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

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


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

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?


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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.