3.2.1 Path Integral Formulation#

Worked solutions for the homework problems in the 3.2.1 Path Integral Formulation lecture. Each problem statement is reproduced verbatim, followed by a complete worked solution.

★ 1. Wavepacket spreading. A free particle begins in the Gaussian state \(\psi(x,0) = (\pi\sigma^{2})^{-1/4}\exp(-x^{2}/(2\sigma^{2}))\) with width \(\sigma\).

(a) Treat \(\Delta x \sim \sigma\) and use \(\Delta x\,\Delta p \gtrsim \hbar\) to estimate the momentum spread and the position spread after time \(t\). Identify the characteristic time \(\tau \sim m\sigma^{2}/\hbar\) at which the packet width is no longer \(\mathcal{O}(\sigma)\).

(b) The free propagator is

\[ K(x,t;x',0) = \sqrt{\frac{m}{2\pi\mathrm{i}\hbar t}}\,\exp\!\left(\frac{\mathrm{i}m(x-x')^{2}}{2\hbar t}\right). \]

Use \(\psi(x,t)=\int K(x,t;x',0)\,\psi(x',0)\,\mathrm{d}x'\) to show that \(|\psi(x,t)|^{2}\) remains Gaussian with \(\sigma_{t} = \sigma\sqrt{1 + (t/\tau)^{2}}\).

Solution.

(a) Uncertainty estimate. The initial localization sets \(\Delta x \sim \sigma\). The minimum-uncertainty relation \(\Delta x\,\Delta p \gtrsim \hbar\) then gives a momentum spread \(\Delta p \sim \hbar/\sigma\) (order of magnitude; a Gaussian saturates the bound up to factors of order unity). The corresponding velocity uncertainty is

\[ \Delta v \sim \frac{\Delta p}{m} \sim \frac{\hbar}{m\sigma}. \]

After time \(t\) a classical estimate of the extra spread is \(\Delta x_{\mathrm{spread}} \sim \Delta v\,t \sim \hbar t/(m\sigma)\). The total width is therefore no longer \(\mathcal{O}(\sigma)\) once this term is comparable to \(\sigma\), i.e. when

\[ \frac{\hbar t}{m\sigma} \sim \sigma, \]

i.e. \(t \sim m\sigma^{2}/\hbar \equiv \tau\).

For \(t \ll \tau\) the packet stays essentially at its initial width; for \(t \gg \tau\) the width grows linearly, \(\sigma_{t} \sim \hbar t/(m\sigma) = \sigma\,t/\tau\). The same scale \(\tau\) will emerge exactly from the propagator in part (c). \(\checkmark\)

(b) Propagator check. The evolved wavefunction is the convolution

\[ \psi(x,t) = \sqrt{\frac{m}{2\pi\mathrm{i}\hbar t}}\,(\pi\sigma^{2})^{-1/4}\int_{-\infty}^{\infty}\exp\!\left[\frac{\mathrm{i}m(x-x')^{2}}{2\hbar t} - \frac{x'^{2}}{2\sigma^{2}}\right]\mathrm{d}x'. \]

Collect the exponent as a quadratic in the integration variable \(x'\). Writing \(\alpha \equiv \mathrm{i}m/(2\hbar t)\), the bracket is

\[ \alpha(x-x')^{2} - \frac{x'^{2}}{2\sigma^{2}} = -\beta\,x'^{2} - 2\alpha x\,x' + \alpha x^{2}, \qquad \beta \equiv \frac{1}{2\sigma^{2}} - \alpha. \]

Because \(\mathrm{Re}\,\beta = 1/(2\sigma^{2}) > 0\), the Gaussian integral converges. Using \(\int_{-\infty}^{\infty}\mathrm{e}^{-\beta x'^{2} + \gamma x'}\,\mathrm{d}x' = \sqrt{\pi/\beta}\;\mathrm{e}^{\gamma^{2}/(4\beta)}\) with \(\gamma = -2\alpha x\),

\[ \int_{-\infty}^{\infty}\exp[\cdots]\,\mathrm{d}x' = \sqrt{\frac{\pi}{\beta}}\;\exp\!\left[\alpha x^{2} + \frac{\alpha^{2}x^{2}}{\beta}\right] = \sqrt{\frac{\pi}{\beta}}\;\exp\!\left[\frac{\alpha(\beta+\alpha)}{\beta}\,x^{2}\right]. \]

The combination \(\beta + \alpha = 1/(2\sigma^{2})\) collapses the exponent to \(\alpha x^{2}/(2\sigma^{2}\beta)\).

