Basic of Group Theory

Definitions

[Definition] : a Group is a set $G$ : with a binary multiplication : $\circ : G\times G\rightarrow G$ defined. satisfying:

Associativity : $f\circ (g\circ h) = (f\circ g)\circ h$ . In the following $\circ$ will be omitted
Existence of identity : $\exists 1 \in G \Rightarrow(\forall g \in G \Rightarrow 1g=g1=g)$
Existence of inverse : $\forall g \in G \exists g^{-1}\in G \Rightarrow g^{-1}g = g g^{-1} =1$

Some concepts about Group are listed following:

Order of group : $|G|$ is the number of finite group elements
Order of an element : $|g|$ is the minimal integer $n$ s.t. $g^n=1$ , or $\infty$ if $n$ does not exist
Abelian group :if $\forall g,h\in G \Rightarrow gh=hg $
subgroup : $H\leq G$ : a subset $H\subset G$ which is also a group under multiplication $\circ$
left(right) coset : $gH(Hg) := \{gh(hg) : h\in H \}$
Normal subgroup : $H \unlhd G$ a subgroup satisfying $\forall g \in G \Rightarrow gH=Hg$
Quotient group : $G/H$ is the group of cosets for normal subgroup $H$ of $G$ : :

$$H\unlhd G \Rightarrow G/H := \{gH : g\in G \} \text{ with } a,b\in G/H \Rightarrow a\circ b :=g_a g_b H \in G/H$$

Conjugacy class : $f,h\in G$ are conjugate if $\exists g \in G \Rightarrow gfg^{-1}=h$
all elements conjugate to $f$ form the conjugacy class of $f$
Direct product of groups : $G\times H$ is the set of $(g,h)$ with $g\in G, h \in H$ , and $(g',h')\circ(g,h)=(g'g,h'h)\in G\times H$

Representation Theory

[Definition] : Linear Representation of group $G$ : on linear space $V$ is a map $R:G\rightarrow \text{GL}(V)$ :

$$\forall g,h\in G \Rightarrow R(g)R(h)=R(gh)$$

This property is usually called as homomorphism

[Note] : one can find that if $R$ is a representation, then $\bm{A}R\bm{A}^{-1}$ with $\bm{A}$ is a non-singular constant matrix. These representation called as equivalent to $R$

[Definition] : Adjoint Representation of Lie groups: the representation space is the Lie Algebra $\text{L}G$ . The representation is defined as:

$$g x g^{-1} = x R(g) \ \ \forall g\in G, x\in \text{L}G$$

Where $gxg^{-1} = -\ti \frac {\td} {\td t} (ge^{\ti t x}g^{-1})|_{t=0}$

[Definition] : Reducible representations : $\exists W \subset V \Rightarrow (\forall g \in G \Rightarrow R(g)V = V)$ . The contrary concept is Irreducible representations (irrep.)

Orthogonality Theorem

[Theorem] : for any two irreps of (finite) group $G$ : , one has:

$$\sum_{g\in G} R(g)^*_{ij}R'(g)_{i'j'} = \frac {|G|} {\text{dim}R} \delta_{ii'}\delta_{jj'}\delta_{R,R'}$$

Where $\delta_{R,R'}=1$ when $R,R'$ are equivalent, it is $0$ otherwise

[Note] : For Compact Lie group, with the definition of Haar measure $\td\mu(g)$ (invariant under action of group elements: $\td\mu(g'g)=\td \mu(g)$ )) , then one has:

$$\int \td \mu(g) R(g)^*_{ij}R'(g)_{i'j'} = \frac {\int \td \mu(g)} {\text{dim}R} \delta_{ii'}\delta_{jj'}\delta_{R,R'}$$

[Definition] : Character : $\chi_R(g) = \text{Tr}R(g)$
Equivalent representations have the same character , and conjugacy class have the same character.

