Things are heating up 🔥
Lecture 4 of our Quantum Mechanics series.
Lecture 2 earned the right to call ρ(x,t) = |ψ(x,t)|² a probability density by showing Schrödinger evolution forces a continuity equation, ∂ρ/∂t ∇·j = 0. Lecture 3 then revealed what actually steers that flow...writing ψ = r exp(iθ) makes the current j become "density × phase-gradient," so phase geometry literally draws the streamlines and quantized vortices show up as phase winding defects.
Lecture 4 is the next link in that same chain...once you accept that probability flows, you can ask whether the bulk of that flowing density has a trajectory...and Ehrenfest says yes, its centroid moves almost classically when the packet is tight and the potential is smooth.
In the 3D panel, the surface height is |ψ(x,t)| and the surface skin is colored by arg(ψ(x,t)). The porcelain bead rides on the surface at the quantum centroid ⟨x⟩(t), the amber bead is a classical centroid integrated under the same V(x,t), and the teal arrow points along ⟨p⟩.
In the 2D panel, the background uses the same encoding (phase as color, amplitude as brightness), but now you can see the current field...the glowing ribbons are streamlines of j, and the small charged defects are vortices where the phase winds by ±2π around a low-density core. Both panels are the same ψ, just two ways of seeing the same objects from Lectures 2 and 3.
The math breakdown
We treat the state as a complex field ψ(x,t) on the plane (x in R²). The Born rule defines the probability density
ρ(x,t) = |ψ(x,t)|²
Schrödinger evolution (ħ = 1 units) is
i ∂ψ/∂t = [ −(1/2m) ∇² V(x,t) ] ψ
From Lecture 2 we have the continuity equation
∂ρ/∂t ∇·j = 0
with probability current
j = (1/2mi) ( ψ* ∇ψ − ψ ∇ψ* )
= (1/m) Im(ψ* ∇ψ)
From Lecture 3, writing ψ = r exp(iθ) turns that into
j = (ρ/m) ∇θ
Now define the centroid and mean momentum:
⟨x⟩(t) = ∫ x ρ(x,t) dx
⟨p⟩(t) = ∫ ψ*(x,t) [ −i ∇ ] ψ(x,t) dx
Ehrenfest’s theorem states
d⟨x⟩/dt = ⟨p⟩/m
d⟨p⟩/dt = −⟨∇V(x,t)⟩
Derive the first identity directly from the continuity equation (so it visibly depends on Lecture 2). Start from ⟨x⟩:
d⟨x⟩/dt
= ∫ x (∂ρ/∂t) dx
= −∫ x (∇·j) dx
Integrate by parts (boundary term vanishes because ρ → 0 at the edges):
−∫ x (∇·j) dx = ∫ j dx
So
d⟨x⟩/dt = ∫ j dx
Now substitute j and rewrite using Im(z) = (z − z*)/(2i):
∫ j dx
= (1/m) ∫ (1/2i)(ψ* ∇ψ − ψ ∇ψ*) dx
= (1/m) ∫ ψ*(−i ∇)ψ dx
= ⟨p⟩/m
Therefore
d⟨x⟩/dt = ⟨p⟩/m
The second identity comes from differentiating ⟨p⟩, using Schrödinger’s equation to replace ∂ψ/∂t and ∂ψ*/∂t, and integrating by parts so the kinetic contributions cancel while the potential contributes a force term:
d⟨p⟩/dt = −⟨∇V⟩
When V is nearly linear across the wavepacket (narrow packet, smooth potential), the mean force is approximately the force at the mean position:
⟨∇V⟩ ≈ ∇V(⟨x⟩)
and you get the Newton-like approximation
m d²⟨x⟩/dt² ≈ −∇V(⟨x⟩)
That’s what the animation is testing live using the same objects from Lectures 2 and 3...ρ, j, and phase.
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