The Generalised Uncertainty Relation
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You should be comfortable with the definition of the standard deviation of an operator, the concept of the commutator \([ \hat{A}, \hat{B} ]\), and the properties of the inner product in Hilbert space.
Why this matters
The uncertainty principle is often misquoted as a limitation of measurement technology, but it is actually a fundamental property of the state vector itself.
In classical mechanics, we can in principle know the position and momentum of a particle to arbitrary precision simultaneously. In quantum mechanics, this is impossible not because our instruments are too crude, but because the mathematical objects representing these quantities do not commute. The Generalised Uncertainty Relation (GUR) provides the rigorous lower bound on the product of the uncertainties of any two observables, linking the spread of measurement outcomes directly to the algebraic structure of the operators. This result is central to understanding why quantum systems behave as they do, from the stability of atoms to the limits of signal processing.
The Intuition Behind the Bound
Imagine you are trying to describe a wave packet. If the wave is very localized in space (small position uncertainty), it must be composed of many different wavelengths (large momentum uncertainty). Conversely, a pure sine wave has a perfectly defined wavelength (zero momentum uncertainty) but is spread out infinitely in space (infinite position uncertainty). This is a Fourier transform property of waves. In quantum mechanics, position and momentum are not just coordinates; they are operators that act on the state vector. When two operators do not commute, no state can give sharp values of both; this is a statement about the spread of outcomes in one prepared state, not about one measurement disturbing the next. The Generalised Uncertainty Relation quantifies this trade-off. It tells us that the product of the "widths" of the probability distributions for two observables cannot be smaller than half the magnitude of the expectation value of their commutator. If the commutator is zero, the observables are compatible, and their uncertainties can both be zero. If the commutator is non-zero, there is a fundamental limit to how precisely we can define both simultaneously.
Mathematical Formulation
Let \(\hat{A}\) and \(\hat{B}\) be two Hermitian operators representing observables. The uncertainty (standard deviation) of \(\hat{A}\) in a state \(|\psi\rangle\) is defined as \(\Delta A = \sqrt{\langle \psi | (\hat{A} - \langle A \rangle)^2 | \psi \rangle}\). Similarly for \(\hat{B}\). We define the deviation operators \(\hat{a} = \hat{A} - \langle A \rangle\) and \(\hat{b} = \hat{B} - \langle B \rangle\). Note that \(\langle \psi | \hat{a} | \psi \rangle = 0\) and \(\langle \psi | \hat{b} | \psi \rangle = 0\). The expectation value of the product of these deviation operators is \(\langle \psi | \hat{a} \hat{b} | \psi \rangle\). This complex number can be decomposed into its real and imaginary parts. The real part is related to the covariance, and the imaginary part is related to the commutator. Specifically, \(\langle \psi | \hat{a} \hat{b} | \psi \rangle = \frac{1}{2} \langle \psi | \{ \hat{a}, \hat{b} \} | \psi \rangle + \frac{1}{2} \langle \psi | [ \hat{a}, \hat{b} ] | \psi \rangle\). Since \(\hat{a}\) and \(\hat{b}\) are Hermitian, the anticommutator \(\{ \hat{a}, \hat{b} \}\) is Hermitian, and the commutator \([ \hat{a}, \hat{b} ]\) is anti-Hermitian. The expectation value of an anti-Hermitian operator is purely imaginary. Thus, \(\text{Im} \langle \psi | \hat{a} \hat{b} | \psi \rangle = \frac{1}{2i} \langle \psi | [ \hat{a}, \hat{b} ] | \psi \rangle\), a real number because \( \langle [\hat{a}, \hat{b}] \rangle \) is purely imaginary. Note that \([ \hat{a}, \hat{b} ] = [ \hat{A}, \hat{B} ]\) because the constants \(\langle A \rangle\) and \(\langle B \rangle\) commute with everything.
Using the Cauchy-Schwarz inequality on the vectors \(\hat{a}|\psi\rangle\) and \(\hat{b}|\psi\rangle\), we have \(\langle \psi | \hat{a}^2 | \psi \rangle \langle \psi | \hat{b}^2 | \psi \rangle \geq |\langle \psi | \hat{a} \hat{b} | \psi \rangle|^2\). The left-hand side is \((\Delta A)^2 (\Delta B)^2\). The right-hand side is the modulus squared of the complex number \(\langle \psi | \hat{a} \hat{b} | \psi \rangle\). For any complex number \(z = x + iy\), \(|z|^2 = x^2 + y^2 \geq y^2\). Therefore, \((\Delta A)^2 (\Delta B)^2 \geq (\text{Im} \langle \psi | \hat{a} \hat{b} | \psi \rangle)^2\). Substituting the expression for the imaginary part, we get \((\Delta A)^2 (\Delta B)^2 \geq \left( \frac{1}{2i} \langle \psi | [ \hat{A}, \hat{B} ] | \psi \rangle \right)^2 = \frac{1}{4} \left| \langle [ \hat{A}, \hat{B} ] \rangle \right|^2\). Taking the square root of both sides yields the Generalised Uncertainty Relation.
