2.6 Expectation values, variance and the uncertainty principle

With $|\Psi|^2$ in hand we have a probability distribution. What remains is statistics: compute means, compute fluctuations.

Expectation values

The definition of an comes straight from probability theory:

\langle x\rangle=\int_{-\infty}^{\infty}x\,|\Psi(x,t)|^2\,\dd x\tag{2.6.1}

Where the momentum operator comes from

The expectation value of position is easy to write. What about momentum? We do not know what “the particle’s momentum distribution” looks like. But there is one clue: $\langle p\rangle$ ought to equal $m\,\dd\langle x\rangle/\dd t$ .

Deriving the momentum operator from d⟨x⟩/dtadvanced~7 min

\frac{\dd\langle x\rangle}{\dd t}=\int x\frac{\partial}{\partial t}|\Psi|^2\,\dd x\tag{2.6.2}

Use the continuity equation $\partial_t|\Psi|^2=-\partial_x j$ from section 2.2:

\frac{\dd\langle x\rangle}{\dd t}=-\int x\frac{\partial j}{\partial x}\dd x\tag{2.6.3}

Integrate by parts; the boundary term vanishes because $\Psi$ is normalisable:

=\int j\,\dd x=\int\frac{\ii\hbar}{2m}\left(\Psi\frac{\partial\Psi^*}{\partial x}-\Psi^*\frac{\partial\Psi}{\partial x}\right)\dd x\tag{2.6.4}

Integrate the first term by parts once more (moving the derivative from $\Psi^*$ onto $\Psi$ ) and combine:

\frac{\dd\langle x\rangle}{\dd t}=-\frac{\ii\hbar}{m}\int\Psi^*\frac{\partial\Psi}{\partial x}\,\dd x\tag{2.6.5}

Hence

\langle p\rangle=m\frac{\dd\langle x\rangle}{\dd t}=\int\Psi^*\left(-\ii\hbar\frac{\partial}{\partial x}\right)\Psi\,\dd x\tag{2.6.6}

and the thing in brackets is the momentum operator:

\hat p=-\ii\hbar\frac{\partial}{\partial x}\tag{2.6.7}

It was not postulated; it was derived — given only the Born rule, the Schrödinger equation, and the classical correspondence “momentum = mass × velocity”.

In general, the operator for a classical quantity $A(x,p)$ is obtained by the substitution $x\to x$ , $p\to-\ii\hbar\partial_x$ , and its expectation value is

\langle \hat A\rangle=\int\Psi^*\,\hat A\,\Psi\,\dd x\tag{2.6.8}

Note that $\hat A$ must sit between $\Psi^*$ and $\Psi$ — the operator acts on the $\Psi$ to its right, and the order cannot be shuffled.

Variance and standard deviation

\sigma_A^2=\langle \hat A^2\rangle-\langle \hat A\rangle^2\tag{2.6.10}

$\sigma_A$ measures the spread of measurement outcomes. In particular $\sigma_A=0$ if and only if $\Psi$ is an eigenstate of $\hat A$ — in which case every measurement returns the same value.

Example: σₓ and σ_p for the infinite-well ground stateadvanced~6 min

The ground state is $\psi_1=\sqrt{2/L}\sin(\pi x/L)$ on $[0,L]$ .

Position: by symmetry $\langle x\rangle=L/2$ .

\langle x^2\rangle=\frac{2}{L}\int_0^L x^2\sin^2\frac{\pi x}{L}\dd x =\frac{2}{L}\int_0^L x^2\cdot\frac{1-\cos(2\pi x/L)}{2}\dd x\tag{2.6.11}

The first term gives $\frac{L^2}{3}$ ; the second, after two integrations by parts, gives $-\frac{L^2}{2\pi^2}$ . Together,

\langle x^2\rangle=L^2\left(\frac13-\frac{1}{2\pi^2}\right)\tag{2.6.12}

\sigma_x^2=\langle x^2\rangle-\frac{L^2}{4}=L^2\left(\frac{1}{12}-\frac{1}{2\pi^2}\right) \approx0.0326\,L^2 \;\Longrightarrow\;\sigma_x\approx0.181\,L\tag{2.6.13}

