Uncertainty Relations

Purpose

Uncertainty relations constrain the spreads of the Born distributions for two observables in the same prepared state. For self-adjoint $A$ and $B$, the Robertson relation is

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

The Robertson–Schrödinger relation is stronger because it also retains the symmetrized covariance. For the canonical pair $X,P$ on the line, Robertson reduces to the familiar position–momentum or Heisenberg relation

$$\Delta X\,\Delta P\geq\frac{\hbar}{2}.$$

These are preparation uncertainty relations. They concern the statistical spread of ideal measurement outcomes. They are not, by themselves, formulas for detector error, estimator uncertainty, or disturbance caused by a prior measurement.

The full derivation and geometric interpretation are at General Uncertainty Relations.

Definitions

For a density operator $\rho$, write

$$\langle A\rangle_\rho \equiv \operatorname{Tr}(\rho A).$$

For a pure state $\rho=\lvert\psi\rangle\langle\psi\rvert$, this becomes $\langle A\rangle_\psi=\langle\psi\rvert A\lvert\psi\rangle$.

Define the centered observables

$$\delta_\rho A \equiv A-\langle A\rangle_\rho I, \qquad \delta_\rho B \equiv B-\langle B\rangle_\rho I.$$

Their variances and standard deviations are

$$(\Delta_\rho A)^2 = \left\langle (\delta_\rho A)^2 \right\rangle_\rho, \qquad \Delta_\rho A\geq0,$$

and likewise for $B$. The real symmetrized covariance is

$$C_\rho(A,B) \equiv \frac12 \left\langle \left\lbrace \delta_\rho A,\delta_\rho B \right\rbrace \right\rangle_\rho.$$

For self-adjoint $A$ and $B$, both $C_\rho(A,B)$ and

$$D_\rho(A,B) \equiv \frac1{2i} \langle[A,B]\rangle_\rho$$

are real. The centered ordered product separates as

$$\left\langle \delta_\rho A\,\delta_\rho B \right\rangle_\rho = C_\rho(A,B)+iD_\rho(A,B).$$

At a glance

Relation	Formula
Centered-overlap bound	$(\Delta_\rho A)^2(\Delta_\rho B)^2\geq\left\lvert\langle\delta_\rho A\,\delta_\rho B\rangle_\rho\right\rvert^2$
Robertson–Schrödinger	$(\Delta_\rho A)^2(\Delta_\rho B)^2\geq C_\rho(A,B)^2+\frac14\left\lvert\langle[A,B]\rangle_\rho\right\rvert^2$
Determinant form	$(\Delta_\rho A)^2(\Delta_\rho B)^2-C_\rho(A,B)^2\geq D_\rho(A,B)^2$
Robertson	$\Delta_\rho A\,\Delta_\rho B\geq\frac12\left\lvert\langle[A,B]\rangle_\rho\right\rvert$
Canonical position–momentum	$[X,P]=i\hbar I\ \Longrightarrow\ \Delta X\,\Delta P\geq\hbar/2$
Angular momentum	$\Delta J_x\,\Delta J_y\geq(\hbar/2)\left\lvert\langle J_z\rangle\right\rvert$
Pure-state equality for the stronger relation	$\delta_\psi A\lvert\psi\rangle=\lambda\,\delta_\psi B\lvert\psi\rangle$ for some $\lambda\in\mathbb C$
Additional condition for Robertson equality	$C_\psi(A,B)=0$

All expectation values and standard deviations in a row refer to the same state. A state-independent commutator produces a state-independent lower bound; a general commutator does not.