[Theorem] :

$$\sum_{g\in G} \chi_R(g)^*\chi_{R'}(g) = |G|\delta_{R,R'}$$

Where $\delta_{R,R'}=1$ when $R,R'$ are equivalent, it is $0$ otherwise

[Theorem] : For finite group $G$ : , let $n=|G|$ , $m$ is the number of conjugacy classes , $r$ is the number of inequivalent irreps , $d_i$ is the dimemsions of irrep. $i$ , then:

$$\begin{aligned} m&=r \\ \sum_i d_i^2 &= n=|G| \\ n &= k_i d_i \ \ k_i\in \mathbb{N} \end{aligned}$$

Projection Operator

Given a possibly reducible representation $R$ of group $G$ : , and the characters of one irrep $\chi_{R'}$ , it is possible to build an irrep. $R'$ on the representation space of $R$ :

Denote the orthonormal basis of representation $R$ by $\ket{e_i} \ \ i=1,\cdots,\text{dim}R$ . The action of group element $g$ , on the basis is:

$$\hat g \ket{e_i} = \sum_j \ket{e_j}R(g)_{ji}$$

Build new basis

$$\ket{\tilde{e}_i} =\sum_{g\in G} \hat g\ket{e_i}\chi_{R'}^* (g) = \sum_{g\in G} \sum_j \ket{e_j}R(g)_{ji}\chi_{R'}^*(g) $$

These are usually `NOT` linearly independent and not orthonormal.

If $R$ contains irrep $R'$ , then $\ket {\tilde{e}_i}$ will span a finite dimensional space (not all $\ket{\tilde{e}_i})$ are zero , then the group $G$ : is represented on this subspace as (several copies of) the irrep $R'$

Conservatin Law & Degeneracy

Symmetry as Unitary Operator: 1-Particle Hilbert Space

Think of a symmetry group $G$ : acting on the coordinate space, and induces the transformation: $\bm{x}\rightarrow g\bm{x}$ , This shall induce unitary transformations let:

$$\ket{\bm{x}}\rightarrow \ket{g\bm{x}}$$

Then we can derive the transformation on arbitary single-particle state:

[Theorem] : For single-particle state $\ket{\psi}$ , when group $G$ : ’s element $g$ , acts on coordinate space, it will be transformed to:

$$(g\psi)(\bm{x}) = J^{-1}\psi(g^{-1}\bm{x})$$

Where $J =|\det \partial_i (g\bm{x})^j| $ is the Jacobian of the transformation.

Proof

for single particle state $\ket{\psi}$ , it shall be transformed as: $$\ket{g\psi} = \int \td\bm{x} \psi(\bm{x})\ket{g\bm{x}} = \int \psi(g^{-1}\bm{x}) \ket{\bm{x}} J^{-1}\td^3 \bm{x}$$ Then we have $$(g\psi)(\bm{x}) \equiv \langle \bm{x}\ket{g\psi} = J^{-1}\psi(g^{-1}\bm{x})$$

Usually, we assume $J=1$ so that we do not need to worry about normalization.
Then we can find that the operation satisfies the Associativity :

$$(h(g\psi))(\bm{x}) = ((hg)\psi)(\bm{x})$$

Moreover , one can define the unitary operator of the operation $g$ , with a orthonormal basis $\ket{e_i}$ :

$$\hat g = \sum_i \ket{ge_i} \bra{e_i}$$

Then one can find that $\ket{g\psi} = \hat g \ket{\psi}$
Then the matrix representation will be a unitary representation of $g$ , :

$$(\bm{U}(g))_{ij} = \bra{e_i} \hat g \ket{e_j} =\langle e_i \ket{ge_j}$$

It is easy to check that:

$$\bm{U}(gh) = \bm{U}(g)\bm{U}(h)$$

Symmetry as Unitary Operator: Fock Space

First of all , we announce an Implicit assumption:
The vacuum is invariant under symmetry. (this shall be carefull to treat!)

