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Write the field $\phi$ and momentum $\pi$ in terms of creation and annihilation operators $a_\mathbf{p}^\dagger, a_\mathbf{p}$

$$ \phi(\mathbf{x}) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_\mathbf{p}}} [a_\mathbf{p}e^{i\mathbf{p}\cdot\mathbf{x}} + a_\mathbf{p}^\dagger e^{-i\mathbf{p}\cdot\mathbf{x}}], $$

$$ \pi(\mathbf{x}) = \int \frac{d^3p}{(2\pi)^3} (-i) \sqrt{\frac{\omega_\mathbf{p}}{2}} [a_\mathbf{p}e^{i\mathbf{p}\cdot\mathbf{x}} - a_\mathbf{p}^\dagger e^{-i\mathbf{p}\cdot\mathbf{x}}]. $$

The goal is to show that

$$ [a_\mathbf{p}, a_\mathbf{q}] = [a_\mathbf{p}^\dagger, a_\mathbf{q}^\dagger] = 0, $$

$$ [a_\mathbf{p}, a_\mathbf{q}^\dagger] = (2\pi)^3 \delta^{(3)}(\mathbf{p}-\mathbf{q}). $$

I have no luck in arriving at the commutation relations.

Take inverse Fourier transform,

$$ \tilde{\phi}(\mathbf{p}) = \int d^3x\ \phi(\mathbf{x}) e^{-i\mathbf{p}\cdot\mathbf{x}} = \frac{1}{\sqrt{2\omega_\mathbf{p}}} a_\mathbf{p} + \frac{1}{\sqrt{2\omega_\mathbf{-p}}} a_\mathbf{-p}^\dagger, $$

$$ \tilde{\pi}(\mathbf{p}) = \int d^3x\ \pi(\mathbf{x}) e^{-i\mathbf{p}\cdot\mathbf{x}} = (-i)\Bigg(\sqrt{\frac{\omega_\mathbf{p}}{2}} a_\mathbf{p} - \sqrt{\frac{\omega_\mathbf{-p}}{2}} a_\mathbf{-p}^\dagger\Bigg). $$

Then

$$ a_\mathbf{p} = \frac{1}{2} \Bigg(\sqrt{2\omega_\mathbf{p}} \tilde{\phi}(\mathbf{p}) + i\sqrt{\frac{2}{\omega_\mathbf{p}}} \tilde{\pi}(\mathbf{p}) \Bigg), $$

$$ a_\mathbf{-p}^\dagger = \frac{1}{2} \Bigg(\sqrt{2\omega_\mathbf{p}} \tilde{\phi}(\mathbf{p}) - i\sqrt{\frac{2}{\omega_\mathbf{p}}} \tilde{\pi}(\mathbf{p}) \Bigg). $$

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

\begin{align} [a_\mathbf{p}, a_\mathbf{q}] &= \frac{1}{4} \int d^3x d^3y\ 2i\sqrt{\frac{\omega_\mathbf{p}}{\omega_\mathbf{q}}} [\phi(\mathbf{x}), \pi(\mathbf{y})] e^{-i\mathbf{x}\cdot\mathbf{p}} e^{-i\mathbf{y}\cdot\mathbf{q}} + 2i\sqrt{\frac{\omega_\mathbf{q}}{\omega_\mathbf{p}}} [\pi(\mathbf{x}), \phi(\mathbf{y})] e^{-i\mathbf{x}\cdot\mathbf{p}} e^{-i\mathbf{y}\cdot\mathbf{q}} \\ &= \frac{1}{4} \int d^3x d^3y\ 2i\sqrt{\frac{\omega_\mathbf{p}}{\omega_\mathbf{q}}} i\delta^{(3)}(\mathbf{x}-\mathbf{y}) e^{-i\mathbf{x}\cdot\mathbf{p}} e^{-i\mathbf{y}\cdot\mathbf{q}} + 2i\sqrt{\frac{\omega_\mathbf{q}}{\omega_\mathbf{p}}} (-i)\delta^{(3)}(\mathbf{y}-\mathbf{x}) e^{-i\mathbf{x}\cdot\mathbf{p}} e^{-i\mathbf{y}\cdot\mathbf{q}} \\ &= -\frac{i}{2} \int d^3x \Bigg(\sqrt{\frac{\omega_\mathbf{p}}{\omega_\mathbf{q}}} - \sqrt{\frac{\omega_\mathbf{q}}{\omega_\mathbf{p}}} \Bigg) e^{-i\mathbf{x}\cdot(\mathbf{p} + \mathbf{q})} \end{align}

Why is this equal to $0$? It's not true that $\omega_\mathbf{p} = \omega_\mathbf{q}$, is it?

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  • $\begingroup$ I've added the homework-and-exercises tag. In the future, please add this tag to this type of problem. This is one of the things that we ask you to do in our homework policy: physics.meta.stackexchange.com/questions/714/… $\endgroup$
    – user4552
    Dec 29, 2019 at 18:59
  • $\begingroup$ Please reference the source of this homework question. This is one of the things that we ask you to do in our homework policy: physics.meta.stackexchange.com/questions/714/… $\endgroup$
    – user4552
    Dec 29, 2019 at 18:59

1 Answer 1

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The only dependence on $\mathbf{x}$ that remains is in the exponent factor. Integrating it we get $\delta$-function, \begin{equation} \int d^3x\, e^{-i\mathbf{x}\cdot(\mathbf{p}+\mathbf{q})}=(2\pi)^3\delta^{(3)}(\mathbf{p}+\mathbf{q}) \end{equation} That means that we can replace $\omega_\mathbf{q}$ with $\omega_{-\mathbf{p}}=\omega_\mathbf{p}$. That results in cancellation.

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  • $\begingroup$ I see, thank you $\endgroup$
    – Bernoulli
    Sep 18, 2019 at 8:35
  • $\begingroup$ Hi @OON Why is it that $\omega_{-p}=\omega_p$? $\endgroup$
    – Lopey Tall
    Feb 23, 2020 at 19:42
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    $\begingroup$ @LopeyTall $\omega_{\mathbf{p}}$ depends only on the module of $\mathbf{p}$ not on its direction. In fact $\omega_{\mathbf{p}}=\sqrt{\mathbf{p}^2+m^2}$. $\endgroup$
    – OON
    Feb 23, 2020 at 19:48
  • $\begingroup$ Hmmm can you elaborate on what "module" means? I was thinking maybe it was because we are working in flat space and these $\omega_{\vec{p}}$ terms are only spatial? $\endgroup$
    – Lopey Tall
    Feb 23, 2020 at 19:50
  • $\begingroup$ ahh "module" $\equiv$ absolute value/magnitude. en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation $\endgroup$
    – Lopey Tall
    Feb 23, 2020 at 19:51

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