Properties and Proofs for Reduced Basis Generation

Projection Operator

Based on the grand orthogonality theorem

gGDαβ(Γ)(g1)Dγδ(Γ)(g)=δΓ,ΓδαδδβγGdimΓ, \sum_{g \in G} D^{(\Gamma)}_{\alpha\beta}(g^{-1}) D^{(\Gamma')}_{\gamma\delta}(g) = \delta_{\Gamma, \Gamma'} \delta_{\alpha\delta} \delta_{\beta\gamma} \frac{|G|}{\dim\Gamma},

We define the following projection operator:

P^α(Γ):=dimΓGgGDαα(Γ)(g1)R^(g) \hat{P}_{\alpha}^{(\Gamma)} := \frac{\dim \Gamma}{|G|}\sum_{g\in G} D_{\alpha\alpha}^{(\Gamma)} (g^{-1}) \hat{R}(g)


Lemma | The operator P^αΓ\hat{P}_{\alpha}^{\Gamma} is Hermitian.

P^α=P^α(Γ) \hat{P}_{\alpha}^{\dagger} = \hat{P}_{\alpha}^{(\Gamma)}

Proof | Both R^\hat{R} and DD are unitary, and therefore

P^α(Γ)=dimΓGgDαα(Γ)(g)R^(g1)=P^α(Γ){\hat{P}_{\alpha}^{(\Gamma)}}^{\dagger} = \frac{\dim \Gamma}{|G|} \sum_{g} D_{\alpha\alpha}^{(\Gamma)}(g) \hat{R}(g^{-1}) = \hat{P}_{\alpha}^{(\Gamma)}

since gg is a dummy variable and thus can be replaced by g1g^{-1}. \square


Lemma | The operator P^α(Γ)\hat{P}_{\alpha}^{(\Gamma)} is a projector, satisfying

P^α(Γ)P^β(Γ)=δΓ,ΓδαβP^α(Γ) \hat{P}_{\alpha}^{(\Gamma)} \hat{P}_{\beta}^{(\Gamma')} = \delta_{\Gamma,\Gamma'} \delta_{\alpha\beta} \hat{P}_{\alpha}^{(\Gamma)}

Proof |

P^α(Γ)P^β(Γ)=dimΓdimΓG2g,hGDαα(Γ)(g1)Dββ(Γ)(h1)R^(gh) \hat{P}_{\alpha}^{(\Gamma)}\hat{P}_{\beta}^{(\Gamma')} = \frac{\dim \Gamma \dim \Gamma'}{|G|^2} \sum_{g,h\in G} D_{\alpha\alpha}^{(\Gamma)} (g^{-1}) D_{\beta\beta}^{(\Gamma')} (h^{-1}) \hat{R}(g \cdot h)

Let m=ghm=g \cdot h. (i.e., g1=hm1g^{-1} = h \cdot m^{-1})

P^α(Γ)P^β(Γ)=dimΓdimΓG2m,hDαα(Γ)(hm1)Dββ(Γ)(h1)R^(m)=γ[dimΓGhDαγ(Γ)(h)Dββ(Γ)(h1)][dimΓGmDγα(Γ)(m1)R^(m)]=γδΓ,Γδαβδαγ[dimΓGmDγα(Γ)(m1)R^(m)]=δΓ,Γδαβ[dimΓGmDαα(Γ)(m1)R^(m)]=δΓ,ΓδαβP^α(Γ).\begin{aligned} \hat{P}_{\alpha}^{(\Gamma)}\hat{P}_{\beta}^{(\Gamma')} &= \frac{\dim \Gamma \dim \Gamma'}{|G|^2} \sum_{m, h} D_{\alpha\alpha}^{(\Gamma)} (h \cdot m^{-1}) D_{\beta\beta}^{(\Gamma')} (h^{-1}) \hat{R}(m)\\ &= \sum_{\gamma} \left[ \frac{\dim \Gamma'}{|G|} \sum_{h} D_{\alpha\gamma}^{(\Gamma)} (h) D_{\beta\beta}^{(\Gamma')} (h^{-1}) \right] \left[ \frac{\dim \Gamma}{|G|} \sum_{m} D_{\gamma\alpha}^{(\Gamma)} (m^{-1}) \hat{R}(m) \right]\\ &= \sum_{\gamma} \delta_{\Gamma, \Gamma'} \delta_{\alpha\beta} \delta_{\alpha\gamma} \left[ \frac{\dim \Gamma}{|G|} \sum_{m} D_{\gamma\alpha}^{(\Gamma)} (m^{-1}) \hat{R}(m) \right] \\ &= \delta_{\Gamma, \Gamma'} \delta_{\alpha\beta} \left[ \frac{\dim \Gamma}{|G|} \sum_{m} D_{\alpha\alpha}^{(\Gamma)} (m^{-1}) \hat{R}(m) \right] \\ &= \delta_{\Gamma, \Gamma'} \delta_{\alpha\beta} \hat{P}_{\alpha}^{(\Gamma)}. \quad \square \end{aligned}