Robertson–Schrödinger relation

Cauchy–Schwarz applied to the centered fluctuation vectors or to their density-operator analogues gives

$$\begin{aligned} (\Delta_\rho A)^2(\Delta_\rho B)^2 &\geq \left\lvert \left\langle \delta_\rho A\,\delta_\rho B \right\rangle_\rho \right\rvert^2 \\ &= C_\rho(A,B)^2 +D_\rho(A,B)^2. \end{aligned}$$

Therefore

$$\begin{aligned} (\Delta_\rho A)^2(\Delta_\rho B)^2 &\geq C_\rho(A,B)^2 \\ &\quad+ \frac14 \left\lvert \langle[A,B]\rangle_\rho \right\rvert^2. \end{aligned}$$

The covariance term is not an optional correction. It is the real part of the same centered overlap whose imaginary part contains the commutator.

In covariance-matrix notation,

$$\Gamma_\rho(A,B) = \begin{pmatrix} (\Delta_\rho A)^2&C_\rho(A,B)\\ C_\rho(A,B)&(\Delta_\rho B)^2 \end{pmatrix},$$

and the relation reads

$$\det\Gamma_\rho(A,B) \geq D_\rho(A,B)^2.$$

For canonical continuous variables, the determinant measures the squared area scale of the uncertainty ellipse. A nonzero covariance tilts that ellipse.

Robertson relation

Dropping the nonnegative covariance term gives

$$(\Delta_\rho A)^2(\Delta_\rho B)^2 \geq \frac14 \left\lvert \langle[A,B]\rangle_\rho \right\rvert^2.$$

Since standard deviations are nonnegative,

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

Robertson is often the fastest useful bound, but it may be weaker than the full relation. In particular,

$$\langle[A,B]\rangle_\rho=0$$

can hold even when $[A,B]\ne0$ as an operator. Robertson then gives only the trivial lower bound zero. This does not make $A$ and $B$ compatible and does not guarantee a common eigenbasis.

Position–momentum specialization

On the real line, with the standard canonical commutation relation

$$[X,P]=i\hbar I,$$

normalization gives

$$\langle[X,P]\rangle_\rho=i\hbar.$$

Robertson therefore yields

$$\Delta_\rho X\,\Delta_\rho P \geq \frac{\hbar}{2}.$$

The lower bound is state independent because the commutator is proportional to the identity. The stronger form is

$$(\Delta_\rho X)^2(\Delta_\rho P)^2 -C_\rho(X,P)^2 \geq \frac{\hbar^2}{4}.$$

An uncorrelated minimum-uncertainty Gaussian has

$$C_\rho(X,P)=0, \qquad \Delta X\,\Delta P=\frac{\hbar}{2}.$$

A chirped or freely evolved pure Gaussian may instead satisfy

$$(\Delta X)^2(\Delta P)^2-C_\rho(X,P)^2 = \frac{\hbar^2}{4}$$

while having $\Delta X\,\Delta P>\hbar/2$. It saturates Robertson–Schrödinger but not Robertson.

The whole-line formula assumes the domains required for $XP$, $PX$, and the second moments. On a finite interval or circle, boundary conditions and the domain of the momentum operator require separate treatment; one cannot simply reuse the whole-line integration-by-parts argument.

Angular-momentum specialization

From

$$[J_x,J_y]=i\hbar J_z,$$

one obtains

$$\Delta_\rho J_x\,\Delta_\rho J_y \geq \frac{\hbar}{2} \left\lvert \langle J_z\rangle_\rho \right\rvert.$$

Cyclic permutations give the other component pairs. Unlike the canonical $X,P$ bound, the right side is state dependent. For a state with $\langle J_z\rangle=0$, this particular Robertson bound is trivial even though $J_x$ and $J_y$ do not commute.

In the state $\lvert j,m\rangle$,

$$(\Delta J_x)^2 = (\Delta J_y)^2 = \frac{\hbar^2}{2} \left[ j(j+1)-m^2 \right].$$

The Robertson inequality becomes

$$j(j+1)-m^2\geq\lvert m\rvert.$$

The extremal states $m=\pm j$ saturate it; generic magnetic sublevels do not.

