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The unitary evolution of a time-dependent hamiltonian is given by the time-ordered matrix exponential $$\begin{aligned} U(t)&=\mathcal T\exp\left[-i\int_0^tH(\tau)d\tau\right]\\ &=I-i\int_0^td\tau\,H(\tau)-\frac12\int_0^td\tau\int_0^t d\tau'\,\mathcal T[H(\tau)H(\tau')]+\ldots\\ &=I-i\int_0^td\tau\,H(\tau)-\int_0^td\tau\int_0^\tau d\tau'\,H(\tau)H(\tau')+\ldots \end{aligned}$$ To implement $U(t)$ on a quantum computer, people often work with a trotterised form, where the integrals are discretized as Riemann sums. One typically encounters expressions such as (arXiv:1805.11568, eq. 1): $$\begin{aligned} U(t)%&\approx\exp[-i\sum_{a=0}^{k-1}H(a\Delta\tau)\Delta\tau]\\ &\approx\prod_{a=0}^{k-1}\exp[-iH(a\Delta\tau)\Delta\tau]~~~~(\Delta\tau=t/k) \end{aligned}$$ However, in this expression, the time ordering symbol $\mathcal T$ has somehow vanished.

What's going on here?

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    $\begingroup$ Well, you might not have the time-ordering symbol in eq. 1 but I would say you still have an implicit time-ordering prescription. The product of piecewise-constant unitary propagators in the right-hand side of eq. 1 is not commutative, so you have to choose a particular order. The correct one is with smaller times on the right, and increasingly larger times as you move to the left, so that you replicate the process described by the time-dependent Hamiltonian H(t). $\endgroup$ – Gianni Mossi Apr 13 at 20:37
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Ok, here's my attempt: take a time-dependent Hamiltonian $H(t)$ and consider its evolution in the time interval $[0,t]$. Discretize this interval in $k$ steps of length $\Delta \tau \equiv t/k$ $$ \tau_{n} \equiv n \Delta \tau, \qquad n = 0,1,\ldots,k-1. $$ Now consider the piecewise-constant product of the propagators $\exp(-i H(\tau_n)\Delta \tau)$ taken in the (right-to-left) time-order: \begin{equation}\tag{1}\label{eq:ordered} e^{-i H(\tau_{k-1}) \Delta \tau} e^{-i H(\tau_{k-2}) \Delta \tau} \cdots e^{-i H(\tau_{0}) \Delta \tau} \end{equation} then expand the exponentials in their power series and multiply the series together. You will get terms of the form

$$ X_{n_1,\ldots,n_{k}} = \frac{H(\tau_{k-1})^{n_1} H(\tau_{k-2})^{n_2} \cdots H(\tau_0)^{n_k}}{n_1! \, n_2! \cdots n_k!} (-i \Delta \tau)^{n_1+\ldots+n_k} $$ for $n_1,\ldots,n_k \in \mathbb{N}$. Now collect together the terms of the same order $m = n_1 + \ldots + n_k$. The first few terms of this series $\sum_{m=0}^{\infty} X_m$ are given by

\begin{align*} X_0 &= I \\ X_1 &= -i \sum_{i=0}^{k-1} H(\tau_i) \Delta \tau \xrightarrow{k \rightarrow \infty} -i \int_0^t H(\tau) \mathrm{d}\tau \\ X_2 &= -\sum_{i>j} H(\tau_i) H(\tau_j) (\Delta \tau)^2 - \sum_i \frac{H(\tau_i)^2}{2!} (\Delta \tau)^2 \end{align*}

Now consider the square $[0,t] \times [0,t] \subseteq \mathbb{R}^2$.

