12.01.05 · quantum / foundations

Spreading of the free Gaussian wave packet: group velocity and dispersion

shipped3 tiersLean: nonepending prereqs

Anchor (Master): Cohen-Tannoudji, Quantum Mechanics, Vol. 1 (1991), Complement G-I

Intuition Beginner

A quantum particle is never perfectly localised. The best you can do is build a wave packet — a lump of wave function centred at some position with some spread. For a free particle (no forces), this lump moves. But it does not stay the same shape. It spreads.

The reason is dispersion. A wave packet is a superposition of many plane waves, each with its own wavelength and therefore its own momentum. In quantum mechanics, each momentum component has a different speed — the dispersion relation $ω = ℏ k^{2} / (2 m)$ is quadratic, not linear. Faster components outrun slower ones. The packet widens.

A concrete number sets the timescale. A Gaussian wave packet of initial width $σ_{0}$ in a particle of mass $m$ doubles its width after a time $t \sim 2 m σ_{0}^{2} /ℏ$ . For a macroscopic object ( $m = 1$ kg, $σ_{0} = 1$ mm) this time exceeds the age of the universe. For an electron ( $m \approx 1 0^{- 30}$ kg, $σ_{0} = 1 0^{- 10}$ m) it is about $1 0^{- 16}$ s — less than a femtosecond. Quantum spreading is real, fast, and inescapable for small particles, but completely negligible for anything you can see.

The centre of the packet travels at the group velocity $v_{g} = p_{0} / m$ , which is exactly the classical particle velocity. The individual plane-wave crests travel at the phase velocity $v_{p h} = p_{0} / (2 m)$ , which is half the group velocity. This mismatch is the hallmark of quantum free-particle dispersion.

Visual Beginner

Draw three snapshots of a Gaussian wave packet at $t = 0$ , $t = t_{1}$ , and $t = 2 t_{1}$ , stacked vertically.

At $t = 0$ : a narrow bell curve centred at $x = 0$ .

At $t = t_{1}$ : a wider, shorter bell curve centred at $x = v_{g} t_{1}$ . The area under each curve is the same (total probability is conserved), but the peak is lower because the packet has spread.

At $t = 2 t_{1}$ : an even wider, even shorter bell curve centred at $x = 2 v_{g} t_{1}$ .

Inside each packet, draw fine oscillations (the de Broglie waves) that move at the phase velocity. Annotate that the crests move at $v_{p h} = p_{0} / (2 m)$ while the envelope moves at $v_{g} = p_{0} / m$ . The crests appear to slide backwards through the envelope.

The width grows as $σ (t) = σ_{0} 1 + ℏ^{2} t^{2} / (4 m^{2} σ_{0}^{4})$ . At short times the packet barely changes. At long times the width grows linearly: $σ (t) \approx ℏ t / (2 m σ_{0})$ .

Worked example Beginner

An electron is localised at $t = 0$ in a Gaussian wave packet of width $σ_{0} = 1 0^{- 10}$ m (about one angstrom) with central momentum $p_{0} = ℏ k_{0}$ corresponding to speed $v_{0} = 1 0^{6}$ m/s.

The group velocity is $v_{g} = p_{0} / m = v_{0} = 1 0^{6}$ m/s — the packet's centre moves at the classical speed.

The spreading time is $t_{s} = 2 m σ_{0}^{2} /ℏ$ . Plugging in:

t_{s} = \frac{2 \times 9.11 \times 1 0 ^{- 31} \times ( 1 0 ^{- 10} ) ^{2}}{1.055 \times 1 0 ^{- 34}} = \frac{2 \times 9.11 \times 1 0 ^{- 51}}{1.055 \times 1 0 ^{- 34}} = 1.73 \times 1 0^{- 16} s .

After one femtosecond ( $1 0^{- 15}$ s), the width is:

σ = 1 0^{- 10} 1 + \frac{( 1 0 ^{- 15} ) ^{2}}{( 1.73 \times 1 0 ^{- 16} ) ^{2}} = 1 0^{- 10} 1 + 33.4 \approx 1 0^{- 10} 34.4 \approx 5.9 \times 1 0^{- 10} m .

The electron's wave packet has spread to nearly six times its original width in one femtosecond. By the time it travels one micrometer ( $1 0^{- 6}$ m), which takes $1 0^{- 12}$ s, the packet is enormously spread out.

Compare with a grain of sand ( $m = 1 0^{- 6}$ kg, $σ_{0} = 1 0^{- 6}$ m): $t_{s} \sim 1 0^{22}$ s, which is roughly $1 0^{14}$ years — a hundred trillion years.

