Ze Chen
 Index Index

## Representation of Finite Groups

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#### Group Representation

##### Definitions
• Representation of a group: a homomorphism from a hroup $$G$$ to a group of operators $$U(G)$$ on a linear vector space $$V$$.
• Dimension: $$\dim V$$.
• Faithful: $$U$$ is a isomorphism.
• Degenerate: not faithful.
• Equivalent representation: two representations related by a similarity transformation, i.e. $U'(G) = S U(G) S^{-1}.$
• Characters of representation: $$\chi(g)$$ defined by $\chi(g) = \Tr U(g) = \sum_i D(g){^i}{_i}.$
• $$D(g)$$: the matrix of $$U(g)$$.
• Invariant subapce: $$V_1 \subset V$$ such that $U(g)V_1 \in V_1,\quad \forall g.$
• Minimal/Proper: $$V_1$$ doesn't contain any non-trivial invariant subspace.
• Irreducible representation: $$U(G)$$ that doesn't admit non-trivial invariant in $$V$$.
• Reducible: not irreducible.
• In matrix form: $D(g) = \begin{pmatrix} D_1(g) & * \\ 0 & D_2(g) \end{pmatrix}.$
• Fully reducible/Decomposible: (if $$V$$ is Hermitian) orthogonal complement of the invariant subspace also invariant.
• Unitary representation:
• $$V$$ is an Hermitian space, and
• $$U(g)$$ are unitary for all $$g$$.
• Direct sum representation: $$V = V_1 \oplus V_2$$ is a sum of two invariant subspaces.
• Notation:
• $$\abs{G}$$: order of $$G$$.
• $$\mu,\nu$$: labels for inequivalent IR of $$G$$.
• $$d_\mu$$: dimension of $$\mu$$-IR.
• $$D^\mu(g)$$: the matrix of $$g$$ in the $$\mu$$-IR, with respect to an orthonormal basis.
• $$\chi^\mu_i$$: character of class $$i$$ in the $$\mu$$-IR.
• $$\abs{C_i}$$: number of elements in the class $$\xi_i$$.
• $$n_C$$: number of conjugacy classes.
• $D{^i}{_j} = \bra{e^i}D\ket{e_j}.$
##### Theorems
• Reducible unitary representations are fully reducible.
• Let $$V$$ be the invariant subspace, and $\Pi^W = U^\dagger \Pi U.$ be the projection onto $$U^{-1} V$$. From $U\Pi^V = \Pi^V U \Pi^V.$ we find $\dim (V^\perp\cap W) = \tr ((1 - \Pi^V) \Pi^W) = 0.$
• Every representation on a Hermitian space is equivalent to a unitary representation.
• We redefine the inner product by $(x,y) = \sum_g \bra{D(g)x}\ket{D(g)y}$ and note that $$D(g)$$ is unitary under the new orthonormal basis.
• Reducible $$\Rightarrow$$ fully reducible.
• Schur's Lemma (I): if $$U(G)$$ is an IR and $AU(g) = U(g)A,\quad \forall A,$ then $$A = \lambda \mathbb{1}$$.
• We prove only for Hermitian $$A$$. It's easily seen that the eigenspace of $$A$$ is an invariant subspace of $$U(g)$$.
• Schur's Lemma (II): if $$U(G)$$ and $$U'(G)$$ are two IRs on $$V$$ and $$V'$$ respectively, and $A: V'\rightarrow V$ satisfies $AU'(g) = U(g)A,\quad \forall g\in G,$ then either
