- #1

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## Homework Statement

I am trying to understand the very last equality for (let me replace the tilda with a hat ) ##\hat{P_{X}(K)}=\hat{P(k_1=k_2=...=k_{N}=k)}##(1)

## Homework Equations

I also thought that the following imaginary exponential delta identity may be useful, due to the equality of the ##k_i##, but see comments below:

##\int dk \exp^{ikx} = \delta(x=0) ##

## The Attempt at a Solution

So it sees to me the goal is something like expressing ##\hat{P_{X}(K)}## in terms of ##P(\hat{k_i})## ?

So these are given by ##\hat{P(k_1)......P(k_n)}= \int d^{N} \vec{x} p(\vec{x}) \exp^{-i \sum\limits_j x_j k_j } ##

I thought I`d first try to look at the simplified case of indepedent random variables to understand (1) but still can`t seem to get it.

So in this case ##p(\vec{x}) = \Pi_{i} p(x_i)##

And then we have

(If I am correct in that the notation is that ##\Pi_{i} dx_i = d^N (\vec{x}) ##)

##\hat{P(k_1)......P(k_n)}= \int \Pi_{i} dx_i p(x_i) \exp^{-i \sum\limits_j x_j k_j } ##

and then you can seperate the integrals and so we have:

##\hat{P(k_1)......P(k_n)}= \hat{P_{x_1}(k_1)} ......\hat{P_{x_n}(k_n)} ## (2)

Now if I consider the independent case in ##\hat{P_{X}(K)}## I have:

##\hat{P_{X}(K)} =\int \Pi_{i} dx_i p(x_i) \exp^{-i k \sum\limits_j x_j }

= \int dx_1 p(x_1) e^{-ix_1 k} \int dx_2 p(x_2) e^{-ix_2 k}...\int dx_N p(x_N) e^{-ikx_N}

= \hat{P_{x_1}(k)} ......\hat{P_{x_n}(k)} ##

So if I compare this to (2), and can reason( I'm not sure you can) that it does matter whether you have ##k_i## or ##k##, this is just the label of the fourier transform, but look at the lower notation that gives the distribution, that ##\hat{P_{x_1}(k)}= \hat{P_{x_1}(k_1)} ## and then I have ##k_1=k## and can do the same for each ##k_i## etc.

Without independence instead i have:

##\int d^{N} \vec{x} p(\vec{x}) \exp^{-i k \sum\limits_j x_j } ##

and I can't see how you can make any conclusions without knowing what ## p(\vec{x}) ## is?

Many thanks