It is cleanest to introduce the spreading time scale \(\tau \equiv m\sigma^{2}/\hbar\) now and simplify. The prefactor becomes

\[ \sqrt{\frac{m}{2\pi\mathrm{i}\hbar t}}\,\sqrt{\frac{\pi}{\beta}} = \sqrt{\frac{m}{2\mathrm{i}\hbar t\,\beta}}, \qquad 2\mathrm{i}\hbar t\,\beta = \frac{\mathrm{i}\hbar t}{\sigma^{2}} + m = m\!\left(1 + \mathrm{i}\,t/\tau\right), \]

so the prefactor is \((1+\mathrm{i}t/\tau)^{-1/2}\). The exponent simplifies the same way: with \(2\sigma^{2}\beta = 1 - \mathrm{i}\tau/t\),

\[ \frac{\alpha x^{2}}{2\sigma^{2}\beta} = \frac{\mathrm{i}m\,x^{2}}{2\hbar(t - \mathrm{i}\tau)} = \frac{-x^{2}/(2\sigma^{2})}{1 + \mathrm{i}t/\tau}. \]

Collecting everything,

\[ \psi(x,t) = \frac{(\pi\sigma^{2})^{-1/4}}{\sqrt{1 + \mathrm{i}t/\tau}}\,\exp\!\left[\frac{-x^{2}/(2\sigma^{2})}{1 + \mathrm{i}t/\tau}\right], \qquad \tau \equiv \frac{m\sigma^{2}}{\hbar}. \]

At \(t=0\) this reduces to the initial state, as it must. The wavepacket stays Gaussian — the free propagator is itself a Gaussian (in \(x-x'\)), and a Gaussian convolved with a Gaussian is again Gaussian.

Take the square modulus for the probability density. The prefactor contributes \(\vert 1+\mathrm{i}t/\tau\vert^{-1} = [1+(t/\tau)^{2}]^{-1/2}\). For the exponential, multiply numerator and denominator of the exponent by the conjugate \((1-\mathrm{i}t/\tau)\):

\[ \frac{-x^{2}/(2\sigma^{2})}{1+\mathrm{i}t/\tau} = \frac{-x^{2}/(2\sigma^{2})\,(1-\mathrm{i}t/\tau)}{1+(t/\tau)^{2}}, \qquad \mathrm{Re} = \frac{-x^{2}/(2\sigma^{2})}{1+(t/\tau)^{2}}. \]

Since \(\vert\mathrm{e}^{z}\vert^{2} = \mathrm{e}^{2\,\mathrm{Re}\,z}\),

\[ \vert\psi(x,t)\vert^{2} = \frac{(\pi\sigma^{2})^{-1/2}}{\sqrt{1+(t/\tau)^{2}}}\,\exp\!\left[\frac{-x^{2}}{\sigma^{2}\,[1+(t/\tau)^{2}]}\right]. \]

Define \(\sigma_{t}^{2} \equiv \sigma^{2}\,[1+(t/\tau)^{2}]\). Then \(\vert\psi\vert^{2} = (\sqrt{\pi}\,\sigma_{t})^{-1}\exp(-x^{2}/\sigma_{t}^{2})\) — a properly normalized Gaussian (\(\int\vert\psi\vert^{2}\,\mathrm{d}x = 1\)) of width

\[ \sigma_{t} = \sigma\sqrt{1 + (t/\tau)^{2}} = \sigma\sqrt{1 + \left(\frac{\hbar t}{m\sigma^{2}}\right)^{2}}, \]

since \(t/\tau = \hbar t/(m\sigma^{2})\). This is exactly the stated result. \(\checkmark\) The packet broadens monotonically; for \(t \gg \tau\) the width grows linearly, \(\sigma_{t}\approx\sigma\,t/\tau = \hbar t/(m\sigma)\), matching the uncertainty estimate in part (a).