• $$A=0$$, or
• $$V$$ is isomorphic to $$V'$$ and $$U(G)$$ is equivalent to $$U'(G)$$.
• Proof:
• $$\operatorname{ker} A$$ is invariant under $$U'$$ and therefore $$A$$ is injective.
• $$\operatorname{im} A$$ is invariant under $$U$$ and therefore $$A$$ is surjective.
• IR of any abelian group is of dimension one.
• Orthonormality of IR: for unitary $$D_\mu$$, $\frac{d_\mu}{\abs{G}}\sum_g D^\dagger_\mu(g){^k}{_i}D^\nu(g){^j}{_l} = \delta^\nu_\mu \delta^j_i \delta^k_l.$
• $D^\dagger_\mu(g){^k}{_i} = \qty[D^\mu(g){^i}{_k}]^*.$
• Proof:
• Let $X_i^j = \sum_g D_\mu^\dagger(g) \ket{e_i}\bra{e^j} D^\nu(g).$
• We have $X_i^j D^\nu(g) = D_\mu^\dagger(g) X_i^j.$ Therefore, either $$X^j_i$$ is $$0$$ or $$D_\mu$$ is equivalent to $$D_\nu$$.
• If $$D_\mu$$ is equivalent to $$D_\nu$$, we have $X_i^j = \lambda \mathbb{1}.$
• Taking trace on both sides we get the desired result.
• The vectors $\ket{\mu,i,j} = \begin{pmatrix} D_\mu(g_1){^i}{_j} & \cdots & D_\mu(g_{\abs{G}}){^i}{_j} \end{pmatrix}{}$ are orthogonal. Each IR contribute $$d_\mu^2$$ such vectors.
• The vectors are complete: $\sum_\mu d_\mu^2 = \abs{G}.$
• The number of IRs equal to the number of conjugacy classes.
• Completeness of IR: $\sum_{\mu,l,k} \frac{d_\mu}{\abs{G}} D^\mu(g){^l}{_k} D^\dagger_\mu(g'){^k}{_l} = \delta_{gg'}.$
• Orthonormality of group characters: $\sum_i \frac{\abs{C_i}}{\abs{G}} \chi^{\dagger i}_\mu \chi^\nu_i = \delta^\nu_\mu.$
• $\chi^{\dagger i}_{\mu} = (\chi^\mu_i)^*.$
• Completeness of group characters: $\sum_\mu \frac{\abs{C_i}}{\abs{G}} \chi^\mu_i \chi^{\dagger j}_\mu = \delta^j_i.$
• Combining the completeness relation of IR with the following lemma:
• Let $$U^\mu(G)$$ be an IR. $\sum_{h\in C_i} U^\mu(h) = \frac{\abs{C_i}}{d_\mu} \chi^\mu_i \mathbb{1}.$
• LHS commutes with any $$U^h(g)$$.
• In the decomposition $U(G) = \bigoplus_{\mu} a_\mu U^\mu(G),$ we have $a_\nu = \sum_i \frac{\abs{C_i}}{\abs{G}} \chi^{\dagger i}_\nu \chi_i.$
• $\mathrm{IR} \quad \Leftrightarrow \quad \sum_i \abs{C_i} \abs{\chi_i}^2 = \abs{G}.$
##### Real, Complex, and Quaternionic Representations
• Real representation: $$\rho$$ equivalent to a real representation.
• In this case, we always have $$\rho \cong \overline{\rho}$$.
• Quaternionic representation: $$\rho \cong \overline{\rho}$$ but $$\rho$$ is not a real representation.
• Complex representation: $$\rho\ncong \overline{\rho}$$.
• Frobenius-Schur indicator: $\frac{1}{\abs{G}} \sum_{g\in G} \chi(g^2) = \begin{cases} 1, & \text{if } g \text{ is real}, \\ 0, & \text{if } g \text{ is complex}, \\ -1, & \text{if } g \text{ is quaternionic}. \end{cases}{}$

