Quantum Field Theory Commutation Relations

Let $$ϕ(x,t)$$ denote the canonical fields and $$π(x,t)$$ denote the canonical momenta where they're given by:

$$$$\phi(x)=\int\frac{d^3\vec{p}}{(2\pi)^3\sqrt{2\omega_{\vec{p}}}}(a_{\vec{p}}e^{-ipx}+a_{\vec{p}}^{\dagger}e^{ipx})$$$$

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

I am trying to prove the following relationships: $$$$[\phi(\vec{x},t),\phi(\vec{y},t)]=0=[\pi(\vec{x},t),\pi(\vec{y},t)].\tag{2.90}$$$$ I have already arrived at: $$$$[\phi(\vec{x},t),\phi(\vec{y},t)]=\int\frac{d^{3}\vec{p}}{(2\pi)^{3}}\frac{1}{2\omega_{p}}(e^{ip(y-x)}-e^{-ip(y-x)})$$$$ I know that the commutator is zero, but I am failed to interpret the above expression. Can someone help me out?

• Peskin and Schroeder have a nice figure 2.4 about when you can Lorentz transform your second term to the first and when not. Here, your separation is spacelike. Mar 24, 2021 at 13:57

Because both field operators are being evaluated at the same time $$t$$, we have that $$y-x = (0, \vec y - \vec x)$$, and so $$p \cdot (y-x) = \vec p \cdot (\vec y- \vec x)$$. From here, note that $$e^{i\vec p \cdot (\vec y - \vec x)}-e^{-i\vec p \cdot (\vec y - \vec x)}=2i\sin\big(\vec p \cdot (\vec y-\vec x)\big)$$, which is an odd function of $$\vec p$$. $$\omega_p$$ is an even function of $$\vec p$$, so the integrand as a whole is odd. What happens when you integrate an odd function over all $$\vec p$$?
$$\int \frac{\mathrm d^3p}{(2\pi)^3} \int \mathrm d^3x\ \int d^3 y\ \frac{1}{2\omega_p} \left(e^{i p \cdot ( y - x)} - e^{-i p\cdot ( y - x)}\right) f(\vec x) g(\vec y)$$
where the integrals over $$x$$ and $$y$$ are performed first. This object is a perfectly well-defined number (which can be straightforwardly shown to be equal to zero).
As a general rule, if you are concerned about the convergence or cancellation of integrals, then you can put your mind at ease by replacing $$\phi(\vec x,t)$$ with the more rigorously defined $$\Phi_t(f) := \int \mathrm d^3 x \ f(\vec x) \phi(\vec x,t)$$, with $$f$$ an aforementioned (arbitrary) test function. It is somewhat inconvenient to carry around these arbitrary test functions, but it has the advantage of making the expressions you write down well-defined. Your mileage may vary, of course.
In this language, we would have $$[\Phi_t(f),\Phi_t(g)] = 0$$ $$[\Phi_t(f),\Pi_t(g) ] = i\int\mathrm d^3 x f(\vec x)g(\vec x)$$ for arbitrary test functions $$f$$ and $$g$$, which imply- and are implied by - the equal-time commutation relations $$[\phi(\vec x,t),\phi(\vec y,t)] = 0$$ $$[\phi(\vec x,t),\pi(\vec y,t)] = i\delta(\vec x - \vec y)$$ for the bare field operators.