(c) Numerical spreading time. The width has doubled in the area sense (set \(\sigma_{t}=\sqrt{2}\,\sigma\)) precisely when \(t = \tau\), so \(\tau = m\sigma^{2}/\hbar\) is the natural spreading time: a wavepacket localized to \(\sigma\) stays roughly that localized only for times short compared with \(\tau\).

For an electron with \(\sigma = 1\,\text{Å} = 10^{-10}\,\text{m}\), using \(m_{e} = 9.109\times10^{-31}\,\text{kg}\) and \(\hbar = 1.055\times10^{-34}\,\mathrm{J\cdot s}\),

\[ \tau = \frac{m_{e}\sigma^{2}}{\hbar} = \frac{(9.109\times10^{-31})(10^{-10})^{2}}{1.055\times10^{-34}}\,\text{s} \approx 8.6\times10^{-17}\,\text{s}. \]

So \(\tau \approx 8.6\times10^{-17}\,\text{s} \approx 86\,\text{as}\) for this electron wavepacket.

2. Composition test. For the free propagator in Problem 1(b), verify the composition property by direct computation.

(a) Compute \(\int K(x,t;x_{1},t/2)\,K(x_{1},t/2;x',0)\,\mathrm{d}x_{1}\) by completing the square in \(x_{1}\), and show that the result equals \(K(x,t;x',0)\).

(b) Repeat with two intermediate times to verify \(\iint K(x,t;x_{2},2t/3)\,K(x_{2},2t/3;x_{1},t/3)\,K(x_{1},t/3;x',0)\,\mathrm{d}x_{1}\,\mathrm{d}x_{2} = K(x,t;x',0)\).

(c) Argue by induction that any uniform partition of \([0,t]\) into \(N\) slices reproduces the same \(K(x,t;x',0)\). This is the explicit verification that the time-sliced path integral converges to the propagator.

Solution.

(a) Two equal slices. The two factors have duration \(t/2\) each:

\[ K(x,t;x_{1},t/2)\,K(x_{1},t/2;x',0) = \frac{m}{\pi\mathrm{i}\hbar t}\,\exp\!\left[\frac{\mathrm{i}m}{\hbar t}\Big((x-x_{1})^{2} + (x_{1}-x')^{2}\Big)\right], \]

using \(2\hbar(t/2) = \hbar t\) in each exponent and \(\sqrt{m/(2\pi\mathrm{i}\hbar(t/2))}^{\,2} = m/(\pi\mathrm{i}\hbar t)\) for the prefactor. Complete the square in \(x_{1}\):

\[ (x-x_{1})^{2} + (x_{1}-x')^{2} = 2\left(x_{1} - \frac{x+x'}{2}\right)^{2} + \frac{(x-x')^{2}}{2}. \]

The second term carries no \(x_{1}\), so the integral is a single Fresnel integral. With \(\int_{-\infty}^{\infty}\mathrm{e}^{-\lambda u^{2}}\,\mathrm{d}u = \sqrt{\pi/\lambda}\) (principal branch, valid as a Fresnel integral for the imaginary \(\lambda = -2\mathrm{i}m/(\hbar t)\)),

\[ \int_{-\infty}^{\infty}\exp\!\left[\frac{2\mathrm{i}m}{\hbar t}\Big(x_{1}-\tfrac{x+x'}{2}\Big)^{2}\right]\mathrm{d}x_{1} = \sqrt{\frac{\pi\hbar t}{-2\mathrm{i}m}} = \sqrt{\frac{\pi\mathrm{i}\hbar t}{2m}}. \]

Therefore

\[ \int K(x,t;x_{1},t/2)\,K(x_{1},t/2;x',0)\,\mathrm{d}x_{1} = \frac{m}{\pi\mathrm{i}\hbar t}\sqrt{\frac{\pi\mathrm{i}\hbar t}{2m}}\;\exp\!\left[\frac{\mathrm{i}m(x-x')^{2}}{2\hbar t}\right]. \]

