Bridge to QFT Capstones

These capstones make the extra input between a wave equation and a field prediction explicit. Each specifies a state, algebra, interaction and observable, then asks for a finite calculation. Use $\hbar=c=1$ and $\eta=(+---)$. The models retain prescribed switching or backgrounds and leading-order approximations where stated.

Helpful background. For the detector, use Propagators to Correlators and First-Order Transition Probability. For the pulse, use Pair Creation, Partial Trace and Entanglement Entropy. For the scattering reduction, use Relativistic QM to QED. The three projects may be attempted separately.

Switched detection from a Wightman function

Capstone 1. In the rest frame of an inertial detector, take a free real scalar field of mass $m\ge0$, initially in the Minkowski vacuum. The detector has ground and excited states separated by $\Omega>0$, with $d=|g\rangle\langle e|$. Its prescribed interaction is

$$\begin{aligned} H_I(t)&=\lambda\chi(t) \left(d e^{-i\Omega t}+d^\dagger e^{i\Omega t}\right) \Phi_f(t),\\ \Phi_f(t)&=\int d^3x\,f(\mathbf x)\phi(t,\mathbf x). \end{aligned}$$

Here $\lambda$ is real, the detector transition matrix element is one, and $\chi,f$ are real smooth rapidly decreasing functions. The spatial smearing makes the specified model well defined; no point-detector ultraviolet limit is assumed. Define

$$\begin{aligned} \widetilde\chi(\omega) &=\int dt\,\chi(t)e^{i\omega t},\\ \widetilde f(\mathbf p) &=\int d^3x\,f(\mathbf x)e^{-i\mathbf p\cdot\mathbf x},\\ d\Pi_p&=\frac{d^3p}{(2\pi)^3\,2E_p}. \end{aligned}$$

Starting in $|g\rangle\otimes|0\rangle$:

1. Derive the order-$\lambda^2$ excitation probability by summing first-order amplitudes over covariantly normalized final field modes.
2. Express the result as a Wightman double integral and explain its frequency sign.
3. For Gaussian switching and spatial smearing, obtain a one-dimensional integral when $m=0$ and verify the closed checkpoint below.
4. Identify the energy source, the perturbative limitation and the difference between a finite switching probability and a stationary rate.

Solution and quantitative checkpoints

The first-order process excites the detector and creates a field quantum. Its coefficient for momentum $\mathbf p$ is $-i\lambda\widetilde f(\mathbf p) \widetilde\chi(\Omega+E_p)$. Consequently

$$P_{\rm exc}^{(2)} =\lambda^2\int d\Pi_p\, |\widetilde f(\mathbf p)|^2 |\widetilde\chi(\Omega+E_p)|^2.$$

The sum $\Omega+E_p$ records the two positive excitation energies. With $W_f(t,t')=\langle0|\Phi_f(t)\Phi_f(t')|0\rangle$, the equivalent expression is

$$\begin{aligned} P_{\rm exc}^{(2)} &=\lambda^2\int dt\,dt'\,\chi(t)\chi(t')\\ &\quad\times e^{-i\Omega(t-t')}W_f(t,t'). \end{aligned}$$

This is an ordered vacuum correlation used in a transition probability. A time-ordered Feynman function is not interchangeable with it.

Take

$$\begin{aligned} \chi_T(t)&=e^{-t^2/(2T^2)},\\ f_L(\mathbf x)&= (2\pi L^2)^{-3/2}e^{-\mathbf x^2/(2L^2)}, \end{aligned}$$

Here $L,T>0$. Their transforms give

$$P_{\rm exc}^{(2)} =2\pi\lambda^2T^2\int d\Pi_p\, e^{-L^2\mathbf p^2-T^2(\Omega+E_p)^2}.$$

For $m=0$, put $A=L^2+T^2$ and $z=T^2\Omega/\sqrt A$. Angular integration gives

$$\frac{P_{\rm exc}^{(2)}}{\lambda^2} =\frac{T^2}{2\pi}e^{-T^2\Omega^2} \int_0^\infty dp\,p\,e^{-Ap^2-2T^2\Omega p}.$$

Complete the square, or differentiate the elementary Gaussian half-line integral, to obtain

$$\begin{aligned} \frac{P_{\rm exc}^{(2)}}{\lambda^2} &=\frac{T^2e^{-T^2\Omega^2}}{4\pi A}\,B(z),\\ B(z)&=1-\sqrt\pi\,z\,e^{z^2}\operatorname{erfc}(z). \end{aligned}$$

At $L=T=\Omega=1$, $P_{\rm exc}^{(2)}/\lambda^2 \approx0.00503997619544$. Check this against the positive radial integral rather than only re-evaluating the closed formula.

At large $z$, computing the two terms of $B$ separately loses precision. Scaled complementary-error functions avoid exponential overflow but do not by themselves remove this subtraction. Useful checks are

$$\begin{aligned} B(z)&=2\int_0^\infty du\,u\,e^{-u^2-2zu}>0,\\ B(z)&\sim\frac1{2z^2}. \end{aligned}$$

Use positive quadrature or a controlled large-$z$ expansion in that regime.

