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?

  • 4
    $\begingroup$ Given the non-commutation of $H_0$ and $V$, why would it be exactly equal? $\endgroup$
    – Ghoster
    Dec 2, 2023 at 21:26
  • $\begingroup$ 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. $\endgroup$
    – Xhorxho
    Dec 2, 2023 at 21:50
  • 3
    $\begingroup$ 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 $\endgroup$ Dec 3, 2023 at 8:10

2 Answers 2


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 $$ \begin{align}e^{tA}e^{2tB}e^{tA}~=~&e^{tA}e^{tB}e^{tB}e^{tA}\cr~\stackrel{(1)+(2)}{=}&~e^{tA+tB}e^{[tA,tB]/2}e^{-[tA,tB]/2}e^{tA+tB}+{\cal O}(t^3)\cr ~=~&e^{2tA+2tB}+{\cal O}(t^3).\end{align}\tag{3}$$

  • $\begingroup$ a really elegant proof, thanku $\endgroup$
    – Xhorxho
    Dec 2, 2023 at 22:12

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.


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