Check your understanding Beginner

Formal definition Intermediate+

The time-dependent Schrodinger equation for a free particle of mass $m$ in one dimension is

i ℏ \frac{\partial ψ}{\partial t} = - \frac{ℏ ^{2}}{2 m} \frac{\partial ^{2} ψ}{\partial x ^{2}} .

The plane-wave solutions are $ϕ_{k} (x, t) = e^{i (k x - ω t)}$ with the dispersion relation

ω (k) = \frac{ℏ k ^{2}}{2 m} .

A general solution is a superposition of plane waves, written as a Fourier integral:

ψ (x, t) = \frac{1}{2 π} \int_{- \infty}^{\infty} \tilde{ψ} (k) e^{i (k x - ω (k) t)} d k .

The initial condition at $t = 0$ determines $\tilde{ψ} (k)$ via the inverse Fourier transform. Time evolution in momentum space is simple: each component simply acquires the phase $e^{- iω (k) t}$ .

Gaussian wave packet. At $t = 0$ the minimum-uncertainty Gaussian is

ψ (x, 0) = (\frac{1}{2 π σ _{0}^{2}})^{1/4} exp (- \frac{x ^{2}}{4 σ _{0}^{2}} + i k_{0} x)

with central wave number $k_{0} = p_{0} /ℏ$ and position-space width $σ_{0}$ . Its Fourier transform is

\tilde{ψ} (k) = (\frac{2 σ _{0}^{2}}{π})^{1/4} exp (- σ_{0}^{2} (k - k_{0})^{2}) .

The momentum-space width is $σ_{k} = 1/ (2 σ_{0})$ , giving $Δ x \cdot Δ p = σ_{0} \cdot ℏ/ (2 σ_{0}) = ℏ/2$ — the minimum-uncertainty product.

Time evolution. Evolving each Fourier component:

ψ (x, t) = \frac{1}{2 π} (\frac{2 σ _{0}^{2}}{π})^{1/4} \int_{- \infty}^{\infty} exp (- σ_{0}^{2} (k - k_{0})^{2} + ik x - i \frac{ℏ k ^{2}}{2 m} t) d k .

Completing the square in $k$ and evaluating the Gaussian integral yields

ψ (x, t) = (\frac{1}{2 π σ ( t ) ^{2}})^{1/4} exp [- \frac{( x - v _{g} t ) ^{2}}{4 σ _{0} σ ( t )} + i k_{0} (x - \frac{v _{g} t}{2}) - i ϕ (t)]

where the group velocity is $v_{g} = ℏ k_{0} / m = p_{0} / m$ , the time-dependent width is

σ (t) = σ_{0} 1 + \frac{ℏ ^{2} t ^{2}}{4 m ^{2} σ _{0}^{4}}

and $ϕ (t) = \frac{1}{2} arctan (ℏ t / (2 m σ_{0}^{2}))$ is a global phase. The probability density remains Gaussian:

∣ ψ (x, t) ∣^{2} = \frac{1}{2 π σ ( t )} exp (- \frac{( x - v _{g} t ) ^{2}}{2 σ ( t ) ^{2}}) .

Group velocity and phase velocity. The centre of the packet moves at the group velocity $v_{g} = d ω / d k ∣_{k_{0}} = ℏ k_{0} / m$ . The wave crests within the packet move at the phase velocity $v_{p h} = ω (k_{0}) / k_{0} = ℏ k_{0} / (2 m) = v_{g} /2$ . The mismatch between envelope motion and crest motion is the hallmark of a dispersive medium.

Uncertainty product in time. The position uncertainty grows as $Δ x (t) = σ (t)$ . The momentum uncertainty is constant: $Δ p = ℏ/ (2 σ_{0})$ at all times, because the free Hamiltonian commutes with $\overset{p}{^}$ . The uncertainty product is

Δ x (t) \cdot Δ p = \frac{ℏ}{2} 1 + \frac{ℏ ^{2} t ^{2}}{4 m ^{2} σ _{0}^{4}} \geq \frac{ℏ}{2} .

The product equals $ℏ/2$ only at $t = 0$ ; for all later times it exceeds the minimum.

Counterexamples to common slips

The packet does not spread because of any force or potential. It spreads because the free-particle dispersion relation is nonlinear ( $ω \propto k^{2}$ ). A medium with linear dispersion ( $ω = c k$ ) would propagate the packet without distortion.
The probability is always conserved: $\int ∣ ψ (x, t) ∣^{2} d x = 1$ for all $t$ . The packet gets wider but lower; the area is unchanged.
The spreading rate does not depend on $k_{0}$ . A stationary packet ( $k_{0} = 0$ ) spreads at exactly the same rate as a moving one. The central momentum sets the velocity of the centre, not the spreading.
The phase velocity $v_{p h} = ℏ k_{0} / (2 m)$ can exceed $c$ for sufficiently large $k_{0}$ . This does not violate relativity because no information travels at the phase velocity. Information and energy travel at the group velocity.

