# Exponential operator approximation: Suzuki-Trotter Expansion

One cannot solve the transition amplitude $$\langle{x}\vert e^{-iHt}\vert{y}\rangle{}$$ with $$H=H_0+V$$ by just applying the operators one after another on the bra/ket, because the free hamiltonian $$H_0$$ doesn’t commute with $$V$$ in general. This gets clear when you expand the exponential operator $$e^{-iH_0t}e^{-iVt}$$, which $$\neq e^{-i(H_0 + V)t}$$.

The book that I’m reading (Quantentheorie, G. Münster, p.364 , only in german) wants to approximate the exponential operator for tiny $$t=\epsilon$$ and suggests the following way:

$$e^{-i(H_0+V)\epsilon} = e^{-iV\frac{\epsilon}{2}}e^{-iH_0\epsilon}e^{-iV\frac{\epsilon}{2}} + \mathcal{O}(\epsilon^3).$$

Does anyone have an idea how this expression is obtained?

• Given the non-commutation of $H_0$ and $V$, why would it be exactly equal? Commented Dec 2, 2023 at 21:26
• At first I thought that the free Hamiltionian is nicely sandwiched between the V on the RHS, so it provides a symmetry just like the LHS does with the powers of the sum. But you are right, that is no proof at all that they are equal. The point is: I don’t know how that expression is obtained, especially the 3rd order approx. Commented Dec 2, 2023 at 21:50
• If you would like to put a name to this expansion, it is called the Suzuki-Trotter expansion. cf. docs.quantum.ibm.com/api/qiskit/qiskit.synthesis.SuzukiTrotter Commented Dec 3, 2023 at 8:10

Hint: Use the BCH formula multiple times: $$e^{tA}e^{tB}~\stackrel{\text{BCH}}{=}~e^{tA+tB+[tA,tB]/2 +{\cal O}(t^3)}~\stackrel{\text{BCH}}{=}~e^{tA+tB}e^{[tA,tB]/2}+{\cal O}(t^3),\tag{1}$$ $$e^{tB}e^{tA}~\stackrel{\text{BCH}}{=}~e^{tB+tA+[tB,tA]/2 +{\cal O}(t^3)}~\stackrel{\text{BCH}}{=}~e^{-[tA,tB]/2}e^{tA+tB}+{\cal O}(t^3).\tag{2}$$ So \begin{align}e^{tA}e^{2tB}e^{tA}~=~&e^{tA}e^{tB}e^{tB}e^{tA}\cr~\stackrel{(1)+(2)}{=}&~e^{tA+tB}e^{[tA,tB]/2}e^{-[tA,tB]/2}e^{tA+tB}+{\cal O}(t^3)\cr ~=~&e^{2tA+2tB}+{\cal O}(t^3).\end{align}\tag{3}
First (as I encourage you to explicitly check), expand the left hand side for small epsilon, keeping track of the ordering since (as you said) $$[H_0, V] \neq 0$$ in general $$\begin{eqnarray} e^{-i \epsilon (H_0+V)} &=& 1 - i\epsilon(H_0 + V) -\frac{\epsilon^2}{2}\left(H_0^2 + H_0 V + V H_0 + V^2\right) \\ && + \frac{i \epsilon^3}{6}\left(H_0^3 + H_0^2 V + H_0 V H_0 + V H_0^2 + H_0 V^2 + V H_0 V + V^2 H_0 + V^3\right) + O(\epsilon^4) \end{eqnarray}$$ Then expand the right hand side for small epsilon (again I encourage you to try this) $$\begin{eqnarray} e^{-\frac{i\epsilon}{2} V} e^{-i \epsilon H_0} e^{-\frac{i \epsilon}{2} V} &=& \left[1 - \frac{i \epsilon}{2} V - \frac{\epsilon^2}{8} V^2 + \frac{i \epsilon^3}{48} V^3 + \cdots \right] \left[1 - i\epsilon H_0 - \frac{\epsilon^2}{2}H_0^2 + \frac{i \epsilon^3}{6}H_0^3 + \cdots \right] \left[1 - \frac{i \epsilon}{2} V - \frac{\epsilon^2}{8} V^2 + \frac{i \epsilon^3}{48} V^3 + \cdots \right] \\ &=& 1 - i \epsilon(H_0 + V) - \frac{\epsilon^2}{2}\left(H_0^2 + H_0 V + V H_0 + V^2\right) \\ && + \frac{i\epsilon^3}{6}\left(H_0^3 + \frac{3}{2}H_0^2 V + \frac{3}{2} V H_0^2 + \frac{3}{4} H_0 V^2 + \frac{3}{2}V H_0 V + \frac{3}{4} V^2 H_0 + V^3 \right) + O(\epsilon^4) \end{eqnarray}$$
Evidently the expressions agree to $$O(\epsilon^2)$$ but not to $$O(\epsilon^3)$$. Indeed if you think about it, there's no way to get a cubic term like $$H_0 V H_0$$ from expanding out the second product of exponentials, but all possible cubic terms appear in the first.