Commutation relations for creation, annihilator operators $a_\mathbf{p}^\dagger, a_\mathbf{p}$

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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– user4552
Dec 29 '19 at 18:59
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Dec 29 '19 at 18:59

1 Answer

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

• I see, thank you Sep 18 '19 at 8:35
• Hi @OON Why is it that $\omega_{-p}=\omega_p$? Feb 23 '20 at 19:42
• @LopeyTall $\omega_{\mathbf{p}}$ depends only on the module of $\mathbf{p}$ not on its direction. In fact $\omega_{\mathbf{p}}=\sqrt{\mathbf{p}^2+m^2}$.
– OON
Feb 23 '20 at 19:48
• 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? Feb 23 '20 at 19:50
• ahh "module" $\equiv$ absolute value/magnitude. en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation Feb 23 '20 at 19:51