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The Double Slit and Which-Path

Send particles one at a time at a pair of slits and each lands as a single dot, yet after enough of them the dots pile up into bright and dark bands: an interference pattern, as if each particle went through both slits and interfered with itself. The intensity is the single-slit envelope times the two-slit fringes, $I = \text{env}\cdot\tfrac12(1 + V\cos 2\alpha)$ with $\alpha = \pi d\sin\theta/\lambda$, and the fringe spacing on a distant screen is $\lambda L/d$. The catch is complementarity: if you install a detector that tells you which slit each particle took, the interference vanishes and you are left with the smooth single-slit hump. You cannot have both the which-path knowledge and the fringes. The trade is quantitative, the visibility $V$ and the path distinguishability $D$ obey $V^2 + D^2 = 1$, so as you turn up the which-path information the bands fade exactly on that circle. Watch the pattern assemble dot by dot, then turn the which-path knob and watch it dissolve.

Figure 1. The double slit and which-path. Top: particles arrive one at a time and accumulate into the fringe pattern on the screen; a which-path detector at the slits erodes the fringes. Bottom: the intensity along the screen (curve) with the detection histogram (bars), and the complementarity bar showing the visibility V and the path distinguishability D summing as V squared plus D squared equals one. Method: rejection sampling from the closed-form intensity. Source: Feynman Lectures, Vol. III, Ch. 1; Englert 1996.
slit sep d4.0
wavelength0.50
which-path0 %

WHAT TO TRY

  • Watch the screen fill dot by dot: each particle lands at a single random spot, but the bright fringes emerge only after many arrive.
  • Turn up the which-path knob: the fringes fade and the pattern collapses to the smooth single-slit hump, and the visibility bar shrinks as the distinguishability grows.
  • Widen the slit separation or shorten the wavelength: the fringes pack closer together (spacing $\lambda L/d$).
  • Set which-path to its half value and see partial fringes, the visibility and distinguishability both about 0.7, riding the $V^2 + D^2 = 1$ circle.