For the quaternion group $Q_8 = \langle \overline{e},i,j,k \vert \overline{e}^2 = e, i^2=j^2=k^2=ijk=\overline{e} \rangle$ that consists of eight elements $\qty{e,i,j,k,\overline{e},\overline{i},\overline{j},\overline{k}},$ the representation given by $\rho(i) = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix},\quad \rho(j) = w = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ is quternionic since $w\rho(g) w^{-1} = \overline{\rho}(g)$ for all $$g$$ but $$\rho$$ could not be sent to a real representation in that way.

#### Typical Representations

##### The Regular Representation
• $$V$$: the group ring.
• Basis: $$\qty{g_j}$$.
• $\mathrm{R}(g_i){^m}_{j} = \delta^{g_m}_{g_i g_j}.$
• $g_i g_j = g_m \Delta^m_{ij}.$
• $\chi^{\mathrm{R}}_i = \begin{cases} \abs{G}, & C_i = \qty{e}, \\ 0, & \text{otherwise}. \end{cases}$
• $\mathrm{R}(G) = \bigoplus_\mu d_\mu U^\mu(G).$
##### Direct Product Representations
• Direct product of vector spaces: $W = U \otimes V$ consists of $\set{\hat{\vb*{w}}_k | k = (i,j); i = 1,\cdots,d_U; j = 1,\cdots,d_V}.$
• $$\hat{\vb*{w}}_k$$ is the formal product $\hat{\vb*{w}}_k = \hat{\vb*{u}}_i \cdot \hat{\vb*{v}}_j.$
• $\bra{w^{k'}}\ket{w_{k}} = \delta^{k'}_{k} = \delta^{i'}_i \delta^{j'}_j.$
• $W = \qty{x^k \ket{w_k}}.$
• $\bra{x}\ket{y} = x^\dagger_k y^k, \quad x^\dagger_k = (x^k)^*.$
• Direct product of operator: $D = A \otimes B.$
• $D\ket{w_k} = \ket{w_{k'}} D^{k'}{_k}.$
• $D^{k'}{_k} = A^{i'}{_i} B^{j'}{_j}.$
• Direct product of representation: $D^{\mu\times \nu}(g) = D^{\mu}(g)\otimes D^{\nu}(g).$
• Character of direct product: $\chi^{\mu\times \nu} = \chi^\mu \chi^\nu.$
• Decomposition $D^{\mu\times \nu}(G) = \bigoplus_{\lambda} a_\lambda D^\lambda.$
• Invariant subspace: $W^\lambda_\alpha,$
• $$\lambda$$: which representation acts on this invariant subspace?
• $$\alpha$$: if there are multiple invariant subspace, distinguish them with this index.
• Basis: new orthonormal basis, $\ket{w^\lambda_{\alpha l}} = \sum_{ij} \ket{w_{i,j}}\langle i,j(\mu\nu)\alpha,\lambda,l \rangle.$
• $$(i,j)$$ as row index.
• $$(\alpha,\lambda,l)$$ as column index.
• The new basis is demanded to be orthornormal (as the old basis is). We have large freedom to mix the basis in each invariant subspace. Therefore, the new basis (and hence the CG coefficients) are not uniquely determined.
• The transformation is unitary.
• Action on old basis: $U(g)\ket{w_{i,j}} = \ket{w_{i',j'}} D^\mu(g)^{i'}{_i} D^\nu(g)^{j'}{_j}.$
• Action on new basis: $U(g)\ket{w^\lambda_{\alpha l}} = \ket{w^\lambda_{\alpha l'}} D^\lambda(g)^{l'}{_l}.$
• Clebsch-Gordan coefficients: $\langle i,j(\mu\nu)\alpha,\lambda,l \rangle.$
• Orthonormality of CG coefficients: $\sum_{\alpha,\lambda,l} \langle i',j'(\mu,\nu)\alpha,\lambda,l\rangle \langle \alpha,\lambda,l (\mu,\nu) i,j\rangle = \delta^{i'}_i \delta^{j'}_j.$
• $\langle \alpha,\lambda,l(\mu\nu) i,j \rangle = \langle i,j(\mu,\nu)\alpha,\lambda,l\rangle^*.$
• Completeness of CG coefficients: $\sum_{i,j} \langle \alpha',\lambda',l'(\mu,\nu) i,j\rangle \langle i,j(\mu,\nu)\alpha,\lambda,l\rangle = \delta^{\alpha'}_{\alpha}\delta^{\lambda'}_{\lambda}$
• Reduction of product representation: \begin{align*} D^\mu(g)^{i'}{_i} D^\nu(g)^{j'}{_j} &= \bra{i'j'}\ket{\alpha,\lambda,l'} D^\lambda(g)^{l'}{_l}\bra{\alpha,\lambda,l}\ket{i,j}, \\ \delta^{\alpha'}_{\alpha}\delta^{\lambda}_{\lambda'} D^\lambda(g)^{l'}{_l} &= \bra{\alpha',\lambda',l'}\ket{i',j'} D^\mu(g)^{i'}{_i} D^\nu(g)^{j'}{_j} \bra{i,j}\ket{\alpha,\lambda,l}. \end{align*}{}
##### Subduced Representations
• If $$T$$ is a representation of a group $$G$$, and
• $$G_1\subset G$$ is a subgroup, then
• $$T$$ subduces the representation $$T_1$$ of $$G_1$$ according to $\color{orange} T_1(g) = T(g),\quad \forall g\in G_1.$
• When this is done for all the IRs of $$G$$, we have a subduction table.
• Subduction of IRs may be reducible.
• Subduced representation denoted $\color{orange} \rho_{G_1} = \rho_G \downarrow G_1.$
##### Induced Representations

Let $$H$$ be a subgroup of $$G$$ and $$\rho: H \rightarrow \mathrm{GL}(V)$$ be a representation of $$H$$. The induced representation $$\mathrm{Ind}^G_H(\pi)$$ acts on $W = \bigoplus_{i=1}^n g_i V,$ where $$g_i \in G$$ are representative elements of the quotient $$G/H$$, by $g\cdot g_i v_i = g_{j(i)} h_i v_i.$ The induced representation is denoted $\color{orange} \rho_G = \rho\uparrow G.$

