Position–Momentum Uncertainty

For a particle on the real line, every normalized state with finite position and momentum variances satisfies

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

This is the canonical position–momentum preparation uncertainty relation. It compares the root-mean-square widths of the position and momentum distributions assigned to the same state. It does not compare the calibration errors of two detectors, and it does not require position and momentum to be measured sequentially on one specimen.

The constant is state independent because the canonical commutator is central:

$$[x,p]=i\hbar I.$$

The same theorem is also a Fourier-analysis statement. Position and momentum wavefunctions are Fourier transforms, and a square-integrable function and its transform cannot both have arbitrarily small variance. Quantum mechanics turns those two mathematical widths into observable probability spreads.

The stronger covariance-sensitive form is

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

where $C_\psi(x,p)$ is the symmetrized position–momentum covariance. An uncorrelated Gaussian saturates the product relation. A chirped Gaussian can saturate the stronger determinant relation while having $\Delta x\,\Delta p>\hbar/2$.

Helpful background. General uncertainty relations supplies Cauchy–Schwarz and its equality condition; momentum-space representation explains the Fourier transform.

Position and Momentum on the Real Line

The standard theorem applies to one Cartesian degree of freedom on the full line. The Hilbert space is

$$\mathcal H=L^2(\mathbb R,dx),$$

with normalized wavefunction

$$\int_{-\infty}^{\infty} |\psi(x)|^2\,dx =1.$$

In the Schrödinger representation,

$$(x\psi)(x)=x\psi(x),$$

and

$$(p\psi)(x) = -i\hbar\frac{d\psi}{dx}(x)$$

on $\mathcal D(p)=H^1(\mathbb R)$, the square-integrable functions with square-integrable weak derivative. Position has domain $\mathcal D(x)=\{\psi:x\psi\in L^2\}$. Finite second moments mean $\psi\in\mathcal D(x)\cap\mathcal D(p)$. The momentum-space wavefunction is

$$\phi(p) = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} e^{-ipx/\hbar}\psi(x)\,dx.$$

The inverse transform is

$$\psi(x) = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} e^{ipx/\hbar}\phi(p)\,dp.$$

With this symmetric convention, Plancherel gives

$$\int_{-\infty}^{\infty}|\phi(p)|^2\,dp=1.$$

Other Fourier conventions move factors of $2\pi$ between the transform and the width inequality. The physical formula remains $\hbar/2$ once momentum is identified as $p=\hbar k$. The transform convention is explained in Fourier Transform Conventions.

The Two Probability Distributions

Position outcomes have density

$$\rho_x(x)=|\psi(x)|^2,$$

with mean

$$x_0 = \langle x\rangle_\psi = \int_{-\infty}^{\infty} x|\psi(x)|^2\,dx,$$

and variance

$$(\Delta_\psi x)^2 = \int_{-\infty}^{\infty} (x-x_0)^2|\psi(x)|^2\,dx.$$

Momentum outcomes have density

$$\rho_p(p)=|\phi(p)|^2,$$

with mean

$$p_0 = \langle p\rangle_\psi = \int_{-\infty}^{\infty} p|\phi(p)|^2\,dp,$$

and variance

$$(\Delta_\psi p)^2 = \int_{-\infty}^{\infty} (p-p_0)^2|\phi(p)|^2\,dp.$$

These are two representations of one state, not two independent probability models. The means $x_0$ and $p_0$ locate the packet in phase space; the uncertainty relation constrains its widths about those means.

A standard deviation is not a hard interval. In a non-Gaussian state, $\Delta x$ does not mean that every position outcome lies between $x_0-\Delta x$ and $x_0+\Delta x$.

Derivation from the Canonical Commutator

The Robertson relation is

$$\Delta_\psi A\,\Delta_\psi B \ge \frac12 \left\lvert \langle[A,B]\rangle_\psi \right\rvert.$$

Set $A=x$ and $B=p$. On a domain where the canonical relation is valid,

$$[x,p]=i\hbar I.$$

Normalization gives

$$\begin{aligned} \langle[x,p]\rangle_\psi &= i\hbar\langle I\rangle_\psi \\ &=i\hbar. \end{aligned}$$

Therefore

$$\begin{aligned} \Delta_\psi x\,\Delta_\psi p &\ge \frac12|i\hbar| \\ &= \frac\hbar2. \end{aligned}$$

Unlike a generic commutator expectation, the right side cannot vanish in a normalized state. The identity operator is what makes this bound universal within the theorem’s domain.

