- #1

Markus Kahn

- 112

- 14

- 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?

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.

$$\partial_t \mathcal{T}\exp\left(-i S(t)\right)\quad \text{where}\quad S(t)\equiv\int_{t_0}^tdu \,H(u) .$$

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.