Let $G = D_{12} = \bra{\rho,\sigma}\ket{\rho^6 = \sigma^2 = 1, \rho\sigma = \sigma \rho^{-1}},$ and $H = \qty{1,\sigma,\rho^3, \sigma\rho^3}.$ A representation of $$H$$ is given by $\rho(\sigma) = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}, \quad \rho(\rho^3) = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \quad \rho(\sigma\rho^3) = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}.$ Note that $$[G:H] = 3$$ and therefore the induced representation consists of $$3\times 3$$ matrices of $$2\times 2$$ blocks. $\rho_G(\rho) = \begin{pmatrix} 0 & 0 & \rho(\rho^3) \\ \mathbb{1} & 0 & 0 \\ 0 & \mathbb{1} & 0 \end{pmatrix},\quad \rho_G(\sigma) = \begin{pmatrix} \rho(\sigma) & 0 & 0 \\ 0 & 0 & \rho(\sigma\rho^3) \\ 0 & \rho(\sigma\rho^3) & 0 \end{pmatrix}.$ We pick representative elements $g_1 = e,\quad g_2 = \rho,\quad g_3 = \rho^2.$ Let's figure out how $$g\in G$$ acts on $$\bigoplus g_i V_i$$, i.e. $g g_i v_i = g_? h_? v_i.$ That is to seek a $$g_?$$ such that $g_?^{-1} g_i v_i = h_? \in H.$ We take $$g = \rho$$ for example, and find that \begin{align*} g_2^{-1} g g_1 &= 1 \in H, \\ g_3^{-1} g g_2 &= 1 \in H, \\ g_1^{-1} g g_3 &= \rho^3 \in H. \end{align*} Therefore, we are now ready to answer the following questions:

1. How does $$g = \rho$$ act on $$g_1 V_1$$?
• $g_1\cdot v_1 \rightarrow g_2\cdot \mathbb{1} v_1.$
2. How does $$g = \rho$$ act on $$g_2 V_2$$?
• $g_2 \cdot v_2 \rightarrow g_3\cdot \mathbb{1} v_2.$
3. How does $$g = \rho$$ act on $$g_3 V_3$$?
• $g_3 \cdot v_3 \rightarrow g_1\cdot \rho(\rho^3) v_3.$

Now we obtain the $$\rho_G$$ above.

This could be written as $[\rho_G(g)]_{\alpha i,\beta j} = [\tilde{\rho}(g_\alpha^{-1} g g_\beta)]_{ij},$ where $[\tilde{\rho}(g)]_{ij} = \begin{cases} [\rho(g)]_{ij}, & \text{if } g\in H, \\ 0, & \text{else}. \end{cases}$

• The induced representation is always in block form, i.e. $$\rho(g)$$ is always a matrix of blocks.
• Big indices, labelling $$g_i\in G/H$$: $$g$$ sends $$g_i V$$ to which $$g_? V$$?
• Small indices, labelling $$v\in V$$: how $$h = g_i^{-1}g$$ acts on $$V$$?
##### Tensor Representations: Symmetry Dictates Selection Rule
• $$U(G)$$ be a representation not necessarily reducible.
• Acting on $V = \bigoplus_{\mu,\alpha} V^\alpha_\mu.$
• Irreducible tensors: a set of operators $\set{O^\mu_i | i = 1,\cdots,d_\mu}$ acting on $$V$$ and transforming as $U(g) O^\mu_i U(g)^{-1} = O^\mu_j D^\mu(g){^j}{_i}.$
• $$O^\mu_i \ket{e^\nu_j}$$ transforms like direct product: $U(g) O^\mu_i \ket{e^\nu_j} = O^\mu_k \ket{e^\nu_j} D^\mu(g){^k}{_i} D^\nu(g){^l}{_j}.$
• New basis $O^\mu_i \ket{e^\nu_j} = \sum_{\alpha,\lambda,l} \langle \alpha,\lambda,l(\mu,\nu)i,j \rangle.$
• Wigner-Eckart theorem: $\bra{e^l_\lambda} O^\mu_i \ket{e^\nu_j} = \sum_\alpha \langle \alpha,\lambda,l(\mu,\nu)i,j \rangle \bra{\lambda} O^\mu \ket{\nu}_\alpha.$
• Reduced matrix element: $\bra{\lambda} O^\mu \ket{\nu}_\alpha = \frac{1}{\abs{d_\lambda}} \sum_k \bra{e^k_\lambda}\ket{w^\lambda_{\alpha k}}.$

#### Obtaining Irreducible Representations

##### Irreducible Vectors
• $$U(G)$$ be a representation not necessarily reducible.
• Acting on $V = \bigoplus_{\mu,\alpha} V^\alpha_\mu.$
• Irreducible set: a set of vectors $\set{\hat{\vb*{e}}^\mu_i | i = 1,\cdots,d_\mu}$ transforming according to $U(g)\ket{e^\mu_i} = \ket{e^\mu_j} D^\mu(g){^j}{_i}.$
• If
• $$\qty{\hat{\vb*{u}}^\mu_i}$$ and $$\qty{\hat{\vb*{v}}^\nu_j}$$ are two sets of irreducible basis, and
• $$\mu$$ and $$\nu$$ labels different IR, then
• the two invariant subspaces spanned by these bases are orthogonal to each other.
• Proof: prove that $\bra{v^j_\nu}\ket{u^\mu_i} = \frac{1}{\abs{G}} \sum_g \bra{v^j_\nu} U^\dagger(g) U(g) \ket{u^\mu_i}$ vanished.