Symmetry operation $g$ , defined on single particle Hilbert space induces symmetry operation of Fock space. As the tensor product restricted in the (anti-)symmetrized many-body Hilbert space.
One can also consider the symmetry operations on the annihilation/creation operators:

[Theorem] : The creation operator $\hat \psi^{\dagger}$ transforms as $\ket{\psi}$ , while annihilation operator $\hat \psi$ transforms as $\bra{\psi}$
With this one can find that the particle number will be invariant under symmetry operation.
For fermion, on can find that:

$$\prod_i \hat{ge_i}^{\dagger} = \det g \prod_i \hat e_i^{\dagger}$$

Symmetry as Unitary Operator: Action on Operators

It is similar to the Heisenberg picture. The action of symmetry operation on state can be transfered to the action of operators:

$$\hat O \rightarrow_g \hat {gO}$$

There are two convention usually used:

Like the Heisenberg picture, let the symmetry operation only acts on operator:

$$\bra {g\psi} \hat O \ket{g\psi} = \bra{\psi} \hat {gO} \ket{\psi} \Rightarrow \hat{gO} = \hat g^{-1} \hat O \hat g$$

Demand the matrix element to be invariant under symmetry operation:

$$\bra{g\psi}\hat {gO} \ket{g\psi} = \bra{\psi}\hat O\ket{\psi} \Rightarrow \hat {gO} = \hat g \hat O \hat g^{-1}$$

The set of linear operators $\hat O_i$ can also form a linear representation of the group:

$$\hat {gO_i} = \sum_j \hat {O_j} R(g)_{ji}$$

Symmetry Generators as Conserved Observables

[Theorem] : The Hamiltonian is invariant under the action of $g$ , , which means that:

$$\bra{g\psi} \hat H \ket{g\phi} = \bra{\psi} \hat H \ket{\phi}$$

for all states $\psi,\phi$ , then for the unitary symmetry $\hat g$ , one have:

$$[\hat H ,\hat g]=0$$

In quantum mechanics, the generators of continuous unitary symmetry corresponds to conserved observables .
For generator $X$ ,, and unitary symmetry $\Phi : g \rightarrow \hat g \equiv \hat \Phi(g)$ , the corresponding observables is:

$$\hat X = -\ti \hat \Phi(e^{\ti \theta X})^{-1} \frac {\td} {\td \theta} \hat \Phi(e^{\ti \theta X})$$

which is independent of $\theta$ value.

[Theorem] : When $\hat \Phi(e^{\ti \theta X})$ is a symmetry unitary operator, which means that $[\hat \Phi(e^{\ti\theta X}),\hat H]=0$ . Then $\hat X$ is an Hermitian, and it is commutative with $\hat H$ , which means that it is a conserved observable.

Proof

Using $\hat \Phi(e^{\ti\theta X})^{-1} \hat H \hat \Phi(e^{\ti\theta X}) = \hat H$ , take $\frac {\td} {\td \theta}$ at $\theta=0$ , and notice that $$\hat X = -\ti \frac {\td} {\td \theta} \hat \Phi(e^{\ti\theta X}) \Big|_{\theta=0} = \ti \frac {\td} {\td \theta} \hat \Phi(e^{\ti \theta X})^{-1} \Big|_{\theta=0}$$ one have: $$\ti \hat X \hat H -\ti \hat H \hat X =0 \Rightarrow [\hat H ,\hat X]=0$$

Symmetry & Level Degeneracy

A symmetry $g$ , satisfies $[\hat H , \hat g]=0$ , which means that $g$ , does not change energy eigenvalues.
Degenerate energy eigenstates form a representation space of the symmetry group. Representation of $g$ , is block diagonalized in energy eigenbasis.
Specially, nondegenerate energy eigenstates are 1-dimension representations.

[Theorem] : The existence of non-commuting symmetry generators:

$$\hat X, \hat Y \Rightarrow ([\hat H,\hat X]=[\hat H,\hat Y]=0 ) \text{ and } ([\hat X,\hat Y]\neq 0)$$

usually implies degeneracy of energy levels.