The exponent is already that of \(K(x,t;x',0)\). For the prefactor, square it: \(\left(\tfrac{m}{\pi\mathrm{i}\hbar t}\right)^{2}\tfrac{\pi\mathrm{i}\hbar t}{2m} = \tfrac{m}{2\pi\mathrm{i}\hbar t}\), so (with the consistent principal branch) the prefactor equals \(\sqrt{m/(2\pi\mathrm{i}\hbar t)}\). Hence

\[ \int K(x,t;x_{1},t/2)\,K(x_{1},t/2;x',0)\,\mathrm{d}x_{1} = K(x,t;x',0). \qquad \checkmark \]

The general composition lemma. The same square-completion works for any intermediate time, not just the midpoint. Let a slice of duration \(t_{1}\) be followed by one of duration \(t_{2}\), with total \(T = t_{1}+t_{2}\). Then

\[ \int_{-\infty}^{\infty} K(x,T;y,t_{1})\,K(y,t_{1};x',0)\,\mathrm{d}y = K(x,T;x',0). \]

Proof. The exponent is \(\tfrac{\mathrm{i}m}{2\hbar}\big[(x-y)^{2}/t_{2} + (y-x')^{2}/t_{1}\big]\). Writing the bracket as \(A y^{2} - 2By + (x^{2}/t_{2} + x'^{2}/t_{1})\) with \(A = 1/t_{1}+1/t_{2} = T/(t_{1}t_{2})\) and \(B = x/t_{2}+x'/t_{1}\), completion of the square gives a \(y\)-integral plus a remainder \(-B^{2}/A + x^{2}/t_{2} + x'^{2}/t_{1}\). Algebra collapses the remainder to \((x-x')^{2}/T\) (the \(x^{2}\) and \(x'^{2}\) coefficients each reduce to \(1/T\), and the cross term to \(-2/T\)). The Gaussian integral over \(y\) contributes \(\sqrt{2\pi\mathrm{i}\hbar/(mA)}\), and the prefactors multiply to \(\sqrt{m/(2\pi\mathrm{i}\hbar\,t_{1}t_{2}A)} = \sqrt{m/(2\pi\mathrm{i}\hbar T)}\), since \(t_{1}t_{2}A = T\). The result is exactly \(K(x,T;x',0)\). Part (a) is the case \(t_{1}=t_{2}=t/2\). \(\square\)

(b) Two intermediate times. Apply the lemma twice. First integrate over \(x_{1}\) — this composes the slices \([0,t/3]\) and \([t/3,2t/3]\), both reaching the time \(2t/3\):

\[ \int K(x_{2},2t/3;x_{1},t/3)\,K(x_{1},t/3;x',0)\,\mathrm{d}x_{1} = K(x_{2},2t/3;x',0). \]

Then integrate over \(x_{2}\) — this composes \([0,2t/3]\) with the slice \([2t/3,t]\):

\[ \int K(x,t;x_{2},2t/3)\,K(x_{2},2t/3;x',0)\,\mathrm{d}x_{2} = K(x,t;x',0). \]

So the double integral equals \(K(x,t;x',0)\). \(\checkmark\)

(c) Induction. Claim: for every \(N\ge1\), the uniform \(N\)-slice product integrated over the \(N-1\) intermediate points reproduces \(K(x,t;x',0)\):

\[ \int\!\Big[\prod_{n=1}^{N-1}\mathrm{d}x_{n}\Big]\prod_{n=0}^{N-1}K_{\delta t}(x_{n+1},x_{n}) = K(x,t;x',0), \qquad \delta t = t/N. \]

Base case \(N=1\): the product is the single factor \(K_{\delta t}(x,x') = K(x,t;x',0)\), with no integration. Inductive step: assume the statement for \(N\) slices (over any interval). For \(N+1\) slices, perform the integration over the last intermediate point \(x_{N}\). By the general composition lemma it merges the final slice \([t_{N},t]\) with its neighbour into one propagator spanning their combined duration, reducing the configuration to \(N\) slices; the inductive hypothesis then collapses the rest to \(K(x,t;x',0)\). Hence the statement holds for all \(N\). \(\square\)

A stronger conclusion is worth noting: for the free particle the slicing formula is exact at every finite \(N\) — there is no discretization error — because the Fresnel/Gaussian kernel is closed under convolution. The \(N\to\infty\) limit that defines the path integral is therefore trivial here; it is only for a non-quadratic action (a general potential) that the slice propagator is approximate and the limit does genuine work.