States of different $$l$$ (living in different spaces of the IRs of $$\mathrm{SO}(3)$$) doesn't overlap.

• Rotates one; kills the rest. $P^j_{\mu i} = \frac{d_\mu}{\abs{G}} \sum_g D^{-1}_\mu(g){^j}{_i} U(g).$
• For any $$\ket{x} \in V$$, the sets of vectors $\set{\ket{e^\mu_i} = P^j_{\mu i}\ket{x}, i = 1,\cdots,d_\mu}.$ transform irreducibly according to $$D^\mu$$ (as long as they are not all null). $U(g)P^j_{\mu i}\ket{x} = P^j_{\mu k}\ket{x} D^\mu(g){^k}{_i}.$
• $$P^j_{\mu i}$$ kills and rotates in the invariant subspace. $P^j_{\mu i} \ket{e^\nu_{\alpha k}} = \ket{e^\nu_{\alpha i}} \delta^\nu_\mu \delta^j_k.$
• Consecutive kill-and-rotate: $P^j_{\mu i} P^l_{\nu k} = \delta^\nu_\mu \delta^j_k P^l_{\mu i}.$
• Inversion $U(g) = \sum_{\mu,i,j} P^{j}_{\mu i} D^\mu(g){^i}{_j}.$
• $U(g) P^l_{\nu k} = \sum_i P^l_{\nu i} D^\nu(g){^i}{_k}.$
• Projection operators:
• Onto the basis vector $$\hat{\vb*{e}}^\mu_i$$ $P_{\mu i} = P^{i}_{\mu i}.$
• Onto the irreducible invariant space $$V_\mu$$ spanned by $$\qty{\hat{\vb*{e}}^\mu_i}$$ $P_\mu = \sum P_{\mu i}.$
• Completeness: $\sum_\mu P_\mu = \mathbb{1}.$

The IRs of one-dimensional discrete translation group are given by $$\qty{e^{ikna}}$$. From a local (Wannier) state $$\phi$$ we generate a one-dimensional invariant (Bloch) basis $\ket{\psi_k} = \sum_n T_n\ket{\phi} e^{ikna}.$

#### Representations of the Permutation Groups

##### One-Dimensional Representations
• Trivial representation: $$E$$.
• Generated by the symmetrizer $$s$$, essentially idempotent and primitive.
• Parity representation: $D(p) = (-1)^{\operatorname{sign} p}.$
• Generated by the anti-symmetrizer $$a$$, essentially idempotent and primitive.
• These are the only one-dimensional representations.
##### Partitions and Young Diagrams
• Partition:
• Partition $\lambda = \qty{\lambda_1,\cdots,\lambda_r}$ of $$n$$:
• $\lambda_i \ge \lambda_{i+1}.$
• $\sum \lambda_i = n.$
• Partition $$\lambda = \mu$$ if $\lambda_i = \mu_i,\quad \forall i.$
• $$\lambda>\mu$$ or $$\lambda<\mu$$ by lexicographic order.
• Parition to Young diagram:
• $4 = 2 + 1 + 1$ represented as follows.
• $4 = 3 + 1$ as follows.
• Partitions of $$n$$ $$\leftrightarrow$$ conjugacy classes of $$S_n$$ $$\leftrightarrow$$ IRs of $$S_n$$.
• Young tableau, normal tableau, and standard tableau:
• A tableau is obtained by filling the squares of a Young diagram with $$1,\cdots,n$$.
• Normal Young tableau: $$1,\cdots,n$$ appear in order from left to right and from the top row to the bottom row.
• Denoted by $$\Theta_\lambda$$.
• Examples as follows.
 1 2 3 4
and
 1 2 3 4
• An arbitrary tableau obtained from $$\Theta_\lambda$$ by permutation $\Theta^p_\lambda = p\Theta_\lambda.$
• Standard Young tableau: number of each row appear increasing to the right, and in each column appear to the bottom.
##### Symmetrizers and Anti-symmetrizers of Young Tableaux

Let $$\Theta_2$$ be given as follows.

