# Properties and Proofs for Reduced Basis Generation

## Projection Operator

Based on the grand orthogonality theorem

$\sum_{g \in G} D^{(\Gamma)}_{\alpha\beta}(g^{-1}) D^{(\Gamma')}_{\gamma\delta}(g) = \delta_{\Gamma, \Gamma'} \delta_{\alpha\delta} \delta_{\beta\gamma} \frac{|G|}{\dim\Gamma},$

We define the following projection operator:

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

### Hermiticity

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

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

Proof | Both $\hat{R}$ and $D$ are unitary, and therefore

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

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

### Projector

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

$\hat{P}_{\alpha}^{(\Gamma)} \hat{P}_{\beta}^{(\Gamma')} = \delta_{\Gamma,\Gamma'} \delta_{\alpha\beta} \hat{P}_{\alpha}^{(\Gamma)}$

Proof |

$\hat{P}_{\alpha}^{(\Gamma)}\hat{P}_{\beta}^{(\Gamma')} = \frac{\dim \Gamma \dim \Gamma'}{|G|^2} \sum_{g,h\in G} D_{\alpha\alpha}^{(\Gamma)} (g^{-1}) D_{\beta\beta}^{(\Gamma')} (h^{-1}) \hat{R}(g \cdot h)$

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

\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 $\Gamma\alpha$ block using $\hat{P}_{\alpha}^{(\Gamma)}$, starting from a basis of the full Hilbert space.

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

$\vert \psi; \Gamma \alpha \rangle := \hat{P}_{\alpha}^{(\Gamma)} \vert \psi \rangle$

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

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

$\vert \tilde{\psi}; \Gamma \alpha \rangle := \frac{1}{\mathcal{N}_{\psi;\Gamma\alpha}} \vert \psi; \Gamma \alpha \rangle$

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

${\mathcal{N}_{\psi;\Gamma\alpha}}^2 := \langle \psi; \Gamma \alpha \vert \psi; \Gamma \alpha \rangle,$

which is

$\mathcal{N}_{\psi;\Gamma\alpha} = \sqrt{ \langle \psi \vert \hat{P} \vert \psi \rangle } = \sqrt{ \langle \psi \vert \psi ; \Gamma\alpha \rangle } = \langle \psi \vert \tilde{\psi} ; \Gamma\alpha \rangle.$

Note that $\mathcal{N}_{\psi;\Gamma\alpha}$ 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 $\vert \psi \rangle$ from the original Hilbert space. Let us introduce the following notation for the transformed states

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

Let $N$ be the the set of group elements that leaves $\vert \psi \rangle$ invariant,

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

The set $N$ is a subgroup of $G$:

$\hat{R}(n_1 n_2) = \hat{R}(n_1) \hat{R}(n_2) = \hat{1}.$

Since

$\hat{R}(g n) \vert \psi \rangle = \hat{R}(g) \hat{R}(n) \vert \psi \rangle = \hat{R}(g) \vert \psi \rangle,$

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

$\vert \psi ; gN \rangle \equiv \vert \psi ; g \rangle.$

Lemma | For all $C_1, C_2 \in G/H$,

$C_1 = C_2 \iff \vert \psi; C_1 \rangle = \vert \psi ; C_2 \rangle$

Proof |
($\Rightarrow$) By definition.
($\Leftarrow$) Assume that $\vert \psi ; C_1 \rangle = \vert \psi ; C_2 \rangle$. Then for $g_1 \in C_1, g_2 \in C_2$,

$\hat{R}(g_1) \vert \psi \rangle = \hat{R}(g_2)\vert \psi \rangle,$

and since $\hat{R}(g^{-1} g_2) = \hat{1}$, $g_1^{-1} g_2 \in N$, and $C_1 = C_2$. $\square$

### Projected state

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

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

We will prove this below.

Consider the following projected state:

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

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

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

After relabeling $g n^{-1}$ by $g$ and factoring out the summation over $N$, we get

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

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

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

$D(g n) = D(g).$

Again, this allows us to define $D$ for every coset:

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

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

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

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

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

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

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

Then the overlap can be written as

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

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

$\langle \psi \vert \hat{P} \hat{P} \vert \psi \rangle = \langle \psi \vert \hat{P} \vert \psi \rangle = \frac{|N| \dim \Gamma}{|G|}.$

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

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

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 $G$ itself may not be an Abelian, the image of the representation $\hat{R}(G)$ acting on the orbit of $\vert\psi\rangle$ 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

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

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

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

$D$ is unitary,

$\sum_{\beta} D_{\alpha\beta}^{(\Gamma)}(g) D_{\gamma\beta}^{(\Gamma)}(g)^{*} = \delta_{\alpha\gamma},$

and thus for $\alpha=\gamma$

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

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

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

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

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

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

## Hamiltonian Elements

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

$\langle \tilde{\phi}; \Gamma \alpha \vert \hat{H} \vert \tilde{\psi} \Gamma \alpha \rangle = \left( \frac{1}{\mathcal{N}_{\phi;\Gamma\alpha}} \langle \phi \vert \hat{P} \right) \hat{H} \left( \frac{1}{\mathcal{N}_{\psi;\Gamma\alpha}} \hat{P} \vert \psi \rangle \right).$

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

$\langle \tilde{\phi} ; \Gamma \alpha \vert \hat{H} \vert \tilde{\psi} ; \Gamma \alpha \rangle = \frac{1}{\mathcal{N}_{\phi;\Gamma\alpha}} \langle \phi \vert \hat{H} \left( \hat{P} \frac{1}{\mathcal{N}_{\psi;\Gamma\alpha}} \vert \psi \rangle \right),$

and thus

$\langle \tilde{\phi}; \Gamma\alpha \vert \hat{H} \vert \tilde{\psi} ; \Gamma \alpha \rangle = \frac{ \langle \phi\vert \hat{H} \vert \tilde{\psi}; \Gamma\alpha \rangle } { \langle \phi \vert \tilde{\phi} ; \Gamma \alpha \rangle}.$

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

$\vert \tilde{\psi} ; \Gamma \alpha \rangle = \sum_{\eta\in B} U_{\tilde{\psi}, \eta} \vert \eta \rangle,$

the matrix element can be written as

$\langle \tilde{\phi}; \Gamma \alpha \vert \hat{H} \vert \tilde{\psi} \Gamma \alpha \rangle = \frac{1}{ U_{\tilde{\phi}\phi} } \sum_{\eta \in B} U_{\tilde{\psi}\eta} \langle \phi \vert \hat{H} \vert \eta \rangle.$