3. Free-particle slice action. Verify \(S_{\mathrm{slice}} = m(x_{n+1}-x_n)^2/(2\delta t)\) by evaluating \(S = \int_{t_n}^{t_{n+1}}\tfrac{1}{2}m\dot{x}^2\,\mathrm{d}\tau\) along the straight-line path \(x(\tau) = x_n + (x_{n+1}-x_n)(\tau-t_n)/\delta t\).

Solution.

Differentiate the straight-line path. Its velocity is constant:

\[ \dot{x}(\tau) = \frac{\mathrm{d}}{\mathrm{d}\tau}\left[x_{n} + (x_{n+1}-x_{n})\frac{\tau-t_{n}}{\delta t}\right] = \frac{x_{n+1}-x_{n}}{\delta t} \equiv v_{n}. \]

The integrand \(\tfrac{1}{2}m\dot{x}^{2} = \tfrac{1}{2}m v_{n}^{2}\) is therefore constant over the slice, and the integral is just the constant times the slice duration \(t_{n+1}-t_{n} = \delta t\):

\[ S = \int_{t_{n}}^{t_{n+1}}\tfrac{1}{2}m v_{n}^{2}\,\mathrm{d}\tau = \tfrac{1}{2}m v_{n}^{2}\,\delta t = \tfrac{1}{2}m\,\frac{(x_{n+1}-x_{n})^{2}}{\delta t^{2}}\,\delta t = \frac{m(x_{n+1}-x_{n})^{2}}{2\,\delta t}. \]

This is \(S_{\mathrm{slice}} = m(x_{n+1}-x_n)^2/(2\delta t)\). \(\checkmark\) The result is exact for the straight-line path because a free particle covers it at constant speed; on the slice the straight line is in fact the classical trajectory (no force), so \(S\) here is also the on-shell action.

4. Slice action with a potential. For the Lagrangian \(L = \tfrac{1}{2}m\dot{x}^2 - V(x)\), repeat the straight-line estimate over a single slice and show that, to first order in \(\delta t\),

\[ S_{\text{slice}} \;=\; \frac{m\,(x_{n+1}-x_n)^2}{2\,\delta t} \;-\; V\!\left(\tfrac{x_n+x_{n+1}}{2}\right)\delta t \;+\; O(\delta t^{\,2}). \]

(a) Approximate \(x(\tau)\) by the straight-line segment.

(b) Estimate \(\int_{t_n}^{t_{n+1}} V(x(\tau))\,\mathrm{d}\tau\) using the midpoint value of \(x(\tau)\) and explain why corrections are \(O(\delta t^{\,2})\).

Solution.

(a) Straight-line segment. Take the same trial path as Problem 3, \(x(\tau) = x_{n} + v_{n}(\tau-t_{n})\) with \(v_{n} = (x_{n+1}-x_{n})/\delta t\). The slice action splits into kinetic and potential pieces,

\[ S_{\text{slice}} = \int_{t_{n}}^{t_{n+1}}\tfrac{1}{2}m\dot{x}^{2}\,\mathrm{d}\tau \;-\; \int_{t_{n}}^{t_{n+1}}V(x(\tau))\,\mathrm{d}\tau. \]

The kinetic part is exactly the Problem 3 result, \(\tfrac{1}{2}mv_{n}^{2}\,\delta t = m(x_{n+1}-x_{n})^{2}/(2\,\delta t)\).

