fits in because $su(2)$ is a lie algebra. It has finite generators but infinite elements. Every element of this set can be found by exponentiating generators. Though there is small subtlety of exponential sending you out of $su(2)$ to $SU(2)$.

The generators of $su(2)$ are $\sigma_x$, $\sigma_y$ and $\sigma_z$ and they are related by $$[\sigma_i, \sigma_j]=2i\epsilon_{ijk} \sigma_k$$ You know two generators using commutator find the third one and then use exponential to find any element of $su(2)$ you want. M.D.Schwartz gives a brief intro to lie algebra in chapter 10 but Wikipedia will suffice for most basic terminology.