The general Cauchy–Schwarz proof and equality criteria belong to General Uncertainty Relations. The algebraic and Weyl forms of the canonical relation belong to Canonical Commutation Relations.

Direct Wavefunction Derivation

First take a Schwartz wavefunction, so all derivatives and boundary steps below are justified. A cutoff or common-core approximation then extends the inequality to every normalized $\psi\in\mathcal D(x)\cap\mathcal D(p)$; the proof uses $x\psi$ and $p\psi$, not their products.

Define the centered fluctuation vectors

$$f(x)=(x-x_0)\psi(x),$$

and

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

Their norms are

$$\|f\|^2=(\Delta_\psi x)^2, \qquad \|g\|^2=(\Delta_\psi p)^2.$$

Cauchy–Schwarz gives

$$(\Delta_\psi x)^2(\Delta_\psi p)^2 \ge |\langle f|g\rangle|^2.$$

Because

$$\int_{-\infty}^{\infty} (x-x_0)|\psi(x)|^2\,dx=0,$$

the $p_0$ term does not contribute, and

$$\langle f|g\rangle = -i\hbar \int_{-\infty}^{\infty} (x-x_0)\psi(x)^*\psi'(x)\,dx.$$

Let

$$I = \int_{-\infty}^{\infty} (x-x_0)\psi^*\psi'\,dx.$$

Then

$$\begin{aligned} I+I^* &= \int_{-\infty}^{\infty} (x-x_0) \frac{d}{dx}|\psi|^2\,dx \\ &= \left[ (x-x_0)|\psi(x)|^2 \right]_{-\infty}^{\infty} - \int_{-\infty}^{\infty}|\psi(x)|^2\,dx. \end{aligned}$$

For a state with a vanishing boundary term, normalization gives

$$I+I^*=-1,$$

so

$$\operatorname{Re}I=-\frac12.$$

Since $\langle f|g\rangle=-i\hbar I$,

$$\operatorname{Im}\langle f|g\rangle = \frac\hbar2.$$

Therefore

$$|\langle f|g\rangle|^2 \ge \frac{\hbar^2}{4},$$

and Cauchy–Schwarz yields

$$\Delta_\psi x\,\Delta_\psi p \ge \frac\hbar2.$$

The real part of $\langle f|g\rangle$ is the symmetrized covariance $C_\psi(x,p)$. Retaining it gives the stronger determinant relation.

Why Fourier Duality Forces the Tradeoff

Under the chosen transform, multiplication by $x$ in position space corresponds to differentiation in momentum space, while differentiation in position space corresponds to multiplication by $p$:

$$x\psi \quad\longleftrightarrow\quad i\hbar\frac{d\phi}{dp},$$

and

$$-i\hbar\frac{d\psi}{dx} \quad\longleftrightarrow\quad p\phi.$$

Localization of $\psi$ requires a broad superposition of Fourier components. Conversely, concentrating $\phi$ near one momentum makes the position-space phase vary almost like a plane wave over a long region.

This is not merely a qualitative analogy. If the wave-number transform is written using $k=p/\hbar$, the variance theorem is

$$\Delta x\,\Delta k \ge \frac12.$$

Multiplying by $\hbar$ gives the physical momentum relation.

Two unchirped Gaussian states with zero means, drawn in dimensionless coordinates $q=x/\ell$ and $k=p\ell/\hbar$. The solid state has $(\Delta q,\Delta k)=(1/2,1)$; the dashed state has $(1,1/2)$. Both have product $1/2$. Brackets span twice the marked standard deviation.

The general transform machinery is developed in Fourier Transform and Wave Packets.

Scaling Makes the Reciprocal Width Exact

Let $a>0$ and define a rescaled normalized state

$$\psi_a(x) = \frac1{\sqrt a} \psi\!\left(\frac{x}{a}\right).$$

Its momentum wavefunction is

$$\phi_a(p) = \sqrt a\,\phi(ap).$$

The means scale as $x_{0,a}=ax_0$ and $p_{0,a}=p_0/a$. After subtracting these means, direct substitution gives

$$\Delta_{\psi_a}x = a\,\Delta_\psi x,$$

and

$$\Delta_{\psi_a}p = \frac1a\,\Delta_\psi p.$$

Thus

$$\Delta_{\psi_a}x\, \Delta_{\psi_a}p = \Delta_\psi x\, \Delta_\psi p.$$

Narrowing the position profile by a factor $a<1$ broadens the momentum profile by exactly $1/a$. Scaling alone does not guarantee saturation; it preserves whatever uncertainty product the original shape had.

