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Quantum Symmetries

Strategy

My strategy is to

• make the quantum symmetries intuitively clear
• show how that is reflected in the Hamiltonians
• explain how the linear complex structures take us through the Hamiltonians
• consider how one Hamiltonian relates to the next Hamiltonian
• explain how the linear complex structures take us through combinations of symmetries

Ideas

• Consider in terms of dimensions, what we are keeping and what we are removing, and how they fit together.
• Think of the symmetries, and the definition of complex conjugation, as depending on whether the complex numbers are expressed intrinsically, algebraically {$a+bi$} or extrinsically as {$2\times 2$} matrices. Perhaps the symmetries takes us back and forth between these two understandings. Whether when we walk around the Bott clock we may be presuming an extrinsic understanding and then running into collapse.
• The Clifford algebras expressing the super division algebras, yielding {$Cl_{\pm $}=M_2(\mathbb{H})$}, seem to match the entries of the Hamiltonian matrices, notably the reals and the quaternions.

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Symmetries

Ludwig explains quantum symmetries in terms of first quantization (wave function) and second quantization (creation {$\hat{\psi}^\dagger_i$} and annihilation {$\hat{\psi}_i$} operators). For fermions, we have the anticommutating relations {$\hat{\psi}_i\hat{\psi}^\dagger_j = -\hat{\psi}^\dagger_j\hat{\psi}_i + \delta_{ij}$} and also {$\hat{\psi}_i\hat{\psi}_j = -\hat{\psi}_j\hat{\psi}_i, \hat{\psi}^\dagger_i\hat{\psi}^\dagger_j = -\hat{\psi}^\dagger_j\hat{\psi}^\dagger_i$}. Here the creation vector {$\hat{\psi}^\dagger=[\hat{\psi}^\dagger_1,\hat{\psi}^\dagger_2,\dots,\hat{\psi}^\dagger_n]$} is a row vector, and the annihilation vector {$\hat{\psi}=[\hat{\psi}_1,\hat{\psi}_2,\dots,\hat{\psi}_n]^T$} is a column vector.

The second quantized Hamiltonian is {$\hat{H}=\sum_{A,B}\hat{\psi}^\dagger_{A}H_{A,B}\hat{\psi}_B=\hat{\psi}^\dagger H\hat{\psi}$}

Here the labels {$A,B$} refer to the {$N$} states (or lattice sites which may be distinguished by extra information such as spin). The Hamiltonian {$H_{A,B}$} is a matrix of numbers, called the first quantization (or single-particle) Hamiltonian. We are interested in large {$N>>1$}, the thermodynamic limit.

Suppose that the Hamiltonian {$H_{A,B}$} is invariant under a group {$G_0$} of symmetries. There exist a representation of {$G$} in terms of {$N\times N$} unitary matrices {$U$}. Unitary means that {$U=U^\dagger$}. These matrices commute with the first quantized Hamiltonian. {$UH=HU$} means that {$UHU\dagger=H$}.

In terms of second quantization, this corresponds to operators {$\hat{U}$} acting on the Fermion Fock space as follows.

{$\hat{\mathcal{U}}\hat{\psi}_A\hat{\mathcal{U}}^{-1}=\sum_B U^\dagger_{A,B}\hat{\psi}_B$}	{$\hat{\mathcal{U}}\hat{\psi}^\dagger_A\hat{\mathcal{U}}^{-1}=\sum_B \hat{\psi}^\dagger_B U_{B,A}$}
the annihilation column vector {$\hat{\psi}$} is converted into a collection of annihilation column vectors with coefficients in the creation matrix {$U^\dagger$}	the creation row vector {$\hat{\psi}$} is converted into a collection of creation row vectors with coefficients in the annihilation matrix {$U={U^\dagger}^{-1}$}

The operators {$\hat{\mathcal{U}}$} and {$\hat{H}$} commute: {$\hat{\mathcal{U}}\hat{H}\hat{\mathcal{U}}^{-1}=\hat{H}$} Can I show this? Consider {$\hat{\mathcal{U}}\hat{H} = \hat{\mathcal{U}}\hat{\psi}^\dagger H\hat{\psi}$}

Three symmetries

Given the annihilation operator {$\Psi=[\psi_1, \dots, \psi_L]^T$} and creation operator {$\Psi^{\dagger}=[\psi_1^\dagger,\dots,\psi_L^\dagger]$}, we may apply a usual symmetry operator {$\mathit{U}_{USL}\Psi^\dagger\mathscr{U}_{USL}^{-1}=\Psi^\dagger\mathbf{U}_{USL}$} or a transposing symmetry operator {$\mathit{U}_{TRN}\Psi^\dagger\mathscr{U}_{TRN}^{-1}=\Psi\mathbf{U}_{USL}^*$}.

