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After reading the previous posts related to the Dyson series, I have decided to open a new thread because there is something that I am still not understanding. It concerns the expression:

$$ ∫_{t_0}^{t}dt^{′}∫_{t_0}^{t^{′}}dt^{''}\hat{T}[\hat{H}(t^{'})\hat{H}(t^{''})] = $$ $$ ∫_{t_0}^{t}dt^{′}∫_{t_0}^{t^{′}}dt^{''}\hat{H}(t^{′})\hat{H}(t^{''}) + ∫_{t_0}^{t}dt^{''}∫_{t_0}^{t^{′}}dt^{'}\hat{H}(t^{''})\hat{H}(t^{′}) $$

that is assumed in many text books. I wonder if it can be derived from the definition of the Time-ordered operator:

$$ \hat{T}[ \hat{H}(t^{′}) \hat{H}(t^{''})]=θ(t^{′}−t^{''}) \hat{H}(t') \hat{H}(t^{''}) + θ(t^{''}−t^{'}) \hat{H}(t^{''}) \hat{H}(t^{'}) $$ and its natural extension to products of integrals: $$ \hat{T}∫_{t_0}^{t}dt^{′}∫_{t_0}^{t^{′}}dt^{''}\hat{H}(t^{'})\hat{H}(t^{''}) = ∫_{t_0}^{t}dt^{′}∫_{t_0}^{t^{′}}dt^{''}\hat{T}[\hat{H}(t^{'})\hat{H}(t^{''})] = $$ $$ = θ(t^{'}−t^{''}) ∫_{t_0}^{t}dt^{′}∫_{t_0}^{t^{′}}dt^{''}\hat{H}(t^{′})\hat{H}(t^{''}) + θ(t^{''}−t^{′}) ∫_{t_0}^{t}dt^{''}∫_{t_0}^{t^{′}}dt^{'}\hat{H}(t^{''})\hat{H}(t^{′}) $$

If I am right, the step-function θ$(t)$ must cancel one of the terms leading to: $$ ∫_{t_0}^{t}dt^{′}∫_{t_0}^{t^{′}}dt^{''}\hat{T}[\hat{H}(t^{'})\hat{H}(t^{''})] =∫_{t_0}^{t}dt^{′}∫_{t_0}^{t^{′}}dt^{''}\hat{H}(t^{′})\hat{H}(t^{''}) \qquad \text{if $t'>t^{''}$} $$ or: $$ ∫_{t_0}^{t}dt^{′}∫_{t_0}^{t^{′}}dt^{''}\hat{T}[\hat{H}(t^{'})\hat{H}(t^{''})] =∫_{t_0}^{t}dt^{''}∫_{t_0}^{t^{′}}dt^{'}\hat{H}(t^{''})\hat{H}(t^{′}) \qquad \text{if $t'< t^{''}$} $$ but in any case it leads to the combination of both: $$ ∫_{t_0}^{t}dt^{′}∫_{t_0}^{t^{′}}dt^{''}\hat{H}(t^{′})\hat{H}(t^{''}) + ∫_{t_0}^{t}dt^{''}∫_{t_0}^{t^{′}}dt^{'}\hat{H}(t^{''})\hat{H}(t^{′}) $$

What I am missing?

Thanks in advance

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up vote 2 down vote accepted

I think you problem lies in wrong integration bounds. The story goes roughly as follows. We would like to evaluate integrals such as $$ \int_{t_0}^{t_1} dt' \int_{t_0}^{t'} dt'' H(t'') H(t').$$ The annoying feature of this integral is that we have to keep track of integration bounds that ensure $t''$ is always greater than $t'$. We can instead rewrite this as $${1 \over 2} \left( \int_{t_0}^{t_1} dt'' \int_{t_0}^{t_1} dt' H(t'') H(t') \Theta (t'' - t') + \int_{t_0}^{t_1} dt'' \int_{t_0}^{t_1} dt' H(t') H(t'') \Theta(t' - t'') \right) = $$ $$ = {1 \over 2} \int_{t_0}^{t_1} dt'' \int_{t_0}^{t_1} dt' T \left[ H(t'') H(t') \right]$$ because both terms are equal. This form is much friendlier since the integration bounds of both integrals are now the same.

In general one has that $$\int_{t_0}^{t_f} dt_{n-1}' \cdots \int_{t_0}^{t_0'} dt_0' H(t_{n-1}') \cdots H(t_{0}') = {1 \over n!} \int_{t_0}^{t_f} dt_{n-1}' \cdots \int_{t_0}^{t_f} dt_0' T \left[ H(t_{n-1}') \cdots H(t_{0}') \right] $$ and finally $$ T \left[ \exp \left( \int_{t_0}^{t_f} dt' H(t') \right) \right] = \sum_{n=0}^{\infty} {1 \over n!} \int_{t_0}^{t_f} dt_{n-1}' \cdots \int_{t_0}^{t_f} dt_0' T \left[ H(t_{n-1}') \cdots H(t_{0}') \right]$$ which is (upto constants) the expansion of an evolution operator in quantum mechanics that can be derived from the Dyson's equation.

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Thanks, indeed I was messed-up with the integration bound. Could you please give some hints about the equality: $$ ∫_{t_0}^{t_1}dt^{′}∫_{t_0}^{t^{′}}dt^{''}\hat{H}(t^{''})\hat{H}(t^{'}) = $$ $$ = \frac{1}{2} \Big( ∫_{t_0}^{t_1}dt^{''}∫_{t_0}^{t_{1}}dt^{'}\hat{H}(t^{''})\hat{H}(t^{'}) θ(t^{''}−t^{′}) + ∫_{t_0}^{t_1}dt^{''}∫_{t_0}^{t_{1}}dt^{'}\hat{H}(t^{'})\hat{H}(t^{''}) θ(t^{'}−t^{''}) \Big) $$ can be formally derived. I have tried by using the diagrams of the previous post, but no way. Thanks again – Carlos Jul 28 '11 at 18:20
@Carlos: you're welcome. As for the derivation, it's essentially nothing else than splitting a rectangle into two right-angled triangles. Does it make sense? Picture would go a long way here :/ – Marek Jul 28 '11 at 18:28
Thanks Marek, after a lot of thinking I got it! – Carlos Jul 30 '11 at 15:46

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