Key theorem with proof Intermediate+

Theorem (Gaussian wave-packet spreading). Let $ψ (x, 0) = (2 π σ_{0}^{2})^{- 1/4} exp (- x^{2} / (4 σ_{0}^{2}) + i k_{0} x)$ be a normalised Gaussian wave packet at $t = 0$ . Under free-particle time evolution governed by $i ℏ \partial_{t} ψ = - (ℏ^{2} /2 m) \partial_{x}^{2} ψ$ , the time-dependent position-space probability density is

∣ ψ (x, t) ∣^{2} = \frac{1}{2 π σ ( t )} exp (- \frac{( x - v _{g} t ) ^{2}}{2 σ ( t ) ^{2}})

with $v_{g} = ℏ k_{0} / m$ and $σ (t) = σ_{0} 1 + ℏ^{2} t^{2} / (4 m^{2} σ_{0}^{4})$ . The packet is Gaussian at all times, with linearly growing width at late times.

Proof. The Fourier transform of the initial state is $\tilde{ψ} (k) = (2 σ_{0}^{2} / π)^{1/4} exp (- σ_{0}^{2} (k - k_{0})^{2})$ . Time evolution in $k$ -space multiplies by $e^{- iω (k) t}$ where $ω (k) = ℏ k^{2} / (2 m)$ :

\tilde{ψ} (k, t) = (\frac{2 σ _{0}^{2}}{π})^{1/4} exp (- σ_{0}^{2} (k - k_{0})^{2} - i \frac{ℏ k ^{2}}{2 m} t) .

The exponent in $k$ is

- σ_{0}^{2} (k - k_{0})^{2} - i \frac{ℏ t}{2 m} k^{2} = - (σ_{0}^{2} + i α) k^{2} + 2 σ_{0}^{2} k_{0} k - σ_{0}^{2} k_{0}^{2}

where $α = ℏ t / (2 m)$ . Completing the square:

= - (σ_{0}^{2} + i α) (k - \frac{σ _{0}^{2} k _{0}}{σ _{0}^{2} + i α})^{2} + \frac{σ _{0}^{4} k _{0}^{2}}{σ _{0}^{2} + i α} - σ_{0}^{2} k_{0}^{2} .

The inverse Fourier transform of a Gaussian $exp (- A (k - B)^{2} + ik x)$ is $π / A exp (- (x - i B \cdot ?)^{2} / (4 A))$ , properly:

\int_{- \infty}^{\infty} e^{- A (k - B)^{2} + ik x} d k = \frac{π}{A} exp (- \frac{x ^{2}}{4 A} + i B x) .

Applying this with $A = σ_{0}^{2} + i α$ and gathering terms, the position-space wave function is a Gaussian with complex width parameter. The probability density $∣ ψ ∣^{2}$ picks out the real part of the inverse width:

\frac{1}{2 Re ( 1/ σ _{0}^{2} + i α )} = σ (t)^{2} = σ_{0}^{2} + \frac{α ^{2}}{σ _{0}^{2}} = σ_{0}^{2} (1 + \frac{ℏ ^{2} t ^{2}}{4 m ^{2} σ _{0}^{4}}) .

The centre of the Gaussian sits at $x = 2 σ_{0}^{2} k_{0} Re (1/ (σ_{0}^{2} + i α)) \cdot (σ_{0}^{2} + i α)$ , which evaluates to $v_{g} t = ℏ k_{0} t / m$ . The probability density is therefore a Gaussian of width $σ (t)$ centred at $v_{g} t$ , as claimed. $□$

Corollary. At late times $t ≫ 2 m σ_{0}^{2} /ℏ$ the width grows linearly: $σ (t) \approx ℏ t / (2 m σ_{0})$ . The spreading rate is $\overset{σ}{˙} \approx ℏ/ (2 m σ_{0}) = Δ p / m$ , i.e. the initial momentum uncertainty divided by the mass.

