Particle Creation by a Classical Source (on-shell mass momenta)

Question

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.

Answer 1

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).

References:

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

Answer 2

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.