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2.10

The free particle and Gaussian wave packets

The simplest potential (no potential at all) turns out to be the most awkward: the stationary states are not normalisable. The fix is to superpose them, and the price is that the packet spreads.

Recommended first

After this section you should be able to

  • Explain why free-particle stationary states are not normalisable, and why they are still useful
  • Distinguish phase velocity from group velocity and say which one is the particle's speed
  • Explain packet spreading from the dispersion relation and estimate the spreading time
V(x)=0(2.10.1)V(x)=0\tag{2.10.1}

It looks like the simplest case. In fact it raises the subtlest question in the chapter.

Stationary states: not normalisable

The stationary equation gives

−ℏ22md2ψdx2=Eψ  ⟹  ψk(x)=Aeikx,k=±2mEℏ(2.10.2)-\frac{\hbar^2}{2m}\frac{\dd^2\psi}{\dd x^2}=E\psi \;\Longrightarrow\; \psi_k(x)=A\ee^{\ii kx},\qquad k=\pm\frac{\sqrt{2mE}}{\hbar}\tag{2.10.2}

With the time factor:

Ψk(x,t)=A ei(kx−ℏk22mt)(2.10.3)\Psi_k(x,t)=A\,\ee^{\ii\left(kx-\frac{\hbar k^2}{2m}t\right)}\tag{2.10.3}

k>0k>0 travels right, k<0k<0 left, and E=ℏ2k2/2mE=\hbar^2k^2/2m is allowed for every real kk — the spectrum is continuous.

The trouble is that

∫−∞∞∣Ψk∣2dx=∣A∣2∫−∞∞dx=∞(2.10.4)\int_{-\infty}^{\infty}|\Psi_k|^2\dd x=|A|^2\int_{-\infty}^{\infty}\dd x=\infty\tag{2.10.4}

Wave packets: superposing plane waves

The equation is linear, so we can superpose. Since kk is continuous, the sum becomes an integral:

Ψ(x,t)=12π∫−∞∞ϕ(k) ei(kx−ℏk22mt)dk(2.10.5)\Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi(k)\,\ee^{\ii\left(kx-\frac{\hbar k^2}{2m}t\right)}\dd k\tag{2.10.5}

ϕ(k)\phi(k) is the momentum-space wavefunction. Given an initial state, it comes from a Fourier transform:

ϕ(k)=12π∫−∞∞Ψ(x,0) e−ikxdx(2.10.6)\phi(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\Psi(x,0)\,\ee^{-\ii kx}\dd x\tag{2.10.6}

The picture

Two representations, one state.

Ψ(x)\Psi(x) and ϕ(k)\phi(k) carry exactly the same information; they differ only in the basis used for the expansion. ∣Ψ(x)∣2|\Psi(x)|^2 gives the position distribution, ∣ϕ(k)∣2|\phi(k)|^2 the momentum distribution (strictly ∣ϕ(k)∣2dk|\phi(k)|^2\dd k, with p=ℏkp=\hbar k).

This is one of the central structures of quantum mechanics: one state can be expanded in different bases. Sections 3.1–3.2 abstract it into “the components of one vector in different coordinate systems”.

The mathematics

Parseval’s theorem guarantees that normalisation agrees in both representations:

∫∣Ψ(x)∣2dx=∫∣ϕ(k)∣2dk=1(2.10.7)\int|\Psi(x)|^2\dd x=\int|\phi(k)|^2\dd k=1\tag{2.10.7}

The basic Fourier property

σxσk≥12(2.10.8)\sigma_x\sigma_k\ge\frac12\tag{2.10.8}

combined with p=ℏkp=\hbar k immediately gives

σxσp≥ℏ2(2.10.9)\sigma_x\sigma_p\ge\frac{\hbar}{2}\tag{2.10.9}

The uncertainty principle is a theorem of Fourier analysis plus the de Broglie relation.

Phase velocity and group velocity

Points of constant phase in a plane wave ei(kx−ωt)\ee^{\ii(kx-\omega t)} move at

vphase=ωk=ℏk2m=p2m(2.10.10)v_\text{phase}=\frac{\omega}{k}=\frac{\hbar k}{2m}=\frac{p}{2m}\tag{2.10.10}

Half the classical speed p/mp/m — if that were the particle’s velocity, quantum mechanics would have been wrong from the start.