Two-level example

Take

$$A=\sigma_x, \qquad B=\sigma_z,$$

and represent a qubit state by

$$\rho = \frac12 \left( I+\mathbf r\cdot\boldsymbol\sigma \right), \qquad \lVert\mathbf r\rVert\leq1.$$

The variances are

$$(\Delta_\rho\sigma_x)^2=1-r_x^2, \qquad (\Delta_\rho\sigma_z)^2=1-r_z^2.$$

Using $\{\sigma_x,\sigma_z\}=0$ and $[\sigma_x,\sigma_z]=-2i\sigma_y$ gives

$$C_\rho(\sigma_x,\sigma_z)=-r_xr_z, \qquad D_\rho(\sigma_x,\sigma_z)=-r_y.$$

Robertson–Schrödinger becomes

$$(1-r_x^2)(1-r_z^2) \geq r_x^2r_z^2+r_y^2.$$

After cancellation, this is exactly

$$1-\lVert\mathbf r\rVert^2\geq0.$$

Every pure qubit state saturates the stronger relation for this pair, whereas Robertson alone is generally not saturated when $r_xr_z\ne0$. This example shows why the covariance term can carry essential information.

Equality conditions

Pure states

For nonzero variances, equality in Robertson–Schrödinger holds precisely when the two centered fluctuation vectors are linearly dependent:

$$\delta_\psi A\lvert\psi\rangle = \lambda\, \delta_\psi B\lvert\psi\rangle$$

for some $\lambda\in\mathbb C$. Equivalently,

$$\left( \delta_\psi A-\lambda\delta_\psi B \right) \lvert\psi\rangle=0.$$

Taking norms gives

$$\lvert\lambda\rvert = \frac{\Delta_\psi A}{\Delta_\psi B}.$$

To saturate Robertson itself, the state must also satisfy

$$C_\psi(A,B)=0.$$

For nonzero variances, the proportionality constant can then be chosen purely imaginary:

$$\delta_\psi A\lvert\psi\rangle = i\kappa\, \delta_\psi B\lvert\psi\rangle, \qquad \kappa\in\mathbb R.$$

The sign and magnitude of $\kappa$ depend on the pair and state. Normalizable solutions need not exist for every pair of observables and every desired mean.

Mixed states

The corresponding Hilbert–Schmidt equality condition is

$$\delta_\rho A\sqrt\rho = \lambda\, \delta_\rho B\sqrt\rho.$$

It must hold on the support of $\rho$. For a full-rank mixed state this is usually much more restrictive than the pure-state condition.

Zero variance

If $\Delta_\rho A=0$, then

$$\delta_\rho A\sqrt\rho=0.$$

Consequently,

$$C_\rho(A,B)=0, \qquad \langle[A,B]\rangle_\rho=0,$$

and both inequalities are saturated trivially. For a pure state, zero variance means that the state is an eigenstate of $A$. It does not imply that $A$ commutes with $B$.

Multiple observables

For self-adjoint observables $X_1,\ldots,X_n$, define

$$\Gamma_{jk} = \frac12 \left\langle \left\lbrace \delta_\rho X_j,\delta_\rho X_k \right\rbrace \right\rangle_\rho$$

and

$$\Omega_{jk} = \frac1{2i} \langle[X_j,X_k]\rangle_\rho.$$

The Gram-matrix uncertainty condition is

$$\Gamma+i\Omega\geq0.$$

Each two-observable Robertson–Schrödinger relation follows from a $2\times2$ principal minor. For three or more observables, the full positive semidefinite matrix can impose constraints not captured by checking pairs independently.

For canonical quadratures $R_j$ satisfying

$$[R_j,R_k]=i\hbar J_{jk}I,$$

the condition becomes

$$\Gamma+\frac{i\hbar}{2}J\geq0.$$

This is the standard covariance-matrix form used for Gaussian states and continuous-variable systems.

