# What is a rotation group and how do we get its unitary representation?

The rotation group is $${\rm SO(3)}$$. It is the group of $$3\times 3$$ orthogonal matrices $$\{g(\theta)\}$$ with unit determinant. So these are already defined in terms of $$3\times 3$$ matrices. But we use unitary representation $$\{U(g(\theta))\}$$ of the rotation group in quantum mechanics. What does that even mean?

How do we define rotation group if not in terms of explicit $${\rm SO(3)}$$ matrices? Is this already not a representation? Isn't the definition of rotation group already in terms of this representation?

Given the elements of the rotation group (i.e., the $${\rm SO(3)}$$ matrices $$g(\theta)$$) how do we get, $$U(g(\theta))$$?

• This is way too broad. Do you know what the formal definition of a representation is? Have you at least read the wikipedia entry? – AccidentalFourierTransform May 29 '19 at 16:20
• @AccidentalFourierTransform yes, I know the definition of representation. Group elements are represented by matrices that obey the group structure. Does it help? – mithusengupta123 May 29 '19 at 17:53
• May or may not be helpful but I wrote some notes on representation theory of SO(3) and QM here: scholar.harvard.edu/files/noahmiller/files/… – user1379857 May 29 '19 at 22:59
• Do you understand that representations can have various dimensions? For example there are $5 \times 5$ and $17 \times 17$ matrices representing 3D rotations, not just $3 \times 3$ ones. – G. Smith May 30 '19 at 0:02
• – Frobenius Dec 26 '19 at 0:36

[about SO(3)] SO(3) is an abstract group with a lot of well-known properties (Lie, compact, topological etc). The representation of SO(3) via 3d, orthogonal, real-valued maricies is one of many possible ones. But this is, by definition, a faithful representation, i.e. every group member has a distinct matrix that corresponds to it.

[about getting unitary reps]

The rotation matricies are orthogonal and real-valued, so they are already unitary, thus, technically, the question is moot.

If your question is how to go from the general orthogonal, 3d, real-valued matrix with unity determinant to its representation as $$R=\exp\left(\dots\right)$$, i would suggest diagonalizing the matrix.

You will find that this does not work over the space $$\mathbb{R}^3$$, but it does work over $$\mathbb{C}^3$$. Quite simple reasoning can show that any

$$R\in \mathbb{R}^3\times\mathbb{R}^3$$ with $$\det R=1$$ and $$R^T=R^{-1}$$ can be diagonalized in the complex space with eigenvalues:

$$\lambda_{1,2,3}=\exp\left(\pm i\phi\right), 1$$ for some $$\phi\in\mathbb{R}$$

Thus $$R=V \exp\left(i\left(\begin{array}\\ \phi &&0 && 0\\ 0 && -\phi && 0\\ 0 && 0 && 0 \end{array}\right)\right) V^{\dagger}$$

Where the $$V$$ is the unitary matrix with eigenvectors. Now simply take the $$V$$ matricies into the exponential and you will have your representation.

The point here is that, if $$\omega\in SO(3)$$, and if $$\omega_1\cdot \omega_2=\omega \in SO(3)$$ is the combination rule on abstract elements, then a representation (by matrices) $$U$$ is a map $$\omega\mapsto U(\omega)$$ so that the rule $$\omega_1\cdot \omega_2=\omega \quad \Rightarrow \quad U(\omega_1)\cdot U(\omega_2)=U(\omega) \tag{1}$$ for any $$\omega_1,\omega_2,\omega\in SO(3)$$ is also satisfied by the matrices $$U(\omega)$$ representing the elements. There is a theorem stating that, for $$SO(3)$$ and a bunch of others, all representation are equivalent to unitary representations, so that $$U(\omega^{-1})=U^{-1}(\omega)=U^\dagger(\omega)$$.

Although the so-called defining representation is in terms of a $$3$$-dimensional space on which $$SO(3)$$ acts "naturally", there may be matrices of dimension other than $$3$$ that satisfy the basic composition law of $$\omega_1\cdot \omega_2=\omega$$ or its matrix version of Eq.(1).

You can "obtain" a representation by larger matrices by tensoring and decomposing the resulting representation. For instance, if $$\{\vert {1}\rangle ,\vert {2}\rangle,\vert {3}\rangle\}$$ are a basis for the $$3$$-dimensional irrep of $$SO(3)$$, then the set$$\{\vert i\rangle\vert j\rangle\}$$ spans a 9-dimensional space with $$U(\omega)\left[\vert i\rangle\otimes\vert j\rangle\right]:= \left[U(\omega)\vert i\rangle\right]\otimes \left[U(\omega)\vert j\rangle\right]$$ will provide you with a $$9$$-dimensional representation, which turns out to be reducible. Note that I'm abusing the notation here because on the left I have $$U$$ as a $$9\times 9$$ matrix but on the right the $$U$$'s are $$3\times 3$$ matrices. In fact, the $$9\times 9$$ representation is reducible: it contains $$L=2,1,0$$, i.e. irreducible pieces of dimensions $$5,3$$ and $$1$$. The $$L=2$$ and $$L=0$$ irreps are spanned by symmetric combinations like $$\vert 1\rangle\vert 2\rangle+ \vert 2\rangle\vert 1\rangle$$ etc, while the $$L=1$$ contains antisymmetric pieces.

In cases other than $$SO(3)$$ (or $$SU(2)$$), one can also obtain inequivalent representations by taking the conjugate. The simplest example would be $$SU(3)$$, where the defining representation ($$3\times 3$$) is often denoted by $$\textbf{3}$$ or $$(1,0)$$ in the Dynkin scheme, and where its (non-equivalent) conjugate is denoted by $$\textbf{3*}$$ or $$(0,1)$$. One can then construct any representation by tensoring a suitable number of copies of $$(1,0)$$ and $$(0,1)$$ and decomposing the result.

Note that $$SO(3)$$ representations (and also $$SU(2)$$ representations) are "self-conjugate" in the sense that taking the conjugate yields the same representation.