# $\rm SU(2)$ symmetry and rotation invariance

I've read in this paper (doi:10.1038/nature07871) that in the context of quantum mechanics, $$\rm SU(2)$$ symmetry leads to the conservation of spin polarization, or in other words invariance with respect to the rotation of the electron’s spin leads to such conservation. I'm very new to the subject, hence, my question is, what exactly does one mean with invariance with respect to the rotation? Most importantly, how can I picture/ draw in my mind ( if possible at all) the statement " invariance with respect to the rotation of the electron’s spin"?

My question is derived from this specific part of the paper.

"According to Noether’s theorem1, for every symmetry in nature there is a corresponding conservation law. For example, invariance with respect to spatial translation corresponds to conservation of momentum. In another well-known example, invariance with respect to rotation of the electron’s spin, or $$\rm SU(2)$$ symmetry, leads to conservation of spin polarization. For electrons in a solid, this symmetry is ordinarily broken by spin–orbit coupling, allowing spin angular momentum to flow to orbital angular momentum. However, it has recently been predicted that $$\rm SU(2)$$ can be achieved in a two-dimensional electron gas, despite the presence of spin–orbit coupling2."

• Please provide details on the contents of the paper especially as it is behind a paywall. Please make your question self-contained so that contributors who do not have access to the paper can clearly understand your question and its context. – ZeroTheHero Jul 21 at 13:19
• Sure, all my questions are derived from this specific paragraph, that I'll add to the question because it can't fit here. – I_R Jul 21 at 13:21
• Link to abstract page? – Qmechanic Jul 21 at 13:56

Let's break down the sentence "Invariance with respect to rotation of the electron’s spin, or SU(2) symmetry, leads to conservation of spin polarization". The invariance means that the Hamiltonian commutes with total spin, $$[H,\vec{S}] =0$$. (You can show this also by requiring that $$H$$ is invariant under arbitrary spin rotation $$U= e^{i \vec{s}\cdot \vec{n} \theta}$$, $$H=UHU^\dagger$$, see also the answer by Undead.) From this it follows that we can simultaneously diagonalize $$H$$ and any component of $$\vec{S}$$. If we diagonalize for example $$S_z$$, you find that any state with $$S_z$$ eigenvalue $$+1/2$$ will keep that eigenvalue, even if it's not an eigenstate of $$H$$. An example of a Hamiltonian that does not conserve spin, is the spin-orbit term $$(\vec{E} \times i \hbar \nabla)\cdot \vec{S}$$ (I believe something like this is considered in the paper you cited). Here $$\vec{E}$$ is some external electric field which intuitively breaks the rotational invariance of the system. As was mentioned by Undead in their answer, even this Hamiltonian has a symmetry: if you rotate both the spin and the angular momentum ($$\vec{E} \times i \hbar \nabla$$), you can keep this term invariant.
If the Hamiltonian commutes with the spin operator $$[H,\vec{S}] = 0$$ and if $$\left | \Omega \right \rangle$$ is an eigenstate of the Hamiltonian (say the ground state) it implies that $$e^{i \vec{n} \cdot \vec{S}} \left | \Omega \right \rangle = e^{i\theta} \left | \Omega \right \rangle$$ so that a spin rotation has no effect other than an unimportant phase factor and the state is invariant.
However, remember that experimentally a spin rotation is not as easy to generate as to just apply $$e^{i \vec{n} \cdot \vec{S}}$$ on a state. What is common to do is a rotation of both angular momentums $$J = L+S$$ with magnetic fields. Spin-orbit coupling commutes with that, in other words, it conserves angular momentum which is expected.