We get four kinds of operator (linear or antilinear vs. usual or transposing). We can show that the operators square to {$+1$} and possibly {$-1$} but note that the relationships for the related unitary matrices are more subtle because they depend on what happens when the matrix is moved outside of the conjugation by the symmetry operator.

• trivial operator
• time reversal operator {$\mathscr{T}$} is usual antilinear. {$\mathscr{T}^2=\pm 1=T, \mathbf{U}_T\mathbf{U}_T^*=T\mathbf{1}$}
• charge conjugation operator {$\mathscr{C}$} is transposing linear. {$\mathscr{C}^2=\pm 1=C, \mathbf{U}_C\mathbf{U}_C^*=C\mathbf{1}$}
• sublattice symmetry operator {$\mathscr{S}$} is transposing antilinear. {$\mathscr{S}^2= 1, \mathbf{U}_S\mathbf{U}_S=\mathbf{1} $}

Note that {$\mathscr{S}=\mathscr{T}\mathscr{C}$} and {$\mathscr{S}^2=\mathscr{I}$}. This means {$\mathscr{T}\mathscr{C}\mathscr{T}\mathscr{C}=\mathscr{I}$} and so {$\mathscr{I}=\mathscr{T}^2\mathscr{T}^{-1}\mathscr{C}\mathscr{T}\mathscr{C}^{-1}\mathscr{C}^2=T\mathscr{T}^{-1}\mathscr{C}\mathscr{T}\mathscr{C}^{-1}C$}. Then {$T^{-1}C^{-1}=\mathscr{T}^{-1}\mathscr{C}\mathscr{T}\mathscr{C}^{-1}=TC$} bęcause {$T,C\in\{-1,+1\}$}. Thus {$\mathscr{C}\mathscr{T}=\mathscr{T}\mathscr{C}TC$} and we can have {$T=C$} (if {$\mathscr{T}$} and {$\mathscr{C}$} commute) or {$T=-C$} (if they anticommute).

• linear {$\mathscr{S}H=-H\mathscr{S}$} sublattice, parity
• antilinear {$\mathscr{C}H\mathscr{C}^{-1}=-H, \mathscr{C}^2=\pm\mathbb{I}$} charge conjugation, particle-hole
• antilinear {$\mathscr{T}H\mathscr{T}^{-1}=H, \mathscr{T}^2=\pm\mathbb{I}$} time reversal
• {$CH^*C^{-1}=-H, C^*C=\pm\mathbb{I}$}
• {$TH^*T^{-1}=H, T^*T=\pm\mathbb{I}$}

Time reversal

Time reversal is a feature of the time-evolution operator {$\hat{\mathcal{U}}(t)=e^{itH}$} which relates the Lie algebra and the Lie group. We have {$\hat{\mathcal{T}}\hat{\mathcal{U}}(t)\hat{\mathcal{T}}^{ž1}=\hat{\mathcal{U}}(-t)$}.

{$T$} is (antilinear? anticommuting, ...?) so it must contain the final linear complex structure {$J_k$}, the one that anticommutes with the elements of {$\frak{m}_k$} whereas the previous {$J_1, \dots J_{k-1}$} commute.

Charge conjugation

{$C$} is based on {$J_2$} because...

The restrictions are based on {$J_1$} grouped with final pairs yielding {$J_1J_2J_3, J_1J_4J_5, J_1J_6J_7$} which one short of the anticommuting linear complex structure.

Math Stack Exchange. Andrius Kulikauskas. How is this antilinear map (charge conjugation) represented as a complex matrix?

Sublattice symmetry {$\mathbf{U}_S$}

The sublattice symmetry {$\mathbf{U}_S$} has eigenvalues restricted to {$\pm 1$} thus can be written at {$\mathbf{1}_{p,q}$}.

Applying the symmetry (by seeing what commutes with operator {$iJ_1$}) breaks up the space into two subspaces. And letting go of the symmetry (by seeing what commutes with a mutually anticommmuting operator {$iJ_2$}) has the subspaces be of the same size and contain the same information.

This relates to the symmetry of counting. There is a distinction between counting forwards and backwards and then the two directions have to be the same size. Similarly, we can think in terms of creation and annihilation operators.

In the real case, with each new operator {$J_i$}, the sublattice symmetry is either arising or disappearing. And when it arises, it is bringing together time reversal {$\mathbf{U}_T$} and charge conjugation {$\mathbf{U}_C$} in one of four ways that the periodicity cycles through.

