# Definition ## Representation Theory Given the [[Basis Vectors of a Representation|basis vectors]] $\ket{\Gamma_n j}$ of an [[irreducible representation]] $\ket{\Gamma_n}$, we can define down a projection operator $ \hat{P}^{(\Gamma_n)}_{k\ell} \ket{\Gamma_n \ell} \equiv \ket{\Gamma_n k}, $ i.e., it transforms the $\ell^\text{th}$ partner into the $k^\text{th}$ partner of the same [[irreducible representation]]. We normally use them to "project out" basis functions for a given partner of a given irreducible representation from an arbitrary function. This allows us to find the basis functions of an irreducible representation if they are not listed in the [[character table]]. ## Explicit Expression Let's start by writing the projection operator as a linear combination of the symmetry operators (i.e., [[Group Element|elements]]) of the symmetry [[group]]: $ \hat{P}^{(\Gamma_n)}_{k\ell} = \sum_{g} A_{k\ell}(g) \hat{P}_g $ where $A_{k\ell}$ are the expansion coefficients to be determined. Plugging into the definition of the projection operator we get: $ \hat{P}^{(\Gamma_n)}_{k\ell} \ket{\Gamma_n \ell} \equiv \ket{\Gamma_{n} k} = \sum_{g} A_{k\ell}(g) \hat{P}_g \ket{\Gamma_n \ell} $ Take the inner product with $\bra{\Gamma_n k}$ $ \braket{\Gamma_{n} k|\Gamma_{n} k} = 1 = \sum_{g} A_{k\ell}(g) \underbrace{\bra{\Gamma_n k}\hat{P}_g \ket{\Gamma_n \ell}}_{D^{(\Gamma_n)}(g)_{k\ell}} $ We can rewrite this as: $ 1 = \sum_g A_{k\ell}(g) D^{(\Gamma_n)}(g)_{k\ell} $ Recall that, by the [[wonderful orthogonality theorem]], we have: $ \frac{\ell_n}{h}\sum_g D^{(\Gamma_n)}(g)_{k\ell}^* D^{(\Gamma_n)}(g)_{k\ell} = 1 $ where $\ell_n$ is the [[Dimension|dimensionality]] of the representation $\Gamma_n$ and $h$ is the [[Group Order|order]] of the symmetry group. So we identify $ A_{k\ell} = \frac{\ell_n}{h} D^{(\Gamma_n)}(g)_{k\ell}^* $ and we end up with an explicit expression for the projection operator $ \hat{P}^{(\Gamma_n)}_{k\ell} = \frac{\ell_n}{h} \sum_{g} D^{(\Gamma_n)}(g)_{k\ell}^* \hat{P}_g $ Thus, if we know *all* the matrix representations $D^{(\Gamma_n)}(g)$ and a *single* basis function, we can use the projection operator to find all the remaining basis functions: $ \ket{\Gamma_n k}= \frac{\ell_n}{h} \sum_{g} D^{(\Gamma_n)}(g)_{k\ell}^* \hat{P}_g \ket{\Gamma_n \ell} $ ## Effect on an Arbitrary Function Expand out an arbitrary function $F$ in the complete set of basis functions $\ket{\Gamma_{n'}j'}$: $ F = \sum_{\Gamma_{n'}} \sum_{j'} f_{j'}^{(\Gamma_{n'})} \ket{\Gamma_{n'}j'} $ Using the explicit expression for the projection operator with $\ell = k$ $ \hat{P}^{(\Gamma_n)}_{kk} = \frac{\ell_n}{h} \sum_{g} D^{(\Gamma_n)}(g)_{kk}^* \hat{P}_g $ we get $ \hat{P}^{(\Gamma_n)}_{kk}F = \frac{\ell_n}{h}\sum_{\Gamma_{n'}} \sum_{j'} \sum_{g} f_{j'}^{(\Gamma_{n'})} D^{(\Gamma_n)}(g)_{kk}^* \hat{P}_g\ket{\Gamma_{n'}j'} $ But recall that: $ \hat{P}_{g}\left|\Gamma_{n'}j'\right\rangle=\sum_{j} D^{\left(\Gamma_{n'}\right)}(R)_{j j'}\left|\Gamma_{n'} j\right\rangle $ so we have $ \hat{P}^{(\Gamma_n)}_{kk}F = \frac{\ell_n}{h}\sum_{\Gamma_{n'}} \sum_{j'} f_{j'}^{(\Gamma_{n'})}\sum_{j} \left[\sum_{g} D^{(\Gamma_n)}(g)_{kk}^* D^{\left(\Gamma_{n'}\right)}(R)_{j j'}\right]\left|\Gamma_{n'} j\right\rangle $ Using the [[wonderful orthogonality theorem]]: $ \sum_{g} D^{\left(\Gamma_{n^{\prime}}\right)}(g)_{j j^{\prime}} D^{\left(\Gamma_{n}\right)}(g)_{k k}^{*}=\frac{h}{\ell_{n}} \delta_{\Gamma_{n} \Gamma_{n}^{\prime}} \delta_{j k} \delta_{j^{\prime} k} $ we end up with $ \hat{P}_{kk}^{(\Gamma_n)} F = f_k^{(\Gamma_n)} \ket{\Gamma_n k} $ Note that the projection operator does not yield normalized basis functions One strategy to find basis functions is to start with an arbitrary function $F$ as shown above then use $\hat{P}_{kk}^{(\Gamma_n)}$ to project out all other partners $\ket{\Gamma_n \ell}$ [[Orthogonality|orthogonal]] to $\ket{\Gamma_n k}$ in the irrep $\Gamma_n$. Again, this requires knowing the matrix representations of all the symmetry operations $g$. ## Characters If we only know the [[Character of a Representation|characters]] of an irrep, then we can still work something out. Define $ \hat{P}^{(\Gamma_n)} \equiv \sum_k \hat{P}_{kk}^{(\Gamma_n)} = \frac{\ell_n}{h} \sum_g \chi^{(\Gamma_n)}(g)^* \hat{P}_g $ We then end up with: $ \hat{P}^{(\Gamma_n)} F = \sum_k f_k^{(\Gamma_n)} \ket{\Gamma_n k} $ i.e., a linear combination of all the partners. So this projects out a function that transforms as $\Gamma_n$ but not as a specific partner of $\Gamma_n$.