Math 4 Wisdom. "Mathematics for Wisdom" by Andrius Kulikauskas.

Hamiltonians and Quantum Symmetries

Two ways of looking at the Lie group embeddings

There is a linear complex structure {$J_i$} and the matrices it commutes with form a subgroup {$H$} of {$G$}.
There is a Lie algebra {$\frak{g}=\frak{h}+\frak{m}$} where {$[\frak{h},\frak{h}]\in\frak{h}, [\frak{h},\frak{m}]\in\frak{m}, [\frak{m},\frak{m}]\in\frak{h}$}, and {$\frak{m}$} are {$i\times$} the Hamiltonians (corresponding to the time evolution operators {$e^{iHt/ \hbar}$} in the unitary group), and {$G$} and {$H$} are the Lie groups corresponding to {$\frak{g}$} and {$\frak{h}$}.

The symmetric spaces form a sequence of order-parameter spaces that arise as we progressively break a large symmetry by introducing more and more symmetry-breaking operators. As a bonus, the matrices representing the symmetry-breaking operators will later serve as the building blocks of model Hamiltonians that yield all possible topologically non-trivial band structures.

I think I have confused the time-reversal and charge conjugation matrices. I think I need to consider how they act on eigenstates {$V_+$} and {$V_-$}. I am not sure!

Michael Stone, Ching-Kai Chiu, Abhishek Roy. Symmetries, Dimensions, and Topological Insulators: the mechanism behind the face of the Bott clock.

{$D≡ O(16r)$}: The coset generators {$m ∈ m_{−1}$} are real skew-symmetric matrices. These are not diagonalizable within the reals. We need to double the Hilbert space to {$R^{32r}$} and tensor with “{$i$}”= {$−iσ_2$} so that the Hamiltonian becomes {$H = −iσ_2 ⊗ m$}. Then taking {$\phi = σ_3 ⊗ I$} to be the real structure that defines complex conjugation, we have {$ϕH = −Hϕ$}. This {$\phi$} therefore defines a particle-hole symmetry that squares to {$+I$}. We should not count the eigenvectors {$v$} and {$iv ≡ (−iσ_2 ⊗ I)v$} as being distinct. (We can regard the {$R^{32r}$} space and the operators {$“i$}” and {$ϕ$} as being inherited from the previous cycle of the Bott clock.)

“{$i$}” {$=\begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix}=-i\sigma_2=-i\begin{pmatrix} 0 & -i \\ i & 0 \\ \end{pmatrix}$}

{$m= \begin{pmatrix} 0 & a \\ -a & 0 \\ \end{pmatrix}$}

{$H = −iσ_2 ⊗ m = \begin{pmatrix} 0 & -m \\ m & 0 \\ \end{pmatrix}=\begin{pmatrix} & & & a \\ & & -a & \\ & -a & & \\ a & & & \\ \end{pmatrix}=akr$} where {$k=\begin{pmatrix} & & & 1 \\ & & 1 & \\ & -1 & & \\ -1 & & & \\ \end{pmatrix}, r=\begin{pmatrix} 1 & & & \\ & -1 & & \\ & & 1 & \\ & & & -1 \\ \end{pmatrix}$}

{$σ_3= \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}$}

{$C=\phi=σ_3 ⊗ I=\begin{pmatrix} I & 0 \\ 0 & -I \\ \end{pmatrix}$}

{$CH^*C^{-1}=CHC^{-1}=\begin{pmatrix} I & 0 \\ 0 & -I \\ \end{pmatrix}\begin{pmatrix} 0 & -m \\ m & 0 \\ \end{pmatrix}\begin{pmatrix} I & 0 \\ 0 & -I \\ \end{pmatrix} = \begin{pmatrix} 0 & -m \\ -m & 0 \\ \end{pmatrix}\begin{pmatrix} I & 0 \\ 0 & -I \\ \end{pmatrix}=\begin{pmatrix} 0 & m \\ -m & 0 \\ \end{pmatrix}=-H$}