Equality Condition and the Gaussian

For nonzero variances, Robertson equality requires the centered fluctuation vectors to be proportional with a purely imaginary coefficient. Write

$$(x-x_0)\psi = -i\lambda (p-p_0)\psi, \qquad \lambda>0.$$

Using $p=-i\hbar d/dx$ gives the first-order equation

$$\frac{d\psi}{dx} = \left[ -\frac{x-x_0}{\lambda\hbar} + \frac{ip_0}{\hbar} \right] \psi.$$

Its normalizable solutions are Gaussians. Writing

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

one obtains

$$\psi(x) = \frac{e^{i\chi}} {(2\pi\sigma_x^2)^{1/4}} \times \exp\!\left[ -\frac{(x-x_0)^2}{4\sigma_x^2} \right] \times \exp\!\left[ \frac{ip_0(x-x_0)}{\hbar} \right],$$

where $\chi$ is an irrelevant global phase.

The probability density is

$$|\psi(x)|^2 = \frac1{\sqrt{2\pi\sigma_x^2}} \exp\!\left[ -\frac{(x-x_0)^2}{2\sigma_x^2} \right],$$

so

$$\Delta x=\sigma_x.$$

Fourier transformation gives a Gaussian centered at $p_0$ with

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

Hence

$$\Delta x\,\Delta p = \frac\hbar2.$$

Translations $x_0$, boosts $p_0$, and a global phase do not affect the uncertainty product. Up to those changes and the width parameter, the uncorrelated Gaussian is the normalized equality state on the line.

The packet’s free evolution and its relation to coherent states belong to Minimum-Uncertainty Wave Packets and Gaussian Wave Packets.

Correlated Gaussians and the Stronger Relation

Consider the chirped Gaussian

$$\psi_\eta(x) = \frac1{(2\pi\sigma_x^2)^{1/4}} \times \exp\!\left[ -\frac{1-i\eta}{4\sigma_x^2}(x-x_0)^2 +\frac{ip_0(x-x_0)}{\hbar} \right],$$

where $\eta\in\mathbb R$. Its position density is unchanged, but the quadratic phase correlates position and momentum. Direct differentiation gives

$$(p-p_0)\psi_\eta = \frac{\hbar(\eta+i)}{2\sigma_x^2} (x-x_0)\psi_\eta.$$

Therefore

$$(\Delta x)^2=\sigma_x^2,$$ $$(\Delta p)^2 = \frac{\hbar^2}{4\sigma_x^2} (1+\eta^2),$$

and

$$C(x,p)=\frac{\eta\hbar}{2}.$$

These quantities satisfy

$$(\Delta x)^2(\Delta p)^2 - C(x,p)^2 = \frac{\hbar^2}{4}.$$

Thus every $\psi_\eta$ saturates Robertson–Schrödinger, while only $\eta=0$ saturates the simpler product relation. A state can be a minimum uncertainty state for the determinant bound without minimizing $\Delta x\,\Delta p$ itself.

Physical Meaning of Narrowing a Packet

A small $\Delta x$ means that repeated ideal position measurements on identically prepared systems have a distribution concentrated near $x_0$ in the root-mean-square sense. It does not mean that the wavefunction has compact support or that every outcome lies in an interval of width $2\Delta x$.

The inequality gives

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

This concerns spread about $p_0$, not the size of $p_0$. A packet can have a large mean momentum and a small momentum spread, or zero mean momentum and a large spread.

For a free particle of mass $m$,

$$\langle T\rangle = \frac{\langle p^2\rangle}{2m} = \frac{p_0^2+(\Delta p)^2}{2m}.$$

Consequently,

$$\langle T\rangle \ge \frac{p_0^2}{2m} + \frac{\hbar^2}{8m(\Delta x)^2}.$$

Localization therefore carries a kinetic-energy cost. For an electron with $p_0=0$ and $\Delta x=1\,\text{Å}$, the lower-bound contribution is about $0.95\,\mathrm{eV}$.

Idealized Sharp States Are Not Counterexamples

Momentum plane waves

An ideal plane wave

$$\psi_p(x) \propto e^{ipx/\hbar}$$

has a sharp generalized momentum but is not square integrable on $\mathbb R$. It has no finite position variance. Normalized packets can approach a sharp momentum distribution only by spreading farther in position.

