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This is based off question 4.30 from Griffith's Introduction to Quanum Mechanics. It asks for the matrix $\textbf{S}_r$ representing the component of spin angular momentum about an axis defined by: $$r = \sin{\theta}\cos{\phi}\hat{i}+\sin\theta\sin\phi\hat{j}+\cos\theta\hat{k}$$

for a spin = $1/2$ particle.The problem is, I can't visualize how the spin vectors relate to spatial coordinates. $$\chi_+ = \begin{pmatrix} 1 \\ 0\end{pmatrix}\ \text{ and } \chi_- = \begin{pmatrix} 0 \\ 1 \end{pmatrix}. $$

So these form a basis of some kind of space, but this is a space I don't understand, for example, why does: $$\chi^{(x)}_+ = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\1 \end{pmatrix} ?$$

I was going to try to use a rotation matrix with angles corresponding to my axis of rotation to try and "rotate" the spin vector onto it, but then I realized that not only do the dimensions not match up, but the angle is π/4, which doesn't make much sense to me. So I guess my question is, how does the 'geometry' of spin work and how can I transform the spins corresponding to a transform I do in space?


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Spin-1/2 representation of rotations involve the Pauli matrices. More details about the physics here. – Michael Brown Feb 22 '13 at 6:55
You may also want to look up the Bloch sphere, which is the canonical geometric incarnation of the state space of a two-level system. – Emilio Pisanty Feb 24 '13 at 19:23
up vote 5 down vote accepted

Great and important question; I hope this response is illuminating and encourages you and others to explore Lie Groups, Lie Algebras, and their representations.

When you want to rotate a vector $\mathbf v$ in three dimensions, then you act on that vector with a rotation matrix $R$ to obtain a rotated vector $\mathbf v'$ related to the original vector by $$ \mathbf v' = R\mathbf v $$ It is a mathematical fact that every rotation (special orthogonal transformation) $R$ is a rotation by some angle $\theta$ about some axis defined by a unit vector $\mathbf n$, and that each such rotation can be written as the matrix exponential of a particular linear combination of certain 3-by-3 matrices $J_i$ called rotation generators. Explicitly $$ R(\theta, \mathbf n) = e^{-i\theta n_i J_i}, \qquad (J_i)_{jk} = i\epsilon_{ijk} $$ So we can write the rotation of a vector as $$ \mathbf v' = e^{-i\theta n_i J_i} \mathbf v $$ It turns out, that we can define the rotation of spinors in an analogous way. Instead of the rotation generators $J_i$ which are 3-by-3 matrices, we choose the pauli matrices which are 2-by-2, and for a given spinor $\xi$, we define a rotated spinor by $$ \chi' = e^{-i\frac{\theta}{2} n_i\sigma_i}\chi $$ Notice how this is basically the same as rotating a vector in 3 dimensions, it's just that we have represented rotations acting on the vector space of spinors in a different way.

The math behind all of this is called representation theory. In particular, when we are talking about spin, the representation theory is related to that of the Lie groups $\mathrm{SO}(3)$ and $\mathrm{SU}(2)$ and their so-called Lie algebras and their representations.

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Thank you for this answer, is the $\theta$ in $X'=e^{-i\frac{\theta}{2}n_i\sigma_i}X$ in "physical" space or "spinor" space? – Mike Flynn Feb 24 '13 at 19:07
@MikeFlynn My Pleasure. That's another great question, and there is more than one way to think of $\theta$. Here's one you may like. Consider the state $\chi_{z,+}$, the eigenvector of $S_z$ corresponding to spin up. If you apply $e^{-i\frac{\theta}{2}n_i\sigma_i}$ to $\chi_{z,+}$ with $\mathbf n = \mathbf e_y$ and $\theta = \pi/2$, then you will get $\chi_{x,+}$, the spin up eigenstate of $S_x$, so in a sense, doing the rotation on the spin state corresponds to doing a physical rotation on the spin the three-dimensional world. There is a lot more one can say here to get more intuition – joshphysics Feb 24 '13 at 19:25
(contd.) for what the rotation angle means and how rotating the spin state relates to the real world; perhaps you should ask another question about this that would address this issue. – joshphysics Feb 24 '13 at 19:26

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