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?

  • $\begingroup$ I am not sure if the bound is as hard as you think. Note that most terms in $A$ commute with most terms in $B$. Calculating $[A,B]$ should not be so challenging. $\endgroup$
    – Jacob
    Apr 7, 2023 at 15:45

1 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.

  • $\begingroup$ Hi Jacob. Thank you for suggestion. Triangle inequality is really what I need. The rest is just a tedious expansion with many terms, as you said. Let me try to work it out. P.S. I have no enough reputation to upvote the answer... $\endgroup$
    – Albert65
    Apr 7, 2023 at 16:54
  • $\begingroup$ @Albert65 I'm glad you found at least one part helpful. No need to upvote, but if you feel that your question has been addressed could you please accept the answer? $\endgroup$
    – Jacob
    Apr 7, 2023 at 17:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.