Time-ordering and Dyson series

Question

1. In Dyson series we use a time-ordered exponential by arguing that Hamiltonians at two different instants of time do not commute. Why is that so?

2. Can anyone explain with an example why should the same Hamiltonians at two different times not commute?

Answer 1

It is so, because Dyson formula was designed for the interaction picture in Quantum Mechanics. Interaction picture may seem strange, especially in comparison with the 'natural' Heisenberg picture, but has been proven useful, especially in perturbative computations.

The basic idea is to split the total Hamiltonian into two parts:

$$ H = H _0 + H _{int} $$

The evolution of operators in interaction picture is then governed by the free theory hamiltonian $H _0$:

$$ a _I (t) = e^{i H _0 (t - t _0)} a _I (t _0) e^{- i H _0 (t - t _0)}, $$

but states also have to evolve (or else observable matrix elements wouldn't be equivalent to the one calculated using the Heisenberg picture):

$$ \left| \Psi _I (t) \right> = U (t, t _0) \left| \Psi (t _0) \right> , $$

where the unitary state-evolution operator is

$$ U(t, t _0) = e^{i H _0 (t - t _0)} e^{-i H (t - t _0)} . $$

The interaction picture is so useful because operators subscripted with $I$ have the same equations of motion as Heisenberg-picture operators in the corresponding free theory. This is why we would like to express everything in terms of subscripted operators.

In particular, take a look at the interaction Hamiltonian in interaction picture:

$$ H _I (t) = e^{i H _0 (t - t _0)} H_{int} e^{- i H _0 (t - t _0)}. $$

This obviously depends on $t$, and therefore, in general,

$$ [ H _I (t _1); H _I (t _2) ] \neq 0. $$

Dyson formula relates the evolution operator $U (t, t _0)$ to the interaction Hamiltonian in interaction picture (which depends on time):

$$ U (t, t _0) = T \exp \left\{ - i \intop _{t _0} ^{t} H _I (t) dt \right\}. $$

Answer 2

Comments to the questions (v2):

1. The Hamiltonian $H(t)$ in QFT often depends on time $t$, e.g. if there are interactions in the interaction picture, or if there are external sources. In particular, Hamiltonians at different times need not commute. For an explicit example see e.g. my Phys.SE answer here.

2. Well, if the Hamiltonian $H(t_1)=H(t_2)$ is the same at two different times $t_1$ and $t_2$, then surely it commutes with itself $[H(t_1),H(t_2)]=0$ at these two time instances.

Answer 3

The Hamiltonian operator in quantum field theory is constructed by using the fields in question. For example, the Hamiltonian for a free scalar field reads \begin{equation} \notag H = \frac{1}{2} \int_V d^3x \left( \pi^2 + ( \partial_i \phi )^2 + m^2 \phi^2 \right) \, . \end{equation} The key observation is now that a given field evaluated at different times does not commute with itself $$ [\phi(t',\vec x) , \phi (t,\vec x)] \neq 0 \, . $$ Therefore, the Hamiltonian evaluated at two different instants of time does not commute: $$[H_i(t) , H_i(t')] \neq 0 . $$

(The field commutator only vanishes for spacelike separations but not for timelike separations. See, for example, page 37 in Tong's notes.)