Derivation: the packet as a whole moves at the group velocityadvanced~6 min

Suppose ϕ(k)\phi(k) is concentrated in a narrow range around k0k_0. Expand the dispersion relation about k0k_0:

ω(k)=ω0+ω′(k0)(k−k0)+12ω′′(k0)(k−k0)2+⋯(2.10.11)\omega(k)=\omega_0+\omega'(k_0)(k-k_0)+\frac12\omega''(k_0)(k-k_0)^2+\cdots\tag{2.10.11}

Keep only the first order for now. With q=k−k0q=k-k_0 and vg≡ω′(k0)v_g\equiv\omega'(k_0):

Ψ(x,t)≈ei(k0x−ω0t)2π∫ϕ(k0+q) eiq(x−vgt)dq(2.10.12)\Psi(x,t)\approx\frac{\ee^{\ii(k_0x-\omega_0t)}}{\sqrt{2\pi}}\int\phi(k_0+q)\,\ee^{\ii q(x-v_gt)}\dd q\tag{2.10.12}

The integral depends on xx and tt only through the combination x−vgtx-v_gt. That is,

∣Ψ(x,t)∣2≈∣Ψ(x−vgt, 0)∣2(2.10.13)|\Psi(x,t)|^2\approx|\Psi(x-v_gt,\,0)|^2\tag{2.10.13}

The envelope translates rigidly at vgv_g with its shape unchanged. And

vg=dωdk∣k0=ddkℏk22m=ℏk0m=p0m(2.10.14)v_g=\left.\frac{\dd\omega}{\dd k}\right|_{k_0}=\frac{\dd}{\dd k}\frac{\hbar k^2}{2m}=\frac{\hbar k_0}{m}=\frac{p_0}{m}\tag{2.10.14}

That is the classical speed. The phase velocity describes the motion of the internal fringes, which carry no particle.

The packet spreads

Now keep the second-order term. ω′′=ℏ/m≠0\omega''=\hbar/m\ne0 means dispersion: different kk components travel at different speeds.

Derivation: the spreading formula for a Gaussian packetadvanced~8 min

Take the initial Gaussian packet

Ψ(x,0)=(2πσ02)−1/4e−x2/4σ02eik0x(2.10.15)\Psi(x,0)=(2\pi\sigma_0^2)^{-1/4}\ee^{-x^2/4\sigma_0^2}\ee^{\ii k_0x}\tag{2.10.15}

Its Fourier transform is again a Gaussian (a special property of Gaussians):

ϕ(k)∝e−σ02(k−k0)2(2.10.16)\phi(k)\propto\ee^{-\sigma_0^2(k-k_0)^2}\tag{2.10.16}

with width σk=1/2σ0\sigma_k=1/2\sigma_0, so σxσp=σ0⋅ℏ2σ0=ℏ2\sigma_x\sigma_p=\sigma_0\cdot\frac{\hbar}{2\sigma_0}=\frac{\hbar}{2} — the Gaussian packet saturates the uncertainty bound exactly. That is what makes it special.

Substituting into the evolution integral (complete the square and do a Gaussian integral: tedious algebra, no tricks) gives

∣Ψ(x,t)∣2=12π σ(t)exp⁡[−(x−vgt)22σ(t)2](2.10.17)|\Psi(x,t)|^2=\frac{1}{\sqrt{2\pi}\,\sigma(t)}\exp\left[-\frac{(x-v_gt)^2}{2\sigma(t)^2}\right]\tag{2.10.17} σ(t)=σ01+(ℏt2mσ02)2 (2.10.18)\boxed{\ \sigma(t)=\sigma_0\sqrt{1+\left(\frac{\hbar t}{2m\sigma_0^2}\right)^2}\ }\tag{2.10.18}

Define the spreading time

τ=2mσ02ℏ(2.10.19)\tau=\frac{2m\sigma_0^2}{\hbar}\tag{2.10.19}

For t≪τt\ll\tau the packet barely changes; for t≫τt\gg\tau, σ≈σ0t/τ=ℏt2mσ0\sigma\approx\sigma_0 t/\tau=\frac{\hbar t}{2m\sigma_0}, growing linearly.

