Higher order product formulas for double bracket rotations in the dbi model

[ashwinieghanesh]

Higher Order Product Formula for DBR in the DBI Model

The third order product formula is given by:
$S_3 := e^{x\hat{A}\frac{\sqrt{5}-1}{2}}e^{x\hat{B}\frac{\sqrt{5}-1}{2}}e^{-x\hat{A}}e^{ -x\hat{B}\frac{\sqrt{5}+1}{2}}e^{x\hat{A}\frac{3-\sqrt{5}}{2}}e^{x\hat{B}} = e^{x^2[\hat{A},\hat{B}]} + \mathcal{O}(x^4)$

where $\hat{A}$ and $\hat{B}$ are arbitrary Hermitian operators.

In the context of DBI, we previously implemented the following second-order product formula which we call GCI:

$S_2 := S_{2}(x):=e^{x \hat{A}} e^{x \hat{B}} e^{-x\hat{A}} e^{-x \hat{B}}=e^{x^{2}[\hat{A}, \hat{B}]}+O\left(x^{3}\right) $

Diagonalization Routine for GCI algorithm

Input

1. $\hat{H}_0$: Initial Hamiltonian
2. $\hat{D}$: Diagonal association (canonical or BHMM operator)
3. $s$: flow duration for a single recursion step $k$
4. $K$: Total number of recursion steps

Output
$\hat{H}_K$: Final (diagonalized) Hamiltonian

Procedure

1. For $k\in[0,K)$, use the reduced group commutator $\hat{U}_k \leftarrow e^{i\sqrt{s_k}\hat{D}_k}e^{i\sqrt{s_k}\hat{H}_k}e^{-i\sqrt{s_k}\hat{D}_k}$
2. $\hat{H}_{k+1} \leftarrow \hat{U^{\dagger}}_k\hat{H}_k\hat{U}_k$
3. Repeat steps 1 to 4 for the next recursion step

Below, I have attached my Jupyter Notebook files in which I studied the higher-order product formula (notably, the 3rd-order product formula) implementation in the DBI algorithm:

1. Multiple DBR steps for product formula orders 2 and 3.

DBR _higher_order_PF.pdf

2. Error scale plots for product formula orders 2 and 3.
Error Scale.pdf

3. DBI implementation for order 3.
DBI _higher_order_PF-copy.pdf

If there are any discrepancies between my typed equations and the manuscript, do let me know.

~Ashwinie

[marekgluza]

Hi @Tellememe thanks for adding. @wrightjandrew cf the updated issue content for your #1342

Ashwinie can you next

• comment below and derive the symmetry relations for the second order by setting $A=\pm iH, B=\pm iD$ or $A=\pm iD, iB=\pm H$, generalize the 1st order https://mathstodon.xyz/@Marekgluza/112279346010080479
• Please add the derivations including the transitions which include $i^2=-1$ (I made mistakes because of that a few times) what happens if the operators are exchanged and what is being approximated
• Finally we want to rotate $S_3^\dagger A S_3$ so please reduce the formula by using the commutation relation like was done for the reduced group commutator in my proposal https://arxiv.org/abs/2206.11772 and propose a worked out formula analogous to DBI 3rd order GCI #1342 for an implementation
• previous point either @Tellememe or @wrightjandrew will implement in a seperate pull request
• Ashwinie, once the derivations are settled here for the future reference, you will be tasked with providing documentation. That will be roughly copy paste of what will be here.

[marekgluza]

https://arxiv.org/pdf/2111.12177 the $S_3$ above comes from this paper (eq 8)

[ashwinieghanesh]

Symmetry Relations for Group Commutators

For $S_2$

1. $V(\hat{A},\hat{B}) \approx e^{[\hat{A},\hat{B}]}$
2. $V(-\hat{A},-\hat{B}) \approx e^{[\hat{A},\hat{B}]}$

For $S_3$

3. $V(\pm i\hat{H},\pm i\hat{D})$ = $e^{i(\pm i\hat{H})\frac{\sqrt{5}-1}{2}}e^{i(\pm i\hat{D})\frac{\sqrt{5}-1}{2}}e^{-i(\pm i\hat{H})}e^{ -i(\pm i\hat{D})\frac{\sqrt{5}+1}{2}}e^{i(\pm i \hat{H})\frac{3-\sqrt{5}}{2}}e^{i(\pm i\hat{D})}$
4. $e^{\mp \hat{H}\frac{\sqrt{5}-1}{2}}e^{\mp \hat{D}\frac{\sqrt{5}-1}{2}}e^{\pm \hat{H}}e^{\pm \hat{D}\frac{\sqrt{5}+1}{2}}e^{\mp \hat{H}\frac{3-\sqrt{5}}{2}}e^{\mp \hat{D}}$

By exchanging the operators,

5. $V(\pm i\hat{D},\pm i\hat{H})$ = $e^{i(\pm i\hat{D})\frac{\sqrt{5}-1}{2}}e^{i(\pm i\hat{H})\frac{\sqrt{5}-1}{2}}e^{-i(\pm i\hat{D})}e^{ -i(\pm i\hat{H})\frac{\sqrt{5}+1}{2}}e^{i(\pm i \hat{D})\frac{3-\sqrt{5}}{2}}e^{i(\pm i\hat{H})}$
6. $e^{\mp \hat{D}\frac{\sqrt{5}-1}{2}}e^{\mp \hat{H}\frac{\sqrt{5}-1}{2}}e^{\pm \hat{D}}e^{\pm \hat{H}\frac{\sqrt{5}+1}{2}}e^{\mp \hat{D}\frac{3-\sqrt{5}}{2}}e^{\mp \hat{H}}$

Compared with the $S_2$ case, for $S_3$, the first three terms and the last term in equations 4 and 6 reflect symmetry upon taking the negative values of both Hermitian operators. The fourth and fifth terms, however, do not follow the symmetry since the operators associated with these terms get changed by a prefactor with a (unique) value $\neq$ 1.