# Virtual terms in the Dyson series (time dependent perturbation theory)

Let the interaction evolution operator in the interaction picture be

$$U_I(t,t_0)=T \exp \Big( -i \int_{t_0}^t dt_1 H_I(t_1) \Big) ,$$

where $$T$$ is the time order operator and $$H_I=H-H_0$$ is the interaction hamiltonian which is supposed to be "small". If $$| i\rangle$$ is a eigenstate of $$H_0$$ at a time $$t_0$$ with eigenvalue $$E_i$$ we can write the evolution of this state at a time $$t$$ as

$$| n \rangle (t) = U_I(t,t_0) | i\rangle= \Big( 1-i \int_{t_0}^t dt_1 H_I(t_1) -\int_{t_0}^t dt_1 H_I(t_1) \int_{t_0}^{t_1} dt_2 H_I(t_2) + \cdots\Big)| i\rangle \\ = | i\rangle - \sum_j | j\rangle \frac{\langle j|H_{int}| i\rangle}{E_j-E_i} + \sum_{f,j} | j\rangle \frac{\langle j|H_{int}| f\rangle \langle f|H_{int}| i\rangle}{(E_j-E_i)(E_f-E_i)} + \cdots ,$$

where $$H_{int}$$ a time independent interaction hamiltonian in the Schrodinger picture (see https://en.wikipedia.org/wiki/Perturbation_theory_(quantum_mechanics)#Time-dependent_perturbation_theory for details)

My question is: Looking at the second order expansion in the r.h.s. of last equation what happens when the initial state and the final state are the same ($$j=i$$), that is,when we have a virtual contribution?

I know that the anwer for this term should be a contribution

$$-\frac{1}{2} \sum_f |i \rangle \frac{\langle i|H_{int}| f\rangle \langle f|H_{int}| i\rangle}{(E_f-E_i)^2}$$

which is what I get if, when doing the expansion of the evolution operator in the second equation, I suppose that $$t_1=t_2$$ and therefore the time order opertor is $$T=1$$. So this implies (talking in terms of particle physics) that, in a virtual term in which the initial and final state are the same, we have to take the time in which the virtual particle is emitted and the time in which it is absorbed to be the same?