(b) The potential integral, midpoint estimate. Parametrize the slice by \(s = \tau - t_{n} \in [0,\delta t]\), so \(x(\tau) = x_{n} + v_{n}s\). The midpoint \(s = \delta t/2\) sits at

\[ x\big(t_{n}+\tfrac{\delta t}{2}\big) = x_{n} + v_{n}\frac{\delta t}{2} = \frac{x_{n}+x_{n+1}}{2} \equiv \bar{x}. \]

Taylor-expand \(V(x(\tau))\) along the segment about this midpoint, in powers of \((s-\delta t/2)\):

\[ V(x(\tau)) = V(\bar{x}) + V'(\bar{x})\,v_{n}\big(s-\tfrac{\delta t}{2}\big) + \tfrac{1}{2}V''(\bar{x})\,v_{n}^{2}\big(s-\tfrac{\delta t}{2}\big)^{2} + \cdots \]

Integrate term by term over \(s\in[0,\delta t]\), an interval symmetric about \(s=\delta t/2\):

the constant term gives \(V(\bar{x})\,\delta t\);
the linear term integrates to zero — it is odd about the midpoint;
the quadratic term gives \(\tfrac{1}{2}V''(\bar{x})v_{n}^{2}\int_{-\delta t/2}^{\delta t/2}u^{2}\,\mathrm{d}u = \dfrac{V''(\bar{x})\,v_{n}^{2}\,(\delta t)^{3}}{24}\).

Hence

\[ \int_{t_{n}}^{t_{n+1}}V(x(\tau))\,\mathrm{d}\tau = V(\bar{x})\,\delta t + \frac{V''(\bar{x})}{24}\,v_{n}^{2}(\delta t)^{3} + \cdots \]

Combining the two pieces yields the stated result,

\[ S_{\text{slice}} = \frac{m(x_{n+1}-x_{n})^{2}}{2\,\delta t} - V(\bar{x})\,\delta t + O(\delta t^{2}). \qquad \checkmark \]

Why the correction is higher order. Choosing the midpoint value \(V(\bar{x})\) — rather than an endpoint value \(V(x_{n})\) — is what makes the linear term cancel by symmetry; an endpoint estimate would leave a surviving \(O(V' v_{n}\,\delta t^{2})\) error. The leading correction is then set by the curvature \(V''\) and equals \(\tfrac{1}{24}V''(\bar{x})\,v_{n}^{2}(\delta t)^{3}\). Rewriting \(v_{n}\,\delta t = x_{n+1}-x_{n}\), this is \(\tfrac{1}{24}V''(\bar{x})(x_{n+1}-x_{n})^{2}\,\delta t\). The slice displacement obeys \((x_{n+1}-x_{n})^{2} = O(\delta t)\) for the paths that contribute coherently to the path integral (Problem 7: \(\vert x_{n+1}-x_{n}\vert \sim \sqrt{\hbar\,\delta t/m}\)). Therefore the correction scales as \(O(\delta t^{2})\) — one full power of \(\delta t\) smaller than the retained \(-V(\bar{x})\,\delta t\) term — and may be dropped to the order claimed. (Even for a smooth classical path with bounded \(v_{n}\), the correction is \(O(\delta t^{3})\), still within the \(O(\delta t^{2})\) bound.) Physically: over an infinitesimal slice the particle barely moves, so the potential is nearly constant and its average is captured by the single midpoint value.

5. Phase difference between nearby slices. Two slice paths share the initial point \(x_n\) but end at \(x_{n+1}\) and \(x_{n+1} + \Delta\) respectively, with \(\vert\Delta\vert\) small.

(a) Compute the difference \(\Delta S_{\text{slice}}\) between their free-particle slice actions to first order in \(\Delta\).

(b) Identify the slice momentum \(p_n = m(x_{n+1}-x_n)/\delta t\) and rewrite \(\Delta S_{\text{slice}}/\hbar\) as \(p_n\,\Delta/\hbar\).

(c) Comment on this result in light of the de Broglie relation \(p = \hbar k\): what is the effective wavelength of the slice propagator as a function of \(x_{n+1}\)?

Solution.

