# Connection to spin 1/2 electron system?

In another Physics stack exchange thread here, Spin matrix for various spacetime fields I obtained the generator of rotations of the SO(2) rotation group for an infinitesimal rotation of 2D vectors. I then tried to relate this to the spin-1/2 electron system, but it appears vectors representing states for that system transform under the Pauli matrices instead. I believe this is because those sets of matrices, when scaled properly, yield the correct eigenvalues for the operators.

However, I also noticed that $$D(\omega) = \text{Id} + \omega \begin{pmatrix} 0&1\\-1&0 \end{pmatrix} = \text{Id} + i \omega \begin{pmatrix} 0&-i \\i&0 \end{pmatrix},$$ so I seemed to have made contact with one of the Pauli matrices, (i.e the scaled form of which the spin states do transform). It looks like the above is the infinitesimal version of $\exp(i\omega \hat n \cdot \sigma)$ with $$\hat n \cdot \sigma = \sigma_2 = \begin{pmatrix} 0&-i \\ i&0 \end{pmatrix}$$ which seems to mean $\hat n = (0,1,0)$. In particular, c.f Sakurai P.159, eqn (3.2.3), he says that the physical spin 1/2 electron system does transform under the operator $D(\phi) = \exp(-iS_z\phi/\hbar)$ which is analogous to what I have above with the replacements $\phi \rightarrow \omega$ and $S_z/\hbar \rightarrow \hat n \cdot \sigma$.

So is there really a connection to the matrix I derived and the spin 1/2 electron system? I do actually think not, since the matrix I derived in that other thread came from analyzing the rotation of vectors in space-time but the spin states live in another space. On the other hand, the analysis above makes me think otherwise at the same time.

Many thanks for clarification!

• The Pauli matrices are the generators of the $\mathfrak{su}(2)$ Lie algebra, which coincides with the Spin Lie algebra since the spin group covers the rotation group, so it is not surprising (in fact, expected) that their infinitesimal versions around the identity be the same. – ACuriousMind Jul 24 '14 at 15:14
• Hi ACuriousMind. Could you elaborate on what you mean by 'the spin group covers the rotation group'? Thanks very much. – CAF Jul 24 '14 at 15:18
• It is a covering space with fiber dimensionality $2$, to be exact. It is known (with a bit of footing in group theory and the like) that the Lie algebras of any Lie groups that are covers of each other coincide. And since the infinitesimal form of a transformation is just $1 + g$, where $g$ is a Lie-algebra element, you have just "rediscovered" that the Spin group and the rotation group have the same Lie algebra – ACuriousMind Jul 24 '14 at 15:24
• Ah, is it related to the isomorphism $SO(3)$ and the factor group $SU(2)/\mathbb{Z}_2$? Okay, so if I then exponentiate my result, is it correct to say that I will obtain an operator that will transform the spin 1/2 states? (at least by doing so, as I said in the OP, I recover the form of one of the operators Sakurai mentions in his book). – CAF Jul 24 '14 at 15:29
• Jup, since the Pauli matrices are the fundamental representation of $\mathfrak{su}(2)$, exponentiating them will yield an element of the fundamental rep of $\mathrm{SU}(2)$, which is the spin-1/2 rep (i.e. the exponentiated forms act naturally on spin-1/2 states, indeed) – ACuriousMind Jul 24 '14 at 15:41

It is not surprising that you found one of the Pauli matrices as the generator of your rotation. Let us see how it can be seen algebraically that it is to be expected:

Observe that 2D rotations embed naturally into 2D unitary matrices, as $\mathrm{SO}(2) \subset \mathrm{SU}(2)$, corresponding to the subgroup of real matrices. Now, as we know, the $\mathrm{SU}(2)$ is generated by the Pauli matrices, and Lie subgroups must be generated by Lie subalgebras. The question is - which Lie subalgebra is generating the $\mathrm{SO}(2)$?

Well, the one generating the real matrices of course! If you examine1 the Pauli matrices, you will see that only the one you found by directly linearizing a 2D rotation is giving real instead of complex matrices when exponentiated. Now, how can we be sure that's it? Because 2D rotations are isomorphic to the $\mathrm{U}(1)$, the one-dimensional Lie group, and so we expect only one generator.

1 Observe that $\mathrm{e}^{\mathrm{i}\phi (\hat n \cdot \vec \sigma)} = \boldsymbol{1}\mathrm{cos}(\phi) + \mathrm{i}(\hat n \cdot \vec \sigma)\mathrm{sin}(\phi)$. Plugging in the explicit form of the Pauli matrices, we find that the summand from $n_y\sigma_y$ is real (since $\mathrm{i}\sigma_y$ is real), while the others are complex.

• Many thanks! The only part I did not understand is this sentence here: 'If you examine the Pauli matrices, you will see that only the one you found by directly linearizing a 2D rotation is giving real instead of complex matrices when exponentiated'. Could you explain this a little more? – CAF Jul 24 '14 at 16:30
• @CAF: Added a footnote – ACuriousMind Jul 24 '14 at 16:36
• Great, thanks. My last question would be when we exponentiate, the result is $D(\omega) = \exp(i\omega S_y/\hbar)$, since $S_i \rightarrow \hbar \sigma_i/2$, so by exponentiating we seem to have obtained an operator that encodes a rotation of a spin 1/2 state by angle $\omega$ around the y-axis. Is this a correct interpretation? I guess if this was correct, a follow-up question would be why physically are we obtaining such a result from exponentiation? – CAF Jul 24 '14 at 16:42
• @CAF: Uhh...I believe that is not a correct interpretation, since the operator $D(\omega)$ we obtained does not act upon a representation of $\mathrm{SU}(2)$, but on a representation of $\mathrm{SO}(2)$. It's just because $\mathrm{SO}(2)\subset \mathrm{SU}(2)$ that they look the same. – ACuriousMind Jul 24 '14 at 16:51
• Ok, but then if $D(\omega)$ acts on a representation of $SO(2)$ then why would it transform the spin 1/2 states? – CAF Jul 24 '14 at 16:55