square

Observe that the first term of $X_2$ above is a Riemann sum of the operator-valued function \begin{equation}\tag{2}\label{eq:F1} F(\tau_1, \tau_2) \equiv H(\tau_1) H(\tau_2) \end{equation} in the interior of the lower-right triangle where $\tau_1 > \tau_2$. You want to extend the definition of $F(\tau_1, \tau_2) $ to the upper-left triangle in such a way that its integral over the upper-left triangle is the same as the integral in the lower triangle. This of course is done by using the time-ordering "operator" $\mathcal{T}$:

\begin{equation}\tag{3}\label{eq:F2} F(\tau_1, \tau_2) \equiv \mathcal{T} \Big[ H(\tau_1) H(\tau_2) \Big] = \begin{cases} H(\tau_1) H(\tau_2) & \text{if $\tau_1 > \tau_2$}\\ H(\tau_2) H(\tau_1) & \text{otherwise} \end{cases} \end{equation}

The operator $\mathcal{T}$ does nothing in the lower-right triangle because there the order $\tau_1 > \tau_2$ is respected (so Eq. \eqref{eq:F2} properly defines an extension of Eq. \eqref{eq:F1}), and "corrects" the order that applies in the other triangle (that is $\tau_2 > \tau_1$) so that the Riemann sums of $F$ over the two triangles are the same. Now you can use this fact $$ -\sum_{i \neq j} \mathcal{T}\Big[ H(\tau_i) H(\tau_j) \Big] (\Delta \tau)^2 = -2\sum_{i>j} \mathcal{T}\Big[ H(\tau_i) H(\tau_j) \Big] (\Delta \tau)^2 $$ and that $\mathcal{T}$ acts trivially if $\tau_i=\tau_j$ to rewrite $X_2$ above as \begin{align} X_2 & = -\sum_{i \neq j} \frac{1}{2!}\mathcal{T}\Big[ H(\tau_i) H(\tau_j) \Big] (\Delta \tau)^2 - \sum_{i} \frac{1}{2!} \mathcal{T}\Big[ H(\tau_i) H(\tau_i) \Big] (\Delta \tau)^2\\ & = -\sum_{i, j=0}^{k-1} \frac{1}{2!}\mathcal{T}\Big[ H(\tau_i) H(\tau_j) \Big] (\Delta \tau)^2 \xrightarrow{k \rightarrow \infty} \frac{(-i)^2}{2!} \int_0^t \int_0^t \mathcal{T} \Big[ H(\tau_1)H(\tau_2) \Big] \,\mathrm{d}\tau_1 \, \mathrm{d}\tau_2 \end{align}

At this point the generic case should be clear, at least in principle: at the $m$-th order you will have an $m$-dimensional cube $[0,t]^m \subseteq \mathbb{R}^m$ split in $m!$ simplices. Each simplex is defined by the specific ordering of the $m$ coordinates $\tau_1,\ldots,\tau_m$ that holds inside of that simplex. The $m$-th term $X_m$ in the series will contain a Riemann sum of $F(\tau_1,\ldots,\tau_m) \equiv H(\tau_1)\cdots H(\tau_m)$ taken over one of these simplices (call it the "fundamental simplex") where the $m$ coordinates $\tau_j$ are ordered according to the order given in Eq. \eqref{eq:ordered}, as well as some boundary terms. Then one has to extend the definition of $F$ over the full hypercube, while using the time-ordering operator $\mathcal{T}$ to "correct back" the order in each simplex to the order in the fundamental simplex. In this way the Riemann sum of the extension of $F$ in each simplex will be the same. After taking care of the additional boundary terms between the simplices, then one will be able to write $X_m$ as the full Riemann sum over $[0,t]^m$ of the operator-valued function $F(\tau_1,\ldots,\tau_m) \equiv \mathcal{T}[ H(\tau_1) \cdots H(\tau_m)]$, modulo the combinatorial factor $1/m!$ that corrects the overcounting obtained from integrating $F$ over all simplices instead of just one. By taking the $k \rightarrow \infty$ limit of the Riemann sum you get $$ X_m \xrightarrow{k \rightarrow \infty} \frac{(-i)^m}{m!} \int_{0}^t \cdots \int_{0}^t \mathcal{T}\Big[ H(\tau_1) \cdots H(\tau_m) \Big] \mathrm{d}\tau_1 \cdots \mathrm{d}\tau_m $$ This is precisely the $m$-th term of the Dyson series, whose sum defines the time-ordered exponential $$ U(t) = \mathcal{T} \exp\Big(-i\int_0^t H(\tau)\, \mathrm{d}\tau \Big) $$ so in the limit of $k \rightarrow \infty$ Eq. \eqref{eq:ordered} (in the right-to-left time ordering) correctly gives the unitary propagator generated by the time-dependent Hamiltonian $H(t)$.

