# The $N$th-order identity of the time-ordered exponential in Quantum Mechanics

In Quantum Mechanics, one often defines the time ordered exponential like e.g. here.

Now my question is how the factor of $N!$ arises. I know the simplex volume as the following integral: \begin{equation} \int_{t_0}^t dt_1 \int_{t_1}^{t} dt_2 \cdots \int_{t_{N-1}}^t dt_N = \frac{(t-t_0)^N}{N!}=\frac{1}{N!}\int_{t_0}^t dt_1 \int_{t_0}^{t} dt_2 \cdots \int_{t_{0}}^t dt_N \end{equation} However, I would like to know how to obtain the identity \begin{equation} \int_{t_0}^t dt_1 \int_{t_1}^{t} dt_2 \cdots \int_{t_{N-1}}^t dt_N~f(t_1)\cdots f(t_N) = \frac{1}{N!} \int_{t_0}^t dt_1\cdots \int_{t_{0}}^{t} dt_N~\mathbb{T}~(f(t_1)\cdots f(t_N)) \end{equation} where $\mathbb{T}$ is the time-ordering operator that acts as follows: \begin{equation} \mathbb{T}~(f(t_1)\cdots f(t_m))=f(t_{\pi(N)})\cdots f(t_{\pi(1)})\qquad\text{with}\qquad t_{\pi(N)}<\cdots< t_{\pi(1)}. \end{equation}

• I think you have a typo in your limits of integration for the time ordering formula. I've fixed the typo in my answer below – Jahan Claes Dec 12 '16 at 20:11
• Thanks. I have changed the time ordering definition now to $t_{\pi(N)}<\cdots< t_{\pi(1)}$, so that it should be correct. – exchange Dec 13 '16 at 9:39

## 1 Answer

We have that $\mathbb{T}~(f(t_1)\cdots f(t_N))=\mathbb{T}~(f(t_{\sigma(1)})\cdots f(t_{\sigma(N)}))$ for every permutation $\sigma$. We also know that if we have some function $F(t_1,...,t_N)$, then

$$\int_{t_0}^t dt_1\cdots \int_{t_{0}}^{t} dt_N~F(t_{\sigma(1)},...,t_{\sigma(N)}) =\sum_\sigma\int_{t_0}^t dt_1 \int_{t_0}^{t_1} dt_2 \cdots \int_{t_0}^{t_{N-1}} dt_N F(t_{\sigma(1)},...,t_{\sigma(N)})$$

We thus have

$$\begin{array}{rcl} \int_{t_0}^t dt_1\cdots \int_{t_{0}}^{t} dt_N~\mathbb{T}~(f(t_1)\cdots f(t_N))&=&\sum_{\sigma}\int_{t_0}^t dt_1 \int_{t_0}^{t_1} dt_2 \cdots \int_{t_0}^{t_{N-1}} dt_N~\mathbb{T}(f(t_{\sigma(1)})\cdots f(t_{\sigma(N)}))\\ &=&\sum_{\sigma}\int_{t_0}^t dt_1 \int_{t_0}^{t_1} dt_2 \cdots \int_{t_0}^{t_{N-1}} dt_N~\mathbb{T}(f(t_1)\cdots f(t_{N}))\\ &=&\sum_{\sigma}\int_{t_0}^t dt_1 \int_{t_0}^{t_1} dt_2 \cdots \int_{t_0}^{t_{N-1}} dt_N~f(t_1)\cdots f(t_{N})\\ &=&N!\int_{t_0}^t dt_1 \int_{t_0}^{t_1} dt_2 \cdots \int_{t_0}^{t_{N-1}} dt_N~f(t_1)\cdots f(t_{N})\\ \end{array}$$ from which the identity immediately follows.

• Thanks for the answer. It would be very nice if you could elaborate on step one, i.e. why can we rewrite $\int_{t_0}^t dt_1\cdots\int_{t_0}^t dt_N F(t_1,\cdots,t_N)=\sum_\sigma \int_{t_0}^t dt_1\int_{t_0}^{t_1} dt_2\cdots \int_{t_0}^{t_{N-1}} F(t_{\sigma(1)},\cdots,t_{\sigma(N)})$?. Thanks. – exchange Dec 13 '16 at 9:43
• Ah, I understood! The N-cube is the union of $N!$ N-simplexes. Now if we let $F(t_{\sigma(1)},\cdots,t_{\sigma(t_N)})$ go through all permutations with ordered times, then every value that the function could take on the N-cube, is taken on one of the Simplexes that make up the cube. And as this union is disjoint because every value that the tuple ($t_1,\cdots,t_n)$ can take only appears in one of these simplexes, the sum of all these permutations with ordered times then is the integral over the whole cube! This was, what I was still missing. – exchange Dec 13 '16 at 10:18