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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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What is $\theta$? – Qiaochu Yuan May 1 '11 at 13:04
@Yuan, sorry $\theta$ is the Heaviside function – wiso May 1 '11 at 13:11
If you check out the answer (…), you might find some inspiration for a proof. – Fabian May 1 '11 at 14:11
Although not a proof, there is a straightforward argument if you consider a not so opaque version of the time ordering operator, i.e $\mathcal{T}[A(t')B(t'')]=A(t')B(t'')$ if $t'>t''$ and $\mathcal{T}[A(t')B(t'')]=B(t'')A(t')$ if $t''>t'$ – Please Delete Account May 2 '11 at 5:16
@Approximist: I don't undertand you comment, you're repeating the definition I wrote above. – wiso May 2 '11 at 20:00
up vote 5 down vote accepted

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 $$\int_{t_o}^tdt_1\int_{t_o}^{t_1}dt_2\cdots\int_{t_o}^{t_{n-1}}dt_n H(t_1)\cdots H(t_n)=\frac{1}{n!} \int_{t_o}^tdt_1\int_{t_o}^{t}dt_2\cdots\int_{t_o}^{t}dt_n \mathcal{T}(H(t_1)\cdots H(t_n))$$


A picture might be worth more:enter image description here

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Thank Approximist, but I don't understand why my last expansion is incorrect, I simply split the inner integral and used the definition of $T$ product. Your first equality is exactly what I want to proof for $n=2$ – wiso May 3 '11 at 20:33
Consider the definition again. Now does this give you a hint? You are confusing the limits, this is an integration over two different domains, for n=2$$\mathop{{\int\!\!\!\!\int}}\limits_{t_0<t'<t''<t}dt'dt''H(t'')H(t') + \mathop{{\int\!\!\!\!\int}}\limits_{t_0<t''<t'<t}dt'dt''H(t')H(t'')$$ Imagine 2D plane with an axis $t'$ and $t''$. Choose two points $t_0$ and $t$ on both axes. This will make up a square divided equally by the diagonal line $t'=t''$. Below the diagonal line is a triangular region with $t_0<t'<t''<t$ and above it is $t_0<t''<t'<t$. – Please Delete Account May 5 '11 at 18:53
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. – Please Delete Account May 5 '11 at 18:59

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