Proving Feynman Identity with Induction

[latentcorpse]

I'm supposed to use the relationship [itex]A^{-1}=\int_0^\infty d \alpha e^{-\alpha A}[/itex] to show that

[itex]\frac{1}{A_1 A_2 \dots A_n}=(n-1)! \int_0^1 dx_1 \dots \int_0^1 dx_n \frac{\delta(1-x_1- \dots - x_n)}{(x_1A_1 + \dots + x_nA_n)^n}[/itex]

I decided that I should try and do this inductively.

So far I have managed to prove the base case n=2 successfully.

Then I made the inductive hypothesis that it will hold for [itex]n=k-1[/itex]

And then I considered [itex]n=k[/itex] and got

[itex]\frac{1}{A_1 \dots A_{k-1}A_k}=\frac{1}{A_1 \dots A_{k-1}} \frac{1}{A_k} = (k-2)! \int_0^1 dx_1 \dots \int_0^1 dx_{k-1} \frac{\delta(1-x_1 - \dots - x_{k-1})}{(x_1A_1 + \dots + x_{k-1}A_{k-1})^{k-1}} \times \int_0^\infty d \alpha e^{-\alpha A_k}[/itex]

However, I don't seem to have any way of going anywhere from here?

Thanks for any help.

 

[fzero]

You don't want to use induction. Write

[tex]
\frac{1}{A_1\cdots A_n} = \int_0^\infty d\alpha_1 \cdots \int_0^\infty d\alpha_n e^{-\sum_i \alpha_i A_i}[/tex]

and change to projective coordinates

[tex]\alpha_i = \alpha x_i, ~~\sum_i x_i=1.[/tex]

 

[latentcorpse]

You don't want to use induction. Write

[tex]
\frac{1}{A_1\cdots A_n} = \int_0^\infty d\alpha_1 \cdots \int_0^\infty d\alpha_n e^{-\sum_i \alpha_i A_i}[/tex]

and change to projective coordinates

[tex]\alpha_i = \alpha x_i, ~~\sum_i x_i=1.[/tex]

So that gives [itex]\frac{1}{A_1 \dots A_n} = \int_0^\infty d \alpha_1 \dots \int_0^\infty e^{-\alpha \sum_i x_i A_i}[/itex]

So I reckon that the sum to 1 gives the 1 in the delta function but I don't know what the connection is.

Also, I tried changing the variable of integration using [itex]\alpha_i = \alpha x_i[/itex] and [itex]d \alpha_i = dx_i[/itex] - whilst this keeps the lower limit of integration at 0, the upper one becomes [itex]\frac{\infty}{\alpha}=\infty \neq 1[/itex] as required - what am i doing wrong here?

 

[fzero]

The [tex]x_i[/tex] range from 0 to 1 from the constraint. Also with the contraint, the measure is

[tex] d\alpha_1 \cdots d\alpha_n = d\alpha dx_1\cdots dx_n \delta\left( \sum_i x_i -1 \right).[/tex]

 

[latentcorpse]

The [tex]x_i[/tex] range from 0 to 1 from the constraint. Also with the contraint, the measure is

[tex] d\alpha_1 \cdots d\alpha_n = d\alpha dx_1\cdots dx_n \delta\left( \sum_i x_i -1 \right).[/tex]

I don't really follow.

Surely the [itex]x_i[/itex] just have to sum to 1. So couldn't we have them being say 4 and -3 (of a n=2 example)? These sum to 1 and are both outside the range 0 to 1. Or are they constrained to being positive since [itex]\alpha_i[/itex] has a positive range?

Secondly, where did that measure come from?

 

[fzero]

I don't really follow.

Surely the [itex]x_i[/itex] just have to sum to 1. So couldn't we have them being say 4 and -3 (of a n=2 example)? These sum to 1 and are both outside the range 0 to 1. Or are they constrained to being positive since [itex]\alpha_i[/itex] has a positive range?

Yes, you have to look at the ranges of the original variables and that [tex]0<\alpha<\infty[/tex].

Secondly, where did that measure come from?

If you compute

[tex] d\alpha_i = x_i d\alpha + \alpha d x_i [/tex]

you can compute the new volume form and obtain the new measure. However, you can reproduce this by integrating over the [tex](n+1)^\text{th}[/tex] variable with a delta function constraint. As a simple example, you could try to show that

[tex] \int_0^\infty dx \int_0^\infty dy \delta(x^2+y^2 -1) [/tex]

gives the correct measure on the unit circle.

 

[latentcorpse]

Yes, you have to look at the ranges of the original variables and that [tex]0<\alpha<\infty[/tex].



If you compute

[tex] d\alpha_i = x_i d\alpha + \alpha d x_i [/tex]

you can compute the new volume form and obtain the new measure. However, you can reproduce this by integrating over the [tex](n+1)^\text{th}[/tex] variable with a delta function constraint. As a simple example, you could try to show that

[tex] \int_0^\infty dx \int_0^\infty dy \delta(x^2+y^2 -1) [/tex]

gives the correct measure on the unit circle.

Why is [itex]0 < \alpha < \infty[/itex]?

And I'm still a bit lost with this delta business. You say I can reproduce the measure using
[tex] d\alpha_i = x_i d\alpha + \alpha d x_i [/tex] - I don't really know how to go about this?

And in your example, why isn't your delta function [itex]\delta(1-x-y)[/itex] like we're trying to get in the final result?

Thanks.

 

[fzero]

Why is [itex]0 < \alpha < \infty[/itex]?

[tex]\alpha[/tex] is like a radial variable. The [tex]x_i[/tex] are affine coordinates.

And I'm still a bit lost with this delta business. You say I can reproduce the measure using
[tex] d\alpha_i = x_i d\alpha + \alpha d x_i [/tex] - I don't really know how to go about this?

You should probably look up what's called the Jacobian matrix, unless you know how to compute the volume form, which encodes the same information.

And in your example, why isn't your delta function [itex]\delta(1-x-y)[/itex] like we're trying to get in the final result?

Thanks.

Because I want to get the standard measure on the circle. It turns out that demanding [tex]x^2+y^2=1[/tex] or [tex]x+y=1[/tex] lead to the same space topologically, but that wasn't what I wanted to demonstrate.

 

[sgd37]

you forgot an [tex] \alpha^{n-1} [/tex] in the measure otherwise there is no [tex] (\sum x_i A_i)^{-n} [/tex]