Part 2 · The postulates

2.5 The uncertainty principle

Source: the Gaussian, the Gaussian wave packet, its momentum distribution and the minimum uncertainty from the 2024/25 board #8; this year’s course has not reached it yet

Objectives

Understand why a sharp position means a spread-out momentum.
Recipe: normalise a Gaussian; find the momentum distribution of a packet by projecting on $u_{p}$ .
Picture: a Gaussian $∣ ψ (x) ∣^{2}$ and its momentum distribution, one widening as the other narrows.
Theorem: the Gaussian packet has $Δ x Δ p_{x} = ℏ/2$ ; in general $Δ x Δ p_{x} \geq ℏ/2$ .
Vocabulary: Gaussian, minimum-uncertainty state.

Theorem 2.5.1Heisenberg’s uncertainty principle2024/25 board #8

Observables that are not compatible cannot be measured simultaneously with arbitrary precision. For position and momentum along one axis, $[\overset{x}{^}, \overset{p}{^}_{x}] = i ℏ$ (Theorem 2.4.3), and $Δ x Δ p_{x} \geq \frac{ℏ}{2} .$

A measurement of $x$ gives $x_{0} \pm Δ x$ and one of $p_{x}$ gives $p_{0 x} \pm Δ p_{x}$ ; the $Δ$ s are the spreads of Definition 2.1.3. ==The sharper the position, the broader the momentum.== It is the bandwidth relation $Δ k Δ x \sim 1$ with $p = ℏ k$ . The board shows it on the one packet where everything can be computed.

The Gaussian

Many measurements of $x$ scatter around $x_{0}$ in a bell shape. The bell is the Gaussian, here with $x_{0} = 0$ : $P (x) = P_{0} e^{- x^{2} /2 (Δ x)^{2}} .$ At $x = \pm Δ x$ it has fallen to $e^{- 1/2} \approx 0.6$ of its peak, and the area between $\pm Δ x$ is $0.683$ of the total. (The board writes “0.66%”, meaning 0.66.) For this $P (x)$ , $⟨ x^{2} ⟩ - ⟨ x ⟩^{2} = (Δ x)^{2}$ exactly (beyond the boards): the width in the formula is the spread.

Recipe 2.5.2Normalising a Gaussian

Ask $P_{0} \int e^{- x^{2} /2 (Δ x)^{2}} d x = 1$ .
Substitute $y = x / (2 Δ x)$ , $d x = 2 Δ x d y$ , and use $\int_{- \infty}^{+ \infty} e^{- y^{2}} d y = π$ : $P_{0} 2 π Δ x = 1$ .
So $P (x) = \frac{1}{2 π Δ x} e^{- x^{2} /2 (Δ x)^{2}} .$ A narrower packet has a taller peak: the area stays 1.

The wave packet. $P = ∣ ψ ∣^{2}$ , so the simplest wavefunction with this distribution is its square root, $ψ (x) = [\frac{1}{2 π Δ x}]^{1/2} e^{- x^{2} / (2Δ x)^{2}} .$ The square root halves the exponent: $\frac{x ^{2}}{2 ( Δ x ) ^{2}} \cdot \frac{1}{2} = \frac{x ^{2}}{( 2Δ x ) ^{2}}$ . This $ψ$ is real. Multiplying it by $e^{i p_{0} x /ℏ}$ leaves $P (x)$ alone and moves the average momentum from 0 to $p_{0}$ : a packet travelling at $p_{0} / m$ (beyond the boards).

Its momentum distribution

Theorem 2.5.3The Gaussian packet’s momentum2024/25 board #8

The momentum distribution of the Gaussian packet is again a Gaussian, $P (p_{x}) = \frac{1}{2 π Δ p _{x}} e^{- p_{x}^{2} /2 (Δ p_{x})^{2}}, Δ p_{x} = \frac{ℏ}{2Δ x} .$

Proof

Project. By Postulate 1, with the continuum-normalised $u_{p}$ of Key observation 1.5.6, $P (p_{x}) = ∣ ⟨ u_{p} ∣ ψ ⟩ ∣^{2}$ . The bra conjugates $u_{p}$ , so its exponent turns negative: $P (p_{x}) = \frac{1}{2 π ℏ} \frac{1}{2 π Δ x} \int e^{- i p_{x} x /ℏ} e^{- x^{2} / (2Δ x)^{2}} d x^{2} .$
Complete the square. $(\frac{x}{2Δ x} + \frac{i p _{x} Δ x}{ℏ})^{2} = \frac{x ^{2}}{( 2Δ x ) ^{2}} + \frac{i p _{x} x}{ℏ} - \frac{p _{x}^{2} ( Δ x ) ^{2}}{ℏ ^{2}},$ so the exponent is $- (\frac{x}{2Δ x} + \frac{i p _{x} Δ x}{ℏ})^{2} - \frac{p _{x}^{2} ( Δ x ) ^{2}}{ℏ ^{2}}$ . The last term does not depend on $x$ and comes out of the integral; squared, it gives $e^{- 2 p_{x}^{2} (Δ x)^{2} / ℏ^{2}}$ .
The remaining integral. With $y = \frac{x}{2Δ x} + \frac{i p _{x} Δ x}{ℏ}$ , $d x = 2Δ x d y$ , it is $2Δ x π$ . Shifting by an imaginary constant does not change it: the integrand has no singularities and vanishes far out (beyond the boards).
Collect. $P (p_{x}) = \frac{4 π ( Δ x ) ^{2}}{2 π ℏ 2 π Δ x} e^{- 2 p_{x}^{2} (Δ x)^{2} / ℏ^{2}} = \frac{2Δ x}{2 π ℏ} e^{- 2 p_{x}^{2} (Δ x)^{2} / ℏ^{2}} .$
Compare with $e^{- p_{x}^{2} /2 (Δ p_{x})^{2}}$ : $\frac{2 ( Δ x ) ^{2}}{ℏ ^{2}} = \frac{1}{2 ( Δ p _{x} ) ^{2}}$ , so $(Δ p_{x})^{2} = \frac{ℏ ^{2}}{4 ( Δ x ) ^{2}}$ . The prefactor is then $\frac{1}{2 π Δ p _{x}}$ , as it must be for a normalised $P (p_{x})$ (that check is beyond the boards).

