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Immersed within the world of Physics (Quantum Theory in particular), I came across the following proof for connecting the starting LHS with the finishing RHS.

$$ \begin{eqnarray} \int_{t_0}^tdt_1 \int_{t_0}^{t_1} dt_2V(t_1)V(t_2) &=& \int_{t_0}^tdt_2 \int_{t_2}^{t} dt_1V(t_1)V(t_2)\\ &=&\int_{t_0}^tdt_1 \int_{t_1}^{t} dt_2V(t_2)V(t_1)\\ &=&\frac{1}{2}\left[\int_{t_0}^tdt_1 \int_{t_1}^{t} dt_2 + \int_{t_0}^tdt_2 \int_{t_0}^{t_1} dt_1\right]V(t_1)V(t_2)\\ &=&\frac{1}{2}\int_{t_0}^tdt_1 \int_{t_0}^{t} dt_2V(t_1)V(t_2)\\ \end{eqnarray} $$

Unfortunately, I've had limited mathematical training (for mathematical physics) and I don't understand the motivation for all of the steps involved. The second line is achieved from the first by change of variables $t_1 \longleftrightarrow t_2$, but intuitively, how would one get from the LHS 1st to RHS 1st and 2nd to 3rd to 4th?

Clearly, I'm missing the fundamentals, so all comments would be welcome.

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2 Answers 2

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I remember first encountering this sort of integral and wondering why there wasn't a more mechanical way to handle them. Then I came across the Iverson bracket. We find $$\begin{eqnarray*} \int_{t_0}^t dt_1 \int_{t_0}^{t_1} dt_2\, V(t_1)V(t_2) &=& \int_{t_0}^t dt_1 \int_{t_0}^t dt_2 \, V(t_1)V(t_2)[t_2 \leq t_1] \\ &=& \int_{t_0}^t dt_1 \int_{t_0}^t dt_2 \, V(t_1)V(t_2) [t_1 \leq t_2] \hspace{5ex}\textrm{(change variables)} \\ &=& \frac{1}{2} \int_{t_0}^t dt_1 \int_{t_0}^t dt_2 \, V(t_1)V(t_2) \left([t_2 \leq t_1] + [t_1 \leq t_2]\right) \\ &=& \frac{1}{2} \int_{t_0}^t dt_1 \int_{t_0}^t dt_2 \, V(t_1)V(t_2). \hspace{5ex}\textrm{(a.e.)} \end{eqnarray*}$$

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+1, for reasons I think you already know. ;) –  J. M. May 1 '12 at 8:30
    
Hey, loving the Iverson bracket notation, thanks. –  TCTopCat May 1 '12 at 9:23
    
@J.M.: I haven't played around with the notation for integrals too much, but so far I like it. It requires a little thought to use the notation "optimally." It seems that constraints that don't change throughout the problem should just be written separately in the simplest way possible to avoid clutter, and the bracket should only be used for the "beef." For example, I originally started with $[t_0\leq t_1\leq t][t_0\leq t_2\leq t_1]$, etc., which is harder to parse. –  user26872 May 1 '12 at 18:18
    
In the same way as with sums, you can effectively treat the integrals with the Iverson brackets as doubly infinite integrals, since the brackets zero out contributions outside the region described within the brackets. –  J. M. May 1 '12 at 18:21
    
@TCTopCat: Glad to help. Here are some other examples of the use of the Iverson bracket on math.stackexchange. –  user26872 May 1 '12 at 18:22

The first line is easiest to see by drawing the domain of integration on the $t_1,t_2$ plane: it is a right triangle. To go from the 2nd to 3rd, observe that the LHS of the first line is equal to the 3rd line, therefore each is equal to the average of the two. To go from the 3rd line to the last line, swap variables on the 2nd term on the 3rd line, and obtain

\begin{eqnarray} & & \frac{1}{2} \left[ \int_{t_0}^t dt_1 \int_{t_1}^t dt_2 + \int_{t_0}^t dt_1 \int_{t_0}^{t_1} dt_2 \right] V(t_1)V(t_2) \\ & = & \frac{1}{2} \left[ \int_{t_0}^t dt_1 \left( \int_{t_1}^t dt_2 + \int_{t_0}^{t_1} dt_2\right) \right] V(t_1) V(t_2) \\ & = & \frac{1}{2} \int_{t_0}^{t_1} dt_1 \int_{t_0}^t dt_2 V(t_1) V(t_2). \end{eqnarray}

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Whenever there is a question about endpoints in a double integral, it is almost always a good idea to sketch the domain. –  Robert Israel May 1 '12 at 2:06
    
I'm sorry, but I'm not quite sure how we can assume that the domain is initially a right triangle? –  TCTopCat May 1 '12 at 8:29
    
Sorry I looked at it again, I see what you mean - I should really think before commenting! –  TCTopCat May 1 '12 at 9:22

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