# Theorem The [[Orthogonality|orthonormality]] relation $ \sum_{g} D_{\mu\nu}^{(\Gamma_j)}(g)D_{\nu'\mu'}^{(\Gamma_{j'})}(g^{-1}) = \frac{h}{\ell_j} \delta_{\Gamma_j, \Gamma_{j'}} \delta_{\mu\mu'} \delta_{\nu\nu'} $ is obeyed for all inequivalent [[Irreducible Representation|irreducible representations]] of a [[group]], where the summation is over all $h$ [[Group Element|group elements]] $g_1, g_2, \ldots, g_h$. $\ell_j$ and $\ell_{j'}$ are the [[Dimension|dimensionalities]] of the representations $\Gamma_j$ and $\Gamma_{j'}$ respectively. If the [[Group Representation|representations]] are [[Theorem on the Unitarity of Representations|unitary]], then the relation becomes: $ \sum_{g} D_{\mu\nu}^{(\Gamma_j)}(g)\left[D_{\mu'\nu'}^{(\Gamma_{j'})}(g)\right]^* = \frac{h}{\ell_j} \delta_{\Gamma_j, \Gamma_{j'}} \delta_{\mu\mu'} \delta_{\nu\nu'} $ Here, $\delta_{\alpha \beta}$ is the [[Kronecker delta|Kronecker delta]]. ## Proof ### Preliminaries Consider the $\ell_{j'}\times\ell_j$ matrix $ M = \sum_{g} D^{(\Gamma_{j'})}(g) X D^{(\Gamma_{j})}(g^{-1}) $ where $X$ is an arbitrary $\ell_{j'} \times \ell_j$. Now let's massage this a little, where in the following $\tilde{g}$ is some arbitrary element in the group. ^8e23fe $ \begin{align} D^{(\Gamma_{j'})}(\tilde{g})M &= \sum_{g} D^{(\Gamma_{j'})}(\tilde{g})D^{(\Gamma_{j'})}(g) X D^{(\Gamma_{j})}(g^{-1}), \\ D^{(\Gamma_{j'})}(\tilde{g})M &= \sum_{g} D^{(\Gamma_{j'})}(\tilde{g} g)X D^{(\Gamma_{j})}(g^{-1}), \\ D^{(\Gamma_{j'})}(\tilde{g})M &= \sum_{g} D^{(\Gamma_{j'})}(\tilde{g} g)X D^{(\Gamma_{j})}(g^{-1})D^{(\Gamma_{j})}(\tilde{g}^{-1}\tilde{g}), \\ D^{(\Gamma_{j'})}(\tilde{g})M &= \sum_{g} D^{(\Gamma_{j'})}(\tilde{g} g)X D^{(\Gamma_{j})}(g^{-1}\tilde{g}^{-1})D^{(\Gamma_{j})}(\tilde{g}), \\ D^{(\Gamma_{j'})}(\tilde{g})M &= \sum_{g} D^{(\Gamma_{j'})}(\tilde{g} g)X D^{(\Gamma_{j})}([\tilde{g} g ]^{-1})D^{(\Gamma_{j})}(\tilde{g}), \\ D^{(\Gamma_{j'})}(\tilde{g})M &= \sum_{g} D^{(\Gamma_{j'})}(g)X D^{(\Gamma_{j})}(g^{-1})D^{(\Gamma_{j})}(\tilde{g}), \\ D^{(\Gamma_{j'})}(\tilde{g})M &= MD^{(\Gamma_{j})}(\tilde{g}), \\ \end{align} $ where in the second to last step we used the [[Group Rearrangement Theorem|rearrangement theorem]], and in the last step we used the definition of $M$. We are now in a state to apply [[Schur's Lemma (Group Theory)#Lemma 2|Schur's Lemma 2]] for the various different cases to the $\ell_{j'} \times \ell_j$ matrix $M$. ### Case 1: $\ell_j \neq \ell_{j'}$ or $\ell_j = \ell_{j'}$ and the representations are inequivalent These are cases [[Schur's Lemma (Group Theory)#^fb3244|(1)]] and [[Schur's Lemma (Group Theory)#^fa8c62|(2a)]], so we have $M = \mathbb{0}$. Thus, from the definition of $M$, we have $ M_{\mu\mu'} = \sum_{g} D^{(\Gamma_{j'})}_{\mu\gamma}(g) X_{\gamma\lambda} D^{(\Gamma_{j})}_{\lambda\mu'}(g^{-1}) = 0. $ Since $X$ is arbitrary, we choose it to be $ X_{\gamma \lambda} = \delta_{\gamma \nu} \delta_{\lambda \nu'}, $ i.e., a matrix with all $0s except at the entry $\nu \nu'$ where it has a $1$. Plugging in, we get: $ M_{\mu\mu'} = \sum_{g} D^{(\Gamma_{j'})}_{\mu\nu}(g) D^{(\Gamma_{j})}_{\nu'\mu'}(g^{-1}) = 0. $ We will come back to this later when we combine it with the second case. ### Case 2: $\ell_j = \ell_{j'}$ and the representations are equivalent If the representations are equivalent, then $\ell_j$ must be equal to $\ell_{j'}$ and by [[Schur's Lemma (Group Theory)#Lemma 1|Schru's Lemma 1]], we have that $M = c I$. Thus: $ M_{\mu\mu'} = \sum_{g} D^{(\Gamma_{j'})}_{\mu\gamma}(g) X_{\gamma\lambda} D^{(\Gamma_{j'})}_{\lambda\mu'}(g^{-1}) = c \delta_{\mu \mu'}. $ Similar to before, we choose $X$, which is arbitrary, to be the following: Since $X$ is arbitrary, we choose it to be $ X_{\gamma \lambda} = c'_{\nu \nu'} \delta_{\gamma \nu} \delta_{\lambda \nu'}, $ i.e., a matrix with all $0s except at the entry $\nu \nu'$ where it has $c'$. Plugging in, we get: $ \sum_{g} D^{(\Gamma_{j'})}_{\mu\nu}(g) D^{(\Gamma_{j'})}_{\nu'\mu'}(g^{-1}) = c''_{\nu\nu'} \delta_{\mu \mu'}. $ where $c''_{\nu \nu'} = c/c'_{\nu \nu'}$. To find the value of $c''_{\nu\nu'}$, we set $\mu = \mu'$ in the equation above and some on $\mu$: $ \begin{align} c''_{\nu\nu'} \ell_{j'} &= \sum_{g} \sum_{\mu} D^{(\Gamma_{j'})}_{\mu\nu}(g) D^{(\Gamma_{j'})}_{\nu'\mu}(g^{-1}) , \\ &=\sum_{g} \sum_{\mu} D^{(\Gamma_{j'})}_{\nu\nu'}(g^{-1}g), \\ &=\sum_{g} \sum_{\mu} D^{(\Gamma_{j'})}_{\nu\nu'}(e), \\ &= \sum_{g} \delta_{\nu\nu'}, \\ &= h \delta_{\nu\nu'} ,\\ c''_{\nu\nu'} &= \frac{h \delta_{\nu\nu'}}{\ell_{j'}}, \\ \end{align} $ Plugging in, we finally get: $ \sum_{g} D^{(\Gamma_{j'})}_{\mu\nu}(g) D^{(\Gamma_{j'})}_{\nu'\mu'}(g^{-1}) = \frac{h}{l_{j'}} \delta_{\nu \nu'} \delta_{\mu \mu'}. $ ### General Case We have the following equation for inequivalent representations: $ \sum_{g} D^{(\Gamma_{j'})}_{\mu\nu}(g) D^{(\Gamma_{j})}_{\nu'\mu'}(g^{-1}) = 0 $ And the following equation for equivalent representations: $\sum_{g} D^{(\Gamma_{j})}_{\mu\nu}(g) D^{(\Gamma_{j})}_{\nu'\mu'}(g^{-1}) = \frac{h}{l_{j}} \delta_{\nu \nu'} \delta_{\mu \mu'}.$ We can combine them into: $ \sum_{g} D_{\mu\nu}^{(\Gamma_j)}(g)\left[D_{\mu'\nu'}^{(\Gamma_{j'})}(g)\right]^* = \frac{h}{\ell_j} \delta_{\Gamma_j, \Gamma_{j'}} \delta_{\mu\mu'} \delta_{\nu\nu'} $ which is the statement of the theore $\square$.