Bridge. This result connects to 12.01.01 where the Heisenberg uncertainty principle is stated in its static form. The spreading theorem is the dynamical consequence: an initially minimum-uncertainty state cannot remain minimum-uncertainty under free evolution, because $Δ p$ is conserved (momentum is a constant of the motion) but $Δ x$ grows. The bridge to 12.10.01 is that the same propagator $K (x, t; 0, 0) = m / (2 π i ℏ t) exp (im x^{2} / (2ℏ t))$ that the path integral computes is the kernel whose convolution with $ψ (x, 0)$ produces the time-evolved wave packet. The bridge to 12.04.01 is that the particle-in-a-box stationary states do not spread (they are energy eigenstates), illustrating that spreading requires a superposition of energy eigenstates with different frequencies — exactly the Fourier decomposition that defines a wave packet.

Exercises Intermediate+

Exercise 4 (medium, symbolic).

At late times $t ≫ 2 m σ_{0}^{2} /ℏ$ , the wave function approaches

ψ (x, t) \approx \frac{1}{( 2 π i ℏ t / m ) ^{1/2}} \tilde{ψ} (\frac{m x}{ℏ t}) e^{im x^{2} / (2ℏ t)} .

Show that $∣ ψ (x, t) ∣^{2}$ at late times is proportional to $∣ \tilde{ψ} (m x / (ℏ t)) ∣^{2}$ — the momentum-space distribution mapped onto position space via $k = m x / (ℏ t)$ .

Hint

The prefactor's modulus is $(m / (2 π ℏ t))^{1/2}$ , independent of $x$ . The $x$ -dependence of the probability density comes entirely from $∣ \tilde{ψ} ∣^{2}$ .

Answer

At late times $σ (t) \approx ℏ t / (2 m σ_{0})$ and the complex exponential's quadratic term dominates. The exact expression from the Fourier inversion is:

ψ (x, t) = \frac{1}{2 π} \int \tilde{ψ} (k) e^{ik x - i ℏ k^{2} t / (2 m)} d k .

For large $t$ , the rapid oscillation of $e^{- i ℏ k^{2} t / (2 m)}$ kills all contributions except near the stationary-phase point $k = m x / (ℏ t)$ . Expanding about this point:

ψ (x, t) \approx \frac{m}{2 π i ℏ t} \tilde{ψ} (\frac{m x}{ℏ t}) e^{im x^{2} / (2ℏ t)} .

Therefore $∣ ψ (x, t) ∣^{2} \approx (m / (2 π ℏ t)) ∣ \tilde{ψ} (m x / (ℏ t)) ∣^{2}$ . Position at late times maps directly to momentum at $t = 0$ via $p = m x / t$ . This is the far-field or Fraunhofer limit: the momentum distribution is displayed on the position screen.

Exercise 6 (hard, symbolic).

A general (non-Gaussian) free wave packet has $\tilde{ψ} (k)$ peaked around $k_{0}$ with width $Δ k$ . Derive the stationary-phase approximation for the propagating packet and show that the centre moves at $v_{g} = d ω / d k ∣_{k_{0}}$ with spreading governed by $d^{2} ω / d k^{2} ∣_{k_{0}}$ . Identify the condition under which the packet does not spread.

Hint

Expand $ω (k)$ to second order about $k_{0}$ . The first derivative gives the group velocity; the second derivative (group velocity dispersion) causes spreading.

Answer

Write $ω (k) = ω_{0} + ω_{0}^{'} (k - k_{0}) + \frac{1}{2} ω_{0}^{''} (k - k_{0})^{2} + \dots$ where $ω_{0}^{'} = d ω / d k ∣_{k_{0}}$ and $ω_{0}^{''} = d^{2} ω / d k^{2} ∣_{k_{0}}$ .

ψ (x, t) \approx \frac{e ^{i (k_{0} x - ω_{0} t)}}{2 π} \int \tilde{ψ} (k) e^{i (k - k_{0}) (x - ω_{0}^{'} t)} e^{- i ω_{0}^{''} (k - k_{0})^{2} t /2} d k .

The first exponential phase vanishes at $x = ω_{0}^{'} t$ — the centre moves at the group velocity $v_{g} = ω_{0}^{'}$ . The second exponential broadens the effective width in $k$ -space, producing spreading in $x$ -space. The spreading rate is proportional to $∣ ω_{0}^{''} ∣$ .

The packet does not spread if and only if $ω_{0}^{''} = 0$ , i.e. $ω (k)$ is linear. For a free non-relativistic particle, $ω_{0}^{''} = ℏ/ m \neq = 0$ , so spreading always occurs. For a photon in vacuum, $ω = c k$ and $ω^{''} = 0$ — light pulses in vacuum propagate without dispersion.

Exercise 7 (hard, symbolic).

