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On p. 692 of 'Quantum Mechanics' by Cohen-Tannoudji, he states that:

Every finite rotation can be decomposed into an infinite number of infinitesimal rotations, since the angle of rotation can vary continuously, and since:

$\begin{equation} \mathcal{R}_{\textbf{u}}(\alpha+d\alpha)=\mathcal{R}_{\textbf{u}}(\alpha)\mathcal{R}_{\textbf{u}}(d\alpha)=\mathcal{R}_{\textbf{u}}(d\alpha)\mathcal{R}_{\textbf{u}}(\alpha), \end{equation}$

where $\mathcal{R}_{\textbf{u}}(d\alpha)$ is an infinitesimal rotation about the axis $\textbf{u}$. Thus, the study of the rotation group can be reduced to an examination of infinitesimal rotations.

Here, $\mathcal{R}_{\textbf{u}}(\alpha)$ represents a geometrical rotation, i.e., it acts on the coordinate space $\Re^{3}$, and with it is associated a rotation operator $R(\alpha)$ which acts on the state space.

In particular, he uses this formulation with infinitesimal rotations to then show that the rotation operator for an infinitesimal rotation about $\textbf{u}$ is:

\begin{equation} R_{\textbf{u}}(d\alpha)=1-\dfrac{i}{\hbar}d\alpha \hspace{0.2em} \textbf{J}\cdot\textbf{u}, \end{equation}

where $\textbf{J}$ is the total angular momentum operator. From this, one can show that the rotation operator for some finite angle is:

\begin{equation} R_{\textbf{u}}(\alpha)=e^{-\frac{i}{\hbar}\alpha \hspace{0.2em} \textbf{J}\cdot\textbf{u}}. \end{equation}

A well known example of such a rotation operator is when $\textbf{J}=\textbf{S}$, i.e., the angular momentum consists of spin only, and when $s$ is allowed to take half-integer values only, such as $\frac{1}{2}$ or $\frac{3}{2}$. In this case, one can show that $R_{\textbf{u}}(2\pi)=-\mathbb{1}$, rather than $+\mathbb{1}$, as one gets in the case of integer spin particles.

Cohen-Tannoudji explains this partly through the fact that we constructed our finite angle rotation operator from a composition of infinitesimal rotation operators, with the footnote:

However, limiting ourselves to infinitesimal rotations, we lose sight of a 'global' property of the finite rotation group: the fact that a rotation through an angle of $2\pi$ is the identity transformation. The rotation operators constructed from infinitesimal operators do not always have this global property. In certain cases (and here he references spin-1/2 particles), the operator associated with a $2\pi$ rotation is not the unit operator but its opposite.

It is not immediately clear to me from the construction he gave why the possible values of $j$ and the fact that we used infinitesimal operators to construct a finite one should be related. How does this relationship come about?

Community
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anon1802
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2 Answers2

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Different Lie groups can have the same (up to isomorphism) Lie algebra. This is the case of, say, $SO(3)$ and $SU(2)$, the latter being the universal 2-cover of the former. When you are given a Lie algebra $\mathfrak g$ and you want to integrate it to a Lie group $G$ having $\mathfrak g$ as a Lie algebra, you will end up with a simply connected group. Hence, if you start with $SO(3)$ and determine its Lie algebra $\mathfrak{so}(3)\cong\mathfrak{su}(2)$, its integration will give you $SU(2)$, which is simply connected and indeed is the 2-cover of $SO(3)$.

Phoenix87
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Say you had didn't know about quantum mechanics and no idea what a spinor is. You're given an "operator" that rotates things. One of the most basic assumptions you will make is that a rotation of $2\pi$ changes nothing. This is really quite reasonable.

As @Phoenix87 said, we can identify $SU(2)$ has having a Lie algebra isomorphic to that of $SO(3)$. We find that for every integer and half-integer $j$ there exists an $SU(2)$ irrep $$R^{(j)}_\mathbf{u}(\alpha)=\exp(-i\alpha\mathbf{u}\cdot\mathbf{J}^{(j)})$$ The physical interpretation of $j$ is the angular momentum of the particle. This is a rotation operator.

Naively one would expect $$R^{(j)}_\mathbf{u}(2\pi)=\mathbb{1}$$ Indeed this is what we find when $j$ is an integer. However, when $j$ is a half integer, we find that a rotation of $2\pi$ is $-\mathbb{1}$. This is more deeply linked to the notion of spinors. One may show that $$R^{(j)}_\mathbf{u}(2\pi)=e^{i2\pi j}$$ holds for any $j$.

From the infinitesimal form of the rotation operator, this is not clear. We need the finite version.

I should mention that we have a topological restriction placed upon rotations: $$R(4\pi)=\mathbb{1}$$

Ryan Unger
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