Position eigenkets

An ideal position eigenket has wavefunction proportional to

$$\delta(x-x_0).$$

It is a distribution rather than a vector in $L^2(\mathbb R)$. Its Fourier amplitude has constant magnitude, so it does not define a normalizable finite-variance momentum distribution.

Infinite-variance states

Some normalized states have heavy tails and infinite $\Delta x$ or $\Delta p$. The variance product is then not a useful finite diagnostic. This does not violate the theorem; it lies outside the finite-moment setting in which the numerical product is defined.

Higher Dimensions and Directional Pairs

For Cartesian components,

$$[x_i,p_j] = i\hbar\delta_{ij}I.$$

Therefore

$$\Delta x_i\,\Delta p_i \ge \frac\hbar2$$

for each matched component. Distinct components commute:

$$[x_i,p_j]=0 \qquad(i\ne j),$$

so this commutator supplies no positive lower bound for their product.

For real vectors $\mathbf u$ and $\mathbf v$, define

$$X_{\mathbf u} = \mathbf u\cdot\mathbf x, \qquad P_{\mathbf v} = \mathbf v\cdot\mathbf p.$$

Then

$$[X_{\mathbf u},P_{\mathbf v}] = i\hbar (\mathbf u\cdot\mathbf v)I,$$

and

$$\Delta X_{\mathbf u}\, \Delta P_{\mathbf v} \ge \frac\hbar2 |\mathbf u\cdot\mathbf v|.$$

The bound depends only on the overlap of the two directions.

Preparation Uncertainty Is Not Error–Disturbance

The theorem can be tested by preparing many copies of $\rho$. Position can be measured on one subensemble and momentum on another. Their empirical widths, after correcting for known detector effects, estimate $\Delta x$ and $\Delta p$.

No first measurement is required to disturb a second measurement in this protocol. A sequential error–disturbance experiment asks different questions about an apparatus, its resolution, and its state-update map.

Heisenberg’s microscope was historically important for motivating operational limits, but the Kennard–Robertson variance inequality is a theorem about prepared-state distributions. The historical development belongs to Uncertainty: Historical Origin.

Domain and Boundary Conditions

Position and momentum are unbounded. A sufficient whole-line setting for the vector proof requires

$$x\psi\in L^2(\mathbb R),$$ $$\psi'\in L^2(\mathbb R),$$

and a representative for which the integration-by-parts boundary term

$$\left[ (x-x_0)|\psi(x)|^2 \right]_{-\infty}^{\infty}$$

vanishes. Schwartz wavefunctions satisfy these conditions comfortably.

The vector proof needs the common operator domain $\mathcal D(x)\cap\mathcal D(p)$, equivalently the common quadratic-form domain of $x^2$ and $p^2$. The vanishing boundary term is automatic for the Schwartz calculation; the inequality extends to this full domain by approximation. Writing

$$\langle\psi|[x,p]|\psi\rangle$$

as an ordinary operator expectation can require stronger product-domain conditions. The weak overlap proof avoids assuming that both $xp\psi$ and $px\psi$ exist as Hilbert-space vectors.

Boundary conditions can alter the story:

• on a finite interval, self-adjoint momentum realizations depend on boundary conditions, and multiplication by $x$ may not preserve their domains;
• on a circle, a globally defined angle observable is not an ordinary Cartesian position operator;
• on a lattice or finite cyclic space, position and quasimomentum have discrete or periodic spectra and obey modified uncertainty relations;
• with gauge fields, canonical momentum and kinetic momentum are different observables.

State the configuration space and the actual observables before using the bound. Regular kinetic momentum also satisfies $[x_i,\Pi_i]=i\hbar I$ on the corresponding common domain; the distinction from canonical momentum matters especially for momentum-component commutators.

Common Pitfalls

Narrow position spread requires broad momentum spread, not large mean momentum. The kinetic-energy bound separates $p_0^2$ from $(\Delta p)^2$ for this reason.

A standard deviation is not a support interval. Even a Gaussian assigns nonzero probability outside $x_0\pm\Delta x$. The theorem constrains second moments, not the largest possible outcomes.

A chirped Gaussian need not minimize the variance product. Its quadratic phase increases $\Delta p$ and introduces covariance while leaving $|\psi(x)|^2$ unchanged. The determinant relation identifies the saturation correctly.

Exercises

Exercise 1: State-independent Robertson bound

Starting from Robertson and $[x,p]=i\hbar I$, derive the position–momentum uncertainty relation. Identify exactly where normalization is used.

