Where are the wavefunctionals in the QFT transition amplitudes?

Struggles

As I’ve previously confessed, there are many seemingly basic issues in physics that I struggle to understand properly. One such issue I’m going to discuss in this post is the path integral representation of the vacuum to vacuum transition in quantum field theory (QFT) and the ensuing $i \epsilon \equiv i 0^+$ prescription for the Feynman propagator.

To begin, let us recall that in quantum mechanics, the transition amplitude from some state $| A \rangle$ to some other state $| B \rangle$ , over the time interval $[t',t]$ , is given by the path integral

(1) $\langle B[t] | A[t'] \rangle \equiv \int_{\mathbb{R}^D} d^D\vec{x} \int_{\mathbb{R}^D} d^D\vec{x}' \psi_B[\vec{x}]^* K[t,t';\vec{x},\vec{x}'] \psi_A[\vec{x}'],$

where

$K[t,t';\vec{x},\vec{x}'] \equiv \int_{\vec{x}'}^{\vec{x}} \mathcal{D}\vec{q} \int_{\prod_\infty \mathbb{R}^{D}} \mathcal{D}\vec{p} \exp\left[ i \int_{t'}^t d\tau L[\vec{q},\vec{p}] \right]$

for some appropriately defined “Lagrangian” $L$ governing the dynamics of the quantum system, that we assume is a function of position $\vec{q}[\tau]$ and momentum $\vec{p}[\tau]$ .

The key observation I wish to highlight in eq. (1) is the need to integrate the path integral $K$ itself against the initial $\psi_A$ and final $\psi_B$ wavefunctions, corresponding respectively to the initial $|A\rangle$ and final $|B\rangle$ states we are interested in. However, for some mysterious reason, when we go on to do QFT, the vacuum-to-vacuum transition amplitude somehow makes no reference whatsoever to the vacuum wavefunctional itself. For instance, the computation of the two-point function (aka the Feynman Green’s function) for a Heisenberg-picture scalar field $\varphi$ is often asserted to be given by

(2) $\langle \text{vac} |T \left\{ \varphi[x] \varphi[y] \right\} | \text{vac} \rangle = \frac{ \int\mathcal{D}\varphi e^{iS[\varphi]} \varphi[x] \varphi[y] }{\int\mathcal{D}\varphi e^{iS[\varphi]} } ;$

where the integrals over field configurations in both the numerator and denominator run over $\mathbb{R}$ at each point in spacetime and — rather crucially — do not seem to contain any explicit quantum-state dependence.

Gaussian Theory in Minkowski: Infinite Spacetime

Now, in infinite flat spacetime — i.e., examining the asymptotic past $t' \to -\infty$ to future $t \to \infty$ transition amplitude — and for a non-interacting theory with mass $m>0$ , namely

(Gaussian) $S \equiv \int_{\mathbb{R}^{D+1}} d^d x \left( \frac{1}{2} (\partial \varphi)^2 - \frac{m^2}{2} \varphi^2 \right),$

the above vacuum expectation value in eq. (2) can readily be computed within the canonical formalism, to yield

(3) $\langle \text{vac} |T \left\{ \varphi[x] \varphi[y] \right\} | \text{vac} \rangle = i \int_{\mathbb{R}^{D+1}} \frac{d\omega d^D\vec{k}}{(2\pi)^{D+1}} \frac{e^{-i\omega(t-t')+i\vec{k}\cdot(\vec{x}-\vec{x}')} }{\omega^2 - \vec{k}^2 - m^2 + i \epsilon} .$

The $i \epsilon \equiv i 0^+$ here tells us the $\omega$ integral’s contour needs to dip below the $\omega = -k$ pole and skirt above the $\omega = +k$ one; i.e., if we viewed the $\omega-$ integral as running over the reals, the poles are located at $\omega = \pm |\vec{k}| (1 - i \epsilon/(2\vec{k}^2))$ . One way to justify this contour prescription, as well as the well-known Feynman propagator result in eq. (3) is to begin with the observation that the ordinary integral

(4) $\int_{\mathbb{R}} dx e^{i\alpha x^2} = \int_{\mathbb{R}} dx e^{i\alpha_\text{R} x^2} e^{-\alpha_\text{I} x^2}$

is well defined only when the imaginary part of $\alpha = \alpha_\text{R} + i \alpha_\text{I}$ is positive; so that the integral itself is damped out in the large $|x|$ region. Next, the action of eq. (Gaussian) in Fourier spacetime,

(Gaussian.2) $S = \int_{\mathbb{R}^{D+1}} \frac{d\omega d^D\vec{k}}{(2\pi)^{D+1}} \frac{1}{2} (\omega^2-\vec{k}^2) | \widetilde{\varphi}[\omega,\vec{k}] |^2,$

tells us the scalar field $\widetilde{\varphi}$ of different momenta are decoupled from one another. Hence, up to an overall normalization, the path integral itself amounts to an infinite product of integrals of the form in eq. (4).