Orders of magnitude: when does spreading matterbasic~4 min
τ=2mσ02ℏ(2.10.20)\tau=\frac{2m\sigma_0^2}{\hbar}\tag{2.10.20}

An electron with σ0=1 nm\sigma_0=1\ \text{nm}:

τ=2×9.11×10−31×(10−9)21.055×10−34≈1.7×10−14 s(2.10.21)\tau=\frac{2\times9.11\times10^{-31}\times(10^{-9})^2}{1.055\times10^{-34}}\approx1.7\times10^{-14}\ \text{s}\tag{2.10.21}

About 17 femtoseconds. Electron packets spread very fast — which is why electron microscopy and quantum-device design have to take it seriously.

A dust grain, m=10−15m=10^{-15} kg, σ0=10−6\sigma_0=10^{-6} m:

τ=2×10−15×10−121.055×10−34≈1.9×107 s(2.10.22)\tau=\frac{2\times10^{-15}\times10^{-12}}{1.055\times10^{-34}}\approx1.9\times10^{7}\ \text{s}\tag{2.10.22}

About seven months. And in practice the grain would have been decohered by air molecules long before that, so the spreading never gets a chance to happen.

Conclusion: quantum spreading is invisible in the macroscopic world not because the formula fails, but because τ∝mσ02\tau\propto m\sigma_0^2 is absurdly large at macroscopic parameters.

Wave-packet evolution and scattering

A Gaussian packet hits a rectangular barrier. The time-dependent Schrödinger equation is solved live by the split-operator method, with ħ = m = 1.

E / V₀
0.80
Reflection R
0.000
Transmission T
0.000
Packet width Δx
3.00

Press play to send the packet into the barrier. R and T lock automatically once the packet has completely left the barrier region. The barrier width on the grid is a = 1.500.

2.00

E = k₀²/2 = 2.00

2.5

positive = a barrier

1.5

T decays exponentially with a: T ~ e^(−2κa)

3.0

larger σ means better-defined momentum, closer to a plane wave

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Try this

  • At the default parameters E ≈ 2.0 < V₀ = 2.5, a classical particle bounces back 100% of the time. Press play: part of the packet gets through. That is tunnelling — the principle behind the scanning tunnelling microscope and alpha decay.
  • Take a from 1.5 to 4 and replay. The transmission collapses — T ~ e^(−2κa) is exponential, so widening the barrier a little costs several orders of magnitude.
  • Set V₀ to 1.0 (now E > V₀, so classically transmission should be 100%) and replay: a clear reflected peak still runs back to the left. A wave partially reflects at any abrupt change of potential, exactly as light does at a glass surface.
  • Switch to the transmission curve and make V₀ negative (a well): at certain energies the curve returns to T = 1. That is resonant transmission, historically seen as the Ramsauer–Townsend effect, where noble gases are almost transparent to slow electrons.
  • Set σ to 8 and scatter again: the measured T lands much closer to the open circle on the analytic curve — the nearer the packet is to a plane wave, the better the single-energy approximation.

Key formulas

  • Free-particle stationary state

    Ψk(x,t)=A ei(kx−ℏk2t/2m)\Psi_k(x,t)=A\,\ee^{\ii(kx-\hbar k^2t/2m)}

    Not normalisable; usable only as an expansion basis

  • Wave-packet expansion

    Ψ(x,t)=12π∫ϕ(k)ei(kx−ω(k)t)dk\Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int\phi(k)\ee^{\ii(kx-\omega(k)t)}\dd k

    φ(k) is the Fourier transform of Ψ(x,0)

  • Group velocity

    vg=dωdk=ℏk0m=p0m=2vphasev_g=\frac{\dd\omega}{\dd k}=\frac{\hbar k_0}{m}=\frac{p_0}{m}=2v_{\text{phase}}

    The envelope's speed is the particle's speed

  • Gaussian packet spreading

    σ(t)=σ01+(ℏt2mσ02)2\sigma(t)=\sigma_0\sqrt{1+\left(\frac{\hbar t}{2m\sigma_0^2}\right)^2}

    Spreading time τ = 2mσ₀²/ħ

Self-check4 questions

  1. 1.

    The plane-wave solution e^{ikx} for a free particle is not normalisable. The correct reading is:

  2. 2.

    For a non-relativistic free particle, the relation between phase and group velocity is:

  3. 3.

    A free Gaussian packet of initial width σ₀ has spreading time τ = 2mσ₀²/ħ. If σ₀ is tripled, τ changes by a factor of:

  4. 4.

    Why does a coherent state in a harmonic oscillator not spread while a free packet does?

What comes next

The last section of the chapter aims a wave packet at a concrete target: a wall it “should not” be able to pass through.

Section 18 of 106 · use ← → to turn the page