Assumptions and domains

The finite-dimensional statement is straightforward for self-adjoint matrices. For unbounded operators, the displayed formulas require more:

• the state must lie in the domains needed for the first and second moments;
• the products $\delta A\,\delta B$ and $\delta B\,\delta A$ must be meaningful in the expectation values used;
• the formal commutator must agree with the operator expression on the relevant vectors;
• boundary terms must respect the chosen self-adjoint domains.

Merely writing a formal differential commutator is not sufficient. A domain failure can invalidate an intermediate derivation even when a related uncertainty statement remains true by another argument.

If a variance diverges, the usual product inequality may be formally true but uninformative. Alternative uncertainty measures, such as entropic uncertainties, can remain meaningful, but they are distinct results with different assumptions.

What the formulas do not imply

No hidden exact values

A small product of standard deviations does not reveal underlying simultaneous values blurred by imperfect apparatus. It summarizes two Born distributions associated with one state.

No automatic error–disturbance law

The symbols $\Delta_\rho A$ and $\Delta_\rho B$ are intrinsic standard deviations. Measurement error and disturbance require an explicit measurement model and different definitions. Robertson should not be relabeled as a universal error–disturbance inequality.

No universal energy–time operator pair

In ordinary nonrelativistic quantum mechanics, time is usually an external parameter rather than a self-adjoint operator satisfying $[T,H]=i\hbar I$ on the physical Hilbert space. Energy–time relations therefore come in several forms, such as characteristic evolution-time bounds, and their assumptions must be stated separately. See Energy–Time Uncertainty.

No compatibility test from one expectation value

The condition

$$\langle[A,B]\rangle_\rho=0$$

for one state does not imply $[A,B]=0$. Compatibility is an operator-level question, while the Robertson bound samples one state-dependent expectation.

Calculation workflow

1. Verify that the state is normalized and the needed second moments exist.
2. Compute $\langle A\rangle$, $\langle A^2\rangle$, and the analogous quantities for $B$.
3. Form the variances using centered operators or $\langle A^2\rangle-\langle A\rangle^2$.
4. Compute $\langle[A,B]\rangle$ without replacing the commutator by a number unless that operator identity is valid.
5. Compute $C_\rho(A,B)$ when Robertson is weak or when testing saturation.
6. Check units, nonnegativity, and special limits.
7. Use the linear-dependence condition only after separating Robertson–Schrödinger equality from Robertson equality.

For numerical work, small negative variances or covariance determinants can arise from roundoff. Use Hermitian symmetrization and a tolerance scaled to the operator norms before interpreting such values as physical violations.

Common mistakes

• Treating $\Delta A$ as an apparatus error bar instead of a Born-distribution standard deviation.
• Omitting the state label when the lower bound is state dependent.
• Inferring $[A,B]=0$ from $\langle[A,B]\rangle_\rho=0$ in one state.
• Calling every state that saturates Robertson–Schrödinger a Robertson minimum-uncertainty state.
• Forgetting the covariance term for correlated Gaussian states.
• Assuming all Gaussian wave packets satisfy $\Delta X\,\Delta P=\hbar/2$ at all times.
• Using the canonical $X,P$ bound on a circle or finite interval without checking operator domains and boundary conditions.
• Applying the same operator proof to energy and time without identifying a valid time observable.
• Dividing by a variance while analyzing a zero-variance equality case.
• Confusing preparation uncertainty with measurement error or disturbance.