{$C^2=I$}

Let {$v=\begin{pmatrix}v_1 \\ v_2 \end{pmatrix}$} then {$Hv=\begin{pmatrix} 0 & -m \\ m & 0 \\ \end{pmatrix}\begin{pmatrix}v_1 \\ v_2 \end{pmatrix}=\begin{pmatrix}-mv_2 \\ mv_1 \end{pmatrix}$}

{$iv ≡ (−iσ_2 ⊗ I)v = \begin{pmatrix} 0 & -I \\ I & 0 \\ \end{pmatrix}\begin{pmatrix}v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix}-v_2 \\ v_1 \end{pmatrix}$}

So set {$\begin{pmatrix}-v_2 \\ v_1 \end{pmatrix}\sim \begin{pmatrix}v_1 \\ v_2 \end{pmatrix}$}

{$DIII ≡ O(16r)/U(8r)$}: The {$m ∈ \frak{m}_0$} ’s are real skew symmetric matrices that anticommute with {$J_1$} . We can keep {$C = ϕ$} as a particle-hole symmetry and take {$T = ϕ ⊗ J_1$} as a time reversal that commutes with {$H = −iσ_2 ⊗ M$} and squares to {$−I$}. The product of {$C$} and {$T$} is {$J_1$} , and this a linear (commutes with {$−iσ_2 ⊗ I$}) “{$P$} ” type symmetry that anticommutes with {$H$}.

{$m=\begin{pmatrix} & & a & b \\ & & b & -a \\-a & -b & & \\ -b & a & & \\ \end{pmatrix}=aj+bk$}

{$H = −iσ_2 ⊗ m = \begin{pmatrix} 0 & -m \\ m & 0 \\ \end{pmatrix}$}

{$C=\phi=\begin{pmatrix} I & 0 \\ 0 & -I \\ \end{pmatrix}$}

{$CH=\begin{pmatrix} I & 0 \\ 0 & -I \\ \end{pmatrix}\begin{pmatrix} 0 & -m \\ m & 0 \\ \end{pmatrix}=\begin{pmatrix} 0 & -m \\ -m & 0 \\ \end{pmatrix}$}

{$HC=\begin{pmatrix} 0 & -m \\ m & 0 \\ \end{pmatrix}\begin{pmatrix} I & 0 \\ 0 & -I \\ \end{pmatrix}=\begin{pmatrix} 0 & m \\ m & 0 \\ \end{pmatrix}=-CH$}

{$T = ϕ ⊗ J_1 = \begin{pmatrix} J_1 & 0 \\ 0 & -J_1 \\ \end{pmatrix} $}

{$TH=\begin{pmatrix} J_1 & 0 \\ 0 & -J_1 \\ \end{pmatrix}\begin{pmatrix} 0 & -m \\ m & 0 \\ \end{pmatrix}=\begin{pmatrix} 0 & -J_1m \\ J_1m & 0 \\ \end{pmatrix}=\begin{pmatrix} 0 & mJ_1 \\ -mJ_1 & 0 \\ \end{pmatrix}=\begin{pmatrix} 0 & -m \\ m & 0 \\ \end{pmatrix}\begin{pmatrix} J_1 & 0 \\ 0 & -J_1 \\ \end{pmatrix}=HT$}

{$P=J_1=CT=\begin{pmatrix} I & 0 \\ 0 & -I \\ \end{pmatrix} \begin{pmatrix} J_1 & 0 \\ 0 & -J_1 \\ \end{pmatrix}=\begin{pmatrix} J_1 & 0 \\ 0 & J_1 \\ \end{pmatrix}$}

{$Pi=\begin{pmatrix} J_1 & 0 \\ 0 & J_1 \\ \end{pmatrix}\begin{pmatrix} 0 & -I \\ I & 0 \\ \end{pmatrix}=\begin{pmatrix} 0 & -J_1 \\ J_1 & 0 \\ \end{pmatrix} = \begin{pmatrix} 0 & -I \\ I & 0 \\ \end{pmatrix}\begin{pmatrix} J_1 & 0 \\ 0 & J_1 \\ \end{pmatrix}=iP$}