Basis Construction

We want to build a basis in the Γα\Gamma\alpha block using P^α(Γ)\hat{P}_{\alpha}^{(\Gamma)}, starting from a basis of the full Hilbert space.

Let ψ\vert\psi\rangle be a basis state in the full Hilbert space. Then we can define

ψ;Γα:=P^α(Γ)ψ \vert \psi; \Gamma \alpha \rangle := \hat{P}_{\alpha}^{(\Gamma)} \vert \psi \rangle

(We will write P^\hat{P} in place of P^α(Γ)\hat{P}_{\alpha}^{(\Gamma)} below for brevity.)

This state, however, is not normalized. Thus when ψ;Γα\vert \psi; \Gamma\alpha \rangle is not null we define the following normalized state,

ψ~;Γα:=1Nψ;Γαψ;Γα \vert \tilde{\psi}; \Gamma \alpha \rangle := \frac{1}{\mathcal{N}_{\psi;\Gamma\alpha}} \vert \psi; \Gamma \alpha \rangle

where Nψ;Γα\mathcal{N}_{\psi;\Gamma\alpha} is the norm of ψ;Γα\vert \psi; \Gamma \alpha \rangle:

Nψ;Γα2:=ψ;Γαψ;Γα, {\mathcal{N}_{\psi;\Gamma\alpha}}^2 := \langle \psi; \Gamma \alpha \vert \psi; \Gamma \alpha \rangle,

which is

Nψ;Γα=ψP^ψ=ψψ;Γα=ψψ~;Γα. \mathcal{N}_{\psi;\Gamma\alpha} = \sqrt{ \langle \psi \vert \hat{P} \vert \psi \rangle } = \sqrt{ \langle \psi \vert \psi ; \Gamma\alpha \rangle } = \langle \psi \vert \tilde{\psi} ; \Gamma\alpha \rangle.

Note that Nψ;Γα\mathcal{N}_{\psi;\Gamma\alpha} is smaller than 1, since it is a norm of a projection of a normalized state.

Orthogonality of the Basis

In order to find a complete orthonormal basis within a given symmetry sector, we need to ensure orthogonality between the basis states that we construct. In general, this requires storing all the generated states states, and checking that a newly generated state is orthogonal to the existing ones.

The process, however, can be optimized in some cases -- one-dimensional irreps, and special choices of higher-dimensional irreps.

Let us start with some definitions. Consider a basis state ψ\vert \psi \rangle from the original Hilbert space. Let us introduce the following notation for the transformed states

R^(g)ψ:=ψ;g \hat{R}(g) \vert \psi \rangle := \vert \psi ; g \rangle

Let NN be the the set of group elements that leaves ψ\vert \psi \rangle invariant,

nN    R^(n)ψ=ψ. n \in N \iff \hat{R}(n) \vert \psi \rangle = \vert \psi \rangle.

The set NN is a subgroup of GG:

R^(n1n2)=R^(n1)R^(n2)=1^. \hat{R}(n_1 n_2) = \hat{R}(n_1) \hat{R}(n_2) = \hat{1}.


