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One of the most important tool in quantum mechanics is the Dyson series because it is the basis of the perturbative theory. There is a step in the derivation that I can't understand.

$\{H(t_i)\}$ are not commuting operators. The $T$ product is defined as follow:

T[H(t)H(t')] = \theta(t-t')H(t)H(t') + \theta(t'-t)H(t')H(t) where $\theta$ is the Heaviside function. You can extend it to $n$ factors, ordering them so that later times ($t$) stand to the left of earlier times.

I need to proof that: $ \int_{-\infty}^{t} dt_1 \int_{-\infty}^{t_1} dt_2 \ldots \int_{-\infty}^{t_{n-1}} dt_n H(t_1)H(t_2)\ldots H(t_n) $ is equal to $\frac{1}{n!}\int_{-\infty}^{t} dt_1 \int_{-\infty}^{t} dt_2 \ldots \int_{-\infty}^{t} dt_n T[H(t_1)H(t_2)\ldots H(t_n)] $

I tried to start with $n=2$, then I think it's easy to use induction, but I'm stuck:

$\int_{-\infty}^{t} dt_1 \int_{-\infty}^{t} dt_2T[H(t_1)H(t_2)] = \int_{-\infty}^{t} dt_1 \int_{-\infty}^{t_1} dt_2 H(t_1)H(t_2) + \int_{-\infty}^{t} dt_1 \int_{t_1}^{t} dt_2 H(t_2)H(t_1)$

but now? I tried to change variables ... can someone help me?

I tried also to visualized it as the integral over a square $(-\infty, t_1=t]\times (-\infty, t_2=t]$ subdivided into two triangles by the diagonal $t_2=t_1$ of an operator $K(t_1,t_2) = K(t_2,t_1)$ because $T[H(t_1)H(t_2)] = T[H(t_2)H(t_1)]$

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    @Approximist: I don't undertand you comment, you're repeating the definition I wrote above.2011-05-02

1 Answers 1

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Your last expansion is incorrect. For the simplest case $\int_{t_o}^tdt'\int_{t_o}^{t'}dt'' H(t')H(t'')=\frac{1}{2}\int_{t_0}^tdt'\int_{t_0}^tdt''\mathcal{T}(H(t')H(t''))$ The domain of integration was the triangle for the former expression while for the latter it is a square.

For three dimensions, the left expression would be integrated over a tetrahedron and the right expression would be integrated over a cube, which is 6 times the volume of the tetrahedron.

In general, the region of integration of the second expression would be a hypercube of $n$ dimensions, which has $n!$ of such simplexes. And so $\int_{t_o}^tdt_1\int_{t_o}^{t_1}dt_2\cdots\int_{t_o}^{t_{n-1}}dt_n H(t_1)\cdots H(t_n)=\frac{1}{n!} \int_{t_o}^tdt_1\int_{t_o}^{t}dt_2\cdots\int_{t_o}^{t}dt_n \mathcal{T}(H(t_1)\cdots H(t_n))$

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A picture might be worth more:enter image description here

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    When you do the integral $\int_{t_o}^tdt'\int_{t_o}^tdt''\mathcal{T}(H(t')H(t''))$ you integrate over the entire square $\{(t',t''):t_0\leq t'\leq t \cap t_0\leq t''\leq t\}$. When you use the definition of the time ordering operator and expand as I have done in previous comment, you integrate over the two triangles.2011-05-05