The board slips twice here: it writes $\overset{p}{^}_{x} = i ℏ \frac{\partial}{\partial x}$ beside $u_{p}$ , without the minus, and drops a minus from the exponent of the factor taken out, restored on the next line.

Loading the demo…

Figure 2.5.1 Drag the width of the

∣ ψ (x) ∣^{2}

packet; the

bar

marks

\pm Δ x

, where the curve has fallen to

e^{- 1/2}

of its peak. Its momentum distribution, below, follows with

Δ p_{x} = ℏ/2Δ x

: halve one and the other doubles. The product stays at the bound

ℏ/2

Minimum uncertainty

Key observation 2.5.4The minimum-uncertainty state2024/25 board #8

$Δ x Δ p_{x} = \frac{ℏ}{2} for the Gaussian, Δ x Δ p_{x} \geq \frac{ℏ}{2} in general .$ ==The Gaussian reaches the bound exactly: it is the minimum-uncertainty state.== Any other shape has a larger product.

The rule behind it, for any two observables (beyond the boards), is $Δ A Δ B \geq \frac{1}{2} ⟨[\hat{A}, \hat{B}]⟩ .$ With $[\overset{x}{^}, \overset{p}{^}_{x}] = i ℏ$ this is $ℏ/2$ . Compatible observables have zero on the right, and nothing stops both being sharp; $x$ and $p_{y}$ commute (Theorem 2.4.3), so there is no bound between them.

Example 2.5.5In numbersbeyond the boards

An electron confined to $Δ x = 1$ nm has $Δ p_{x} \geq \frac{1.055 \times 1 0 ^{- 34}}{2 \times 1 0 ^{- 9}} = 5.3 \times 1 0^{- 26}$ kg m/s, so $Δ v_{x} \geq 5.8 \times 1 0^{4}$ m/s, and a kinetic energy $\frac{( Δ p _{x} ) ^{2}}{2 m} \approx 10$ meV, comparable with thermal energy at room temperature, 25 meV. ==Confinement on the nanometre scale costs energy that matters==; it is why the lowest level of the infinite well is above zero.
A 1 g bead located to 1 µm has $Δ v \geq 5 \times 1 0^{- 26}$ m/s. Unmeasurable: classical mechanics is safe for everyday objects.

Exercises

Exercise 2.5.1Sample questions, Q2 (iv)

What is Heisenberg’s uncertainty principle? How does it relate to wave–particle duality and de Broglie?

Solution

Position and momentum cannot both be sharp, $Δ x Δ p_{x} \geq ℏ/2$ . A particle of definite $p$ is a plane wave of one $k = p /ℏ$ , spread over all space; localising it needs a packet of many $k$ , so a spread of $p$ : $Δ k Δ x \sim 1$ times $ℏ$ . In operator terms, $[\overset{x}{^}, \overset{p}{^}_{x}] = i ℏ \neq = 0$ .

Exercise 2.5.2Sample questions, Q2 (vii)

Discuss how a Gaussian wavepacket satisfies the uncertainty principle in the best possible case.

Solution

$ψ \propto e^{- x^{2} / (2Δ x)^{2}}$ has a Gaussian position distribution of width $Δ x$ . Projected on $u_{p} = e^{i p_{x} x /ℏ} / 2 π ℏ$ it gives a momentum distribution that is again Gaussian, of width $Δ p_{x} = ℏ/2Δ x$ (Theorem 2.5.3). So $Δ x Δ p_{x} = ℏ/2$ , the equality case; any other shape gives more. Narrowing the packet in $x$ widens it in $p_{x}$ in proportion.

Exercise 2.5.3beyond the boards

An electron is in a Gaussian packet with $Δ x = 0.1$ nm, about an atom’s size. Smallest $Δ p_{x}$ , and the kinetic-energy scale $(Δ p_{x})^{2} /2 m$ ?

Solution

$Δ p_{x} = \frac{1.055 \times 1 0 ^{- 34}}{2 \times 1 0 ^{- 10}} = 5.3 \times 1 0^{- 25}$ kg m/s; $\frac{( 5.3 \times 1 0 ^{- 25} ) ^{2}}{2 \cdot 9.11 \times 1 0 ^{- 31}} = 1.5 \times 1 0^{- 19}$ J $\approx 1$ eV. Atomic confinement alone costs electron-volts: the energy scale of chemistry.

Exercise 2.5.4beyond the boards

A packet in a box has $ψ = 2/ L cos (2 π x / L)$ (Theorem 1.5.1). What does a momentum measurement give? Find $⟨ \overset{p}{^}_{x} ⟩$ and $Δ p_{x}$ .

Solution

$ψ = \frac{1}{2} (u_{1} + u_{- 1})$ : $\pm \frac{2 π ℏ}{L}$ , 50% each. $⟨ \overset{p}{^}_{x} ⟩ = 0$ , $⟨ \overset{p}{^}_{x}^{2} ⟩ = (2 π ℏ/ L)^{2}$ , $Δ p_{x} = 2 π ℏ/ L$ . A standing wave is two travelling waves going opposite ways.