1

I have to proove that $$\hat{U}(t,t_{0})=T\bigg\{\mathrm{exp}\bigg (\int_{t_{0}}^{t}\hat{H}_{\mathrm{int}}^{I}(\tau)\,\mathrm{d}\tau\bigg )\bigg\}\tag{1}$$

is a solution of the Schrödinger equation:

$$i\partial_{t}\hat{U}(t,t_{0})=\hat{H}_{\mathrm{int}}^{I}\hat{U}(t,t_{0})\tag{2}$$

The proof that this is the case is very simple, when I am allowed to interchange the time-ordering operator with the time derivative....but why am I allowed to do this?

Qmechanic
  • 220,844

3 Answers3

3

  1. OP asks a valid question. It is presumably spurred by the well-known fact that changing the order of time differentiation and time ordering may produce contact terms.

  2. So in order to evaluate the derivative $$\frac{\partial }{\partial t_2}U(t_2,t_1)\tag{A}$$ of the time evolution operator $$ U(t_2,t_1)~=~T\exp\left[-\frac{i}{\hbar}\int_{t_1}^{t_2}\! \mathrm{d}t~H(t)\right],\tag{B}$$ we are not just differentiating under the time ordering symbol $T$.

  3. Instead we are considering the difference quotient $$\frac{U(t_2+\delta t,t_1) - U(t_2,t_1)}{\delta t} ~\stackrel{(D)}{=}~ \frac{U(t_2+\delta t,t_2) - {\bf 1} }{\delta t}U(t_2,t_1), \tag{C}$$ where we used the group property $$ U(t_3,t_1)~=~U(t_3,t_2)U(t_2,t_1) \tag{D} $$ of the time evolution operator.

  4. Now taking the limit $\delta t\to 0$ in eq. (C) yields OP's sought-for time-dependent Schrödinger equation (TDSE) $$\frac{\partial }{\partial t_2}U(t_2,t_1) ~\stackrel{(B)+(C)}{=}~ -\frac{i}{\hbar}H(t_2)U(t_2,t_1).\tag{E}$$

  5. For more details, see e.g. my related Phys.SE answer here.

Qmechanic
  • 220,844
0

The formal expression of the Dyson series is:
$\hat U(t, t_0) = T \{ exp [ -i \int_{t_0}^t dt' \hat H_{int}(t') ] \}$
where $T\{\}$ is the time-ordering operator.

To demonstrate that it is the solution to the Schroedinger equation you are allowed to take the derivative as the time-ordering just places the operators within the brackets in order regardless of whether they commute or not.

Michele Grosso
  • 3,874
0

A pedestrian way is simply solving the equation by the method of successive approximations: $$ i\partial_t U(t,t_0)=HU(t,t_0)\Leftrightarrow U(t,t_0)=-i\int_{t_0}^tdt_1 H(t_1)U(\tau,t_0),\\ U^{(0)}(t,t_0)=1,\\ U^{(1)}(t,t_0)=-i\int_{t_0}^tdt_1 H(t_1)U^{(0)}(\tau,t_0)=-i\int_{t_0}^tdt_1 H(t_1),\\ U^{(2)}(t,t_0)=-i\int_{t_0}^tdt_2 H(t_2)U^{(1)}(t_2,t_0)= (-i)^2\int_{t_0}^td t_2\int_{t_0}^{t_2}d t_1 H(t_2)H(t_1),\\ ...,\\ U^{(n)}(t,t_0)=-i\int_{t_0}^tdt_n H(t_n)U^{(n-1)}(t_n,t_0)=\\ (-i)^n\int_{t_0}^tdt_n \int_{t_0}^{t_n}d t_{n-1}...\int_{t_0}^{t_3}d t_2 \int_{t_0}^{t_2}d t_1 H(t_n)H(t_{n-1})...H(t_2)H(t_1),\\ U(t,t_0)=\sum_{n=0}^{\infty}U^{(n)}(t,t_0) $$

Now we can transform every term, e.g., $$ U^{(2)}(t,t_0)= (-i)^2\int_{t_0}^td t_2\int_{t_0}^{t_2}d t_1 H(t_2)H(t_1)=\\ (-i)^2\int_{t_0}^td t_1\int_{t_1}^{t}d t_2 H(t_2)H(t_1)= (-i)^2\int_{t_0}^td t_2\int_{t_2}^{t}d t_1 H(t_1)H(t_2)=\\ \frac{(-i)^2}{2}\int_{t_0}^td t_2\int_{t_0}^{t_2}d t_1 H(t_2)H(t_1) + \frac{(-i)^2}{2}\int_{t_0}^td t_2\int_{t_2}^{t}d t_1 H(t_1)H(t_2)=\\ \frac{(-i)^2}{2}\int_{t_0}^td t_2\int_{t_0}^{t_2}d t_1 T\left[H(t_2)H(t_1)\right] + \frac{(-i)^2}{2}\int_{t_0}^td t_2\int_{t_2}^{t}d t_1 T\left[H(t_2)H(t_1)\right]= \\ \frac{(-i)^2}{2}\int_{t_0}^td t_2\int_{t_0}^{t}d t_1 T\left[H(t_2)H(t_1)\right]$$ (I leave proving it for an arbitrary order as a homework exercise... surely a tedious one, but not scary to a physicist.) The full evolution operator then collapses to the exponent: $$ U^{(n)}(t,t_0)= \frac{(-i)^n}{n!}\int_{t_0}^tdt_n \int_{t_0}^td t_{n-1}...\int_{t_0}^td t_2 \int_{t_0}^ttd t_1 T\left[H(t_n)H(t_{n-1})...H(t_2)H(t_1)\right],\\ U(t,t_0)=\sum_{n=0}^{\infty}U^{(n)}(t,t_0)=T\exp\left[-i\int_{t_0}^t dt_1 H(t_1)\right] $$

Roger V.
  • 68,984