All in all this procedure is the same as the one used in deriving the Dyson series. The only difference is that here we have to handle Riemann sums instead of intergals, and so we have some pesky boundary terms (and their combinatorics) to take care of but nothing serious.

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  • $\begingroup$ Aahh, I see how this works now. Thanks for the elaborate answer, this makes it completely clear! $\endgroup$ – Dyon J Don Kiwi van Vreumingen Apr 21 at 13:45
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The point of time-ordering is to take into account that a Hamiltonian may not commute with itself at different times, $[ H(t), H(t') ] \neq 0$. But the time-ordering operator is really just a reminder to keep track of the correct ordering when expanding the exponential as a Taylor-series (see the Dyson-series).

The whole point of the Trotter-Suzuki approximation, on the other hand, is to split an exponential of a sum of non-commuting operators into a product of exponentials, $e^{A+B} \approx (e^{A/n} e^{B/n})^n$.

Now, if you approximate integration as a Riemann sum, $$\int_0^t dt' H(t') \approx \sum_l H(l \Delta t) \Delta t \ ,$$ what you get is a series of non-commuting operators, $[H(l\Delta t), H((l+1)\Delta t)] \neq 0$. This is now exactly what you can use your Trotter approximation for: \begin{align} \mathcal{T}\exp\Bigl(-i \int_0^t dt' H(t') \Bigr) \approx \exp\Bigl(-i \sum_l H(l \Delta t) \Delta t \Bigr) \\ \approx \Biggl[ \prod_l \exp\Bigl(-i H(l \Delta t) \frac{\Delta t}{n} \Bigr) \Biggr]^n \ . \end{align}

EDIT:

I want to highlight that, generally, choosing the order in which you trotterize your exponential, i.e., whether you write $e^{A+B} \approx (e^{A/n} e^{B/n})^n$ or $e^{A+B} \approx (e^{B/n} e^{A/n})^n$, is arbitrary. However, this does not mean that the order is irrelevant. In fact, the accuracy of the approximation might vary quite significantly depending on the chosen order of factors.

That said, from a physical point of view, time-ordering should be implied in an expansion like the above and is well-motivated, as @Gianni Mossi pointed out in the comments with an intuitive example. In the above case, ordering should be with times increasing from right to left (i.e., operator at the smallest time applied to the state first).

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  • $\begingroup$ Hi @Durd3nT, I do agree that in the Suzuki-Trotter formula the order is irrelevant but then I would like to propose a thought experiment. Think of a time-dependent Hamiltonian H(t) where some potential term V is turned on continuously from zero to a fixed intensity as you go from time t=0 to time t=1. Now think of the inverse process, where the full potential V is present at t=0 and is turned off continuously as you go to time t=1. $\endgroup$ – Gianni Mossi Apr 14 at 14:34
  • $\begingroup$ These are clearly two distinct physical processes but if you take the time-ordering (in eq. 1 of the OP) to be arbitrary, then their propagators can approximated to arbitrary accuracy by the same operator (the product of piecewise-constant propagators with a large enough k, and opposite orderings). $\endgroup$ – Gianni Mossi Apr 14 at 14:35
  • $\begingroup$ Thanks, @Gianni Mossi. I agree, ordering a Trotter-Suzuki expansion may be arbitrary but it certainly makes sense to imply a time-ordering from a physical viewpoint! I will add this to my answer. $\endgroup$ – Durd3nT Apr 23 at 9:21

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