 1 2 3

The irreducible symmetrizer is given by $e_2 = \frac{[e+(12)][e-(31)]}{4}.$ Let $r_2 = (23) e_2.$ Then $$\qty{e_2,r_2}$$ span an invariant subspace of the group ring, giving rise to the two-dimensional IR of $$S_3$$.

##### Symmetry Classes of Tensors
• Horizontal and vertical permutations: $$\Theta^p_\lambda$$ given,
• Horizontal permutations: $$\qty{h^p_\lambda}$$ leave invariant the sets of numbers appearing in the same row of $$\Theta^p_\lambda$$.
• $$\qty{h_\lambda}$$ form a subgroup of $$S_n$$.
• Vertical permutations: $$\qty{v^p_\lambda}$$ leave invariant the sets of numbers appearing in the same column of $$\Theta^p_\lambda$$.
• $$\qty{v_\lambda}$$ form a subgroup of $$S_n$$.
• Symmetrizer, anti-symmetrizer, and irreducible/Young symmetrizer: associated with the Young tableau $$\Theta^p_\lambda$$,
• Symmetrizer: summation over horizontal permutations, $s_\lambda^p = \sum_{h} h^p_\lambda.$
• Idempotent.
• Anti-symmetrizer: summation over vertical permutations, $a^p_\lambda = \sum_v (-1)^{v_\lambda} v_\lambda^p.$
• Idempotent.
• Irreducible/Young symmetrizer: sum over horinzontal and vertical permutations: $e^p_\lambda = \sum_v (-1)^{v_\lambda} h^p_\lambda v^p_\lambda.$
• Primitive idempotent.
• Properties of symmetrizers: superscript $$p$$ omitted.
• For $$\lambda$$ in the group ring $$\mathbb{C}S_n$$, these exists $$\xi\in\mathbb{C}$$ such that $s_\lambda r a_\lambda = \xi e_\lambda.$
• There exists $$\xi \in \mathbb{\eta}$$ such that $e_\lambda^2 = \eta e_\lambda.$
• $$e^p_\lambda$$ associated with $$\Theta_\lambda^p$$ is a primitive idempotent. It generates an IR of $$S_n$$ on the group ring.
• The representations generated by $$e^p_\lambda$$ and $$e_\lambda$$ are equivalent.
• Two irreducible symmetrizers $$e_\lambda$$ and $$e_\mu$$ generate inequivalent IR if $$\lambda \neq \mu$$.
• If $$\lambda \neq \mu$$, then $e^p_\mu e^q_\lambda=0$ for all $$p,q\in S_n$$.
• The left ideals generated by irreducible symmetrizers $$\qty{e_\lambda}$$ associated with normal Young tableaux $$\qty{\Theta_\lambda}$$ are all IRs of $$S_n$$.
• The direct sum (yes they are lineraly-independent) of the left ideals generated by all standard tableaux is the group ring $$\mathbb{C}S_n$$.
##### Irreducible Representations of the Permutation Groups
• $$\qty{i}$$ denotes a collection of indices. We work on the space $$V^n_m$$, i.e. $$n$$ copies of a $$m$$-dimensional space $$V$$.

• Direct product of operator acting on direct product of spaces: $g\ket{\qty{i}} = \ket{\qty{j}} D(g)^{\qty{j}}{_{\qty{i}}}.$

• $(gx)^{\qty{j}} = D(g)^{\qty{j}}{_{\qty{i}}} x^{\qty{i}}.$
• $(Dg)^{\qty{j}}{_{\qty{i}}} = g^{{j_1}}{_{i_1}} \cdots g^{{j_n}}{_{i_n}}.$
• Action of permutation: $p\ket{\qty{i}} = \ket{p^{-1}\qty{i}}.$

• $(px)^{\qty{i}} = x^{p\qty{i}}.$
• $D(p)^{\qty{j}}{_{\qty{i}}} = \delta^{p\qty{j}}{_{\qty{i}}}.$
• The two set of matrices $$\qty{D(p)}$$ of where $$p\in S_n$$ and $$\qty{D(g)}$$ where $$g\in \mathrm{GL}(m)$$ commute with each other.