{$PH=\begin{pmatrix} J_1 & 0 \\ 0 & J_1 \\ \end{pmatrix}\begin{pmatrix} 0 & -m \\ m & 0 \\ \end{pmatrix}=\begin{pmatrix} 0 & -J_1m \\ J_1m & 0 \\ \end{pmatrix}$}

{$HP=\begin{pmatrix} 0 & -m \\ m & 0 \\ \end{pmatrix}\begin{pmatrix} J_1 & 0 \\ 0 & J_1 \\ \end{pmatrix}=\begin{pmatrix} 0 & -mJ_1 \\ mJ_1 & 0 \\ \end{pmatrix}=\begin{pmatrix} 0 & J_1m \\ -J_1m & 0 \\ \end{pmatrix}=-PH$}

Define {$J_1\equiv P$}

{$AII≡ U(8r)/Sp(4r)$}: The generators {$m ∈ \frak{m}_1$} are real skew matrices that commute with {$J_1$} and anticommute with {$J_2$} . They can be regarded as skew-quaternion-hermitian matrices with complex entries. We no longer need set {$i → −iσ_2 ⊗ I$} as the matrices no longer have elements coupling between the artificial copies. We instead use {$J_1$} as the surrogate for “{$i$}.” Now {$H = J_1 m$} is real symmetric, and commutes with {$T = J_2$} . This {$T$} acts as a time reversal operator squaring to {$−I$}.

{$m=\begin{pmatrix} & -b & & -d \\ b & & d & \\ & d & & -b \\ -d & & b & \\ \end{pmatrix}=bi+dkr$} where {$k=\begin{pmatrix} & & & 1 \\ & & 1 & \\ & -1 & & \\ -1 & & & \\ \end{pmatrix}, r=\begin{pmatrix} 1 & & & \\ & -1 & & \\ & & 1 & \\ & & & -1 \\ \end{pmatrix}$}

{$H=J_1m=\begin{pmatrix} & -1 & & \\ 1 & & & \\ & & & -1 \\ & & 1 & \\ \end{pmatrix}m=\begin{pmatrix} -b & & -d & \\ & -b & & -d \\ d & & -b & \\ & d & & -b \\ \end{pmatrix} = -bI -djr$} (NOT real symmetric!)

{$T=J_2=\begin{pmatrix} & & 1 & \\ & & & -1 \\-1 & & & \\ & 1 & & \\ \end{pmatrix}$}

{$T^2=-I$}

{$HT=J_1mJ_2=J_1(-J_2m)=J_2J_1m=TH$}

Define {$J_2\equiv T$}

{$CII≡ Sp(4r)/Sp(2r) × Sp(2r)$}: The matrices {$m ∈ \frak{m}_2$} commute with {$J_1$} and {$J_2$} but anti- commute with {$J_3$}. Again {$H = J_1 m$}. We can set {$T = J_3$} as this commutes with with H and squares to {$−I$}. {$P = J_2 J_3$} anticommutes with {$H$} but commutes with {$J_1$} (and so is a linear map) while {$C = J_2$} anticommutes with {$H$}, is antilinear and squares to {$−I$}.

If a matrix commutes with {$J_1$} then its {$2\times 2$} blocks are complex linear (they match with complex numbers). And if it furthermore commutes with {$J_2$} then its {$4\times 4$} blocks have the form {$s=\begin{pmatrix} z_1 & z_2 \\ -\bar{z_2} & \bar{z_1} \\ \end{pmatrix}$}.

