Annihilation and creation operator - $\phi$ and $\pi$ for Klein-Gordon Field

Question

Introduction and Notation

Let $\phi(\vec{x})$ be the real Klein-Gordon (quantum) field, written as:

$$\phi(\vec{x})=\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{p}}}\left(a_{\vec{p}}+a^{\dagger}_{-\vec{p}} \right)e^{i\vec{p}\cdot\vec{x}} $$

where $\omega_{p}=\sqrt{|\vec{p}|^2+m^2}$, $a_{\vec{p}}$, $a^{\dagger}_{-\vec{p}}$ the annihilation and creation operators, and let $\pi(\vec{x})$ the momentum density conjugate to $\phi(\vec{x})$, given by

$$\pi(\vec{x})=\int\frac{d^3p}{(2\pi)^3}(-i)\sqrt{\frac{\omega_{p}}{2}}\left(a_{\vec{p}}-a^{\dagger}_{-\vec{p}} \right)e^{i\vec{p}\cdot\vec{x}} $$

The question

The only non trivial equal-time commutator is

$$[\phi(\vec{x}),\pi(\vec{y})]=i\hbar\delta^{(3)}(\vec{x}-\vec{y}) $$

As the relation between $\phi(\vec{x})$ and the $a,a^{\dagger}$ is linear, and so is between $\pi(\vec{x})$ and them, I'm going to express the commutator obeyed by $a$ and $a^{\dagger}$. I'm failing to derive the inverse Fourier transform

$$ \int d^3x'\phi(\vec{x})e^{i\vec{p}'\cdot\vec{x}'}=\int\int d^3x'\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{p}}}\left(a_{\vec{p}}+a^{\dagger}_{-\vec{p}} \right)e^{i\vec{p}\cdot\vec{x}}e^{i\vec{p}'\cdot\vec{x}'}=\\=\int d^3x'e^{i\vec{p}'\cdot\vec{x}'} \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{p}}}\left(a_{\vec{p}}+a^{\dagger}_{-\vec{p}} \right)e^{i\vec{p}\cdot\vec{x}}=\\=(2\pi)^3\delta^{3}(\vec{p}')\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{p}}}\left(a_{\vec{p}}+a^{\dagger}_{-\vec{p}} \right)e^{i\vec{p}\cdot\vec{x}}$$

What I want is a Dirac delta $\delta^{3}(\vec{p}-\vec{p}')$, so where is my procedure failing? I know this is a math question, but given the physical context, it may fit here better.

Thanks for your time, any hint will be appreciated

EDIT 1

EDIT 2 EDIT 1 Now as an answer

Answer 1

For the field:

$$ \int d^3x\phi(\vec{x})e^{-i\vec{p}'\cdot\vec{x}}=\int\int d^3x\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{p}}}\left(a_{\vec{p}}+a^{\dagger}_{-\vec{p}} \right)e^{i\vec{p}\cdot\vec{x}}e^{-i\vec{p}'\cdot\vec{x}}=\\=\int d^3x e^{-i(\vec{p}'-{p})\cdot\vec{x}}\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{p}}}\left(a_{\vec{p}}+a^{\dagger}_{-\vec{p}} \right)=\\=\int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{p}}}\left(a_{\vec{p}}+a^{\dagger}_{-\vec{p}} \right)(2\pi)^3\delta(\vec{p}'-\vec{p})=\frac{1}{\sqrt{2\omega_{p}'}}\left(a_{\vec{p}'}+a^{\dagger}_{-\vec{p}'} \right)\equiv\tilde{\phi}(\vec{p}')$$