Solution

Robertson gives

$$\Delta_\psi x\,\Delta_\psi p \ge \frac12 \left\lvert \langle[x,p]\rangle_\psi \right\rvert.$$

The canonical relation implies

$$\langle[x,p]\rangle_\psi = i\hbar\langle I\rangle_\psi.$$

Normalization is used in

$$\langle I\rangle_\psi = \langle\psi|\psi\rangle =1.$$

Hence

$$\Delta_\psi x\,\Delta_\psi p \ge \frac12|i\hbar| = \frac\hbar2.$$

Exercise 2: The integration-by-parts constant

Assume $\psi$ is normalized, differentiable, and decays sufficiently rapidly. Show that

$$\operatorname{Re} \int_{-\infty}^{\infty} (x-x_0)\psi^*(x)\psi'(x)\,dx = -\frac12.$$

Explain how this fixes the constant $\hbar/2$.

Solution

Let the integral be $I$. Then

$$\begin{aligned} 2\operatorname{Re}I &= I+I^*\\ &= \int_{-\infty}^{\infty} (x-x_0) \frac{d}{dx}|\psi(x)|^2\,dx. \end{aligned}$$

Integration by parts gives

$$\begin{aligned} 2\operatorname{Re}I &= \left[ (x-x_0)|\psi(x)|^2 \right]_{-\infty}^{\infty} - \int_{-\infty}^{\infty}|\psi(x)|^2\,dx \\ &=-1. \end{aligned}$$

Thus $\operatorname{Re}I=-1/2$. The centered position–momentum overlap is $-i\hbar I$, whose imaginary part is $\hbar/2$. Cauchy–Schwarz must therefore bound its modulus by at least $\hbar/2$.

Exercise 3: Gaussian saturation

For

$$\psi(x) = \frac1{(2\pi\sigma^2)^{1/4}} \exp\!\left[ -\frac{(x-x_0)^2}{4\sigma^2} \right] \times \exp\!\left[ \frac{ip_0(x-x_0)}{\hbar} \right],$$

compute $\Delta x$ and $\Delta p$ without performing the full Fourier transform.

Solution

The density is a normal distribution of variance $\sigma^2$, so

$$\Delta x=\sigma.$$

Differentiate the wavefunction:

$$\psi' = \left[ -\frac{x-x_0}{2\sigma^2} + \frac{ip_0}{\hbar} \right]\psi.$$

Therefore

$$(p-p_0)\psi = \frac{i\hbar(x-x_0)}{2\sigma^2}\psi.$$

Taking the norm gives

$$\begin{aligned} (\Delta p)^2 &= \frac{\hbar^2}{4\sigma^4} \langle(x-x_0)^2\rangle \\ &= \frac{\hbar^2}{4\sigma^2}. \end{aligned}$$

Thus $\Delta p=\hbar/(2\sigma)$ and the product is $\hbar/2$.

Exercise 4: Fourier scaling

For $a>0$, let

$$\psi_a(x) = a^{-1/2}\psi(x/a).$$

Derive $\phi_a(p)=a^{1/2}\phi(ap)$ and show how the two variances scale.

Solution

Substitute $x=ay$ into the Fourier transform:

$$\begin{aligned} \phi_a(p) &= \frac1{\sqrt{2\pi\hbar}} \times \int e^{-ipx/\hbar} a^{-1/2}\psi(x/a)\,dx \\ &= a^{1/2} \frac1{\sqrt{2\pi\hbar}} \times \int e^{-i(ap)y/\hbar}\psi(y)\,dy \\ &= a^{1/2}\phi(ap). \end{aligned}$$

Changing variables in the second moments gives

$$\Delta_{\psi_a}x=a\Delta_\psi x,$$

and

$$\Delta_{\psi_a}p=\frac1a\Delta_\psi p.$$

Their product is unchanged.

Exercise 5: A chirped Gaussian

For the state $\psi_\eta$ defined in the text, verify

$$(p-p_0)\psi_\eta = \frac{\hbar(\eta+i)}{2\sigma_x^2} (x-x_0)\psi_\eta,$$

then compute $\Delta p$ and $C(x,p)$.

