# Poisson brackets of pre-quantized annihilation and creation operators localized in $p$-space

Let $\phi(\vec{x},t)$ be real classical scalar field and $\pi(\vec{x},t)$ its conjugate momentum. It can be written as Fourier Transform $$\phi(\vec{x},t)=\int \frac{d^3p}{(2\pi)^3}e^{i\vec{p}\vec{x}}\phi(\vec{p},t).$$ Then it is put into Klein-Gordon equation to show that free field can be viewed as system of infinite many harmonic oscillators. In order to quantize free field one introduces symbols (pre-quantized a/c operators): \begin{equation} a(\vec{p},t)=\sqrt{\frac{\omega_{\vec{p}}}{2}}\phi(\vec{p},t)+i\sqrt{\frac{1}{2\omega_{\vec{p}}}}\pi(\vec{p},t) \end{equation} \begin{equation} a^{+}(\vec{-p},t)=\sqrt{\frac{\omega_{\vec{p}}}{2}}\phi(\vec{p},t)-i\sqrt{\frac{1}{2\omega_{\vec{p}}}}\pi(\vec{p},t) \end{equation} To proceed further in canonical quantization one has to calculate poisson brackets $$\{a(\vec{p},t),a^{+}(\vec{q},t)\}.$$ My lecturer's notes give $$\{a(\vec{p},t),a^{+}(\vec{q},t)\}=(2\pi)^{3}i\delta(\vec{p}-\vec{q})$$ without any calculation or explanation. Poison brackets are defined as (ommiting $t$ for clarity): $$\{A,B\}=\int d^{3}z \frac{\delta{A}}{\delta\phi(\vec{z})}\frac{\delta{B}}{\delta\pi(\vec{z})}-\frac{\delta{B}}{\delta\phi(\vec{z})}\frac{\delta{A}}{\delta\pi(\vec{z})}.$$ Where $\delta$ denotes functional derivative. Problem with my calculation is that I have to calculate poisson bracket of those pre-quantized annihilation and creation operators localized in $p$-space but poisson brackets are defined in $x$-space and thus I have to take Fourier transform $a(\vec{x},t)$ of $a(\vec{p},t)$, but it doesn't make much sense to me because it immediately destroys sharp localization in $p$-space (because I integrate over whole $p$-space). I would be grateful for consistent explanation.

I think I figured it out. Let \begin{equation} \phi(\vec{p},t)=\int d^3x \:\, e^{-i\vec{p}\vec{x}}\phi(\vec{x},t) \end{equation} and the same goes for $\pi(\vec{p},t)$. One can then view these Fourier integrals $\phi(\vec{p},t)$ as functionals on $\phi(\vec{x},t)$. Calculating functional derivatives (ommiting $t$): \begin{equation} \frac{\delta \phi(\vec{p})}{\delta \phi(\vec{z})}=\lim_{\epsilon \rightarrow 0} \frac{\int d^3x \; \, e^{-i\vec{p}\vec{x}}(\phi(\vec{x})+\epsilon \delta(\vec{x}-\vec{z})) \; \; -\int d^3x \; \, e^{-i\vec{p}\vec{x}}\phi(\vec{x})}{\epsilon}=e^{-i\vec{p}\vec{z}} \end{equation} and analogously for $\phi(\vec{q})$,$\pi(\vec{p})$,$\pi(\vec{q})$. Calculation of poisson bracket is now straightforward - just calculating fucntional derivative of $a(\vec{p})$, $a^{+}(\vec{q})$ using above formula. Then it yields: \begin{equation} \{a(\vec{p}),a^{+}(\vec{q})\}=\int d^3z \;\, \frac{-i}{2}\left( \sqrt{\frac{\omega_p}{\omega_q}}e^{-i(\vec{p}-\vec{q})\vec{z}} + \sqrt{\frac{\omega_q}{\omega_p}}e^{-i(\vec{p}-\vec{q})\vec{z}} \right) \end{equation} And this integral is equal to \begin{equation} -i(2\pi)^3\delta(\vec{p}-\vec{q}) \end{equation} factors in squate root cancel out because formula is nonzero for $\vec{p}=\vec{q}$ only. Sign is different from the one in lecture notes but I checked it and there is used another convention (different in sign) for poisson bracket as I am used to.