The sublattice symmetry {$S$}

• there may be no symmetries
• there may only be the sublattice symmetry and {$S^2=+1$}
• {$S$} may not exist but either {$C$} or {$T$} does
• the sublattice symmetry is one of three symmetries and {$S^2=+1 = T^2C^2$}
  • {$T^2=+1, C^2=+1$}
  • {$T^2=-1, C^2=-1$}
• the sublattice symmetry is one of three symmetries and {$S^2=-1 = T^2C^2$}
  • {$T^2=+1, C^2=-1$}
  • {$T^2=-1, C^2=+1$}

The Hamiltonians express what is left free by the symmetries. They are the freedom of the self. They express how the freedom expresses the constraints. The freedom is channeled in perspectives.

Twofold periodicity is driven by the symmetry {$SHS^{-1}=-H$} which acts on {$H$} directly, thus bringing out the symmetry of the Hermitian matrix, yielding a unitary matrix upon exponentiation {$e^{iH}$}.

God

There is no additional symmetry but there is simply the fact that {$H=H^\dagger$} is Hermitian.

Human

There is the symmetry {$1_{p,q}H1_{p,q}=-H$}.

Eightfold periodicity is driven by the conjugate {$H*$}, through two symmetries {$TH^*T^{-1}=H$} and {$CH^*C^{-1}=-H$}. Consider what it means to have unmarked-marked opposites {$H$} and {$-H$} and twin unmarked-unmarked opposites, imaginary numbers {$i$} and {$j$} with {$\bar{i}=j$} and {$\bar{j}=i$}. Note that known (raw, direct, unreflected experience) and unknown (indirect, reflected) are unmarked and marked.

Note that {$H$} and {$H^\dagger$} are twin opposites. One is the context of the other. But neither is intrinsically primary, as regards the underlying norm. However, one is then chosen as the inhabited one, the resident one, the being. Note also, however, that the dual vector space has a notion of functional which is secondary and brings in a map into the field, which is a marker, making the dual vector space marked.

Note also that {$+1$} and {$-1$} are unmarked and marked only as the secondary operation (multiplication in a ring). In an additive group, such as {$\mathbb{Z}$} they are the same, they both serve as generators, because the identity is zero. Note also the difference between zero and a blank space, as opposites, here marked and unmarked.

Nullsome

{$C=\phi\otimes\mathbb{I}$}

{$\mathbf{H=-H^*}$}

{$\frak{o}\textrm{(L)}$}

The symmetry {$\mathbf{1_{p,q}H=-H1_{p,q}}$} can be written {$1_{p,q}H1_{p,q}= H^* = -H$}.

Onesome

{$C=\phi, T=\phi\otimes J_1$}

{$\begin{pmatrix} & h_{mm} \\ -h_{mm}^* & \end{pmatrix}$}

{$\frak{o}\textrm{(2L)} \backslash \frak{u}\textrm{(L)}$}

{$H$} and {$-H^*$} are separated, identified with going back and forth between subspaces, and real numbers are now allowed.

Twosome

{$T=J_2$}

{$\begin{pmatrix} h_{aa} & h_{ab} \\ -h_{ab}^* & h_{aa}^* \end{pmatrix}$}

{$\frak{u}\textrm{(2L)} \backslash \frak{sp}\textrm{(L)}$}

{$\mathbf{\begin{pmatrix} & 1_{mm} \\ -1_{mm} & \end{pmatrix}H^*=H\begin{pmatrix} & 1_{mm} \\ -1_{mm} & \end{pmatrix}}$}

The symmetry can be written

{$\begin{pmatrix} & -1_{mm} \\ 1_{mm} & \end{pmatrix}\mathbf{H}\begin{pmatrix} & 1_{mm} \\ -1_{mm} & \end{pmatrix}= \mathbf{H}^*$}

{$\begin{pmatrix} & -1_{mm} \\ 1_{mm} & \end{pmatrix}\begin{pmatrix} h_{aa} & h_{ab} \\ h_{ba} & h_{bb} \end{pmatrix}\begin{pmatrix} & 1_{mm} \\ -1_{mm} & \end{pmatrix}= \begin{pmatrix} h_{aa}^* & h_{ab}^* \\ h_{ba}^* & h_{bb}^* \end{pmatrix}$}

{$=\begin{pmatrix} -h_{ba} & -h_{bb} \\ h_{aa} & h_{ab} \end{pmatrix}\begin{pmatrix} & 1_{mm} \\ -1_{mm} & \end{pmatrix}= \begin{pmatrix} h_{bb} & -h_{ba} \\ -h_{ab} & h_{aa} \end{pmatrix} = \begin{pmatrix} h_{aa}^* & h_{ab}^* \\ h_{ba}^* & h_{bb}^* \end{pmatrix}$}

Thus {$h_{bb}=h_{aa}^*, h_{ba}=-h_{ab}^*$}.

