Let $\phi(\vec{x},t)$ denote the canonical fields and $\pi(\vec{x},t)$ denote the canonical impulses 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}) \tag{2.78/81} \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}) \tag{2.91} \end{equation}

where (I'll omit the vectors in the subscripts now because I can't align them properly)

\begin{equation} [a_{p},a_{p'}^{\dagger}]=(2\pi)^{3}\delta^{3}(\vec{p}-\vec{p}'),\qquad [a_{p},a_{p'}]=0=[a_{p}^{\dagger},a_{p'}^{\dagger}].\tag{2.69} \end{equation}

I am trying to prove the commutation relations:

\begin{equation} [\phi(\vec{x},t),\phi(\vec{y},t)]=0=[\pi(\vec{x},t),\pi(\vec{y},t)].\tag{2.90} \end{equation}

With a straightforward calculation I arrive at: \begin{equation} [\phi(\vec{x},t),\phi(\vec{y},t)]=\int\frac{d^{3}\vec{p}}{(2\pi)^{3}}\frac{1}{\omega_{p}}(e^{ip(y-x)}-e^{ip(x-y)}).\tag{2.89} \end{equation}

And I must admit I see no way how this is zero. I then tried to consult books on QFT and I came across a "proof" in Schwartz' Quantum Field Theory and the Standard Model, on the bottom of p. 24:

Since the integral measure and $\omega_{p}=\sqrt{\vec{p}^{2}+m^{2}}$ are symmetric under $\vec{p}\to-\vec{p}$ we can flip the sign on the exponent of one of the terms to see that the commutator vanishes.

But to me this just doesn't seem right. How can we flip the sign on only one part of the function? This comes out just as wishful thinking. Is there an alternative way to prove these relations? Or did we "observe" that these commutation relations are correct so we just go with the flow?

  • 1
    $\begingroup$ Separate the integrals into two parts - one with $e^{ip(y-x)}$ and another with $e^{ip(x-y)}$. Then, flip $p \to - p$ in the second one. $\endgroup$
    – Prahar
    Mar 18, 2018 at 15:06

1 Answer 1


There's no "wishful thinking" here, just calculus. As an example let's consider the integral $$I = \int_{-1}^1 \cos(x) - \cos(-x) \, dx.$$ We evaluate it as $$I = \int_{-1}^1 \cos(x) \, dx - \int_{-1}^1 \cos(-x) \, dx = \int_{-1}^1 \cos(x) \, dx + \int_{1}^{-1} \cos(u) \, du$$ where we substituted $u = -x$. By exchanging the limits of integration, we get $$I = \int_{-1}^1 \cos(x) \, dx - \int_{-1}^{1} \cos(u) \, du.$$ Finally, by renaming $u$ to $x$, we have $$I = \int_{-1}^1 \cos(x) \, dx - \int_{-1}^{1} \cos(x) \, dx = \int_{-1}^1 \cos(x) - \cos(x) \, dx = \int_{-1}^1 0 \, dx = 0.$$ Schwartz's reasoning is the same, just replace $dx$ with $d^3\vec{p}$ and the cosine with the exponential. Since this is usually taught in calculus class, Schwartz didn't both filling in the intermediate steps.

  • 2
    $\begingroup$ Thank you for your quick reply. While what you wrote is mathematically correct, it narrows the application of the theory, in my opinion. This applies only to symmetric intervals which needs not be the case but this way of calculating leads to things like: $2i\int_{-\infty}^{\infty}sinxdx=\int_{-\infty}^{\infty}(e^{ix}-e^{-ix})dx=\int_{-\infty}^{\infty}e^{ix}dx-\int_{-\infty}^{\infty}e^{ix}dx=0$ which is obviously false. So in that sense, it seems to me that it not only applies to symmetric situations but also only to finite spaces. Please correct me if I'm wrong, I am still new to QFT. $\endgroup$
    – Kandrax
    Mar 18, 2018 at 12:21
  • $\begingroup$ @Kandrax Of course, integrals in QFT need to be regularized, and a QFT without regularization is meaningless. If you start with creation and annihilation operators, then indeed there are plenty of strange regularization procedures that would make this cancellation false, e.g. if you defined $\int_{-\infty}^\infty dp$ to be $\int_{-2\Lambda}^{\Lambda} dp$. So part of the definition of a QFT in the first place is that it comes with a regulator that preserves appropriate symmetries. $\endgroup$
    – knzhou
    Mar 18, 2018 at 13:19
  • $\begingroup$ The particular $p \to -p$ symmetry in this question is simple enough that it's preserved by every reasonable regulator I can think of, but other symmetries might not be preserved and lead to quantum anomalies, as you see later in Schwartz. $\endgroup$
    – knzhou
    Mar 18, 2018 at 13:20
  • 2
    $\begingroup$ $\phi(x)$ is, strictly speaking, an operator valued distribution. If you are worried about these sort of problems, then you can work with the normalized operators $\phi(f) = \int d^4x f(x) \phi(x)$ where $f(x)$ is taken to be an $L^2$-normalizable function. Then, you can freely take $p \to - p$ without having to worry about convergence of integrals issues. Your example with $\int_{-\infty}^\infty sin x dx$" fails precisely because the integral does not converge to begin with. $\endgroup$
    – Prahar
    Mar 18, 2018 at 15:25

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