The external switching apparatus can supply energy; a vacuum click is not an inventory of pre-existing particles. Smooth adiabatic stretching of this Gaussian switching suppresses a positive-gap vacuum response. An eternally stationary inertial positive-gap excitation rate is zero. Growing only a plateau with fixed switching edges is a different limit and need not erase its edge probability. The switching distinction is discussed by Fewster, Juárez-Aubry and Louko (2016).

Require $P_{\rm exc}^{(2)}\ll1$ and control higher perturbative orders before treating this as an accurate probability. Gaussian profiles are not compactly supported: they cannot be inserted into the exact spacelike support argument in Derivation Problems.

Extension. For a homogeneous momentum-diagonal occupation distribution $n_{\mathbf p}$ with no anomalous pair coherence, show that the vacuum frequency factor becomes

$$\begin{aligned} &n_{\mathbf p} |\widetilde\chi(\Omega-E_p)|^2\\ &\quad +(n_{\mathbf p}+1) |\widetilde\chi(\Omega+E_p)|^2. \end{aligned}$$

The first term permits absorption. At a discrete-mode resonance a perturbative $T^2$ growth eventually invalidates the small-probability approximation; it is not automatically a finite continuum transition rate.

A coherent pulse and a mixed one-species state

Capstone 2. Consider a complex scalar with a spatially uniform, externally prescribed positive mass parameter. For one particle–antiparticle channel, let

$$\omega_j=\sqrt{\mathbf p^2+m_j^2}>0 \quad (j=0,1).$$

The paired operators are $a_{\mathbf p}$ and $b_{-\mathbf p}$; their momentum subscripts are suppressed below. The frequency is $\omega_0$ before $t=0$, $\omega_1$ during $0<t<T$, and $\omega_0$ after $T$. This neutral scalar-parameter pump preserves global U(1) charge. It is not an electric-field pulse. Begin with the incoming vacuum and first treat one channel or a regulated finite set of momenta.

1. Match a normalized incoming mode and its derivative at both jumps. Find the final occupation and its reversible durations.
2. Convert the mode coefficients into the operator convention and find the pair-number distribution.
3. Trace over the antiparticle and compute the particle’s entropy.
4. Determine whether the continuum mean number and produced energy are ultraviolet finite for an abrupt pulse in $3+1$ dimensions.

Solution and quantitative checkpoints

Let the incoming numerical mode be $q_{\rm in}(t)=e^{-i\omega_0t}/\sqrt{2\omega_0}$ before the first jump. Write the outgoing basis using its local time origin:

$$q_{\rm out}(t)= \frac{e^{-i\omega_0(t-T)}}{\sqrt{2\omega_0}}, \qquad t>T.$$

Define the mode coefficients by $q_{\rm in}=\alpha_m q_{\rm out} +\beta_m q_{\rm out}^*$ at late times. With $r=\omega_1/\omega_0$ and $\theta=\omega_1T$, continuity of $q,\dot q$ gives

$$\begin{aligned} \alpha_m&=\cos\theta -\frac{i}{2}(r+r^{-1})\sin\theta,\\ \beta_m&=\frac{i}{2}(r-r^{-1})\sin\theta. \end{aligned}$$

They obey $|\alpha_m|^2-|\beta_m|^2=1$. Equality of the two field expansions gives, in this convention,

$$a_{\rm out}=\alpha_m a_{\rm in} +\beta_m^*b_{\rm in}^\dagger.$$

Thus the off-diagonal operator coefficient is $\beta_m^*$, not $\beta_m$. The phase convention in Pair Creation must be translated before comparing complex coefficients. Their moduli give the same in-vacuum mean occupation:

$$N=|\beta_m|^2 =\frac{(\omega_1^2-\omega_0^2)^2} {4\omega_0^2\omega_1^2} \sin^2(\omega_1T).$$

For $r=2$, $N=(9/16)\sin^2\theta$. At $\theta=\pi$, the outgoing occupation is zero despite mixing at both interfaces. At $\theta=\pi/2$, $N=9/16$. A direct integration of $\ddot q+\omega_1^2q=0$ between the jumps is an independent check of the interference formula.

The incoming vacuum expressed in the outgoing number basis is a pure two-mode squeezed state. With a phase $\varphi$ fixed by the Bogoliubov convention, its Schmidt form can be written

$$\begin{aligned} |\Psi\rangle&= \frac1{\sqrt{1+N}} \sum_{n=0}^{\infty}c_n|n\rangle_a|n\rangle_b,\\ c_n&=\left(e^{i\varphi}\sqrt{\frac{N}{1+N}}\right)^n. \end{aligned}$$

For $N>0$, this phase satisfies $e^{i\varphi}\sqrt{N/(1+N)} =\beta_m^*/\alpha_m^*$ in the operator convention above. Consequently

$$P_n=\frac1{1+N} \left(\frac{N}{1+N}\right)^n.$$

For $N=9/16$, $P_0=16/25$ and $P_1=144/625$. Every term has equal particle and antiparticle number, hence exactly zero total charge. The mean number of quanta is $2N$, while the mean number of pairs is $N$.