The coherent state of the harmonic oscillator $∣ α ⟩$ is a minimum-uncertainty Gaussian that does not spread. Prove that the free-particle Gaussian wave packet is a coherent state of the harmonic oscillator with $ω \to 0$ in the appropriate sense, and explain why setting $ω = 0$ restores spreading.

Hint

The harmonic-oscillator coherent state oscillates at frequency $ω$ with constant width. The free-particle limit $ω \to 0$ makes the oscillation period infinite.

Answer

The harmonic-oscillator coherent state in position representation is

ψ_{α} (x, t) \propto exp (- \frac{( x - x _{0} cos ω t ) ^{2}}{4 σ _{0}^{2}} + phase terms)

with $σ_{0} = ℏ/ (2 mω)$ and $x_{0}$ set by $α$ . The width is constant because the harmonic potential provides a restoring force that refocuses the spreading.

The free-particle limit $ω \to 0$ gives $σ_{0} \to \infty$ — the minimum-uncertainty width diverges, which is correct (the free-particle ground state is not normalisable). For any finite $σ_{0}$ the free particle lacks the refocusing mechanism, and the packet spreads irreversibly. The spreading time $t_{s} = 2 m σ_{0}^{2} /ℏ = 1/ (2 ω σ_{0}^{2} \cdot σ_{0}^{2} / σ_{0}^{2}) = 1/ ω$ — the spreading timescale is exactly the inverse oscillator frequency that was sent to zero.

The oscillator prevents spreading by curving the classical trajectories back together (all classical paths are periodic with the same period). The free particle has no such refocusing.

Exercise 8 (hard, conceptual).

Explain why the spreading of a free wave packet does not violate the conservation of energy. The wave function spreads to occupy an ever-larger region of space; does this not require energy?

Hint

Compute $⟨ H ⟩ = ⟨ p^{2} ⟩ / (2 m)$ as a function of time. Is it conserved?

Answer

The expectation value of energy is $⟨ H ⟩ = ⟨ p^{2} ⟩ / (2 m)$ . Since $[H, p] = 0$ for a free particle, $⟨ p^{2} ⟩$ is constant in time. The energy is exactly conserved.

The confusion arises from conflating "spreading in position space" with "requiring energy." The spreading costs no energy because it is driven by the momentum that was already present in the initial state. The kinetic energy stored in the initial momentum uncertainty $Δ p$ is the "fuel" for the spreading. More precisely, $⟨ p^{2} ⟩ = p_{0}^{2} + (Δ p)^{2}$ is constant; the centrepiece $p_{0}^{2} / (2 m)$ is the "classical" kinetic energy of the packet's centre, and $(Δ p)^{2} / (2 m)$ is the zero-point kinetic energy that drives the spreading. No external energy input is required.

An analogy: an initially collimated beam of classical particles, each moving at a slightly different speed, spreads out over time with no energy cost. The quantum spreading is the wave-mechanical version of the same kinematic fact.

Exercise 9 (hard, symbolic).

Two Gaussian wave packets of widths $σ_{1}$ and $σ_{2}$ are superposed at $t = 0$ with their centres well separated. Describe the time evolution of the interference pattern between them and show that the fringe spacing scales as $Δ x_{fringe} = 2 π ℏ t / (m d)$ where $d$ is the initial separation.

Hint

The interference pattern comes from the cross term $2 Re (ψ_{1}^{*} ψ_{2})$ . At late times each packet maps its momentum distribution to position space.

Answer

Write $ψ = ψ_{1} + ψ_{2}$ where $ψ_{j} (x, 0) \propto exp (- (x - x_{j})^{2} / (4 σ_{j}^{2}) + i k_{j} x)$ with $x_{1} = - d /2, x_{2} = d /2$ .

At late times, $ψ_{j} (x, t) \approx m / (2 π i ℏ t) \tilde{ψ}_{j} (m x / (ℏ t)) e^{im x^{2} / (2ℏ t)}$ . The cross term in $∣ ψ ∣^{2}$ is

2 Re (ψ_{1}^{*} ψ_{2}) \propto cos (\frac{m ( k _{1} - k _{2} ) x}{ℏ} \cdot t + \frac{( k _{1}^{2} - k _{2}^{2} ) ℏ t}{2 m} + \frac{( x _{1} - x _{2} ) m x}{ℏ t}) .

For $k_{1} = k_{2} = k_{0}$ (same central momentum) the dominant oscillation comes from the $x_{1} - x_{2}$ term:

fringe spacing = \frac{2 π ℏ t}{m ∣ x _{1} - x _{2} ∣} = \frac{2 π ℏ t}{m d} .

This is the quantum-mechanical double-slit formula in the time domain: two separated sources produce fringes whose spacing grows linearly with time. The fringe pattern is a momentum-space measurement of the initial separation projected onto position space.

