Dyson's series and the time derivative

[Markus Kahn]

Homework Statement

First define $$U(t):=\mathcal{T} \exp\left(-i\int_{t_0}^t du\, H(u)\right),$$
where $\mathcal{T}$ is the time-ordering operator. Show that it satisfies the differential equation
$$i\partial_t U(t)= H(t)U(t)\quad U(t_0)=1.$$

Relevant Equations

Given above.

I'm having a hard time understanding how exactly to evaluate the expression}
$$\partial_t \mathcal{T}\exp\left(-i S(t)\right)\quad \text{where}\quad S(t)\equiv\int_{t_0}^tdu \,H(u) .$$
The confusing part for me is that if we can consider the following:
$$\partial_t \mathcal{T}\exp\left(-i S(t)\right) = \partial_t \mathcal{T}\left[ \sum_k \frac{(-i)^k}{k!}S^k(t)\right] =\partial_t \sum_k \frac{(-i)^k}{k!} \mathcal{T}\left[S^k(t)\right],$$
but since all the $S(t)$ are happening at the same time, the time ordering isn't doing anything, which seems quite wrong (in that case we wouldn't need it in the first place...). So where am I going wrong here?P.S. Since I had trouble understanding how exactly the time-ordering and the time-derivative influence each other I played around a bit and ended up proving the following:
Lemma
Let $A(t)$ and $B(u)$ be two, non-commuting Operators, than we have
$$\partial_t \mathcal{T}(A(t)B(u)) = \mathcal{T}(\partial_t A(t) B(u)) + \delta(t-u)[A(t),B(u)].$$
Not sure if it is useful for the exercise at hand.

 

[Antarres]

The definition you've given is missinterpreted. The exponential form of the time evolution operator is just formal, the definition is the Dyson series, where the product of interaction Hamiltonians is defined at different times, not at the same time. So the definition you started from is the final formal form of the Dyson's formula.
Expanding that you find:

$$\mathcal{T}\exp{\left(-i\int_{t_0}^t duH(u)\right)} = \sum_{k=0}^\infty \frac{(-i)^k}{k!}\int_{t_0}^t dt_1\dots dt_n \mathcal{T}\{H(t_1)\dots H(t_n)\}$$

So time ordering here is ordering the interaction Hamiltonians with their own times of interaction inside, integration variable can't be $t$ which you get in the final dependence when the integral is calculated, when you integrate something(definitely) the final function depends on the parameter/variable that's left, not on the integration variable.

Since you're going backwards from the formal exponent to the differential equation that is usually used as the beginning of the approach, it will be useful for you to show:
$$\int_{t_0}^t dt_1 \int_{t_0}^{t_1}dt_2 \dots \int_{t_0}^{t_{n-1}}dt_{n}H(t_1)\dots H(t_n)= \frac{1}{n!} \int_{t_0}^t dt_1dt_2\dots dt_n \mathcal{T}\{H(t_1)...H(t_n)\}$$

From that point, it should be straightforward to derive the differential equation you're looking for.