R^(gn)ψ=R^(g)R^(n)ψ=R^(g)ψ, \hat{R}(g n) \vert \psi \rangle = \hat{R}(g) \hat{R}(n) \vert \psi \rangle = \hat{R}(g) \vert \psi \rangle,

two group elements related by an element in NN have the same representation. This allows us to define R^\hat{R} on ψ\vert \psi \rangle for every left coset gNgN:

ψ;gNψ;g. \vert \psi ; gN \rangle \equiv \vert \psi ; g \rangle.

Lemma | For all C1,C2G/HC_1, C_2 \in G/H,

C1=C2    ψ;C1=ψ;C2 C_1 = C_2 \iff \vert \psi; C_1 \rangle = \vert \psi ; C_2 \rangle

Proof |
(\Rightarrow) By definition.
(\Leftarrow) Assume that ψ;C1=ψ;C2\vert \psi ; C_1 \rangle = \vert \psi ; C_2 \rangle. Then for g1C1,g2C2g_1 \in C_1, g_2 \in C_2,

R^(g1)ψ=R^(g2)ψ, \hat{R}(g_1) \vert \psi \rangle = \hat{R}(g_2)\vert \psi \rangle,

and since R^(g1g2)=1^\hat{R}(g^{-1} g_2) = \hat{1}, g11g2Ng_1^{-1} g_2 \in N, and C1=C2C_1 = C_2. \square

Projected state

Now let us first consider a one-dimensional irrep D(Γ)D^{(\Gamma)} (which we denote DD for brevity). We will show that any pair of states constructed using P^α(Γ)\hat{P}_{\alpha}^{(\Gamma)} is mutually either orthogonal, or parallel (i.e. equal up to an overall phase). This can be expressed as

ϕP^P^ψ=P^ψ2 or 0. \left\vert \langle \phi \vert \hat{P} \hat{P} \vert \psi \rangle \right\vert = \left\vert \hat{P} \vert \psi \rangle \right\vert^2 \text{ or } 0.

We will prove this below.

Consider the following projected state:

P^ψ=dimΓGgD(g)R^(g)ψ \hat{P} \vert \psi \rangle = \frac{\dim \Gamma}{|G|} \sum_{g} D(g)^{*} \hat{R}(g) \vert \psi \rangle

We can insert R^(n1)\hat{R}(n^{-1}) for nNn \in N, since it does not change ψ\vert\psi\rangle:

P^ψ=dimΓGgD(g)R^(g)(1NnNR^(n1)ψ) \hat{P} \vert \psi \rangle = \frac{\dim \Gamma}{|G|} \sum_{g} D(g)^{*} \hat{R}(g) \left( \frac{1}{|N|} \sum_{n \in N} \hat{R}(n^{-1}) \vert \psi \rangle \right)

After relabeling gn1g n^{-1} by gg and factoring out the summation over NN, we get

P^ψ=dimΓGgD(g)R^(g)ψ(1NnD(n)) \hat{P} \vert \psi \rangle = \frac{\dim \Gamma}{|G|} \sum_{g} D(g)^{*} \hat{R}(g) \vert \psi \rangle \left( \frac{1}{|N|} \sum_{n} D^{*}(n) \right)

The summation over nn can only be one or zero: It is zero if any D(n)1D(n) \neq 1, and P^ψ\hat{P}\vert\psi\rangle is a null vector. This means that, given an irrep DD, any basis state (from the original Hilbert space) which is invariant under element nn whose representation D(n)1D(n) \neq 1 should be thrown away in constructing the reduced basis.

If, on the other hand, D(n)=1D(n)=1 for all nNn \in N, then any elements related by nn have the same representation:

D(gn)=D(g). D(g n) = D(g).

Again, this allows us to define DD for every coset:

D(gN)D(g). D(g N) \equiv D(g).