• Tensors of symmetry class $$\lambda$$: $T_\lambda(\alpha) = \set{re_\lambda \ket{\alpha} | r\in \mathbb{C}S_n},$

• $$\lambda$$ the Young diagram given.
• $$T_\lambda$$ is an irreducible invariant subspace with respect to $$S_n$$.
• If $$T_\lambda(\alpha)$$ is not empty, then the realization of $$S_n$$ on $$T_\lambda(\alpha)$$ coincides with the IR generated by $$e_\lambda$$.
• $T'_\lambda = \set{e_\lambda\ket{\alpha} | \ket{\alpha} \in V^n_m}.$
• With Young symmetrizers, tensors may be completely decomposed into irreducible tensors $\ket{\lambda, \alpha, a},$ where
• $$\lambda$$: the symmetry class (Young diagram),
• $$\alpha$$: distinct set of tensors $$T_\lambda(\alpha)$$ invariant under $$S_n$$,
• $$a$$: basis element whtin each $$T_\lambda(\alpha)$$.
• Indices of basis chosen in a way such that $p\ket{\lambda,\alpha,a} = \ket{\lambda,\alpha,b} D_\lambda(p){^{b}}{_{a}}.$
• Tensors of symmetry $$\Theta^p_\lambda$$: $\set{e^p_\lambda\ket{\alpha}}.$

• $$\Theta^p_\lambda$$ the Young tableau given.
• Two tensor subspaces irreducibly invariant with respect to $$S_n$$ and belonging to the same symmetry class either overlap completely or are disjoint.

• Two irreducible invariant tensor subspaces corresponding to two distinct symmetry classes are necessarily disjoint.

• If $$g\in \mathrm{GL}_m$$ and $$\qty{\ket{\lambda,\alpha,a}}$$ is a set of basis tensors, then the subspaces $$T'\lambda(a)$$ spanned by $$\qty{\ket{\lambda,\alpha,a}}$$ with fixed $$\lambda$$ and $$a$$ are invariant under $$\mathrm{GL}_m$$. The representations of $$\mathrm{GL}_m$$ on $$T'_\lambda(a)$$ are independent of $$a$$. $g\ket{\lambda,\alpha,a} = \ket{\lambda,\beta,a} D_\lambda(g){^\beta}{_\alpha}.$

• The representations of $$\mathrm{GL}_m$$ on the subspace $$T'_\lambda(a)$$ of $$V^n_m$$ are irreducible.

Let $$\Theta_{\lambda=s}$$ be the Young tableau associated with the total symmetrizer, then $$T_\lambda(\alpha)$$ is a totally symmetric tensor, $$S_n$$ acts on which irreducibly (and trivially).

There is one and only one independent totally anti-symmetric tensor of rank $$n$$ in $$n$$-dimensional space.

For second-rank tensors on $$m$$-dimensional $$V$$, we have $e_s\ket{ij} = \frac{\ket{ij} + \ket{ji}}{2},$ while $e_a\ket{ij} = \frac{\ket{ij} - \ket{ji}}{2}.$

Let $$\Theta_m$$ be given as follows.

 1 2 3

The irreducible symmetrizer is given by $e_m = \frac{[e+(12)][e-(31)]}{4}.$ With \begin{align*} \ket{\alpha} &= \ket{++-}, \\ \ket{\beta} &= \ket{--+}. \end{align*} we obtain \begin{align*} e_m\ket{\alpha} = \ket{m,\alpha,1} &= \frac{2\ket{++-} - \ket{-++} - \ket{+-+}}{4}, \\ (23) e_m\ket{\alpha} = \ket{m,\alpha,2} &= \frac{2\ket{+-+} - \ket{-++} - \ket{++-}}{4}; \\ e_m\ket{\beta} = \ket{m,\beta,1} &= \frac{2\ket{--+} - \ket{+--} - \ket{-+-}}{4}, \\ (23) e_m\ket{\beta} = \ket{m,\beta,2} &= \frac{2\ket{-+-} - \ket{+--} - \ket{--+}}{4}. \end{align*}{} The action of $$S_3$$ on the subspace generated by these two vectors gives rise to its two-dimensional representation.

#### Application: Degeneracy in Central Potential

An Example of Symmetry Group Acting on Wavefunctions

• The states $\set{\ket{nl,m}|m = -l,\cdots,l}$ could be related by operators in $$\mathrm{O(3)}$$.