Solution

Differentiation gives

$$\psi_\eta' = \left[ -\frac{1-i\eta}{2\sigma_x^2}(x-x_0) + \frac{ip_0}{\hbar} \right] \psi_\eta.$$

Applying $-i\hbar d/dx-p_0$ yields the stated relation. Its squared norm is

$$\begin{aligned} (\Delta p)^2 &= \frac{\hbar^2|\eta+i|^2}{4\sigma_x^4} \langle(x-x_0)^2\rangle \\ &= \frac{\hbar^2(1+\eta^2)}{4\sigma_x^2}. \end{aligned}$$

The centered overlap is

$$\begin{aligned} \langle(x-x_0)(p-p_0)\rangle &= \frac{\hbar(\eta+i)}{2\sigma_x^2} \times \langle(x-x_0)^2\rangle \\ &= \frac\hbar2(\eta+i). \end{aligned}$$

Its real part is the symmetrized covariance:

$$C(x,p)=\frac{\eta\hbar}{2}.$$

Substitution verifies saturation of the determinant relation.

Exercise 6: Localization energy scale

For a free particle with $p_0=0$, show that

$$\langle T\rangle \ge \frac{\hbar^2}{8m(\Delta x)^2}.$$

Estimate the bound for an electron localized to $\Delta x=1\,\text{Å}$.

Solution

Since $p_0=0$,

$$\langle T\rangle = \frac{(\Delta p)^2}{2m}.$$

The uncertainty relation gives

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

so

$$\langle T\rangle \ge \frac{\hbar^2}{8m(\Delta x)^2}.$$

Using $m_e\simeq9.11\times10^{-31}\,\mathrm{kg}$ and $\Delta x=10^{-10}\,\mathrm m$ gives

$$\langle T\rangle \gtrsim 1.53\times10^{-19}\,\mathrm J \simeq0.95\,\mathrm{eV}.$$

Exercise 7: Directional uncertainty in three dimensions

Let $\mathbf u$ and $\mathbf v$ be unit vectors. Derive

$$\Delta(\mathbf u\cdot\mathbf x)\, \Delta(\mathbf v\cdot\mathbf p) \ge \frac\hbar2|\mathbf u\cdot\mathbf v|.$$

What does Robertson give when the directions are orthogonal?

Solution

Using $[x_i,p_j]=i\hbar\delta_{ij}I$,

$$\begin{aligned} [\mathbf u\cdot\mathbf x, \mathbf v\cdot\mathbf p] &= \sum_{i,j}u_iv_j[x_i,p_j] \\ &= i\hbar \sum_i u_iv_i I \\ &= i\hbar(\mathbf u\cdot\mathbf v)I. \end{aligned}$$

Robertson gives the stated bound. If the directions are orthogonal, the commutator vanishes and this particular relation gives only the trivial lower bound zero. It does not require either variance to vanish.

Exercise 8: Why the boundary term matters

Repeat the integration-by-parts step on an interval $[0,L]$. Show that

$$\begin{aligned} 2\operatorname{Re}I &= \left[ (x-x_0)|\psi(x)|^2 \right]_{0}^{L} -1. \end{aligned}$$

Explain why the whole-line proof cannot simply be copied to arbitrary boundary conditions. Check the constant wavefunction $\psi(x)=1/\sqrt L$ with periodic momentum boundary conditions.

Solution

The calculation is identical except that the endpoints are finite:

$$\begin{aligned} I+I^* &= \int_0^L (x-x_0) \frac{d}{dx}|\psi(x)|^2\,dx \\ &= \left[ (x-x_0)|\psi(x)|^2 \right]_0^L - \int_0^L|\psi(x)|^2\,dx. \end{aligned}$$

Normalization makes the last integral one, yielding the stated result. The endpoint contribution need not vanish for periodic or other self-adjoint boundary conditions. In addition, multiplication by $x$ may not preserve the domain of the chosen momentum operator. For $\psi=1/\sqrt L$, $x_0=L/2$, $\Delta x=L/\sqrt{12}$, and $p\psi=0$, so $\Delta p=0$. The boundary contribution is exactly one, giving $2\operatorname{Re}I=0$, not $-1$. Multiplication by $x$ spoils the periodic endpoint condition, so the formal $[x,p]=i\hbar I$ cannot be applied to this vector as a product identity. There is no contradiction with the real-line theorem.

References

• G. B. Folland and A. Sitaram, “The Uncertainty Principle: A Mathematical Survey,” Journal of Fourier Analysis and Applications 3, 207–238 (1997). DOI.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• E. H. Kennard, “Zur Quantenmechanik einfacher Bewegungstypen,” Zeitschrift für Physik 44, 326–352 (1927). DOI.