The state projection can then be written as a sum over cosets:

P^ψ=dimΓGgD(g)ψ;g=NdimΓGCG/ND(C)ψ;C. \hat{P} \vert \psi \rangle = \frac{\dim \Gamma}{|G|} \sum_{g} D(g)^{*} \vert \psi; g \rangle = \frac{|N| \dim \Gamma}{|G|} \sum_{C \in G/N} D(C)^{*} \vert \psi; C \rangle.

Now, the overlap between P^ϕ\hat{P} \vert \phi \rangle and P^ψ\hat{P} \vert \psi \rangle is

ϕP^P^ψ=ϕP^ψ=NdimΓGCG/ND(C)ϕψ;C. \langle \phi \vert \hat{P} \hat{P} \vert \psi \rangle = \langle \phi \vert \hat{P} \vert \psi \rangle = \frac{|N| \dim \Gamma}{|G|} \sum_{C \in G/N} D(C)^{*} \langle \phi \vert \psi; C \rangle.

If the inner product is zero for all CC (i.e., ϕ\vert\phi\rangle and ψ\vert\psi\rangle are not related by any symmetry operation), then P^ψ\hat{P}\vert\psi\rangle and P^ϕ\hat{P}\vert\phi\rangle are also orthogonal. It can be nonzero for at most one coset. Let us say that it is nonzero for a coset CG/NC' \in G/N:

ϕψ;C=eiθ \langle \phi \vert \psi; C' \rangle = e^{i \theta}

Then the overlap can be written as

ϕP^ψ=NdimΓGD(C)eiθ \langle \phi \vert \hat{P} \vert \psi \rangle = \frac{|N| \dim \Gamma}{|G|} D(C')^{*} e^{i \theta}

while the the norm of P^ψ\hat{P} \vert \psi \rangle is

ψP^P^ψ=ψP^ψ=NdimΓG. \langle \psi \vert \hat{P} \hat{P} \vert \psi \rangle = \langle \psi \vert \hat{P} \vert \psi \rangle = \frac{|N| \dim \Gamma}{|G|}.

Since D(C)=1|D(C')|=1, we find that ϕ\vert \phi \rangle and ψ\vert \psi \rangle are parallel:

ϕ=D(C)ψ. \vert \phi \rangle = D(C') \vert \psi \rangle.

The implication is the following. When constructing the reduced basis, we only need to check the orbit and see if a basis state exists in any of the orbits of the existing basis states. Also, this means that an original basis state has overlap with only one of the reduced basis state. This greatly reduces the computational complexity, and also the memory size required to store the information.

This nice property stems from the Abelian structure of the representation. Even if the group GG itself may not be an Abelian, the image of the representation R^(G)\hat{R}(G) acting on the orbit of ψ\vert\psi\rangle is. We will see below that this Abelian structure can be applied to higher dimensional (non-Abelian) irreps, depending on the choice of the form of the matrices of the irrep.

Extension to Higher Dimensional Irreps

Now how do we extend this to higher dimensional irreps?

Of course because of non-Abelian nature of the group and its representation, the optimization cannot be applied in general. It does, however, work when the irreps satisfy a special property.

The projection operator for higher dimensional irrep is written as

P^α(Γ):=dimΓGgDαα(Γ)(g)R^(g) \hat{P}_{\alpha}^{(\Gamma)} := \frac{\dim \Gamma}{|G|}\sum_{g} {D_{\alpha\alpha}^{(\Gamma)}}^{*}(g) \hat{R}(g)

Now let us assume that all the elements of all the representation matrices are either 0 or unit length.

Dαβ(Γ)(g)=0 or 1. \left\vert D_{\alpha\beta}^{(\Gamma)}(g) \right\vert = 0 \text{ or } 1.

DD is unitary,

βDαβ(Γ)(g)Dγβ(Γ)(g)=δαγ, \sum_{\beta} D_{\alpha\beta}^{(\Gamma)}(g) D_{\gamma\beta}^{(\Gamma)}(g)^{*} = \delta_{\alpha\gamma},

and thus for α=γ\alpha=\gamma

βDαβ(Γ)(g)2=1, \sum_{\beta} \left\vert D_{\alpha\beta}^{(\Gamma)}(g) \right\vert^2 = 1,

implying that only one element per row(column) is nonzero.