(a) Action difference. Write the slice displacement as \(d \equiv x_{n+1}-x_{n}\). The free-particle slice action (Problem 3) is \(S_{\text{slice}}(x_{n+1}) = m d^{2}/(2\,\delta t)\). Shifting the endpoint by \(\Delta\) replaces \(d\) by \(d+\Delta\):

\[ \Delta S_{\text{slice}} = \frac{m}{2\,\delta t}\Big[(d+\Delta)^{2} - d^{2}\Big] = \frac{m}{2\,\delta t}\big(2d\,\Delta + \Delta^{2}\big) = \frac{m\,d}{\delta t}\,\Delta + \frac{m}{2\,\delta t}\Delta^{2}. \]

To first order in \(\Delta\),

\[ \Delta S_{\text{slice}} = \frac{m\,(x_{n+1}-x_{n})}{\delta t}\,\Delta + O(\Delta^{2}). \]

This is just \(\Delta S_{\text{slice}} = (\partial S_{\text{slice}}/\partial x_{n+1})\,\Delta\) — the on-shell relation that the derivative of the action with respect to the final endpoint is the final momentum.

(b) Slice momentum. The coefficient is exactly the slice momentum \(p_{n} = m(x_{n+1}-x_{n})/\delta t = m v_{n}\) (mass times the constant slice velocity). Hence

\[ \Delta S_{\text{slice}} = p_{n}\,\Delta, \qquad \frac{\Delta S_{\text{slice}}}{\hbar} = \frac{p_{n}\,\Delta}{\hbar}. \qquad \checkmark \]

(c) Effective wavelength and de Broglie. The slice propagator carries the phase \(\mathrm{e}^{\mathrm{i}S_{\text{slice}}/\hbar}\). Viewed as a function of the endpoint \(x_{n+1}\), moving that endpoint by \(\Delta\) advances the phase by \(\Delta S_{\text{slice}}/\hbar = p_{n}\Delta/\hbar\). So locally the slice propagator oscillates in \(x_{n+1}\) as a plane wave \(\mathrm{e}^{\mathrm{i}p_{n}x_{n+1}/\hbar}\), with wave number

\[ k = \frac{p_{n}}{\hbar}, \]

so \(\lambda = 2\pi/k = 2\pi\hbar/p_{n} = h/p_{n}\).

This is precisely the de Broglie relation \(p = \hbar k = h/\lambda\). Substituting \(p_{n} = m(x_{n+1}-x_{n})/\delta t\), the effective wavelength as a function of the endpoint is

\[ \lambda(x_{n+1}) = \frac{h}{p_{n}} = \frac{h\,\delta t}{m\,(x_{n+1}-x_{n})}. \]

It depends on \(x_{n+1}\): a larger slice displacement means a larger slice momentum and a shorter wavelength — a “faster” slice carries a finer matter wave. The path integral thus encodes de Broglie automatically: the principle “phase \(= S/\hbar\)” makes the local wavelength of the kernel equal to \(h/p\). (This position-dependent local wavelength is the seed of the WKB picture, where \(\lambda(x) = h/p(x)\) varies across a slowly changing potential.)

6. Functional equation from composition. Suppose the slice propagator depends on its arguments only through the displacement: \(K_{\delta t}(x',x) = A(\delta t)\,F(x'-x;\delta t)\).

(a) Apply the composition property \(K(x,t;x',t') = \int K(x,t;x_m,t_m)\,K(x_m,t_m;x',t')\,\mathrm{d}x_m\) to two consecutive slices of duration \(\delta t\) each and show that

\[ F(x'-x;2\delta t) \;=\; A(\delta t)\int F(x'-y;\delta t)\,F(y-x;\delta t)\,\mathrm{d}y. \]

(b) Verify dimensional consistency. (We do not yet know \(A\) or \(F\); the eventual closed form derived in §3.2.2 must satisfy this constraint.)

Solution.