Exercise 10 (hard, symbolic).

Prove that the free-particle propagator $K (x, t; 0, 0) = m / (2 π i ℏ t) exp (im x^{2} / (2ℏ t))$ acting on the Gaussian initial state reproduces the time-evolved Gaussian derived above. That is, verify that

ψ (x, t) = \int_{- \infty}^{\infty} K (x, t; x^{'}, 0) ψ (x^{'}, 0) d x^{'}

gives the correct Gaussian with width $σ (t)$ and centre $v_{g} t$ .

Hint

Write out the integral explicitly. The exponent is quadratic in $x^{'}$ — complete the square and evaluate the Gaussian.

Answer

ψ (x, t) = \frac{m}{2 π i ℏ t} (\frac{1}{2 π σ _{0}^{2}})^{1/4} \int d x^{'} exp (\frac{im ( x - x ^{'} ) ^{2}}{2ℏ t} - \frac{x ^{'2}}{4 σ _{0}^{2}} + i k_{0} x^{'}) .

The exponent in $x^{'}$ is

- (\frac{1}{4 σ _{0}^{2}} - \frac{im}{2ℏ t}) x^{'2} + (i k_{0} - \frac{im x}{ℏ t}) x^{'} + \frac{im x ^{2}}{2ℏ t} .

Let $A = 1/ (4 σ_{0}^{2}) - im / (2ℏ t)$ and $B = i k_{0} - im x / (ℏ t)$ . Completing the square: $- A (x^{'} - B / (2 A))^{2} + B^{2} / (4 A) + im x^{2} / (2ℏ t)$ .

The Gaussian integral over $x^{'}$ gives $π / A$ . The result is

ψ (x, t) = N exp (\frac{B ^{2}}{4 A} + \frac{im x ^{2}}{2ℏ t})

with $N$ a normalisation. Computing $B^{2} / (4 A)$ and separating real and imaginary parts reproduces a Gaussian of width $σ (t) = ∣ A ∣^{- 1/2} / 2$ centred at $ℏ k_{0} t / m = v_{g} t$ , with the correct phase factor. The algebra is lengthy but mechanical; the key step is that $A^{- 1} = 4 σ_{0}^{2} / (1 + i ℏ t / (2 m σ_{0}^{2}))$ and the real part of $1/ A$ gives $1/ (2 σ (t)^{2})$ .

Coherence length and the classical limit Master

The spreading formula $σ (t) = σ_{0} 1 + t^{2} / t_{s}^{2}$ with $t_{s} = 2 m σ_{0}^{2} /ℏ$ defines a coherence length $L_{c} = v_{g} t_{s} = 2 m σ_{0}^{2} v_{g} /ℏ = 2 σ_{0}^{2} p_{0} /ℏ$ . Over distances shorter than $L_{c}$ the packet maintains its shape; over distances longer than $L_{c}$ the packet has spread far beyond its initial width and the probability density reflects the momentum distribution rather than the initial position distribution. The coherence length is the quantum-mechanical distance scale on which wave behaviour gives way to classical ballistic propagation.

For a macroscopic object ( $m = 1$ kg, $σ_{0} = 1 0^{- 6}$ m, $v = 1$ m/s): $L_{c} \approx 2 \times 1 0^{- 12} \times 1/ (1 0^{- 34}) \approx 1 0^{22}$ m — light-years. The packet is coherent over astronomical distances; for all practical purposes the object follows a classical trajectory.

For an electron at the same speed ( $m = 1 0^{- 30}$ kg, same $σ_{0}$ ): $L_{c} \approx 2 \times 1 0^{- 12} \times 1 0^{- 30} \times 1/ (1 0^{- 34}) \approx 1 0^{- 8}$ m — tens of nanometres. The electron delocalises after travelling a few atomic diameters.

The classical limit of quantum mechanics is the regime $L_{c} ≫ L_{observation}$ , where $L_{observation}$ is the length scale of the measurement apparatus. In this regime the wave packet is effectively a point particle throughout the experiment. The condition is $m σ_{0} v_{0} ≫ ℏ$ , i.e. the action $m σ_{0} v_{0}$ associated with the packet's phase-space volume greatly exceeds $ℏ$ . This is the same condition that appears in the path-integral stationary-phase analysis of 12.10.01: classical mechanics is the $ℏ \to 0$ limit, but for fixed $ℏ$ the effective small parameter is $ℏ/ (m σ_{0} v_{0})$ .

