In the Time Evolution module, we learned about Trotterization, which is based on the Trotter-Suzuki decomposition $$ e^{\hat{A} + \hat{B}} = \left(e^{\frac{\hat{A}}{n}}e^{\frac{\hat{B}}{n}}\right)^n + O\left(\frac{1}{n}\right). $$ This method allows us to approximate the time evolution for Hamiltonians that can be expressed as a sum of Hermitian operators. We relied on approximating to order $\tfrac{1}{n},$ whose coefficient we did not specify. We would like to see we can make better approximations, so we will now analyze this coefficient explicitly using a powerful tool called the **Baker-Campbell-Hausdorff formula** (in **Zassenhaus** form), $$ e^{\hat{A}+\hat{B}} = e^{\hat{A}} e^{\hat{B}}\exp\left(-\frac{1}{2}[\hat{A}, \hat{B}]\right)\exp\left(\frac{1}{2}[\hat{B}, [\hat{A}, \hat{B}]]+ \frac{1}{6}[\hat{A}, [\hat{A}, \hat{B}]]\right)\cdots , \tag{1} $$ where $[\hat{A}, \hat{B}] = \hat{A}\hat{B}- \hat{B}\hat{A}$ is the **commutator** of $\hat{A}$ and $\hat{B}$ and the $\cdots$ on the right denotes a product of exponentials of higher-order nested commutators. In essence, the commutator $[\hat{A}, \hat{B}]$ distinguishes the difference in applying $\hat{A}$ then $\hat{B}$ versus the opposite order. If $\hat{A}$ and $\hat{B}$ *commute*, the order in which they are applied doesn't change anything, and $[\hat{A}, \hat{B}] = 0$. As an illustration of how this works, let us apply it to the Trotter-Suzuki formula. First, because an operator $A$ commutes with itself, $[\hat{A}, \hat{A}] = 0$, $$ \left(e^{\hat{A}}\right)^n = e^{n\hat{A}}. $$ Then, $$ \begin{align*} e^{\hat{A}+\hat{B}} & = e^{\tfrac{\hat{A}+\hat{B}}{n} n} \\ & = \left(e^{\frac{\hat{A}}{n}} e^{\frac{\hat{B}}{n}}\exp\left(-\frac{1}{2n^2}[\hat{A}, \hat{B}]\right)\cdots\right)^n, \end{align*} $$ where we used $(1)$ on the second line. When $n$ is large, the first two terms in this product dominate, and we recover the Trotter-Suzuki formula $$ e^{\hat{A}+\hat{B}} = \lim_{n\to\infty} \left(e^{\frac{\hat{A}}{n}}e^{\frac{\hat{B}}{n}}\right)^n. $$ When we go to the next order in $1/n,$ the result will be the **second-order Trotter formula**, $$ e^{\hat{A} + \hat{B}} = \left(e^{\frac{\hat{A}}{2n}}e^{\frac{\hat{B}}{2n}}e^{\frac{\hat{B}}{2n}}e^{\frac{\hat{A}}{2n}}\right)^n + O\left(\frac{1}{n^2}\right). $$ This provides a Trotterization circuit that yields the time evolution with better precision, in exchange for using more gates. **Exercise HS.1.1.** Use the Zassenhaus expansion $(1)$ to confirm the second order Trotter formula $$ e^{\hat{A} + \hat{B}} = \left(e^{\frac{\hat{A}}{2n}}e^{\frac{\hat{B}}{2n}}e^{\frac{\hat{B}}{2n}}e^{\frac{\hat{A}}{2n}}\right)^n + O\left(\frac{1}{n^2}\right). $$

Hint.

Exploit the fact that $$ [\hat{A}, \hat{B}] = \hat{A}\hat{B}- \hat{B}\hat{A}= \hat{B}\hat{A}- \hat{A}\hat{B}= -[\hat{B}, \hat{A}]. $$

Solution.

