# How to verify the Klein-Gordon field commutation relations, Peskin and Schroeder Equation (2.30)

I am trying to verify the commutation relation given in Peskin and Schroeder. In particular, I don't know how to go between these two lines:

$$[\phi(\textbf{x}), \pi(\textbf{x}')] = \int \frac{d^3p d^3p'}{(2\pi)^6} \frac{-i}{2}\sqrt{\frac{\omega_{p'}}{\omega_p}}\left([a^\dagger_{-p}, a_{p'}] - [a_p, a^\dagger_{-p'}] \right)e^{i(p\cdot{}x+p'\cdot{}x')}$$ $$[\phi(\textbf{x}), \pi(\textbf{x}')] = i\delta^{(3)}(\textbf{x}-\textbf{x}') \hspace{10mm}(2.30)$$

Using equations (2.27) and (2.28) for $$\phi$$ and $$\pi$$: $$\phi(\textbf{x}) = \int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}(a_p + a^\dagger_{-p})e^{ip \cdot{} x} \hspace{10mm}(2.27)$$

$$\pi(\textbf{x}) = \int \frac{d^3p}{(2\pi)^3} (-i)\sqrt{\frac{\omega_p}{2}}(a_p - a^\dagger_{-p})e^{ip \cdot{} x} \hspace{6mm}(2.28)$$

And the ladder operator commutation relation: $$[a_p, a^\dagger_{p'}] = (2\pi)^3\delta^{(3)}(\textbf{p} - \textbf{p}') \hspace{10mm}(2.29)$$

My Attempt

Using the commutation relation, I sub in for the two ladder operator commutators: $$1) \hspace{5mm}[a^\dagger_{-p}, a_{p'}] = -[a_{p'},a^\dagger_{-p}] = -(2\pi)^3\delta^{(3)}(\textbf{p}'- (-\textbf{p)}) = -(2\pi)^3\delta^{(3)}(\textbf{p}' +\textbf{p})$$ Where I have used a negative $$\textbf{p}$$ inside the dirac delta, since the commutator is $$a_{-p}$$ (I am unsure whether this is correct).

$$2) \hspace{5mm}[a_{p}, a^\dagger_{-p'}] = (2\pi)^3\delta^{(3)}(\textbf{p}-(-\textbf{p}')) = (2\pi)^3\delta^{(3)}(\textbf{p} + \textbf{p}')$$ Using the same thinking as before. Subbing this into the integral: $$[\phi(\textbf{x}), \pi(\textbf{x}')] = \int \frac{d^3p d^3p'}{(2\pi)^6} \frac{-i}{2}\sqrt{\frac{\omega_{p'}}{\omega_p}}\left(-2(2\pi)^3\delta^{(3)}(\textbf{p} + \textbf{p}')\right)e^{i(p\cdot{}x+p'\cdot{}x')}$$

Dealing with the minus and cancelling terms: $$[\phi(\textbf{x}), \pi(\textbf{x}')] = \int \frac{d^3p d^3p'}{(2\pi)^3} i \sqrt{\frac{\omega_{p'}}{\omega_p}}\delta^{(3)}(\textbf{p} + \textbf{p}')e^{i(p\cdot{}x+p'\cdot{}x')}$$

Here I am stuck: I do not know hot to deal with the dirac-delta in the integral, and I'm unsure whether I'm even right up to here. Any help on how to proceed or corrections thus far are appreciated! I'm told it's important to understand this part for the upcoming chapters.

Well you're nearly done. That delta function $$\delta^3(\mathbf{p}+\mathbf{p}^\prime)$$ just sets $$\mathbf{p}^\prime = -\mathbf{p}$$ when you integrate in $$\mathbf{p}^\prime$$ (or the opposite if you integrate in $$\mathbf{p}$$, but it's the same). What you get at this point is what follows
\begin{align} [\phi(\textbf{x}), \pi(\textbf{x}')] &= \int \frac{d^3p d^3p'}{(2\pi)^3} i \sqrt{\frac{\omega_{p'}}{\omega_p}}\delta^{(3)}(\textbf{p} + \textbf{p}')e^{i(p\cdot{}x+p'\cdot{}x')}\\ &=\int\frac{d^3p}{(2\pi)^3}i\sqrt{\frac{\omega_p}{\omega_p}}e^{ipx-ipx^\prime} = i\int\frac{d^3p}{(2\pi)^3}e^{ip(x-x^\prime)} \end{align}
Now by definition the last integral is just a delta function $$\delta^3(\mathbf{x}-\mathbf{x}^\prime)$$ since it's just the Fourier transform of $$1$$. At this point you just get the result you search for. The fact that $$\omega_{-p}=\omega_p$$ just comes from the fact that the energy is quadratic in $$p$$ so the sign does not matter.
• This solves my question, thank you! One thing: does the $\delta ^{(3)}$ mean a delta function in 3 dimensions? But I can just treat it with the same rules as a 1D one? May 31, 2020 at 9:32
• Yes, exactly! You may find in next chapter even $\delta^{(4)}$ which is the $\delta$ function in $4$ dimension, whether Euclidean of Minkowski depends on the situation, I'll know when you see it. And yes, you can treat it exactly in the same way as the one dimensional one. May 31, 2020 at 9:53