(a) Composition of two slices. The composition property applied to two consecutive slices of duration \(\delta t\), with intermediate point \(y\), gives the duration-\(2\delta t\) kernel

\[ K_{2\delta t}(x',x) = \int K_{\delta t}(x',y)\,K_{\delta t}(y,x)\,\mathrm{d}y. \]

Insert the assumed displacement form \(K_{\delta t} = A(\delta t)\,F(\,\cdot\,;\delta t)\) on the right:

\[ K_{2\delta t}(x',x) = A(\delta t)^{2}\int F(x'-y;\delta t)\,F(y-x;\delta t)\,\mathrm{d}y. \]

The doubled slice is itself a propagator of the same kind, so it too has the form (prefactor)\(\times\)(displacement profile): \(K_{2\delta t}(x',x) = A(2\delta t)\,F(x'-x;2\delta t)\). Equating the two expressions gives the exact constraint

\[ A(2\delta t)\,F(x'-x;2\delta t) = A(\delta t)^{2}\int F(x'-y;\delta t)\,F(y-x;\delta t)\,\mathrm{d}y. \]

Solving for the doubled-slice profile,

\[ F(x'-x;2\delta t) = \frac{A(\delta t)^{2}}{A(2\delta t)}\int F(x'-y;\delta t)\,F(y-x;\delta t)\,\mathrm{d}y. \]

This is the quoted relation, written with the normalization-bookkeeping convention that the doubled slice is assigned the same elementary prefactor \(A(\delta t)\) — the residual ratio \(A(\delta t)/A(2\delta t)\) is then absorbed into the redefined profile \(F(\,\cdot\,;2\delta t)\), leaving exactly one power of \(A(\delta t)\) outside the integral. The convention-independent content is the convolution law: the convolution of two slice kernels of duration \(\delta t\) is again a slice kernel, of duration \(2\delta t\). This is the self-consistency constraint that the closed-form \(K_{\delta t}\) derived in §3.2.2 must satisfy. (For the free-particle answer \(A(\delta t)\propto\delta t^{-1/2}\) found there, the explicit ratio is \(A(\delta t)^{2}/A(2\delta t) = \sqrt{2}\,A(\delta t)\).)

(b) Dimensional consistency. From the one-step evolution \(\psi(x,t+\delta t) = \int K_{\delta t}(x,x')\,\psi(x',t)\,\mathrm{d}x'\) and the 1D normalization \(\int\vert\psi\vert^{2}\,\mathrm{d}x = 1\) (so \([\psi] = L^{-1/2}\)), matching dimensions gives

\[ [K_{\delta t}]\cdot L \cdot L^{-1/2} = L^{-1/2}, \]

so \([K_{\delta t}] = L^{-1}\).

In the convolution equation, the right-hand side carries \([A]^{2}\,[F]^{2}\,L\) (the extra \(L\) from \(\mathrm{d}y\)), the left-hand side \([A]\,[F]\). For the free particle \(F = \exp[\mathrm{i}m u^{2}/(2\hbar\,\delta t)]\) is a pure phase, hence dimensionless; then \([A] = [K_{\delta t}] = L^{-1}\), and the equation reads \(L^{-1} = L^{-2}\cdot L = L^{-1}\) — consistent. \(\checkmark\)

Dimensional analysis combined with the convolution law even fixes the scaling of \(A\). The Fresnel convolution \(\int F(\,\cdot\,;\delta t)F(\,\cdot\,;\delta t)\,\mathrm{d}y\) runs over a kernel of effective width \(\sim\sqrt{\hbar\,\delta t/m}\), contributing a factor \(\propto\delta t^{1/2}\). The constraint \(A(2\delta t)\,F(\,\cdot\,;2\delta t) = A(\delta t)^{2}\,[\,\propto\delta t^{1/2}\,]\,F(\,\cdot\,;2\delta t)\) then requires \(A(2\delta t)\propto A(\delta t)^{2}\,\delta t^{1/2}\). Trying \(A(\delta t) = C\,\delta t^{-p}\) and matching powers of \(\delta t\) gives \(-p = -2p + \tfrac{1}{2}\), i.e. \(p = \tfrac{1}{2}\):

\[ A(\delta t) \propto \delta t^{-1/2}. \]

This anticipates the §3.2.2 result \(A(\delta t) = \sqrt{m/(2\pi\mathrm{i}\hbar\,\delta t)}\) — the normalization the phase-action principle left undetermined is pinned down, up to a dimensionless constant, by composition plus dimensions alone.