Quantum Experiments · II · Paths, information, and time · Chapter 08
Can a movable slit record a path without changing interference?
At the \(1927\) Solvay conference, Einstein proposed mounting the first slit screen of a double-slit interferometer on delicate springs so that the transverse momentum recoil imparted as a particle deflects toward the upper or lower slit would register which slit the particle traversed. Could an experimenter read this mechanical recoil record after each trial while still observing high-contrast two-slit interference fringes on the detection screen? Bohr answered by showing that the movable slit plate must itself be treated as a quantum degree of freedom subject to position–momentum uncertainty.
When the particle’s two path alternatives \(\lvert L\rangle\) and \(\lvert R\rangle\) kick the movable support into pointer states \(\lvert d_L\rangle\) and \(\lvert d_R\rangle\), the combined particle–apparatus system forms an entangled state. If \(\lvert d_L\rangle=\lvert d_R\rangle\), the support remains unentangled with the path and stores zero path information; if \(\langle d_R\vert d_L\rangle=0\), an optimal readout of the support distinguishes the two paths without error. What governs interference loss is not the macroscopic mass of the slit plate by itself, but the inner product \(\gamma=\langle d_R\vert d_L\rangle\) of its resulting quantum states.
In the interactive diagram below, the movable support is represented by two Gaussian pointer wave packets centered at \(\pm\Delta/2\) along a dimensionless readout coordinate \(q\) with width \(\sigma\). Increasing the scaled separation \(\Delta/\sigma\) pulls the two pointer wave packets apart and continuously suppresses their overlap integral \(\gamma\).
Visual animation · Section 8.1
Coupling path alternatives to a mechanical pointer
For a balanced two-path preparation coupled unitarily to a pointer initially in \(\lvert d_0\rangle\), the joint state after the slit interaction is
\[\lvert\Psi\rangle = \frac{1}{\sqrt{2}}\bigl(\lvert L\rangle\lvert d_L\rangle + e^{i\phi}\lvert R\rangle\lvert d_R\rangle\bigr).\]
When the interference screen measures only the particle without conditioning on the pointer, we trace over the pointer Hilbert space. The off-diagonal coherence between \(\lvert L\rangle\) and \(\lvert R\rangle\) is multiplied by the pointer overlap \(\gamma=\langle d_R\vert d_L\rangle\), so the unconditional fringe visibility is simply \(V=|\gamma|=|\langle d_R\vert d_L\rangle|\).
Suppose the two pointer wavefunctions in coordinate \(q\) are normalized real Gaussians whose probability densities have standard deviation \(\sigma\) and centers separated by \(\Delta\):
\[\psi_{L,R}(q) = \frac{1}{(2\pi\sigma^2)^{1/4}}\exp\left[-\frac{(q\pm\Delta/2)^2}{4\sigma^2}\right] \quad\Longrightarrow\quad \gamma = \int_{-\infty}^{\infty}\psi_R^*(q)\psi_L(q)\,dq = \exp\left(-\frac{\Delta^2}{8\sigma^2}\right).\]
As the dimensionless separation \(\Delta/\sigma\) increases from \(0\) to \(6\), the fringe visibility \(V=\exp[-\Delta^2/(8\sigma^2)]\) decays smoothly from \(1\) toward \(0\) without any discontinuous jump requiring a human observer to inspect the pointer.
How well can any measurement on the pointer distinguish \(\lvert d_L\rangle\) from \(\lvert d_R\rangle\)? By the Helstrom minimum-error discrimination bound for two equally likely pure states, the path distinguishability is \(D=\sqrt{1-|\langle d_R\vert d_L\rangle|^2}=\sqrt{1-|\gamma|^2}\) (so the maximum probability of guessing the path correctly is \((1+D)/2\)). Combining \(V=|\gamma|\) and \(D=\sqrt{1-|\gamma|^2}\) yields the exact duality relation
\[D^2 + V^2 = 1 \qquad (\text{pure, balanced two-path state}),\]
which generalizes to the Englert–Greenberger–Yasin inequality \(D^2+V^2\le 1\) when the preparation or environment introduces mixedness.
