For a particle with a spin $S$, the rotation operator is given by $$ e^{iJ_i\theta/\hbar} $$ where $J_i$ is the component of the total angular momentum along the direction of the rotation axis. The total angular momentum itself is given by $$ \mathbf J = \mathbf L \otimes I_S + I_L\otimes \mathbf S $$ here $I_S$ and $I_L$ are the identity operators in $S$ and $L$ spaces, respectively. The form of total angular momentum above arises from the fact that combining spatial with spin degree of freedom is realized by a tensor product between the respective spaces and from the Lie algebra of the rotation group $SO(3)$.

My first question is can I write the following $$ e^{iJ_i\theta/\hbar} = e^{iL_i\theta/\hbar} \otimes e^{iS_i\theta/\hbar} \hspace{5mm}?$$ The RHS is not something I personally came up with, I found it in my textbook but the author doesn't pertain its relationship with the more common rotation operator which is the LHS.

The second question is about the representation of the rotation group $SO(3)$ in this tensor product space, which eventually leads to the RHS of the above equation for the rotation operator. From what I have read, a tensor product representation space is a tensor product between two different representations spaces of the same group, here the $SO(3)$ group. The tensor product representation space for a particle with spin $S$ is then given by $\mathcal H = \Pi_L(R(\theta)) \otimes \Pi_S(R(\theta))$, where $R(\theta) \in SO(3)$. $\Pi_L(R(\theta))$ is the representation space of $SO(3)$ in $L^2(\mathbb R^3)$, which is easy to describe - in this space $\mathbf L = \mathbf r \times \mathbf p$. My second question is the representation of $SO(3)$ in the spin space which is $\mathbb C^{2S+1}$. How does it look like and how to describe it? I am at lost here because the standard representation of $SO(3)$ is $3\times 3$ orthogonal matrix while any operators in $\mathbb C^{2S+1}$ must be $(2S+1)\times(2S+1)$.

  • 2
    $\begingroup$ To your first question $[L_i, S_i] = 0$ (follows directly from your second equation) so breaking up the exponential in this way is perfectly OK, just treat them like they were regular numbers. $\endgroup$
    – qgp07
    Commented Jun 12, 2016 at 3:48

1 Answer 1


The issue is that the "spin representation of $SO(3)$" is not a representation of $SO(3)$ at all, but a representation of its double cover $SU(2)$ (see https://en.wikipedia.org/wiki/Spin_group). Since we sometimes write down representations in terms of infinitesimal generators (in other words, as a representation of the Lie algebra of the Lie group in consideration) and since $SO(3)$ and $SU(2)$ are the same at the infinitesimal level (their Lie algebras are the same), the distinction between the groups does not produce any confusion when writing the spin representation in terms of infinitesimal generators of the rotation group, which you wrote as $J_i$. However, if you try to exponentiate the infinitesimal generators $S_i$ of any of the spin representations to a representation of $SO(3)$ you will run into issues! In particular, rotation by 360 degrees, which should be the same as no rotation at all, will act by $-1$ on the representation, which doesn't make sense.

In general, there is no reason to expect $SO(3)$ to have any interesting representations on the even-dimensional vector spaces of the spin representations. In fact, you can show that any such representation will decompose as a direct sum of representations on odd-dimensional vector spaces. For instance, when $s = 1/2$, the only representation of $SO(3)$ on the 2-dimensional vector space $\mathbb{C}^2$ is the trivial representation where every element of $SO(3)$ acts by the identity 2 by 2 matrix. However, $SU(2)$ has a natural representation on $\mathbb{C}^2$, which is the spin representation.

  • $\begingroup$ The thing is in my book (and probably in any other books), the author discusses a tensor product between two different representations of the same group. Now if I consider a particle in Euclidean space having spin, the rotation operator (SO(3)) on the corresponding Hilbert space will be given by $e^{iJ_i\theta} = e^{iL_i\theta} \otimes e^{iS_i\theta}$. Again, according to my book the two operators in the RHS are actually two different representations of SO(3), one in $L^2(\mathbb R^3)$ and another one in the spin space. $\endgroup$
    – nougako
    Commented Jun 18, 2016 at 22:37
  • $\begingroup$ By the way how could you come up with the representation of SO(3) on $\mathbb C^2$ being the trivial representation? Is this definition or something one can prove? $\endgroup$
    – nougako
    Commented Jun 18, 2016 at 22:39
  • 1
    $\begingroup$ The only thing I can think of is that the notation $e^{iS_i\theta}$ means the exponentiated representation, which in this case will be a representation of $SU(2)$. The usual representation $e^{iL_i\theta}$ also defines a representation of $SU(2)$ since there is the double cover map $SU(2)\to SO(3)$. The representations of $SO(3)$ can be decomposed into a direct cum of irreducible representations, all of which are odd-dimensional. In fact, the $l$-dimensional representation comes from the action of $SO(3)$ on the spherical harmonics with eigenvalue $-l(l+1)$. $\endgroup$ Commented Jun 18, 2016 at 23:43
  • $\begingroup$ Sorry, the irreducible representations I mention above are $2l+1$-dimensional. $\endgroup$ Commented Jun 18, 2016 at 23:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.