Momentum: $\langle p\rangle=0$ (the wavefunction is real, and by symmetry the integral vanishes; physically, left- and right-moving components are equally likely). And

\langle p^2\rangle=2m\langle T\rangle=2mE_1=\frac{\pi^2\hbar^2}{L^2} \;\Longrightarrow\;\sigma_p=\frac{\pi\hbar}{L}\tag{2.6.14}

The product:

\sigma_x\sigma_p\approx0.181L\cdot\frac{\pi\hbar}{L}=0.568\,\hbar\approx1.14\cdot\frac{\hbar}{2}\tag{2.6.15}

Larger than $\hbar/2$ , and fairly close to the bound. Note that $L$ cancels completely: shrink the box and $\sigma_x$ shrinks proportionally while $\sigma_p$ grows proportionally, leaving the product untouched. You can see this directly in the simulation below.

The uncertainty principle

\sigma_x\,\sigma_p\ \ge\ \frac{\hbar}{2}\tag{2.6.16}

The rigorous proof needs the generalised uncertainty relation (section 3.6). What matters here is what it actually says.

This is not about measurement disturbance

The popular story runs: “to see where the electron is you must shine short-wavelength light on it, the photon kicks momentum into the electron, and so the momentum becomes uncertain.”

That story — Heisenberg’s microscope — has historical value, but as an explanation of the inequality it is wrong, for a simple reason.

$\sigma_x$ and $\sigma_p$ are two statistical properties of the same wavefunction. You can compute both from $\Psi$ with pen and paper without measuring anything at all — which is exactly what the derivation above did. The inequality holds before you decide what to measure.

The correct statement: a wavefunction with small $\sigma_x$ (narrow in space) necessarily contains a broad range of wave numbers in its Fourier expansion, hence a large $\sigma_p$ . That is a theorem of Fourier analysis, equally true for sound waves and light pulses — it is why an ultrashort laser pulse necessarily has a broad spectrum. The distinctively quantum ingredient is only $p=\hbar k$ , which translates “spread in wave number” into “uncertainty in momentum”.

The picture

How to read the inequality

It is a statement about states, not about instruments.

It says: there is no quantum state whose position and momentum distributions are both arbitrarily narrow. However good your apparatus, and whether or not you measure at all, the constraint stands.

The states that saturate it are Gaussian wave packets, which is why Gaussians occupy a special place in quantum mechanics (section 2.10).

The mathematics

\sigma_x\sigma_p\ge\frac{\hbar}{2}\tag{2.6.17}

The standard order-of-magnitude estimate:

\Delta p\sim\frac{\hbar}{\Delta x},\qquad E\sim\frac{(\Delta p)^2}{2m}\sim\frac{\hbar^2}{2m(\Delta x)^2}\tag{2.6.18}

Confine a particle to a length $L$ and the zero-point kinetic energy is $\sim\hbar^2/2mL^2$ . The infinite well’s $E_1=\pi^2\hbar^2/2mL^2$ differs from that estimate only by $\pi^2$ .

The infinite square well

Drag the width L and the quantum number n and watch the energy, the waveform and the probability distribution move together. Units: ħ = m = 1.

Energy Eₙ (n=1): 4.935
relative to E₁(L=1): 1.00 ×
⟨x⟩ / L: 0.500
Δx / L: 0.180

Well width L1.00

Shrink the box and every level is pushed up together: E ∝ 1/L²

Quantum number n1

nodes = n − 1 = 0 (endpoints excluded)

State

Show ψShow |ψ|²

Try this

Drag L from 2.5 down to 0.6 and keep your eye on the dashed line in the level diagram (pinned at the reference energy E₁(L=1)): it sinks all the way to the bottom, meaning every level has risen far above it. The "relative to E₁(L=1)" readout climbs from 0.16 to 2.78 — "the tighter you confine a particle, the more kinetic energy it has", a direct consequence of ΔxΔp ≥ ħ/2.
Change n in the stationary mode and note that |ψ|² never changes with time (the curve stands still); switch to a superposition, press play, and |ψ|² immediately starts sloshing. That is exactly what "stationary" refers to.
Leave only c₁ in the superposition (drag the rest to 0) and press play — the probability density stops moving again. However long a single eigenstate evolves, it only picks up an overall phase e^(−iEₙt/ħ), which no observable can see.
Push n above 10 and look at |ψ|²: the fringes get so fine that the distribution is nearly uniform — the classical picture of a particle equally likely to be anywhere in the box. The correspondence principle, in view.