Cross-links

• General Uncertainty Relations
• Position–Momentum Uncertainty
• Energy–Time Uncertainty
• Commutators
• Correlations and Covariance
• Variance and Standard Deviation
• Minimum-Uncertainty Wave Packets
• Canonical Commutation Relations

References

• H. P. Robertson, “The uncertainty principle,” Physical Review 34, 163–164 (1929).
• E. Schrödinger, “Zum Heisenbergschen Unschärfeprinzip,” Sitzungsberichte der Preussischen Akademie der Wissenschaften, Physikalisch-mathematische Klasse, 296–303 (1930).
• E. H. Kennard, “Zur Quantenmechanik einfacher Bewegungstypen,” Zeitschrift für Physik 44, 326–352 (1927).
• J. Hilgevoord and J. Uffink, “The uncertainty principle,” in The Stanford Encyclopedia of Philosophy, substantive revision 2016.
• P. Busch, P. Lahti, and R. F. Werner, “Colloquium: Quantum root-mean-square error and measurement uncertainty relations,” Reviews of Modern Physics 86, 1261–1281 (2014).
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• A. S. Holevo, Probabilistic and Statistical Aspects of Quantum Theory, 2nd English ed., Edizioni della Normale, 2011.

Exercises

1. A spin-$1/2$ state is an eigenstate of $\sigma_z$. Evaluate $\Delta\sigma_x$, $\Delta\sigma_y$, and the Robertson lower bound for the pair $\sigma_x,\sigma_y$.

Solution

In a $\sigma_z$ eigenstate with eigenvalue $s=\pm1$,

$$\langle\sigma_x\rangle = \langle\sigma_y\rangle = 0,$$

and $\sigma_x^2=\sigma_y^2=I$. Therefore

$$\Delta\sigma_x=\Delta\sigma_y=1.$$

Since

$$[\sigma_x,\sigma_y]=2i\sigma_z,$$

Robertson gives

$$\Delta\sigma_x\,\Delta\sigma_y \geq \frac12 \left\lvert 2i\langle\sigma_z\rangle \right\rvert = 1.$$

The product is exactly one, so the relation is saturated.

2. Show that the Robertson–Schrödinger inequality is invariant under the replacements $A'=aA+bI$ and $B'=cB+dI$ for real $a,b,c,d$, apart from the common scale factor $a^2c^2$.

Solution

Centering removes the additive constants:

$$\delta A'=a\,\delta A, \qquad \delta B'=c\,\delta B.$$

Hence

$$(\Delta A')^2=a^2(\Delta A)^2, \qquad (\Delta B')^2=c^2(\Delta B)^2,$$

while

$$C(A',B')=ac\,C(A,B)$$

and

$$\langle[A',B']\rangle = ac\,\langle[A,B]\rangle.$$

Every term in the squared relation therefore acquires the same factor $a^2c^2$. If either scale vanishes, both sides vanish consistently.

3. Let a pure state satisfy

$$\delta_\psi A\lvert\psi\rangle = \lambda\, \delta_\psi B\lvert\psi\rangle.$$

Explain why this saturates Robertson–Schrödinger, and identify the additional condition needed to saturate Robertson.

Solution

The two fluctuation vectors are linearly dependent, so equality holds in the Cauchy–Schwarz inequality from which Robertson–Schrödinger is derived. Therefore

$$(\Delta_\psi A)^2(\Delta_\psi B)^2 = C_\psi(A,B)^2 +\frac14 \left\lvert \langle[A,B]\rangle_\psi \right\rvert^2.$$

Robertson discards $C_\psi(A,B)^2$. It is saturated only if this discarded term vanishes:

$$C_\psi(A,B)=0.$$

For nonzero variances this is equivalent to choosing the proportionality constant purely imaginary.

4. For a normalized state with $\langle P\rangle=0$, use $\Delta X\,\Delta P\geq\hbar/2$ to bound the mean kinetic energy $\langle P^2/(2m)\rangle$ in terms of $\Delta X$.

Solution

Because $\langle P\rangle=0$,

$$\langle P^2\rangle=(\Delta P)^2.$$

The uncertainty relation implies

$$(\Delta P)^2 \geq \frac{\hbar^2}{4(\Delta X)^2}.$$

Therefore

$$\left\langle \frac{P^2}{2m} \right\rangle \geq \frac{\hbar^2}{8m(\Delta X)^2}.$$

Localization carries a kinetic-energy cost proportional to $1/(\Delta X)^2$.