#### Application: Crystal Field Theory

An Example of Compatibility Relations

• My free atom has five $$d$$-orbitals. Without an external potential, the atom enjoys the full $$\mathrm{O}(3)$$ symmetry. Therefore, its five $$d$$-orbitals are degenerate since $$\mathrm{O}(3)$$ connects them.
• Now my atom is placed in a crystal of $$\mathrm{T_d}$$ symmetry. The symmetries that $$d$$-orbitals enjoy are reduced. Therefore, they are not guaranteed to be degenerate since two orbitals may not be be connected by an operation in $$\mathrm{T_d}$$.
• In the worst case, when there is no symmetry present, we no longer have any degeneracy.
• The case has not come to the worst. We still have a non-trivial symmetry group and we wanna know if there could be any degenracy. We note the following one-to-one correspondence: $\text{degeneracy} \Leftrightarrow \text{invariant subspace} \Leftrightarrow \text{IR}.$
• We take the following procedures to determine the possible degeneracies:
1. Determine how $$\mathrm{T_d}$$ acts on our $$d$$-orbitals. Find out the character. From this we get a representation of $$\mathrm{T_d}$$.
2. With the character of IRs of $$\mathrm{T_d}$$, we could easily decompose the above representation into IRs and know the invariant subspaces, hence the degeneracy.
• We know how $$\mathrm{O}(3)$$ acts on $$d$$-orbitals. The action of $$\mathrm{T_d}$$ is obtained directly by subduction, and the characters are obtained just as if we are treating $$\mathrm{O}(3)$$ operations.
• The only question is, to each $$l$$ we have two IRs of $$\mathrm{O}(3)$$ of dimension $$2l+1$$. Which one should we use?
• Since $$d$$-orbitals are invariant under parity, we should therefore use the IR where parity has positive character.
• Character table of $$\mathrm{T_d}$$: where $$6S_4$$ denotes the rotoreflections by $$90^\circ$$ and $$6\sigma_{\mathrm{d}}$$ the reflections.
$$T_d$$ $$E$$ $$8C_3$$ $$3C'_2$$ $$6S_4$$ $$6\sigma_{\mathrm{d}}$$
$$A_1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$
$$A_2$$ $$1$$ $$1$$ $$1$$ $$-1$$ $$-1$$
$$E$$ $$2$$ $$-1$$ $$2$$ $$0$$ $$0$$
$$T_2$$ $$3$$ $$0$$ $$-1$$ $$1$$ $$-1$$
$$T_1$$ $$3$$ $$0$$ $$-1$$ $$-1$$ $$1$$
• We determine the characters of $$\mathrm{O}(3)$$ on the above conjugacy classes with ($$j=2$$) $\chi = \frac{\sin (j+1/2)\psi}{\sin (\psi/2)}.$
• Character table of $$\mathrm{O}(3)$$.
$$\mathrm{O}(3)$$ $$E$$ $$8C_3$$ $$3C'_2$$ $$6S_4$$ $$6\sigma_{\mathrm{d}}$$
$$D_2^+$$ $$5$$ $$-1$$ $$1$$ $$-1$$ $$1$$
• The decomposition is given by $D_2^+ = E\oplus T_1.$
• The five $$d$$-orbitals splits into a two-fold level and a three-fold level.
###### Wrap Up: Compatibility Relations
• Originally, the atom has a higher symmetry $$\mathrm{O}(3)$$.
• As we move the atom into a crystal, the symmetry is reduced and therefore the originally irreducible representation ($$d$$-orbitals) $$D_2^+$$ becomes reducible.
• The IR $$D_2^+$$ is decomposed into a direct sum. The possible decompositions are those decompositions into IRs of subgroups of $$\mathrm{O}_3$$. The decomposition will be different if the reduced symmetry is given by $$\mathrm{O}_h$$. These possible decompositions are compatibility relations.
• Compatibility relations $$\Leftrightarrow$$ patterns of crystal field splitting.

#### Application: Band Representations

##### Point Group Action on Bloch Functions
• The actions that carry a point $$\vb*{k}$$ in the reciprocal space to its equivalent $\vb*{k}' = \vb*{k} + \vb*{K}$ form the little group at point $$\vb*{k}$$.
• If $$g$$ is an element of the point group, then $g \psi_{\vb*{k}}(\vb*{r}) = \psi_{\vb*{k}}(g^{-1}\vb*{r}) = \psi_{g\vb*{k}}(\vb*{r}).$
• If, in particular, $$h$$ is an element of the little group of $$\vb*{k}$$, then the set of wavefunctions $\set{h \psi_{\vb*{k}}(\vb*{r})}$ are degenerate, just like what we have in a central potential.
##### Compatibility Relations: Moving Away From Points of High Symmetry
• The symmetry group of points on the line $$XY$$ joining two points of high symmetry $$X$$ and $$Y$$ is a subgroup of the symmetry group thereof. Therefore, the representations decomposes along the line into direct sums. The decomposition should satisfy the compatibility relations, just like what we have in crystal field splitting.

#### Application: Stark Effect

• We ignore the spin-orbit interaction.
• When $$E=0$$, the $$l$$-orbitals ($$2l+1$$-fold degenerate) transforms according to $$D^l$$.
• For $$E=E\hat{\vb*{z}}$$, the $$l$$-orbitals split, each of which transforms according to the $$m$$-th representation of $$\mathrm{SO}(2)$$. Therefore, the eigenkets are of the form $$\ket{m}$$.
• That's why we could directly apply the perturbation theory to $$\ket{lm}$$.
 Index Index

2021/4/27 18:56:43

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