# Theorem Every [[Group Representation|representation]] with [[Matrix|matrices]] having *nonvanishing* [[Determinant|determinants]] can be brought into [[Unitary Operator|unitary]] form by a [[similarity transformation]]. # Proof Let $D_1, D_2, \ldots, D_h$ be the matrices of the representation. Consider the matrix $ H = \sum_{\alpha}^h D_\alpha D_\alpha^\dagger $ $H$ is a [[Hermitian Operator|Hermitian matrix]] as one can easily verify: $ H^\dagger = \sum_{\alpha}^h (D_\alpha D_\alpha^\dagger)^\dagger = \sum_{\alpha}^h D_\alpha D_\alpha^\dagger = H $ Thus, we can [[Diagonalizable Matrix|diagonalize]] it via a suitable [[similarity transformation]]. Let $U$ be a [[Unitary Operator|unitary matrix]] made up of the orthonormal [[eigenvectors]] which diagonalize $H$ to give the diagonal matrix $d$: $ \begin{align} d &= U^{-1} H U, \\ &= \sum_{\alpha}^h U^{-1} D_\alpha D_\alpha^\dagger U,\\ &= \sum_{\alpha}^h U^\dagger D_\alpha U U^\dagger D_\alpha^\dagger U,\\ &= \sum_{\alpha}^h \hat{D}_\alpha \hat{D}_\alpha^\dagger, \end{align} $ where in the last step we defined $ \hat{D}_\alpha = U^\dagger D_\alpha U $ and its adjoint. Note that $ \begin{align} d_{kk} &= \sum_{\alpha} \sum_j (\hat{D}_\alpha )_{kj}(\hat{D}_\alpha^\dagger)_{jk},\\ &= \sum_{\alpha} \sum_j (\hat{D}_\alpha )_{kj}(\hat{D}_\alpha)^*_{jk},\\ &= \sum_{\alpha} \sum_j |(\hat{D}_\alpha )_{kj}|^2 > 0 \end{align} $ Thus, $d$ is a diagonal matrix with only real, positive entries. We can then define some secondary matrices: $ \begin{align} d^{1/2} &\equiv \text{diag}(\sqrt{d_{11}}, \sqrt{d_{22}}, \ldots), \\ d^{-1/2} &\equiv \text{diag}(1/\sqrt{d_{11}}, 1/\sqrt{d_{22}}, \ldots) \end{align} $ Note that the generation of $d^{-1/2}$ from $d^{1/2}$ requires none of the $d_{kk}$ to vanish. ==How do we guarantee they don't?== Note the following properties: $ \begin{align} (d^{1/2})^\dagger &= d^{1/2}, \\ (d^{-1/2})^\dagger &= d^{-1/2}, \\ (d^{1/2})(d^{1/2}) &= d, \\ d^{1/2}d^{-1/2} &= d^{-1/2}d^{1/2} = I \end{align} $ Now we define $ \begin{align} \hat{\hat{D}}_\alpha &= d^{-1/2} \hat{D}_\alpha d^{1/2}, \\ \hat{\hat{D}}_\alpha^\dagger &= (d^{-1/2} \hat{D}_\alpha d^{1/2})^\dagger, \\ &= d^{1/2} \hat{D}^{\dagger}_\alpha d^{-1/2} \end{align} $ Now we prove that $\hat{\hat{D}}_\alpha^\dagger \hat{\hat{D}}_\alpha = I$, i.e. the $\hat{\hat{D}}$ matrices are unitary. $ \begin{align} \hat{\hat{D}}_\alpha \hat{\hat{D}}_\alpha^\dagger &= (d^{-1/2} \hat{D}_\alpha d^{1/2})(d^{1/2} \hat{D}_\alpha^\dagger d^{-1/2}), \\ &=d^{-1/2} \hat{D}_\alpha d \hat{D}_\alpha^\dagger d^{-1/2}, \\ &=d^{-1/2} \sum_{\beta}(\hat{D}_\alpha \hat{D}_\beta \hat{D}_\beta^\dagger\hat{D}_\alpha^\dagger) d^{-1/2}, \\ &=d^{-1/2} \sum_{\beta}[(\hat{D}_\alpha \hat{D}_\beta)(\hat{D}_\alpha \hat{D}_\beta)^\dagger] d^{-1/2}, \\ &=d^{-1/2} \sum_{\gamma}[\hat{D}_\gamma \hat{D}_\gamma^\dagger] d^{-1/2}, \\ &=d^{-1/2} d d^{-1/2} = I. \\ \end{align} $ In the third and last equality, we used the definition of $d$. In the second-to-last equality, we used the [[group rearrangement theorem|rearrangement theorem]]. It is valid here because the matrices $\hat{D}_\alpha$ are [[Group Representation|representations]] of [[group]] [[Group Element|elements]]. # Conclusion Given representation matrices $D_\alpha$ with non-vanishing [[determinant]], we can transform them into a unitary representation $ \hat{\hat{D}}_\alpha = d^{-1/2} U^{-1} D_\alpha U d^{1/2} $ where $U$ is the [[Unitary Operator|unitary matrix]] that [[Diagonalizable Matrix|diagonalizes]] the [[Hermitian Operator|Hermitian]] matrix $H$ $ H = \sum_{\alpha} D_\alpha D_\alpha^\dagger $ and $ d = \sum_\alpha \hat{D}_\alpha \hat{D}_\alpha^\dagger $ where $ \hat{D}_\alpha = U^{-1} A_\alpha U. $ # Notes Not all physical operators can be represented by a unitary matrix, see, e.g., the [[time-reversal operator]], which is an [[anti-unitary operator]].