This means that, under the symmetry operation, one basis state transforms into another basis state, not to a superposition many. This greatly simplifies the process for the construction of the basis.

Remember that

P^α(Γ):=dimΓGgGDαα(Γ)(g)R^(g) \hat{P}_{\alpha}^{(\Gamma)} := \frac{\dim \Gamma}{|G|}\sum_{g \in G} {D_{\alpha\alpha}^{(\Gamma)}}^{*}(g) \hat{R}(g)

where Dαα(Γ)D_{\alpha\alpha}^{(\Gamma)} has length one or zero. Let GG' be the set of elements where Dαα(g)0D_{\alpha\alpha}(g) \neq 0. Then the projection operator can be written as a sum only over GG':

P^α(Γ)=dimΓGgGDαα(Γ)(g)R^(g) \hat{P}_{\alpha}^{(\Gamma)} = \frac{\dim \Gamma}{|G|}\sum_{g\in G'} {D_{\alpha\alpha}^{(\Gamma)}}^{*}(g) \hat{R}(g)

Note that GG' is a subgroup of GG, and Dαα(Γ)D_{\alpha\alpha}^{(\Gamma)} is a one-dimensional irrep of GG'. Therefore, the basis generation algorithm for one-dimensional irrep can be applied (up to normalization).

Hamiltonian Elements

Consider two states in the reduced basis, ϕ~;Γα,ψ~Γα\vert \tilde{\phi}; \Gamma \alpha \rangle, \vert \tilde{\psi} \Gamma \alpha \rangle. The matrix element of the Hamiltonian is

ϕ~;ΓαH^ψ~Γα=(1Nϕ;ΓαϕP^)H^(1Nψ;ΓαP^ψ). \langle \tilde{\phi}; \Gamma \alpha \vert \hat{H} \vert \tilde{\psi} \Gamma \alpha \rangle = \left( \frac{1}{\mathcal{N}_{\phi;\Gamma\alpha}} \langle \phi \vert \hat{P} \right) \hat{H} \left( \frac{1}{\mathcal{N}_{\psi;\Gamma\alpha}} \hat{P} \vert \psi \rangle \right).

Since H^\hat{H} and P^\hat{P} commute,

ϕ~;ΓαH^ψ~;Γα=1Nϕ;ΓαϕH^(P^1Nψ;Γαψ), \langle \tilde{\phi} ; \Gamma \alpha \vert \hat{H} \vert \tilde{\psi} ; \Gamma \alpha \rangle = \frac{1}{\mathcal{N}_{\phi;\Gamma\alpha}} \langle \phi \vert \hat{H} \left( \hat{P} \frac{1}{\mathcal{N}_{\psi;\Gamma\alpha}} \vert \psi \rangle \right),

and thus

ϕ~;ΓαH^ψ~;Γα=ϕH^ψ~;Γαϕϕ~;Γα. \langle \tilde{\phi}; \Gamma\alpha \vert \hat{H} \vert \tilde{\psi} ; \Gamma \alpha \rangle = \frac{ \langle \phi\vert \hat{H} \vert \tilde{\psi}; \Gamma\alpha \rangle } { \langle \phi \vert \tilde{\phi} ; \Gamma \alpha \rangle}.

Therefore, when the symmetrized basis state is written in terms of the states in the original basis state BB as

ψ~;Γα=ηBUψ~,ηη, \vert \tilde{\psi} ; \Gamma \alpha \rangle = \sum_{\eta\in B} U_{\tilde{\psi}, \eta} \vert \eta \rangle,

the matrix element can be written as

ϕ~;ΓαH^ψ~Γα=1Uϕ~ϕηBUψ~ηϕH^η. \langle \tilde{\phi}; \Gamma \alpha \vert \hat{H} \vert \tilde{\psi} \Gamma \alpha \rangle = \frac{1}{ U_{\tilde{\phi}\phi} } \sum_{\eta \in B} U_{\tilde{\psi}\eta} \langle \phi \vert \hat{H} \vert \eta \rangle.