Why should I take the time order to solve this differential equation?

Question

In a textbook on quantum field theory I come across the following differential equation.

$$ i \partial_t U(t) = H(t) U(t) $$

I would say that the solution to this equation would be

$$U(t) = e^{-i \int_0^t dt H(t) }$$

Since

$$ \partial_t e^{-i \int_0^t dt H(t) } = \partial_t ( -i\int_0^t dt H(t))e^{-i\int_0^t dt H(t) } = -i H(t) e^{-i\int_0^t dt H(t) } = -i H(t) U(t)$$

exactly as desired.

However the solution in the textbook states:

$$ T(e^{-i\int_0^t dt H(t) })$$

where T stands for the time order parameter. In other words the solution should be

$$ U(t) = 1 -i \int_0^t dt_1 H(t_1) + \frac{1}{2}(-i)^2 \int_0^{t} \int_0^{t} dt_t dt_2 T(H(t_1)H(t_2)) + \cdots $$

where $T(H(t_1)H(t_2))$ equals $H(t_1)H(t_2)$ if $t_1<t_2$ and $H(t_2)H(t_1)$ otherwise.

So why is this the correct solution? What goes wrong in the reasoning above?

Answer 1

Your solution is only correct if the ODE is scalar, that is $U$ is one-dimensional, or all matrix values of $H(t)$ commute with each other.

In the more general non-commuting case the exponential formula is wrong and you need to apply the time order operation.

Answer 2

Of course I see.

we have

$$ \frac{d}{dt} e^{\int_0^t dt_1 H(t_1)} = \frac{d}{dt} ( 1 + \int_0^t dt_1 H(t_1) + \frac{1}{2} \int_0^t dt_1 \int_0^{t} dt_2 H(t_1) H(t_2) = 0 + H(t) + \frac{1}{2} ( (\int_0^t dt_1 H(t_1)) H(t) + H(t) (\int_0^t dt_2 H(t_2)) ) $$

In the comuting case the last two terms would be equal making sure that

$$ \frac{d}{dt} e^{\int dt H(t)} = H(t) e^{\int dt H(t)} $$

But in this case I cannot pull the $H(t)$ in $(\int_0^t dt_1 H(t_1)) H(t) $ to the left.

Enforcing the time ordering solves this.