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

\begin{equation} \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}) \end{equation}

\begin{equation} \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}) \end{equation}

I am trying to prove the following relationships: \begin{equation} [\phi(\vec{x},t),\phi(\vec{y},t)]=0=[\pi(\vec{x},t),\pi(\vec{y},t)].\tag{2.90} \end{equation} I have already arrived at: \begin{equation} [\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)}) \end{equation} I know that the commutator is zero, but I am failed to interpret the above expression. Can someone help me out?

  • $\begingroup$ 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. $\endgroup$ Mar 24, 2021 at 13:57

1 Answer 1


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 $\vec p$?

Note that it is reasonable to object on the grounds that this integral does not actually converge in the sense of ordinary improper integrals. We must remember that, like the field operators themselves, this object is meant to be integrated against smooth and rapidly-decaying test functions. In other words, the well-defined quantity is

$$\int \frac{\mathrm d^3p}{(2\pi)^3} \int \mathrm d^3x\ \int d^3 y\ \frac{1}{2\omega_p} \left(e^{i p \cdot ( y - x)} - e^{-i p\cdot ( y - x)}\right) f(\vec x) g(\vec y) $$

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.


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