Applying the partial trace gives $\rho_a=\sum_n P_n|n\rangle\langle n|$. Its entropy, using natural logarithms, is

$$S(\rho_a)=(N+1)\ln(N+1)-N\ln N.$$

For $N=9/16$ it is approximately $1.0209659294$ nats; at $N=0$ take the continuous limit $S=0$. This is bipartite entanglement entropy of a pure channel state, not an entropy increase of the full closed channel. Discarding the partner loses information that can matter during a later coherent pulse.

For large momentum, with $\Delta m^2=m_1^2-m_0^2$,

$$N_{\mathbf p}= \frac{(\Delta m^2)^2}{4p^4}\sin^2(pT) +O(p^{-5}).$$

For a nontrivial abrupt pulse with $T>0$, the number-density integral $\int d^3p\,N_{\mathbf p}/(2\pi)^3$ has an integrable ultraviolet tail. The produced-energy density includes $2\omega_0N_{\mathbf p}$ and has a logarithmic ultraviolet divergence. A finite momentum cutoff does not remove that physical limitation. Smooth the time profile before claiming a finite continuum energy prediction.

Finally, $\omega_1$ depends on momentum. One duration generally cannot undo all channels. The per-mode cancellation does not imply a reversed continuum pulse, and this prescribed-pump model does not include backreaction.

From dynamical exchange to a fixed potential

Capstone 3. Use two distinct massive charged Dirac species in rationalized natural units and the covariantly normalized tree exchange checkpoint from Relativistic QM to QED:

$$\mathcal M= \frac{q_1q_2\,j_1\cdot j_2}{k^2+i0}.$$

The currents are $j_i^\mu=\bar u_i'\gamma^\mu u_i$ with the accepted covariant spinor normalization. The on-shell currents satisfy $k\cdot j_1=k\cdot j_2=0$. Explain how this amplitude can yield an electrostatic potential, then identify the additional limits required to compare it with fixed-source scattering. Address gauge dependence, the force sign, external-state factors, recoil and omitted photon processes.

Solution checkpoints

Longitudinal propagator terms are proportional to $(k\cdot j_1)(k\cdot j_2)$ and vanish for the stated external on-shell currents. On each elastic Dirac line, $k_\mu\bar u(p')\gamma^\mu u(p)=0$ follows from the two Dirac equations. This is the current-conservation checkpoint, not a general proof about arbitrary off-shell currents.

In the static low-momentum limit $k^0=0$, $k^2=-\mathbf K^2$ and $j_1\cdot j_2\simeq4m_1m_2$ for leading spin-preserving matrix elements in a common canonical-spin basis. More generally the leading term includes each line’s Pauli-spinor overlap; spin-flip elements vanish at this order. The leading potential therefore acts as the identity on spin. The potential matching is

$$\begin{aligned} \widetilde V(\mathbf K) &=-\frac{\mathcal M}{4m_1m_2} =\frac{q_1q_2}{\mathbf K^2},\\ V(r)&=\frac{q_1q_2}{4\pi r}. \end{aligned}$$

Equal signed charges repel and opposite charges attract. The overall amplitude sign cannot be recovered from its squared modulus alone. The factor $4m_1m_2$ comes from covariantly normalized external currents; it is part of the matching, not optional.

A dynamical $2\to2$ matrix element also includes a target external state, its current and recoil, and a four-momentum conservation delta. A prescribed static source has no normalized target external leg and conserves only the matter energy. Its reduced amplitude and flux must be obtained consistently before using the Mott formula.

The static matching above takes both species to low momentum. A relativistic projectile scattering from a heavy target is a different reduction: keep its full current and reduce the target. Use the recoil bound from Derivation Problems; for a massless projectile it depends on $(E/M)(1-\cos\theta)$. Small Born coupling does not make that parameter small.

Dynamical photon emission needs photon final states and the corresponding amplitudes; radiative corrections need loop calculations and their renormalized and infrared-safe interpretation. They are absent from a fixed classical potential. The distinguishable-species assumption also matters: electron–positron scattering is not this single exchange channel with no annihilation contribution. The LSZ discussion states the charged-asymptotic-state qualifications.

References

• Fewster, Christopher J., Benito A. Juárez-Aubry, and Jorma Louko. “Waiting for Unruh.” Classical and Quantum Gravity 33, 165003 (2016). doi:10.1088/0264-9381/33/16/165003. Switched detector response and the distinction between switching limits.
• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge University Press, 2014. doi:10.1017/9781139540940. Free-field normalization, scattering and QED.
• Tong, David. Lectures on Quantum Field Theory. University of Cambridge, 2006–2007. Free scalar fields and QED.
• Weinberg, Steven. The Quantum Theory of Fields, Volume I: Foundations. Cambridge University Press, 1995. doi:10.1017/CBO9781139644167. Particle states, fields and scattering conventions.