Minimum-Uncertainty Wave Packets

A minimum-uncertainty wave packet is a normalized state whose position and momentum spreads reach the lower limit allowed by the uncertainty relation. For one-dimensional position and momentum,

$$\Delta x\,\Delta p \ge \frac{\hbar}{2}.$$

The canonical example is a Gaussian wave packet. It is the most localized possible state in position for a given momentum spread, and the most sharply peaked momentum distribution possible for a given position spread. This is why Gaussians appear so often in free-particle motion, semiclassical approximations, and coherent states.

The Standard Gaussian

At an initial time, take

$$\psi(x) = \left(\frac{1}{2\pi\sigma_x^2}\right)^{1/4} \exp\left[ -\frac{(x-x_0)^2}{4\sigma_x^2} +\frac{i}{\hbar}p_0(x-x_0) \right].$$

Its probability density is

$$\lvert\psi(x)\rvert^2 = \frac{1}{\sqrt{2\pi}\sigma_x} \exp\left[ -\frac{(x-x_0)^2}{2\sigma_x^2} \right],$$

so

$$\Delta x=\sigma_x.$$

With the momentum convention used in this volume, the momentum-space wavefunction is also Gaussian and has standard deviation

$$\Delta p = \frac{\hbar}{2\sigma_x}.$$

Therefore

$$\Delta x\,\Delta p = \sigma_x\frac{\hbar}{2\sigma_x} = \frac{\hbar}{2}.$$

This is exact saturation of the position-momentum uncertainty relation.

What Saturation Means

Saturating the uncertainty relation does not mean the particle has a hidden exact position and exact momentum. It means the probability distributions predicted by the state are as jointly narrow as quantum mechanics permits for this pair of observables.

The tradeoff is reciprocal:

$$\Delta p_{\min} = \frac{\hbar}{2\Delta x}.$$

If the packet is squeezed to half its initial spatial width, the minimum possible momentum spread doubles. The mean momentum $\langle p\rangle=p_0$ need not become large; the spread around the mean becomes large.

Equality Condition

The equality case of the Robertson derivation occurs when the centered momentum action is proportional to the centered position action:

$$(\hat p-p_0)\lvert\psi\rangle = i\lambda(\hat x-x_0)\lvert\psi\rangle, \qquad \lambda\gt0.$$

In position representation this becomes

$$\left( -i\hbar\frac{d}{dx} -p_0 \right)\psi(x) = i\lambda(x-x_0)\psi(x).$$

Solving gives

$$\psi(x) \propto \exp\left[ -\frac{\lambda}{2\hbar}(x-x_0)^2 +\frac{i}{\hbar}p_0x \right].$$

Normalizability requires $\lambda\gt0$. Writing

$$\lambda=\frac{\hbar}{2\sigma_x^2}$$

recovers the standard Gaussian. Thus the Gaussian is not merely a convenient example; it is the equality solution.

Wave-Packet Interpretation

A narrow position-space packet requires many momentum components whose phases interfere constructively near $x_0$ and destructively away from it. The Gaussian is special because its Fourier transform is again Gaussian, with the reciprocal width required to meet the lower bound.

In momentum space the same state has the schematic form

$$\phi(p) \propto \exp\left[ -\frac{\sigma_x^2(p-p_0)^2}{\hbar^2} -\frac{i}{\hbar}x_0(p-p_0) \right].$$

The phase records the packet center $x_0$; the width records the momentum uncertainty. The probability density $\lvert\phi(p)\rvert^2$ is centered at $p_0$.

Free Evolution And Correlations

An initially unchirped Gaussian saturates

$$\Delta x\,\Delta p=\frac{\hbar}{2}$$

at that instant. Under free evolution, the packet remains Gaussian, but its width grows:

$$\Delta x(t) = \sigma_x \sqrt{ 1+\left( \frac{\hbar t}{2m\sigma_x^2} \right)^2 }, \qquad \Delta p(t)=\frac{\hbar}{2\sigma_x}.$$

For $t\ne0$, the simple product becomes

$$\Delta x(t)\Delta p(t) = \frac{\hbar}{2} \sqrt{ 1+\left( \frac{\hbar t}{2m\sigma_x^2} \right)^2 },$$

which is larger than $\hbar/2$.

This does not mean the Gaussian stopped being special. Free evolution creates position-momentum correlation: points farther from the center tend to have phases corresponding to different local momenta. The stronger covariance-aware inequality is

$$(\Delta x)^2(\Delta p)^2-C_{xp}^2 \ge \frac{\hbar^2}{4},$$

where

$$C_{xp} = \frac12 \left\langle (\hat x-\langle x\rangle)(\hat p-\langle p\rangle) + (\hat p-\langle p\rangle)(\hat x-\langle x\rangle) \right\rangle.$$

A freely spreading pure Gaussian saturates this stronger inequality even when $\Delta x\,\Delta p\gt\hbar/2$. The extra product comes from correlation, not from a non-Gaussian shape.