Furthermore the {$8\times 8$} matrices {$m$} anti-commmute with {$J_3$}. Thus they have the form {$m=\begin{pmatrix} 0 & s_1 \\ s_2 & 0 \\ \end{pmatrix}$}

{$H=J_1m=\textrm{diag}[\begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix}]m= \begin{pmatrix} 0 & i_4s_1 \\ i_4s_2 & 0 \\ \end{pmatrix}$} where {$i_4=\begin{pmatrix} & -1 & & \\ 1 & & & \\ & & & -1 \\ & & 1 & \\ \end{pmatrix}$}

Note that {$i_4s$} has the form {$is=\begin{pmatrix} y_1 & y_2 \\ \bar{y_2} & -\bar{y_1} \\ \end{pmatrix}$}.

{$T=J_3= \begin{pmatrix} & & & -1 & & & & \\ & & -1 & & & & & \\ & 1 & & & & & & \\ 1 & & & & & & & \\ & & & & & & & 1 \\ & & & & & & 1 & \\ & & & & & -1 & & \\ & & & & -1 & & & \\ \end{pmatrix}$}

{$TH=J_3J_1m=J_1mJ_3=HT$}

{$C=J_2$}

{$CH=J_2J_1m=-J_1mJ_2$}

{$P=CT=J_2J_3$}

{$PJ_1=J_2J_2J_1=J_1J_2J_3=J_1P$}

{$PH=J_2J_3J_1m=J_1J_2J_3m=-J_1J_2mJ_3=-J_1mJ_2J_3=-HP$}

Define {$J_3\equiv T$}

{$C≡ {Sp(2r) × Sp(2r)}/Sp(2r) ≃ Sp(2r)$}: The matrices {$m ∈ \frak{m}_3$} commute with {$J_1 , J_2 , J_3$} , and anticommute with {$J_4$} , and we can restrict ourselves to the subspace in which {$K = J_1 J_2 J_3$} takes a definite value, say {$+1$}. The Hamiltonian {$J_1 m$} commutes with {$J_4$} – but {$J_4$} does not commute with {$J_1 J_2 J_3$} and so is not allowed as an operator on our subspace. Indeed no product involving {$J_4$} is allowed. But {$C = J_2$} commutes with {$J_1 J_2 J_3$} and still anticommutes with {$H$}. Thus we still have a particle-hole symmetry squaring to {$−1$}. The old time reversal {$J_3$} now anticommutes with {$H$} and looks like another particle- hole symmetry, but is not really an independent one as in this subspace {$J_3 = J_2 J_1$} and {$J_1$} is simply multiplication by “{$i$}.”

{$m=\begin{pmatrix} s_1 & \\ & -s_1 \\ \end{pmatrix}$}

{$H=J_1m=\begin{pmatrix} i_4s_1 & \\ & -i_4s_1 \\ \end{pmatrix}$}

Restrict to {$4$} dimensional {$V_+$} for operator {$J_1J_2J_3$}.

{$m=\begin{pmatrix} s_1 \\ \end{pmatrix}$}

{$H=J_1m=\begin{pmatrix} i_4s_1 \\ \end{pmatrix}$}

{$HJ_4=J_1mJ_4=J_4J_1m=J_4H$}

{$J_4K=J_4J_1J_2J_3=-J_1J_2J_3J_4=-KJ_4$}

{$C=J_2$}

{$CK=J_2J_1J_2J_3=-J_1J_2J_2J_3=J_1J_2J_3J_2=KC$}

{$CH=J_2J_1m=-J_1J_2m=-J_1mJ_2=-HC$}

{$CI≡ Sp(2r)/U(2r)$}: The {$m ∈ \frak{m}_4$} anticommute with {$J_5$} . Now {$J_4 J_5$} commutes with {$J_1 J_2 J_3 $}, and so is an allowed operator. It anticommutes with {$H = J_1 m$} and commutes with {$J_1$} . It is therefore a “{$P$} ”-type linear map. The map {$T = J_2 J_4 J_5$} is antilinear (anticommutes with {$J_1$} ), commutes with {$H$} and {$T^2 = +I$}. We can take {$C = J_2$} again.