$\int\mathcal{D}\varphi e^{ i S[\varphi] } = \mathcal{N} \prod_{\omega,\vec{k}} \int d\widetilde{\varphi}[\omega,\vec{k}] \exp[ \frac{i}{2} (\omega^2-\vec{k}^2 + i \epsilon) | \widetilde{\varphi}[\omega,\vec{k}] |^2 ] .$

where we have now introduced an $i \epsilon$ — this amounts to rendering $\text{Im}[\alpha] > 0$ in eq. (4) — for otherwise the integral would be ill defined for each and every fixed $(\omega,\vec{k})$ . Up to a factor of 2 that I’ve not been able to hunt down thus far, I was able to show using such a Fourier spacetime calculation that eq. (2) does indeed lead to eq. (3); Peskin and Schroeder (in their path integral Chapter) does a similar calculation by placing the quantum system in a box, by doing the path integral over discrete Fourier spacetime.

To sum: For non-interacting theories, it appears to be possible to justify the apparent lack of vacuum state dependence within the path integral representation of the vacuum-to-vacuum transition amplitude, albeit in a round-about manner, if one takes the infinite past to infinite future limits.

Finite Time Propagation

Why should we, though, restrict our attention only to the asymptotic past and future? Surely we may learn more physics by demanding that the transition take place over a finite interval $[t',t]$ ? For instance, in cosmological applications, physicists have become interested in whether the perturbations imprinted in the cosmic microwave sky are due to an initial quantum state other than the vacuum one. This ‘initial state’ is usually released at a finite time, not the infinite past.

To this end, let us return to eq. (Gaussian) but take this finite range into account:

(Gaussian.3) $\int_{t'}^t d\tau \int_{\mathbb{R}^D} \frac{ d^D\vec{k} }{(2\pi)^D} \frac{1}{2} \left( |\dot{\widetilde{\varphi}}[\tau,\vec{k}|^2 + E_{\vec{k}}^2 |\widetilde{\varphi}[\tau,\vec{k}|^2 \right) ;$

with the positive energy defined as

$E_{\vec{k}} \equiv \sqrt{\vec{k}^2+m^2}.$

By expressing the action in mixed frequency-real space, we see that non-interacting field theories in infinite space are really a continuous infinity collection of simple harmonic oscillators (SHOs), with oscillation angular frequency $\Omega = E_{\vec{k}}$ , described by the Lagrangian $(1/2)(\dot{q}^2 - \Omega^2 q^2)$ . From quantum mechanics, we already know that its ground state is the Gaussian

$\langle x | E_0 \equiv \Omega/2 \rangle = (\Omega/\pi)^{1/4} \exp[-(\Omega/2) x^2]$ .

This immediately informs us that our non-interacting massive field theory has a vacuum wavefunctional

$\langle \Psi | \text{vac} \rangle = \mathcal{N}' \exp\left[ -\int_{\mathbb{R}^D} \frac{d^D\vec{k}}{(2\pi)^D} \frac{E_{\vec{k}}}{2} |\widetilde{\Psi}[\vec{k}]|^2 \right].$

Now, what we actually need to compute in QFT involves insertions of operators; namely,

$\left( \int\mathcal{D}\varphi e^{iS[\varphi]} \varphi[x] \varphi[y] \varphi[z] \dots \right)/\int\mathcal{D}\varphi e^{iS[\varphi]} .$

The well-known trick to achieve this, which facilitates perturbation theory for interacting fields, is to introduce a source in the action. Here, I shall simply sketch the construction for the quantum mechanical SHO:

$\langle E_0[t] | E_0[t'] \rangle_J \equiv \sqrt{\Omega/\pi}\int_{\mathbb{R}} dx \int_{\mathbb{R}} dx' e^{-\frac{\Omega}{2}(x^2+x'^2)} K_{\text{SHO}}[t,t';x,x'] ;$

where the SHO propagator — i.e., the path integral — is

$K[t,t';x,x'] = \mathcal{N} \int_{x'}^x \mathcal{D}q \exp\left[ i \int_{t'}^t \left( \frac{1}{2} \dot{q}^2 - \frac{\Omega^2}{2} q^2 + J \cdot q \right) d\tau \right] .$

This introduction of $J$ allows the insertion of $q[t_1] q[t_2] \dots q[t_N]$ by taking functional derivatives:

$\sqrt{\Omega/\pi}\int_{\mathbb{R}} dx \int_{\mathbb{R}} dx' e^{-\frac{\Omega}{2}(x^2+x'^2)} \mathcal{N} \int_{x'}^x \mathcal{D}q \exp\left[ i \int_{t'}^t \left( \frac{1}{2} \dot{q}^2 - \frac{\Omega^2}{2} q^2 \right) d\tau \right] q[t_1] \dots q[t_N] \\ = \frac{1}{i^N} \left. \frac{\delta^N \langle E_0[t] | E_0[t'] \rangle_J}{\delta J[t_1] \dots \delta J[t_N]} \right\vert_{J=0}.$

