I am reading Peskin & Schroeder Ch9 and am stuck on a calculation going from equation 9.36. The problem is essentially a change of variable of a Klein-Gordon field.

Beginning, we have an integral of the KG Lagrangian and a source term

\begin{align} \int d^4 x\: [\mathcal{L}_0(\phi) + J\phi] = \int d^4 x \left[\frac{1}{2} \phi (-\partial^2 - m^2 + i \epsilon)\phi + J\phi\right]. \end{align} Now we want to shift the field so introduce \begin{align} \phi' = \phi - i\int d^4 y\: D_F(x-y)J(y). \end{align}

Substitution is then supposed to give \begin{align} \int d^4 x\: [\mathcal{L}_0(\phi) + J\phi] = \int d^4 x \left[\frac{1}{2} \phi' (-\partial^2 - m^2 + i \epsilon)\phi' \right] - \int d^4x d^4y\; \frac{1}{2}J(x)[-iD_F(x-y)]J(y). \end{align} To do this, they use that $D_F$ is a Green's function of the KG operator, which I have been using as \begin{align} (-\partial^2 - m^2 + i \epsilon)i\int d^4 y\: D_F(x-y)J(y) = -J(x). \end{align} So I find

\begin{align} \int d^4 x\: [\mathcal{L}_0(\phi) + J\phi] = &\int d^4 x \left[\frac{1}{2} \left(\phi' + i\int d^4 y\: D_F(x-y)J(y)\right) (-\partial^2 - m^2 + i \epsilon)\left(\phi' + i\int d^4 y\: D_F(x-y)J(y)\right) + J(x)\left(\phi' + i\int d^4 y\: D_F(x-y)J(y)\right)\right] \\ =& \int d^4 x \left[\frac{1}{2} \phi'(-\partial^2 - m^2 + i \epsilon)\phi' - \frac{1}{2} \phi' J(x) + \frac{1}{2} i\int d^4 y\: D_F(x-y)J(y) (-\partial^2 - m^2 + i \epsilon)\phi' \\ - \frac{1}{2} i\int d^4 y\: D_F(x-y)J(y) J(x) + J(x)\phi' + J(x)i\int d^4 y\: D_F(x-y)J(y)\right] \\= &\int d^4 x \left[\frac{1}{2} \phi' (-\partial^2 - m^2 + i \epsilon)\phi' \right] - \int d^4x d^4y\; J(x)[-iD_F(x-y)]J(y) \\ &+ \frac{1}{2}\int d^4x \: J(x)\phi' +\frac{1}{2} i\int d^4x d^4 y\: D_F(x-y)J(y) (-\partial^2 - m^2 + i \epsilon)\phi' \\= &\int d^4 x \left[\frac{1}{2} \phi' (-\partial^2 - m^2 + i \epsilon)\phi' \right] - \int d^4x d^4y\; J(x)[-iD_F(x-y)]J(y) \\ &+ \frac{1}{2}\int d^4x \: J(x)\phi' -\frac{1}{2} i\int d^4x d^4 y\: (-\partial^2 - m^2 + i \epsilon)D_F(x-y)J(y) \phi' + \text{boundary terms} \\= &\int d^4 x \left[\frac{1}{2} \phi' (-\partial^2 - m^2 + i \epsilon)\phi' \right] - \int d^4x d^4y\; J(x)[-iD_F(x-y)]J(y) \\ &+ \int d^4x \: J(x)\phi' , \end{align} where the boundary terms come from integration by parts. I'm off by a factor of $\frac{1}{2}$ in the second term, and I also have an additional term, but I can't see where it's going wrong.

  • $\begingroup$ When you do integration by parts leading to a new term with additional boundary terms it seems like you changed the sign, however, we actually apply integration by parts twice since we have a second derivative. When you apply integration by parts twice, the sign remains the same (see en.wikipedia.org/wiki/Integration_by_parts section 'repeated integration by parts'). Does this explain your additional term? $\endgroup$ May 29, 2023 at 7:15

1 Answer 1


$\newcommand{\ip}[1]{\left<#1\right>}\newcommand{\d}{\mathrm{d}}\newcommand{\inv}[1]{{#1}^{-1}}$What happens here is a simple completion of a square. To see it in its cleanest form, without potentially error-inducing calculations, remember that, in the space where $\phi$ lives1 $$\ip{\bullet,\tilde\bullet}:= \int \mathrm{d}^4 x\ \bullet(x)\,\tilde{\bullet}(x),$$ is an inner product. Moreover, let $\Omega := -\partial^2-m^2+\mathrm{i}\epsilon$ be the Klein-Gordon operator that you are interested in.

The scenario you have, then, is that \begin{align} \frac12\ip{\phi,\Omega\phi}+\ip{J,\phi}&=\frac12\ip{(\phi'-\inv{\Omega}J),\Omega\,(\phi'-\inv{\Omega}J)}+\ip{J,(\phi'-\inv{\Omega}J)} = \\ &= \frac12\Big(\ip{\phi',\Omega\phi'}-\ip{J,\phi'}-\ip{\phi',J}+\ip{J,\inv{\Omega}J}\Big)+\ip{J,\phi'}-\ip{J,\inv{\Omega}J} = \\ &= \frac12\ip{\phi',\Omega\phi'} - \frac12\ip{J,\inv{\Omega}J}, \end{align} where I used that the invertible operator $\Omega$ is self-adjoint, i.e. $\ip{\bullet,\Omega\tilde\bullet}=\ip{\Omega\bullet,\tilde\bullet}$, and hence so is its inverse and further that the inner product defined above is symmetric, so $\ip{\phi,J}=\ip{J,\phi}$.

The missing factor of $\frac12$ comes in your second-leading-to-third line where you have $$-\frac12\mathrm{i}\int\d^4 x J(x) D(x-y) J(y) + \mathrm{i}\int\d^4 x J(x) D(x-y) J(y) = \color{red}{\frac12} \mathrm{i}\int\d^4 x J(x) D(x-y) J(y) $$ but you've missed the red $\frac12$. As for the $J\phi'$, there are two terms that give this contribution from the kinetic part of the action, while you've only written one of them. If you write both, they exactly cancel the term that comes from the $J\phi$ part (cf. my equations above).

1Recall: $\phi:\mathbb{R}^{4}\to\mathbb{R}$, with boundary conditions $\phi(x)\overset{\Vert x\Vert\to\infty}{\longrightarrow}0$.

  • $\begingroup$ Saying that the Klein-Gordon operator is invertible would suggest that it has a two sided inverse, however, it seems to have only a right inverse. As any solution to the homogeneous Klein-Gordon equation is mapped to 0 by the Klein-Gordon operator, hence the Klein-Gordon operator is not injective, hence it does not admit a left-inverse. Does your argument still work if $\Omega^{-1}$ is only a right inverse? $\endgroup$ May 29, 2023 at 7:24
  • $\begingroup$ And could you please elaborate on why $\Omega$ is self-adjoint? This seems to follow from a repeated integration by parts, which would give two boundary terms. Why would these boundary terms vanish? $\endgroup$ May 29, 2023 at 7:24
  • 2
    $\begingroup$ @JensWagemaker in every qft setup on a non-compact space the fields are taken to vanish at infinity. This takes care of both of your worries. The plane waves are discarded from the spectrum, hence the Klein-Gordon operator admits no zero-modes and has both left- and right-inverses. It is also self-adjoint since the boundary terms vanish. $\endgroup$ May 29, 2023 at 11:44

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