For the momentum:

$$ \int d^3x\pi(\vec{x})e^{-i\vec{p}'\cdot\vec{x}}=\int d^3x\int\frac{d^3p}{(2\pi)^3}(-i)\sqrt{\frac{\omega_{p}}{2}}\left(a_{\vec{p}}-a^{\dagger}_{-\vec{p}} \right)e^{i\vec{p}\cdot\vec{x}}e^{-i\vec{p}'\cdot\vec{x}}=\\=\int d^3x e^{-i(\vec{p}'-{p})\cdot\vec{x}}\int\frac{d^3p}{(2\pi)^3}(-i)\sqrt{\frac{\omega_{p}}{2}}\left(a_{\vec{p}}-a^{\dagger}_{-\vec{p}} \right)=\\=\int \frac{d^3p}{(2\pi)^3}(-i)\sqrt{\frac{\omega_{p}}{2}}\left(a_{\vec{p}}-a^{\dagger}_{-\vec{p}} \right)(2\pi)^3\delta(\vec{p}'-\vec{p})=(-i)\sqrt{\frac{\omega_{p'}}{2}}\left(a_{\vec{p}'}-a^{\dagger}_{-\vec{p}'} \right)\equiv\tilde{\pi}(\vec{p}')$$

So

$$\boxed{\frac{1}{2}\left(\sqrt{2w_{p'}}\tilde{\phi}(\vec{p}')+i\sqrt{\frac{2}{w_{p'}}}\tilde{\pi}(\vec{p}')\right)=a_{\vec{p}'}}$$

$$\boxed{\frac{1}{2}\left(\sqrt{2w_{p'}}\tilde{\phi}(\vec{p}')-i\sqrt{\frac{2}{w_{p'}}}\tilde{\pi}(\vec{p}')\right)=a^{\dagger}_{-\vec{p}'}}$$

Now we can use the fact that the CCR for the momentum and the field is almost the same as the CCR for the Fourier transforms:

$$[\phi(\vec{x}),\pi(\vec{y})]-i \delta^{(3)}(\vec{x}-\vec{y})=0$$

Multiplying by $e^{-i(\vec{p}\cdot\vec{x})}$ and $e^{-i(\vec{p}'\cdot\vec{y})}$ integrating with respect to $\vec{x}$ and $\vec{y}$

$$ \int d^3xd^3y\left( e^{-i(\vec{p}\cdot\vec{x})}\phi(\vec{x})e^{-i(\vec{p}'\cdot\vec{y})}\pi(\vec{y})-\pi(\vec{y})e^{-i(-\vec{p}'\cdot\vec{y})}\phi(\vec{x})e^{-i(\vec{p}\cdot\vec{x})}-i\delta^{(3)}(\vec{x}-\vec{y})\right)=0$$

We get the commutator in terms of the fields and momentum in the momentum basis:

$$[\tilde{\phi}(\vec{p}),\tilde{\pi}(\vec{p}')]=(2\pi)^3i\delta^{(3)}(\vec{p}-\vec{p}') $$

so that

$$\left[a_{\vec{p}},a^{\dagger}_{-\vec{p}'}\right]=\left[\frac{1}{2}\left(\sqrt{2w_{p}}\tilde{\phi}(\vec{p})+i\sqrt{\frac{2}{w_{p}}}\tilde{\pi}(\vec{p})\right),\frac{1}{2}\left(\sqrt{2w_{p'}}\tilde{\phi}(-\vec{p}')-i\sqrt{\frac{2}{w_{p'}}}\tilde{\pi}(-\vec{p}')\right)\right]=\frac{1}{2}\left(-i[\tilde{\phi}(\vec{p}),\tilde{\pi}(\vec{p}')]+i[\tilde{\pi}(-\vec{p}),\tilde{\phi}(-\vec{p}')] \right)=(2\pi)^3\delta({\vec{p}+\vec{p}'})$$

Final Remarks

1)I think there are some $\hbar$ missing and 2) I'm not sure about the symmetry between $\vec{p}$ and $-\vec{p}$ for the real KG field

Edit

I set $\hbar=1$, but the commutator still looks a little ugly with that $+$ sign

Edit

Given that the field is real may I say $\tilde{\phi}(\vec{p})=\tilde{\phi}^*(-\vec{p})=\tilde{\phi}(-\vec{p})$