The $\langle E_0[t] | E_0[t'] \rangle_J$ itself can be tackled by first shifting the integration variables

$q[t' \leq s \leq t] = q_c[s] + \xi[s] + \int_{t'}^t G_s[s,\tau] J[\tau] d\tau,$

where $q_c[s]$ is the classical trajectory of the SHO that begins at $q_c[t'] = x'$ and ends at $q_c[t] = x$ ; $\xi[s]$ is the quantum trajectory that now needs to be integrated over all trajectories joining $\xi[t']=0$ to $\xi[t]=0$ , namely $\int_{x'}^x \mathcal{D}q = \int_0^0 \mathcal{D}\xi$ ; whereas $G_s[t,t']$ is the Green’s function of the SHO operator obeying Dirichlet boundary conditions. (These boundary conditions on $q_c$ , $\xi$ , and $G_s$ ensure the total $q[s]$ obeys the boundary condition $q[s=t'] = x'$ and $q[s=t] = x$ , as required by the definition of $K$ itself.) After quite a bit of work, I find

$\langle E_0[t] | E_0[t'] \rangle_J = e^{-i\frac{\Omega}{2} (t-t')} \exp\left[ \frac{1}{2} \int_{t'}^t d\tau \int_{t'}^t d\tau' (iJ[\tau]) \widetilde{G}_\text{F}[\tau-\tau'] (iJ[\tau']) \right] ,$

where

$\widetilde{G}_{\text{F}}[s] \equiv \frac{1}{2\Omega} \left( \Theta[s] e^{-i\Omega\cdot s} + \Theta[-s] e^{+i\Omega \cdot s} \right) .$

The reader already familiar with QFT would recognize this to be intimately related to the Feynman Green’s function. As we have already previously identified, going from the quantum mechanical SHO to the QFT of a massive scalar amounts to replacing $\Omega \to E_{\vec{k}}$ followed by multiplying all the relevant transition amplitudes over all momenta $\vec{k}$ . The result is

$\langle \text{vac} | T \left\{ \varphi[x] \dots \varphi[x_N] \right\} | \text{vac} \rangle = \frac{1}{i^N} \frac{\delta^N}{\delta J[x_1] \dots \delta J[x_N]} \left. \exp\left[ \frac{1}{2} \int_{\mathbb{R}^D} \frac{d^D\vec{k}}{(2\pi)^D} \int_{t'}^t d\tau \int_{t'}^t d\tau' (i \widetilde{J}[\tau,\vec{k}]) \widetilde{G}_{\text{F}}[\tau-\tau'] (i \widetilde{J}[\tau',\vec{k}]) \right] \right\vert_{J=0} .$

If we re-write the above exponent in real space and time, for e.g., $x \equiv (t,\vec{x})$ and $x' \equiv (t',\vec{x}')$ ,

$\int_{\mathbb{R}^D} \frac{d^D\vec{k}}{(2\pi)^D} \int_{t'}^t d\tau \int_{t'}^t d\tau' (i \widetilde{J}[\tau,\vec{k}]) \widetilde{G}_{\text{F}}[\tau-\tau'] (i \widetilde{J}[\tau',\vec{k}]) \\ = \int_{\mathbb{R}^{D+1}} d^d x \int_{\mathbb{R}^{D+1}} d^d x' (i J[x]) G_{\text{F}}[x-x'] (i J[x']) ,$

we uncover the Feynman Green’s function

(Feynman) $G_{\text{F}}[x-x'] = \int_{\mathbb{R}^{D}} \frac{d^D\vec{k}}{(2\pi)^D} \frac{e^{i\vec{k}\cdot(\vec{x}-\vec{x}')}}{2 E_{\vec{k}}} \left( \Theta[t-t'] e^{-iE_{\vec{k}}(t-t')} + \Theta[t'-t] e^{+iE_{\vec{k}}(t-t')} \right) .$

This result would also be obtained by performing the $\omega-$ integral in eq. (3) while ensuring the $i \epsilon$ prescription is taken into account; thereby justifying the latter by explicitly taking into account the vacuum state dependence at the ends of the time interval $[t',t]$ .

Closing Remarks

I was quite happy to see Richard Woodard recently paying tribute to the amazing Steven Weinberg, where he explained how Weinberg was one of the few field theorists who did carefully take into account the vacuum state dependence in the path integral formulation.

References

R.P. Woodard, “Big Steve and the State of the Universe,” Symmetry 15, no.4, 856 (2023) doi:10.3390/sym15040856 [arXiv:2303.05111 [hep-th]].
M.E.Peskin and D.V.Schroeder, “An Introduction to quantum field theory,” Addison-Wesley, 1995, ISBN 978-0-201-50397-5

Author: Yi-Zen Chu

I am a theoretical physicist, with research interests spanning gravitation and field theory, particle cosmology and Mathematica software development. View all posts by Yi-Zen Chu

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