Chirped Gaussians

A Gaussian with a quadratic phase,

$$\psi(x) \propto \exp\left[ -\frac{1-ib}{4\sigma_x^2}(x-x_0)^2 +\frac{i}{\hbar}p_0(x-x_0) \right], \qquad b\in\mathbb R,$$

is called chirped in wave-packet language. Its probability density has the same position width $\sigma_x$, but its local phase gradient varies with $x$. This increases the ordinary momentum spread:

$$\Delta p = \frac{\hbar}{2\sigma_x}\sqrt{1+b^2}.$$

Thus

$$\Delta x\,\Delta p = \frac{\hbar}{2}\sqrt{1+b^2}.$$

For $b\ne0$, the packet no longer saturates the simple product relation, but it is still a correlated Gaussian. This is the form naturally generated by free spreading.

Relation To Coherent States

The harmonic oscillator ground state is a minimum-uncertainty Gaussian. A coherent state is a displaced version of that ground-state packet. It has nonzero mean position and momentum, but the same uncertainty product:

$$\Delta x\,\Delta p=\frac{\hbar}{2}.$$

The important difference from a free Gaussian is dynamical. A free minimum-uncertainty packet spreads because the free dispersion relation is quadratic in momentum. A harmonic-oscillator coherent state keeps its shape because the oscillator potential refocuses the packet. This is why coherent states are the oscillator analogue of classical phase-space points.

The oscillator construction is treated in Coherent States.

Common Mistakes

• Saying a minimum-uncertainty packet has exact position and exact momentum.
• Thinking every Gaussian at every time satisfies $\Delta x\,\Delta p=\hbar/2$.
• Confusing a larger momentum spread with a larger mean momentum.
• Forgetting that a quadratic phase can increase $\Delta p$ without changing $\lvert\psi(x)\rvert^2$.
• Treating plane waves as minimum-uncertainty states; they are not normalizable finite-variance states.
• Assuming only harmonic-oscillator coherent states can saturate the position-momentum bound.

Where This Is Used

• Gaussian Wave Packets gives the explicit free-particle packet and spreading formula.
• Position-Momentum Uncertainty gives the core uncertainty relation.
• Wave Packet Spreading explains how the free packet broadens after the initial minimum-product instant.
• Coherent States gives the oscillator version, where the Gaussian packet remains shape-preserving.
• Squeezed States: First Encounter shows minimum-uncertainty Gaussian states with unequal conjugate widths.
• Wave Packets gives the Fourier-analysis viewpoint.

References

• E. H. Kennard, “Zur Quantenmechanik einfacher Bewegungstypen,” Zeitschrift fur Physik 44, 326-352, 1927.
• H. P. Robertson, “The Uncertainty Principle,” Physical Review 34, 163-164, 1929.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• J. R. Klauder and B.-S. Skagerstam, Coherent States, World Scientific, 1985.

Exercises

1. A normalized packet has $\Delta x=a$. What is the smallest possible $\Delta p$?

Solution

The uncertainty relation gives

$$a\,\Delta p \ge \frac{\hbar}{2}.$$

Thus

$$\Delta p \ge \frac{\hbar}{2a}.$$

A suitable unchirped Gaussian attains this lower bound.

2. Derive the Gaussian form from the equality condition.

Solution

Use

$$(\hat p-p_0)\psi = i\lambda(x-x_0)\psi.$$

In position space,

$$\left( -i\hbar\frac{d}{dx} -p_0 \right)\psi = i\lambda(x-x_0)\psi.$$

Solving the first-order equation gives

$$\psi(x) \propto \exp\left[ -\frac{\lambda}{2\hbar}(x-x_0)^2 +\frac{i}{\hbar}p_0x \right].$$

Normalizability requires $\lambda\gt0$, so the equality state is Gaussian.

3. Why does a freely spreading Gaussian have $\Delta x(t)\Delta p(t)\gt\hbar/2$ for $t\ne0$?

Solution

Free evolution leaves $\Delta p$ unchanged but increases $\Delta x(t)$:

$$\Delta x(t) = \sigma_x \sqrt{ 1+\left( \frac{\hbar t}{2m\sigma_x^2} \right)^2 }.$$

Therefore the ordinary product becomes larger than its initial value. The state remains a correlated Gaussian, but position and momentum are no longer uncorrelated.

4. How is a harmonic-oscillator coherent state related to a minimum-uncertainty Gaussian?

Solution

The oscillator ground state is a Gaussian that saturates $\Delta x\,\Delta p=\hbar/2$. A coherent state is a displaced version of that ground state, so its mean position and momentum can be nonzero while its uncertainty product remains minimal. Unlike a free Gaussian, it does not spread because the oscillator dynamics refocuses the packet.