Minimum-uncertainty packets and squeezing. The initial Gaussian saturates $Δ x \cdot Δ p = ℏ/2$ . A squeezed state has $Δ x < σ_{0}$ and $Δ p > ℏ/ (2 σ_{0})$ , still saturating the bound. A squeezed free-particle wave packet spreads faster than an unsqueezed one: reducing $σ_{0}$ by a factor $s$ shortens the spreading time by $s^{2}$ . Squeezing the position uncertainty below the ground-state width accelerates the inevitable delocalisation. This is a direct consequence of the uncertainty principle: you pay for localisation in faster spreading.

The Kennard wave packet. The explicit time-dependent Gaussian derived above was first written down by Kennard in 1927 ^{[Kennard 1927]}, in the same year as the Davisson-Germer experiment confirmed de Broglie's hypothesis. Kennard's construction showed that the minimum-uncertainty Gaussian maintains its Gaussian shape under free evolution, with a complex width parameter whose modulus grows as $σ (t)$ . The phase factor $exp (- i arctan (ℏ t / (2 m σ_{0}^{2})) /2)$ that accompanies the spreading is the Gouy phase of the quantum wave packet — the analogue of the phase shift that a focused laser beam acquires as it passes through its waist. In optics the Gouy phase produces a $π$ shift from $- \infty$ to $+ \infty$ ; in quantum mechanics the same phase appears in the wave-packet time evolution and contributes to the phase of the propagator.

Connections to classical mechanics and optics Master

The wave-packet spreading formula has precise analogues in two classical domains.

Classical dispersion. A pulse of light in a dispersive medium (refractive index $n (ω)$ ) broadens as it propagates. The group velocity is $v_{g} = c / n_{g}$ where $n_{g} = n + ω d n / d ω$ is the group index. The pulse broadens at a rate proportional to the group-velocity dispersion $β_{2} = d^{2} k / d ω^{2}$ . The quantum wave packet is the matter-wave version of this phenomenon, with the dispersion relation $ω = ℏ k^{2} / (2 m)$ playing the role of the refractive-index curve. The quantum free-particle GVD is $β_{2} = m /ℏ$ , which is always positive — there is no anomalous dispersion for non-relativistic free particles.

Diffraction. A beam of light passing through a slit of width $a$ diffracts. The far-field diffraction angle is $θ \sim λ / a$ , giving a beam width $w \sim λ L / a$ at distance $L$ . The quantum analogue: a particle localised to width $σ_{0}$ has de Broglie wavelength $λ \sim h / p_{0}$ , and after propagation time $t$ the packet width is $σ (t) \sim ℏ t / (2 m σ_{0}) = λ t / (4 π σ_{0})$ . Substituting $L = v_{g} t = p_{0} t / m$ gives $σ (t) \sim λ L / (4 π σ_{0})$ — exactly the diffraction formula. Wave-packet spreading is diffraction of the matter wave.

Liouville's theorem. In classical statistical mechanics, a phase-space distribution evolves under Hamilton's equations while preserving its phase-space volume (Liouville's theorem). The quantum analogue is that the Wigner quasi-probability distribution of the free Gaussian wave packet evolves with constant phase-space area but changing shape: the initially circular distribution in $(x, p)$ space shears into an elongated ellipse. The area $Δ x \cdot Δ p = ℏ/2$ is constant (the minimum-uncertainty value at $t = 0$ ), but this is the area of the uncertainty ellipse, not $σ (t) \cdot Δ p$ (which grows). The distinction is that the Wigner distribution includes correlations between $x$ and $p$ that the simple product $Δ x \cdot Δ p$ does not capture.

Connections Master

12.01.01 (Wave-particle duality): The double-slit experiment uses the spreading of the electron's wave function through the slits to produce interference. The slit width sets $σ_{0}$ and the screen distance sets $t$ ; the fringe pattern is the far-field momentum distribution displayed by a spread wave packet.
12.04.01 (Particle in a box): Stationary states of the infinite square well are energy eigenstates and do not spread. A superposition of box eigenstates does spread, reflecting against the walls. The free-particle spreading formula applies only between reflections.
12.10.01 (Path integral): The propagator $K (x, t; 0, 0)$ that produces the time-evolved Gaussian via convolution is the same object computed by the path integral as a sum over all paths. The spreading formula is the Gaussian packet's response to this kernel.
12.10.02 (Euclidean path integrals): Under Wick rotation $t \to - i τ$ , the spreading wave packet becomes a diffusing Gaussian in the Euclidean time variable $τ$ . The width $σ (τ) = σ_{0} 1 + ℏ^{2} τ^{2} / (4 m^{2} σ_{0}^{4})$ is the Euclidean version of the heat-kernel spreading.
11.04.01 (Statistical mechanics): The coherence length $L_{c}$ connects to the thermal de Broglie wavelength $λ_{th} = 2 π ℏ^{2} / (m k_{B} T)$ : at temperatures where $λ_{th}$ exceeds the inter-particle spacing, wave-packet spreading causes quantum degeneracy (Bose-Einstein condensation or Fermi degeneracy).