{$m=\begin{pmatrix} s_2 & \\ & s_2 \\ \end{pmatrix}$} where {$s_2=\begin{pmatrix} d & & & -c \\ & d & c & \\ & -c & d & \\ c & & & d \\ \end{pmatrix}$} is made of complex linear {$2\times 2$} blocks.

{$H = J_1 m = \begin{pmatrix} i_4s_2 & \\ & i_4s_2 \\ \end{pmatrix}$} where {$i_4s_2=\begin{pmatrix} & -d & -c & \\ d & & & -c \\ -c & & & -d \\ & -c & d & \\ \end{pmatrix}$}

{$C=J_2$}

{$T=J_2J_4J_5=\begin{pmatrix} & & & 1 & & & & \\ & & 1 & & & & & \\ & 1 & & & & & & \\ 1 & & & & & & & \\ & & & & & & & -1 \\ & & & & & & -1 & \\ & & & & & -1 & & \\ & & & & -1 & & & \\ \end{pmatrix}$}

{$P=J_4J_5=-CT=\begin{pmatrix} & -1 & & & & & & \\ 1 & & & & & & & \\ & & & 1 & & & & \\ & & -1 & & & & & \\ & & & & & 1 & & \\ & & & & -1 & & & \\ & & & & & & & -1 \\ & & & & & & 1 & \\ \end{pmatrix}$}

{$PK=KP$}

{$AI ≡ U(2r)/O(2r)$}: The {$m ∈ \frak{m}_5 n$} anticommute with {$J_6$} , and we are to restrict ourselves to the subspace on which {$K = J_1 J_2 J_3 = +1$} and {$M = J_1 J_4 J_5 = +1$}. The map {$T = J_3 J_4 J_6$} commutes with {$K$} and {$M$} and commutes {$H = J_1 m$}. We have {$T^2 = +I$}. We could equivalently take {$T = J_2 J_4 J_6$}.

{$H=J_1m$}

{$T=N=J_2J_4J_6=\textrm{diag}[-1,1,-1,1,-1,1,-1,1]$}

{$BDI≡ O(2r)/O(r) × O(r)$}: The {$m ∈ \frak{m}_6$} anticommute with {$J_7$} , and we are to restrict our- selves to the eigenspaces of {$K = J_1 J_2 J_3 , M = J_1 J_4 J_5$} . The {$m$} also commute with the antilinear operator {$N = J_2 J_4 J_6$} which we can regard as our real structure {$ϕ$}. We therefore set {$C = ϕ = N$}. Now {$J_3 J_4 J_7 , J_3 J_5 J_6 , J_2 J_5 J_7$} and {$J_1 J_6 J_7$} all commute with {$K$}, {$M$} and {$N$}, each squaring to {$+I$}. In the restricted subspace {$J_3 J_4 J_7 ∝ J_2 J_5 J_7 ∝ J_1 J_6 J_7$} and {$J_3 J_5 J_6 ∝ I$}. The problem here is what to take for “{$i$},” as the current {$H → J_1 m$} will take us out of the eigenspace of {$N$}. But this is the problem we started with. We need to double the space and keep “{$i$}”= {$J_1$} and the real structure {$ϕ = N$}. We are therefore retaining the {$R^{2r}$} Hilbert space. With {$H = J_1 m$} we have that {$T = J_1 J_6 J_7$} commutes with {$H$} and squares to {$+I$}.

{$D≡ {O(r) × O(r)}/O(r) ≃ O(r)$}: Now the {$m ∈ \frak{m}_7$} anticommute with {$J_8$} and, except for the factor “{$i$}”= {$J_1$} , we should stay in the space where {$K = J_1 J_2 J_3 , M = J_1 J_4 J_5 , N = J_2 J_4 J_6$} and {$P = J_1 J_6 J_7$} take the value {$+1$} With {$H = J_1 m$} we have that {$C = ϕ = N$} anticommutes with {$H$}, and brings us full circle.