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In the book Cosmology by Daniel Baumann, when talking about the two-point correlation function of the density fluctuation $\delta(\vec{x})=\delta(\vec{x})$ for a fixed time $t$, the author states that this correlation function in Fourier space is the following:

\begin{aligned} \left\langle\delta(\mathbf{k}) \delta^{*}\left(\mathbf{k}^{\prime}\right)\right\rangle &=\int \mathrm{d}^{3} x \mathrm{~d}^{3} x^{\prime} e^{-i \mathbf{k} \cdot \mathbf{x}} e^{i \mathbf{k}^{\prime} \cdot \mathbf{x}^{\prime}}\left\langle\delta(\mathbf{x}) \delta\left(\mathbf{x}^{\prime}\right)\right\rangle \\ &=\int \mathrm{d}^{3} r \mathrm{~d}^{3} x^{\prime} e^{-i \mathbf{k} \cdot \mathbf{r}} e^{-i\left(\mathbf{k}-\mathbf{k}^{\prime}\right) \cdot \mathbf{x}^{\prime}} \xi(r) \\ &=(2 \pi)^{3} \delta_{\mathrm{D}}\left(\mathbf{k}-\mathbf{k}^{\prime}\right) \int \mathrm{d}^{3} r e^{-i \mathbf{k} \cdot \mathbf{r}} \xi(r) \\ & \equiv(2 \pi)^{3} \delta_{\mathrm{D}}\left(\mathbf{k}-\mathbf{k}^{\prime}\right) \mathcal{P}(k), \end{aligned}

where $\mathbf{r}=\mathbf{x}-\mathbf{x'}$. I don't know why this correlation function needs to include the complex conjugate of $\delta(\mathbf{k'})$ instead of just being $\left\langle\delta(\mathbf{k}),\delta(\mathbf{k'})\right\rangle$, but apart from that, I understand the first and second lines. My question arises in the step between the second and third lines. I know that the Fourier transform of the Dirac delta $\delta_D$ is:

$$\mathcal{F}[\delta(\mathbf{x}-\mathbf{x_0})]=\int_{\mathbb{R}^3}d^3xe^{-i\mathbf{k}\cdot\mathbf{x}}\delta(\mathbf{x}-\mathbf{x_0})=e^{-i\mathbf{k}\cdot\mathbf{x_0}}$$

Therefore, if I use the inverse Fourier transform on this result, I get:

$$\delta(\mathbf{x}-\mathbf{x_0})=\mathcal{F}^{-1}(e^{-i\mathbf{k}\cdot\mathbf{x_0}})=\int\dfrac{d^3k}{(2\pi)^3}e^{i\mathbf{k}\cdot\mathbf{x}}e^{-i\mathbf{k}\cdot\mathbf{x_0}}=\int\dfrac{d^3k}{(2\pi)^3}e^{i\mathbf{k}\cdot(\mathbf{x}-\mathbf{x_0})}$$

This means that, renaming $\mathbf{k}$ as $\mathbf{x'}$, $\mathbf{x}$ as $\mathbf{k}$ and $\mathbf{x_0}$ as $\mathbf{k'}$, we have:

$$(2\pi)^3\delta(\mathbf{k}-\mathbf{k'})=\int d^3x\ e^{i(\mathbf{k}-\mathbf{k'})\mathbf{x'}}$$

But what the book seems to have used to get the third line from the second one is this formula instead:

$$(2\pi)^3\delta(\mathbf{k}-\mathbf{k'})=\int d^3x\ e^{-i(\mathbf{k}-\mathbf{k'})\mathbf{x'}}$$

So, where does the minus in the exponent come from? Or is there an errata in the book?

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1 Answer 1

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The delta function is an even distribution: i.e. $\delta(x)=\delta(-x)$ in general.

Also note that $\langle \delta(k) \delta^*(k') \rangle$ is an expectation value here, $\bf{not}$ an inner product (although this could be written differently using an inner product).

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  • $\begingroup$ Wow, it was so easy, I don't how I didn't see it. Thank you! I'm curious, is it very difficult to see how to write the expectation value in terms of the inner product? I've never seen it that way. $\endgroup$ May 8, 2023 at 21:02

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