Calculation of Trotter-Suzuki error bound

Question

Suppose we are given Hamiltonian in the form: $$ H = -\sum_{k=0}^{n-1} \alpha\sigma^x_k\sigma^x_{k+1} + \beta\sigma^y_k\sigma^y_{k+1} + \gamma\sigma^z_k\sigma^z_{k+1}, $$ where $n$ is the number of qubits, and $\alpha$, $\beta$, $\gamma$ $\in \mathbb{R}^+$. By splitting Hamiltonian terms into even/odd subsets: $$ A = -\sum_{k=0} \alpha\sigma^x_{2k}\sigma^x_{2k+1} + \beta\sigma^y_{2k}\sigma^y_{2k+1} + \gamma\sigma^z_{2k}\sigma^z_{2k+1}, $$ $$ B = -\sum_{k=0} \alpha\sigma^x_{2k+1}\sigma^x_{2k+2} + \beta\sigma^y_{2k+1}\sigma^y_{2k+2} + \gamma\sigma^z_{2k+1}\sigma^z_{2k+2}, $$ it is possible to write down the upper bound on 2nd order Trotter-Suzuki approximation $S_2(t)$ for evolution $e^{-i t H} = e^{-i t (A + B)}$: $$ S_2(t) = e^{-i (t/2) A} e^{-i t B} e^{-i (t/2) A}, $$ which had been derived in this paper as follows: $$ \left\|S_2(t) - e^{-i t H}\right\| \le \frac{t^3}{12}\left\|[B,[B,A]]\right\| + \frac{t^3}{24}\left\|[A,[A,B]]\right\|, $$ where $\left\|.\right\|$ stands for operator norm.

The problem: it is very difficult to obtain analytical expression for the right-hand side of the last equation for arbitrary $\alpha$, $\beta$, $\gamma$. On the other hand, its direct computation is impossible for large number of qubits, say, $n > 16$.

The question: does there exist any analytical approximation, asymptotic ($n \rightarrow \infty$), etc., which is easy to compute?

Answer 1

There are two aspects of your question that I will address separately: the computational and analytical. I believe the underlying issue is the same, but I will nonetheless address them separately.

Computational: You claim in your question that the computation is "impossible" for the number of qubits $n > 16$. I have to imagine your reasoning for the claim is thinking about the pauli strings involved as $2^{16} \times 2^{16}$ matrices.

Yet, there is another, often much better way to think of them: as abstract mathematical entities that obey certain algebraic rules. For example, to calculate $(\sigma_x \otimes \sigma_y \otimes \sigma_z)^2$, you could do the appropriate $8 \times 8$ matrix multiplication. But isn't it better to just note that $\sigma_i ^2 = I$?

Whenever there is a better conceptual way to do something, there is usually a better way to code it too. Indeed, instead of representing the operations involved as matrices, we could store $H$ as a list of paulis, with corresponding weights. Whenever we perform, say, commutators in our code, we can simply define it according to the algebraic rules of the paulis. This will generally be much faster than doing everything in terms of matrices, because for reasonable models in physics there will typically only be polynomially many pauli terms with respect to the number of qubits

Once you've computed the nested commutators this way, you'll have yet another list of paulis with associated weights. At this point don't try to compute the norm directly. A good, worst-case but generally tight upper bound can be achieved with the triangle inequality. $$ \left\| \sum_i \alpha_i P_i\right\| \leq \sum_i \vert \alpha_i\vert $$

Analytical: Though the computational procedure I described above will work, I think in your particular case we can get quite far analytically. For simplicity, I am going to assume $n$ is even and you are taking periodic boundaries, so that $(n-1) + 1 = 0$. Let's write $A = \sum_{k=0}^{n-1} A_k$ and similar for $B$, where $A_k$ is the term acting on qubits $k$ and $k+1$. We have $$ [A, B] = \sum_{k=0, \, \text{even}} [A_k, B]. $$ Crucially, the significant locality of the terms involved means there are only a couple terms in $B$ that do not commute with $A_k$. In fact, $$ [A, B] = \sum_{k=0, \, \text{even}} [A_k, B_{k-1}] + [A_k, B_{k+1}]. $$ We can evaluate these commutators straightforwardly using the algebraic properties of the paulis. Here is an example: $$ [\sigma_1^x \sigma_2^x, \sigma_2^y \sigma_3^y] = \sigma_1^x [\sigma_2^x, \sigma_2^y] \sigma_3^y = i \sigma_1^x \sigma_2^z \sigma_3^y $$ It's admittedly a bit tedious, but perhaps less so than you were expecting.