Consequently, {$H=\begin{pmatrix} h_{aa} & h_{ab} \\ -h_{ab}^* & h_{aa}^* \end{pmatrix}$}

The absolute space and relative space are distinguished and allowed to have self-relations.

Threesome

{$C=J_2, T=J_3$}

{$\begin{pmatrix} & & h_{aa} & h_{ab} \\ & & -h_{ab}^* & h_{aa}^* \\ h_{aa}^\dagger & -h_{ab}^T & & \\ h_{ab}^\dagger & h_{aa}^T & & \end{pmatrix}$}

{$\frak{sp}\textrm{(2L)} \backslash \frak{sp}\textrm{(L)}\oplus\frak{sp}\textrm{(L)}$}

{$H$} and {$H^\dagger$} are separated.

Foursome

{$C=J_2$}

{$\begin{pmatrix} h_{aa} & h_{ab} \\ h_{ab}^* & -h_{aa}^* \end{pmatrix}$}

{$\frak{sp}\textrm{(L)}$}

{$\mathbf{\begin{pmatrix} & 1_{mm} \\ -1_{mm} & \end{pmatrix}H^*=-H\begin{pmatrix} & 1_{mm} \\ -1_{mm} & \end{pmatrix}}$}

The symmetry can be written

{$\begin{pmatrix} & -1_{mm} \\ 1_{mm} & \end{pmatrix}\mathbf{H}\begin{pmatrix} & 1_{mm} \\ -1_{mm} & \end{pmatrix}= -\mathbf{H}^*$}

{$\begin{pmatrix} & -1_{mm} \\ 1_{mm} & \end{pmatrix}\begin{pmatrix} h_{aa} & h_{ab} \\ h_{ba} & h_{bb} \end{pmatrix}\begin{pmatrix} & 1_{mm} \\ -1_{mm} & \end{pmatrix}= \begin{pmatrix} -h_{aa}^* & -h_{ab}^* \\ -h_{ba}^* & -h_{bb}^* \end{pmatrix}$}

{$=\begin{pmatrix} -h_{ba} & -h_{bb} \\ h_{aa} & h_{ab} \end{pmatrix}\begin{pmatrix} & 1_{mm} \\ -1_{mm} & \end{pmatrix}= \begin{pmatrix} h_{bb} & -h_{ba} \\ -h_{ab} & h_{aa} \end{pmatrix} = \begin{pmatrix} -h_{aa}^* & -h_{ab}^* \\ -h_{ba}^* & -h_{bb}^* \end{pmatrix}$}

Thus {$h_{bb}=-h_{aa}^*, h_{ba}=h_{ab}^*$}.

Consequently, {$H=\begin{pmatrix} h_{aa} & h_{ab} \\ h_{ab}^* & -h_{aa}^* \end{pmatrix}$}

The complementary space is allowed and united, allowing for diagonal elements. The negative sign of the bottom row has been swapped.

Fivesome

{$C=J_2, T=J_2J_4J_5$}

{$\begin{pmatrix} & h_{mm} \\ h_{mm}^* & \end{pmatrix}$}

{$\frak{sp}\textrm{(L)} \backslash \frak{u}\textrm{(L)}$}

The diagonal elements are made zero.

Sixsome

{$T=J_2J_4J_6$}

{$\mathbf{H=H^*}$}

{$\frak{u}\textrm{(L)} \backslash \frak{o}\textrm{(L)}$}

{$\mathbf{1_{p,q}H=H1_{p,q}}$}

The symmetry {$\mathbf{1_{p,q}H=H1_{p,q}}$} can be written {$1_{p,q}H1_{p,q}= H^* = H$}.

The off-diagonal blocks {$h_{mm}$} and {$h_{mm}^*$} are identified and united, making them real.

Sevensome

{$C=J_2J_4J_6=\phi, T=J_1J_6J_7$}

{$\begin{pmatrix} & h_{pq} \\ h_{pq}^* & \end{pmatrix}$}

{$\frak{o}\textrm{(2L)} \backslash \frak{o}\textrm{(L)}\oplus\frak{o}\textrm{(L)}$}

{$H$} and {$H^*$} are distinguished, and allowed to be non-square.

Eightsome

{$C=J_2J_4J_6=\phi$}

{$\mathbf{H=-H^*}$}

{$\frak{o}\textrm{(L)}$}

{$\mathbf{1_{p,q}H=-H1_{p,q}}$}

{$h_{pq}$} and {$h_{pg}^*$} are identified as negatives of each other. The negative sign of the bottom row has been swapped (from the zero onto {$h_{pq}^*$}).