[Note] :

If $[\hat X,\hat Y]=\ti z$ is a non-zero c-number, there must be degeneracy. For unitary operator $e^{\ti \hat X}$ changes eigenvalue of $\hat Y$ by $z$ (Landau level)
If $[\hat X,\hat Y]$ is not a c-number, there may be a non-degeneracy energy level. IF the state is vanished by commutators of all order:

$$0=[\hat X,\hat Y]\ket{E} = [\hat X,[\hat X,\hat Y]]\ket{E}=[\hat Y,[\hat X,\hat Y]]\ket{E}=\cdots$$

(ground state pf electron in hydrogen atom(ignore spin) with angular momentum   $L=0$   , take   $\hat X =\hat L_x, \hat Y=\hat L_y$ ))

Examples of Symmetry

Translation

Continuous translation in 1D open space: $T_a : x\rightarrow x+1$ for all $a\in\mathbb{R}$ form a group: $T_aT_{a'}=T_{a+a'}$ ,, which is isomorphic to $\mathbb{R}$
The related unitary operator is:

$$\hat T(a) = \int \ket{x+a}\bra{x}\td x$$

Or use momentum basis: $\hat T(a) = \exp^{-\ti a \hat p}$ .
This shows that the generator of translation is $\hat p$ .
The momentum eigenstates $\ket{p}$ are 1D irreps, for $\hat T(a)\ket{p} = e^{-\ti a p}\ket{p}$

Discrete Translation

Discrete translation(lattice translation): $T: x\rightarrow x+a$ for a constant $a\neq 0$ .. The discrete translation group is the cyclic group generated by $T$ : $\{T^n\}$ for all $n\in \mathbb{Z}$

The unitary operator is still $\hat T = \exp(-\ti a \hat p)$
There is no associated Lie algebra and conserved observables(discrete group)
Unitary irreps are 1D , $\hat T = e^{\ti \theta}$ with real $\theta$ mod $2\pi$
Momentum eigenstates $\ket{p+\frac 2\pi n/a}$ for $n\in\mathbb{Z}$ belong to the same 1D irrep.
Bloch's Theorem : for system with the above translation symmetry, $[\hat H,\hat T]=0$ , the m-th energy eigenstate of irrep $\hat T = e^{-\ti a \hat p}$ will be a superposition of $\ket{p+nG}$ , $\ket{E_m,p}=\sum_n u_{mn}\ket{p+nG}$ Where $n\in\mathbb{Z}$ and $G=2\pi/a$ (the reciprocal lattice vector) And the Bloch function $u_m(x)=\sum_{n}u_{mn}e^{2\pi \ti n(x/a)}$ is periodic: $u_m(x+a)=u_m(x)$
The Bloch function : $u_m(x)$ (or Fourier coefficients $u_{mn}$ )) and the crystal momentum $p$ in first Brillouin zone : $-\pi/a\leq p \leq \pi/a$ in 1D case, define the eigenstates.

Selection Rule

Symmetry Constraints on Matrix Elements

In general we want to consider the matrix element: $(f_k)_{ij}=\bra{\phi_i}\hat f_k\ket{\psi_j}$ . Where $i(j)$ indicates that $\phi_i(\psi_j)$ is one of the degenerate energy levels of irrep. $R_{\phi}(R_{\psi})$ And $k$ means that $\hat f_k$ belongs to a set of operators forming irrep. $R_f$ . Then these matrix elements shall form a tensor product representation $R_f\otimes R_{\phi}^{*}\otimes R_{\psi}$

$\hat f_k$ form a representation, in the sense that $\hat g \hat f_k \hat g^{-1}=\sum_{k'}\hat f_{k'}R_f(g)_{k'k}$
$R_{\phi}^{*}$ is the conjugate representation of $R_{\phi}$ , $R_{\phi}^{*}(g)=[R_{\phi}(g)]^{*}$ it is easy to check it is also a representation. More over: $\bra{g\phi_i}=\bra{\phi_i}\hat g^{\dagger}=\sum_{i'}R_{\phi}^{*}(g)_{i'i}\bra{\phi_{i'}}$
$(f_k)_{ij}$ form a tensor represetntation in the sense that:

$$\begin{aligned} (f_k)_{ij}&=\bra{\phi_i}\hat f_k \ket{\psi_j} \equiv \bra{\phi_i}\hat g^{-1}\hat g \hat f_k \hat g^{-1}\hat g \ket{\psi_j} \\ &= \sum_{i',j',k'} R_{\phi}^{*}(g)_{i'i}R_f(g)_{k'k}R_{\psi}(g)_{j'j} \bra{\phi_{i'}} \hat f_{k'} \ket{\psi_{j'}} \\ &= \sum_{i',j',k'} (f_{k'})_{i'j'} R_f(g)_{k'k}R_{\phi}^{*}(g)_{i'i}R_{\psi}(g)_{j'j} \end{aligned}$$

Sum over $g$ , , by the orthogonality theorem, we have:

[Theorem(Selection Rule)] : the matrix element $(f_k)_{ij}$ will vanish if the tensor representation $R_f\otimes R_{\phi}^{*}\otimes R_{\psi}$ , after decomposed into direct sum of irreps, does NOT contain trivial representation.