Historical notes Master

The Gaussian wave packet was introduced by Kennard (1927) ^{[Kennard 1927]} and independently by Darwin (1927), as one of the first exact time-dependent solutions of the Schrodinger equation. The result resolved a conceptual puzzle of early quantum mechanics: if electrons are described by waves, why do they seem to follow definite trajectories? The answer is that for macroscopic masses the spreading time is astronomically long, and for electrons the spreading is real but fast enough that the "trajectory" picture never applies at the atomic scale.

Heisenberg's 1927 uncertainty paper ^{[Heisenberg 1927]} used the wave-packet spreading phenomenon as a heuristic argument for the uncertainty principle: localising a particle more precisely requires a broader momentum distribution, which produces faster spreading. The rigorous proof (via the Fourier-transform Cauchy-Schwarz argument given in 12.01.01) came slightly later.

The concept of group velocity versus phase velocity for matter waves was clarified by de Broglie himself, who noted that the phase velocity $v_{p h} = c^{2} / v$ for a relativistic particle exceeds $c$ , but the group velocity equals the particle velocity $v$ . The non-relativistic limit $v_{p h} = v /2$ is the version encountered in this unit. The relativistic analysis shows that $v_{p h} \cdot v_{g} = c^{2}$ , which for $v ≪ c$ reduces to $v_{g} \approx v$ and $v_{p h} \approx c^{2} / v ≫ c$ .

The observation of single-electron wave-packet dynamics was achieved by Tonomura et al. (1989) ^{[Tonomura 1989]} in their celebrated double-slit experiment with electrons, where the buildup of the interference pattern one electron at a time directly visualises the probabilistic nature of the wave-packet detection.

Bibliography Master

Cohen-Tannoudji, C., Diu, B. & Laloë, F. Quantum Mechanics, Vol. 1, Complement G-I (Wiley, 1991). Comprehensive treatment of the Gaussian wave packet including coherent-state connection.
Griffiths, D. J. & Schroeter, D. F. Introduction to Quantum Mechanics, 3e (Cambridge, 2018), Ch. 2.4. Accessible derivation with worked examples.
Sakurai, J. J. & Napolitano, J. Modern Quantum Mechanics, 3e (Cambridge, 2021), §2.1. Time evolution via propagator methods.
Kennard, E. H. "Zur Quantenmechanik einfacher Bewegungstypen," Z. Phys. 44, 326 (1927). Original derivation of the Gaussian wave-packet solution.
Heisenberg, W. "Über den anschaulichen Inhalt der quantentheoretischen Kinematik und Mechanik," Z. Phys. 43, 172 (1927). The uncertainty principle paper, using wave-packet arguments.
Tonomura, A. et al. "Demonstration of single-electron buildup of an interference pattern," Am. J. Phys. 57, 117 (1989). Experimental visualisation of electron wave behaviour.

Prerequisites

12.01.01 pending
12.04.01 pending

Tier anchors

beginner: Griffiths & Schroeter, Introduction to Quantum Mechanics, 3e (2018), Ch. 2.4
intermediate: Sakurai & Napolitano, Modern Quantum Mechanics, 3e (2021), §2.1
master: Cohen-Tannoudji, Quantum Mechanics, Vol. 1 (1991), Complement G-I

References

Griffiths & Schroeter — Introduction to Quantum Mechanics, 3e (Cambridge, 2018) · Ch. 2.4 The free particle
Sakurai & Napolitano — Modern Quantum Mechanics, 3e (Cambridge, 2021) · §2.1 Time evolution
Cohen-Tannoudji, Diu & Laloë — Quantum Mechanics, Vol. 1 (Wiley, 1991) · Complement G-I: One-dimensional Gaussian wave packet
Heisenberg, W. — Über den anschaulichen Inhalt der quantentheoretischen Kinematik und Mechanik · Z. Phys. 43, 172 (1927)
Kennard, E. H. — Zur Quantenmechanik einfacher Bewegungstypen · Z. Phys. 44, 326 (1927)
Tonomura et al. — Demonstration of single-electron buildup of an interference pattern · Am. J. Phys. 57, 117 (1989)

Estimated time

beginner: 18m
intermediate: 40m
master: 55m