This is an interactive simulation (infinite-well) and needs JavaScript to run. The derivations and conclusions in the text do not depend on it, so feel free to read on.

Drag the well width $L$ and watch $\Delta x/L$ in the readout: it is a constant, about 0.181. So $\sigma_x\propto L$ , while energy $\propto1/L^2$ means $\sigma_p\propto1/L$ . The product is independent of $L$ — the inequality is equally tight for a box of any size.

Ehrenfest’s theorem: the return of classical mechanics

⟨x⟩ and ⟨p⟩ obey Newton's equationsadvanced~5 min

We already have the first relation:

\frac{\dd\langle x\rangle}{\dd t}=\frac{\langle p\rangle}{m}\tag{2.6.19}

A similar computation for $\langle p\rangle$ (differentiate in time, substitute the Schrödinger equation, integrate by parts) gives

\frac{\dd\langle p\rangle}{\dd t}=\left\langle-\frac{\partial V}{\partial x}\right\rangle=\langle F\rangle\tag{2.6.20}

Together these are Newton’s second law for expectation values.

But look carefully at the right-hand side: it is $\langle F(x)\rangle$ , not $F(\langle x\rangle)$ . In general these differ. Expanding $F$ about $\langle x\rangle$ :

\langle F(x)\rangle=F(\langle x\rangle)+\frac12 F''(\langle x\rangle)\,\sigma_x^2+\cdots\tag{2.6.21}

Only when the force is approximately linear across the width of the packet ( $F''\approx0$ ), or the packet is narrow enough ( $\sigma_x$ small), do the two agree and the expectation value genuinely follow a classical trajectory.

This explains when classical mechanics is a good approximation: for macroscopic objects $\sigma_x$ is absurdly small and everyday force fields are extremely smooth on that scale. It also explains when it is not: for an electron crossing a double slit the packet width is comparable to the slit separation, and the $F''$ term is not remotely negligible.

Key formulas

Expectation value
$\langle \hat A\rangle=\int\Psi^*\hat A\Psi\,\dd x$
The operator sits in the middle and acts on the Ψ to its right
Momentum operator
$\hat p=-\ii\hbar\frac{\partial}{\partial x}$
Derived from ⟨p⟩ = m d⟨x⟩/dt, not assumed
Standard deviation
$\sigma_A=\sqrt{\langle \hat A^2\rangle-\langle \hat A\rangle^2}$
σ_A = 0 ⟺ Ψ is an eigenstate of Â
Uncertainty principle
$\sigma_x\sigma_p\ge\frac{\hbar}{2}$
A property of the state, independent of measurement; Gaussians saturate it
Ehrenfest's theorem
$m\frac{\dd^2\langle x\rangle}{\dd t^2}=\left\langle-\frac{\partial V}{\partial x}\right\rangle$
Note ⟨F(x)⟩, not F(⟨x⟩)

Self-check4 questions

What comes next

The tools are in place. The next five sections apply them to specific potentials, starting from the simplest: one particle and two infinitely high walls.

Expectation values, variance and the uncertainty principle

Recommended first

After this section you should be able to

Expectation values

Where the momentum operator comes from

Variance and standard deviation

The uncertainty principle

The picture

The mathematics

The infinite square well

Ehrenfest’s theorem: the return of classical mechanics

Key formulas

Self-check4 questions

What comes next

Expectation values

Where the momentum operator comes from

Variance and standard deviation

The uncertainty principle

◑The picture

∑The mathematics

Ehrenfest’s theorem: the return of classical mechanics

∎Key formulas

What comes next

The picture

The mathematics

Key formulas