Examples: Selection Rule

Parity selection rule Consider group $\{1,I\}$ generated by spatial inversion $I$ . It has only two irreps., trivial(even) one and odd representation: $\{1,-1\}$
- States & Operators are classified into parity odd (u) and parity even (g) classes
- Atomic orbitals of even(odd) angular momentum are parity even(odd)
- The matrix element is nonzero only for $\bra{g}\hat O_g\ket{g'}$ , $\bra{g}\hat O_u\ket{u}$ , $\bra{u}\hat O_g\ket{u'}$ , $\bra{u}\hat O_u\ket{g}$
- Main contribution to optical transitions (absorption/emission of one photon) is from $\bra{\phi}\bm{E}\cdot \bm{r}\ket{\psi}$ , the operator is parity odd. So initial and final states should have opposite parity
- Pseudo-vector : vectors that are even under inversion.
- Pseudo-scalar : scalars that are odd under inversion.
Raman selection rule absorb a photon of polarization $\bm{E}_{\text{in}}$ and emit a photon of polarization $\bm{E}_{\text{out}}$ , the relevant matrix element is $\bra{\text{final}}(\bm{E}_{\text{out}}\cdot \bm{r})(\bm{E}_{\text{in}\cdot \bm{r}})\ket{\text{initial}}$ .
- If the system have $C_{4v}$ symmetry, the initial state is of trivial representation $A_1$ , $\bm{E}_{\text{in}}$ , $\bm{E}_{\text{out}}$ are along $x,y$ directions respectively, then the final state must have the symmetry of function $xy$ , or $B_2$ representation.

Examples: Symmetry-allowed Hamiltonian

In many cases the symmetry of the system is known, but exact Hamiltonian is not. The goal is to write down a Hamiltonian consistent with the symmetry.

Then general rule: find out representation $\hat g$ of all symmetry generators, and demand that $\hat g^{-1}\hat H \hat g = \hat H$
For continuous symmetry, $\hat H$ should be commutative with all symmetry generators, or carry vanishing symmetry quantum number.

There are some examples:

Translation symmetry For each monomial of operators, the sum of momenta of the factors must be zero. That means that Hamiltonian can only contain those items like: $\hat \psi^{\dagger}(p)\hat \psi(p)$ , $\hat \psi(-p)\hat \psi(p)$ , $\hat \psi^{\dagger}(p_1)\hat \psi^{\dagger}(p_2)\hat \psi(p_3)\hat \psi(p_1+p_2-p_3)$
Point group symmetry Generally, free particle Hamiltonian can be written as:

$$\hat H = \int \hat {\bm{\psi}}^{\dagger}(\bm{k}) \bm{H}(\bm{k})\hat {\bm{\psi}}(\bm{k}) \td^3 \bm{k}$$

Where   $\hat{\bm{\psi}}(\bm{k})$   is a column vector of annihilation operators.

Under              $g$   ,          there is :

$$\begin{aligned} \hat{\bm{\psi}}(\bm{k}) &\rightarrow_g R(g) \hat{\bm{\psi}}(g\bm{k}) \\ \hat H &\rightarrow_g \int \hat {\bm{\psi}}^{\dagger}(g\bm{k})R(g)^{-1} \bm{H}(\bm{k})R(g)\hat {\bm{\psi}}(\bm{k}) \td^3 \bm{k} \end{aligned}$$

If              $g$   ,          is symmetry transformation, Hamiltonian should be invariant. Then   $H(g\bm{k})=R(g)^{-1}H(\bm{k})R(g)$ 

This will be one of the constraints of the Hamiltonian.

Leonard