From Zassenhaus and the Taylor expansion (using similar logic to part (a)), we immediately have $$ \begin{align*} e^{\frac{\hat{A} + \hat{B}}{2n}} & = e^{\frac{\hat{A}}{2n}}e^{\frac{\hat{B}}{2n}}e^{-[\hat{A}, \hat{B}]/8n^2}\cdots \\ & = e^{\frac{\hat{A}}{2n}}e^{\frac{\hat{B}}{2n}}\left(I - \frac{[\hat{A}, \hat{B}]}{8n^2}\right) + O\left(\frac{1}{n^3}\right). \end{align*} $$ Swapping the roles of $\hat{A}$ and $\hat{B}$, $$ e^{\frac{\hat{B} + \hat{A}}{2n}} = e^{\frac{\hat{B}}{2n}}e^{\frac{\hat{A}}{2n}}\left(I + \frac{[\hat{A}, \hat{B}]}{8n^2}\right) + O\left(\frac{1}{n^3}\right), $$ where we used the hint that $[\hat{B}, \hat{A}] = -[\hat{A}, \hat{B}]$. Taking the product and raising to the power $n$, we then obtain $$ \begin{align*} e^{\hat{A}+\hat{B}} & = e^{\frac{1}{2n}(\hat{A} + \hat{B}) n}e^{\frac{1}{2n}(\hat{B} + \hat{A}) n} \\ & = \left(e^{\frac{\hat{A}}{2n}}e^{\frac{\hat{B}}{2n}}\left(I - \frac{[\hat{A}, \hat{B}]}{8n^2}\right)e^{\frac{\hat{B}}{2n}}e^{\frac{\hat{A}}{2n}}\left(I + \frac{[\hat{A}, \hat{B}]}{8n^2}\right) + O\left(\frac{1}{n^3}\right)\right)^n \\ & = \left(e^{\frac{\hat{A}}{2n}}e^{\frac{\hat{B}}{2n}}e^{\frac{\hat{B}}{2n}}e^{\frac{\hat{A}}{2n}} + e^{\frac{\hat{A}}{2n}}e^{\frac{\hat{B}}{2n}}e^{\frac{\hat{B}}{2n}}e^{\frac{\hat{A}}{2n}} \frac{[\hat{A}, \hat{B}]}{8n^2} - e^{\frac{\hat{A}}{2n}}e^{\frac{\hat{B}}{2n}}\frac{[\hat{A}, \hat{B}]}{8n^2} e^{\frac{\hat{B}}{2n}}e^{\frac{\hat{A}}{2n}} + O\left(\frac{1}{n^3}\right)\right)^n. \end{align*} $$ Using the fact that $e^{\frac{\hat{A}}{2n}}$ and $e^{\frac{\hat{B}}{2n}}$ commute up to order $\tfrac{1}{n}$ and $[\hat{A}, \hat{B}]$ has a commutator with $e^{\frac{\hat{A}}{n}}$ and $e^{\frac{\hat{B}}{n}}$ of order $\tfrac{1}{n}$, we can move the $[\hat{A}, \hat{B}]/8n^2$ coefficient in the second and third terms to the front as follows: $$ e^{\hat{A}+\hat{B}} = \left(e^{\frac{\hat{A}}{2n}}e^{\frac{\hat{B}}{2n}}e^{\frac{\hat{B}}{2n}}e^{\frac{\hat{A}}{2n}} - \frac{[\hat{A}, \hat{B}]}{8n^2}(e^{\frac{\hat{B}}{n}}e^{\frac{\hat{A}}{n}} - e^{\frac{\hat{A}}{n}}e^{\frac{\hat{B}}{n}}) + O\left(\frac{1}{n^3}\right)\right)^n, $$ where we also moved factors of $e^{\frac{\hat{A}}{2n}}$ and $e^{\frac{\hat{B}}{2n}}$ around to combine them with the same factors, creating factors of $e^{\frac{\hat{A}}{n}}$ and $e^{\frac{\hat{B}}{n}}$. All resulting $O\left(\tfrac{1}{n}\right)$ contributions from every move are multiplied by $\tfrac{1}{n^2}$ (from $[\hat{A}, \hat{B}]/8n^2$), moving all non-commuting remnants into the $O\left(\tfrac{1}{n^3}\right)$ term. The middle term in the smaller brackets is simply $[ e^{\frac{\hat{B}}{n}}, e^{\frac{\hat{A}}{n}} ]$, which we know is $O\left(\tfrac{1}{n}\right)$. This also is multiplied by $\tfrac{1}{n^2}$, which moves the entire term into $O\left(\tfrac{1}{n^3}\right)$. We therefore find that $$ \begin{align*} e^{\hat{A}+\hat{B}} & = \left(e^{\frac{\hat{A}}{2n}}e^{\frac{\hat{B}}{2n}}e^{\frac{\hat{B}}{2n}}e^{\frac{\hat{A}}{2n}} + O\left(\frac{1}{n^3}\right)\right)^n \\ & = \left(e^{\frac{\hat{A}}{2n}}e^{\frac{\hat{B}}{2n}}e^{\frac{\hat{B}}{2n}}e^{\frac{\hat{A}}{2n}}\right)^n + O\left(\frac{1}{n^2}\right). \end{align*} $$ Note that the second line is obtained by using the binomial expansion of the first line, where the result is changing the multiplicative correction $O\left(\tfrac{1}{n^3}\right)$ to an additive correction $O\left(\tfrac{1}{n^2}\right)$.

--- This second order formula turns out to be just one in a family of product formulas. Let's see some higher order formulas.

Codercise HS.1.1 - Trotter error

Codercise HS.1.2 - Second order Trotter formula

Codercise HS.1.3 - Generalized Trotterization

Codercise HS.1.4 - Optimizing steps

Codercise HS.1.5 - PennyLane's TrotterProduct

What is Hamiltonian Simulation?

Product formulas

Higher-order Trotterization