Visual animation · Section 8.2
Pointer overlap and the visibility–distinguishability tradeoff
Because the reduced density matrix of the particle depends only on the inner product \(\langle d_R\vert d_L\rangle\), storing a distinguishing recoil in the support suppresses the unconditional interference fringes immediately upon interaction—whether or not anyone ever reads the support’s position. Conversely, observing \(V<1\) in an experiment does not prove that \(D=\sqrt{1-V^2}\) is stored in a single controllable pointer, because classical phase noise also yields \(V<1\) with \(D=0\) (\(D^2+V^2<1\)).
If the pointer remains isolated and coherent, measuring it in the superposition basis \((\lvert d_L\rangle\pm\lvert d_R\rangle)/\sqrt{2(1\pm\gamma)}\) rather than the coordinate basis \(q\) allows conditional fringes to be sorted out in coincidence—the quantum eraser protocol of Chapter 10.
Modern realizations of Einstein’s recoiling slit replace mechanical springs with trapped ions, Bose–Einstein condensates, or dissociating molecules whose momentum recoil and internal states are independently calibrated.
Visual animation · Section 8.3
Entanglement with the apparatus versus classical noise
Sweep the dimensionless pointer separation \(\Delta/\sigma\) from \(0\) to \(6\). Watch the two Gaussian pointer wave packets separate while \(V=\exp[-\Delta^2/(8\sigma^2)]\) falls and \(D=\sqrt{1-V^2}\) rises, maintaining \(D^2+V^2=1\) everywhere.
Interactive model · Idealized predictions
On a small screen, swipe the diagram to inspect its components. Moving marks illustrate the model; they are not measured trajectories.
When the pointer separation equals twice the packet width (\(\Delta=2\sigma\)), what are the fringe visibility \(V\) and the path distinguishability \(D\)?
Substituting \(\Delta/\sigma=2\) into the Gaussian overlap formula gives \(V=\exp[-2^2/8]=e^{-1/2}\approx 0.6065\). The optimal path distinguishability is then \(D=\sqrt{1-V^2}=\sqrt{1-e^{-1}}\approx 0.7950\). At this intermediate coupling, both partial interference (\(V\approx 0.61\)) and partial path information (\(D\approx 0.80\)) coexist, with \(D^2+V^2=e^{-1}+(1-e^{-1})=1\).
What are the visibility \(V\) and distinguishability \(D\) when the two pointer states coincide (\(\Delta/\sigma=0\), so \(\lvert d_L\rangle=\lvert d_R\rangle\))?
Use \(\langle d_R\vert d_L\rangle=1\) for normalized identical states.
When \(\lvert d_L\rangle=\lvert d_R\rangle\), the overlap is \(\gamma=\langle d_R\vert d_L\rangle=1\), so \(V=1\) and \(D=\sqrt{1-1^2}=0\): full interference visibility is retained and the pointer carries zero path information.
Suppose the particle interaction leaves the pointer in two orthogonal states (\(\langle d_R\vert d_L\rangle=0\)), but the experimenter chooses never to look at the pointer. Do unconditional fringes reappear on the screen?
Evaluate the off-diagonal term of the particle’s reduced state, which is proportional to \(\langle d_R\vert d_L\rangle\).
No. Tracing over the pointer multiplies the interference cross term by \(\langle d_R\vert d_L\rangle=0\) regardless of whether a human observer reads the pointer output.
In a noisy interferometer where random thermal phase fluctuations reduce the visibility to \(V=0.50\) without entangling the particle with any readable path marker (\(D=0\)), does the equality \(D^2+V^2=1\) hold?
Distinguish the pure-state equality \(D^2+V^2=1\) from the general mixed-state inequality \(D^2+V^2\le 1\).
No: here \(D^2+V^2=0^2+0.50^2=0.25<1\). The equality \(D^2+V^2=1\) holds only for pure balanced two-path states without classical phase averaging; mixedness reduces \(V\) without increasing accessible path distinguishability \(D\).
Whenever a path-dependent interaction entangles a particle with an apparatus pointer, the unconditional fringe visibility is scaled by the pointer state overlap \(V=|\langle d_R\vert d_L\rangle|\), enforcing \(D^2+V^2\le 1\).
These readings provide historical context, experimental implementations, or deeper derivations. The interactive display above is an educational model, not a reproduction of their measured data.