It is noted in Peskin and Schroeder's QFT text that the momenta used in the evaluation of the field operator $\phi(x)$ are "on mass-shell": $p^2=m^2$. Specifically, this is in relation to the expression: $$\phi(x)=\int{d^3p\over (2\pi)^3}{1\over\sqrt{2E_p}}\bigg\{\bigg(\mathbf{a_p}+{i\over\sqrt{2E_p}}\tilde{j}(p)\bigg)e^{-ip\cdot x}+\;\text{h.c.}\bigg\}\quad \tag{2.64}.$$ Why is this the case? My educated guess is that the particle momenta must be "on shell" because without this condition, $\phi(x)$ will not be guaranteed to have Lorentz invariance.

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    $\begingroup$ Isn't it because you already integrated out the $dp_0$ component, which if i recall correctly gives you a delta for the on-shell relation? $\endgroup$
    – LolloBoldo
    Commented May 18 at 13:11

2 Answers 2


TL;DR: The on-shellness of the Fourier decomposition (2.64) comes from the source-full/inhomogeneous Klein-Gordon equation (2.61).

P&S argue this in 2 steps:

  1. First P&S consider the Fourier-transformed source-free/homogeneous Klein-Gordon equation, and argue that its solution $\phi_0$ is on-shell.

  2. Next P&S introduce in eq. (2.63) a classical source $j$ via the retarded Greens function, which can also be brought on an on-shell form: $$\begin{align} D_R(x-y)~\stackrel{(2.58)}{=}~& \int\!\frac{d^4p}{(2\pi)^4}\left. \frac{i}{p^2-m^2}\right|_{p^0\to p^0+i\epsilon} e^{-ip\cdot(x-y)}\cr ~=~~~&\int\!\frac{d^4p}{(2\pi)^4}\frac{i}{(p^0+i\epsilon)^2-E_{\bf p}^2}e^{-ip\cdot(x-y)}\cr ~=~~~&-\int\!\frac{d^3p}{2E_{\bf p}(2\pi)^3} \frac{dp^0}{2\pi i}\left(\frac{1}{p^0+i\epsilon-E_{\bf p}}-\frac{1}{p^0+i\epsilon+E_{\bf p}}\right)e^{-ip\cdot(x-y)}\cr ~\stackrel{\text{Res.Thm.}}{=}&\theta(x^0-y^0)\int\!\frac{d^3p}{2E_{\bf p}(2\pi)^3} \left(\left. e^{-ip\cdot(x-y)}\right|_{p^0=E_{\bf p}} -\left. e^{-ip\cdot(x-y)}\right|_{p^0=-E_{\bf p}} \right)\cr ~=~~~&\theta(x^0-y^0)\int\!\frac{d^3p}{2E_{\bf p}(2\pi)^3} \left.\left( e^{-ip\cdot(x-y)} - e^{ip\cdot(x-y)} \right)\right|_{p^0=E_{\bf p}}, \end{align}\tag{A} $$ where the above $i\epsilon$ prescription is consistent with the integration contour in the figure on p.30 between eqs. (2.54-55). The above eq. (A) can alternatively be extracted from eqs. (2.54-55).


  1. M.E. Peskin & D.V. Schroeder, An Intro to QFT, 1995; section 2.4.

You can view the mass shell condition as a constraint on the Hilbert space that comes from the action in the path integral formalism. More specifically, there is an equality of Lorentz invariant measures $$\int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_p} = \int \frac{d^4p}{(2\pi)^4} \delta(p^2-m^2) \Theta(p^0) $$

You should check that these really are equal. The RHS is Lorentz invariant because the sign of $p^0$ can't change under Lorentz transformations that are connected to the identity, and even if we included time reversal we integrate over all the momenta anyways. Anyways, hopefully this helps explain where the form of the Fourier decomposition $$\phi(x) = \int \frac{d^3p}{(2\pi)^3 2E_p} \phi(p) e^{ipx} + h.c.$$ comes from. If you want to use the more covariant looking measure, you'll find that the term I wrote comes from one Theta function, and the +h.c. gives the opposite theta function, so that if we defined $\phi^\dagger(p) := \phi(-p)$, then $$\phi(x) = \int \frac{d^4p}{(2\pi)^4} \delta(p^2-m^2) \phi(p) e^{ipx} $$

Finally, we can rewrite the delta function as $$\phi(x) = \int \frac{d^4p d\tau}{(2\pi)^4} \phi(p) e^{ipx + i\tau (p^2 - m^2 +i \epsilon)} $$

This is called Schwinger's proper time formalism. In fact, by noting that $\langle 0| \phi(p) \phi(-p) |0\rangle \sim \frac{i}{p^2-m^2+i\epsilon}$, we can see that $$\langle 0| \phi(x) \phi(y) |0\rangle = \int \frac{d^4p}{(2\pi)^4}\int_0^\infty d\tau e^{ip(x-y) + i\tau (p^2 - m^2 +i \epsilon)}$$ you can interpret $\tau$ as the proper time of a particle traveling from the endpoints $y \to x$, which as a Lagrange multiplier on the mass shell constraint.

  • $\begingroup$ So, does Schwinger's proper time formalism lend itself well to path integral approaches to QFT? $\endgroup$ Commented May 18 at 15:03
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    $\begingroup$ Yes, and a good reference to get you started is the relevant chapter in Schwartz. $\endgroup$
    – 11zaq
    Commented May 18 at 16:15

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