Application to Position and Momentum
The most famous application is the Heisenberg Uncertainty Principle for position \(x\) and momentum \(p\). The canonical commutation relation is \([ \hat{x}, \hat{p} ] = i\hbar\). This is a constant, independent of the state. Therefore, the expectation value is simply \(i\hbar\). The magnitude is \(\hbar\). Substituting this into the GUR, we get \(\Delta x \cdot \Delta p \geq \frac{1}{2} \hbar\). This result holds for any quantum state. It is not a statement about the disturbance caused by measurement, but about the intrinsic spread of the wavefunction. If a particle is in a momentum eigenstate, \(\Delta p = 0\), which implies \(\Delta x \to \infty\). If a particle is localized in a small region, \(\Delta x\) is small, which forces \(\Delta p\) to be large. This has profound implications for bound states, such as the electron in a hydrogen atom. If the electron were to collapse into the nucleus, \(\Delta x\) would become very small, causing \(\Delta p\) (and thus the kinetic energy) to become very large, preventing the collapse.
Saturation and Minimum Uncertainty States
The inequality becomes an equality when the vectors \(\hat{a}|\psi\rangle\) and \(\hat{b}|\psi\rangle\) are linearly dependent and, in addition, the real (covariance) part of \( \langle \hat{a}\hat{b} \rangle \) vanishes; together these require \( \hat{b}|\psi\rangle = i\lambda\, \hat{a}|\psi\rangle \) with \( \lambda \) real. Such states are called minimum uncertainty states. For position and momentum, the minimum uncertainty states are Gaussian wave packets. These states have the property that \(\Delta x \cdot \Delta p = \frac{\hbar}{2}\). They are the closest quantum analogues to classical particles. Under free evolution, however, a Gaussian spreads and stops being a minimum uncertainty state; in a harmonic oscillator the ground state, and the coherent states built from it, keep \( \Delta x \cdot \Delta p = \hbar/2 \) for all time. For \( x \) and \( p \), the saturation condition is the first-order equation \( (\hat{p} - \langle p \rangle)\psi = i\lambda(\hat{x} - \langle x \rangle)\psi \), whose normalisable solutions are exactly these Gaussians. This connection between the algebraic structure of the commutator and the specific form of the wavefunction is a key insight in quantum mechanics.
The derivation
The Generalised Uncertainty Relation
Derives the Robertson uncertainty bound for any two observables from the Cauchy-Schwarz inequality, recovering the Heisenberg position-momentum limit.
A common misconception is that the uncertainty principle sets a lower limit on \(\Delta x\) or on \(\Delta p\) separately. It does not: either one alone can be as small as you like, provided the other is large enough that the product stays at or above \(\hbar/2\). What is forbidden is making both small at once. Another subtle point is that the GUR applies to the standard deviations of the probability distributions in a given state, not to the precision of a single measurement. A single measurement yields a definite value, but the spread of many repeated measurements on identically prepared systems is what is constrained by the uncertainty relation. C
Worked example
Problem. Consider a particle in a state where the position uncertainty is \(\Delta x = 1.0 \times 10^{-10} \ \mathrm{m}\). Calculate the minimum possible momentum uncertainty \(\Delta p\). Use \(\hbar = 1.055 \times 10^{-34} \ \mathrm{J \cdot s}\).
- Identify the relevant uncertainty relation: \(\Delta x \cdot \Delta p \geq \frac{\hbar}{2}\).
- Rearrange to solve for the minimum \(\Delta p\): \(\Delta p_{\min} = \frac{\hbar}{2 \Delta x}\).
- Substitute the given values: \(\Delta p_{\min} = \frac{1.055 \times 10^{-34} \ \mathrm{J \cdot s}}{2 \times 1.0 \times 10^{-10} \ \mathrm{m}}\).
- Perform the calculation: \(\Delta p_{\min} = \frac{1.055}{2} \times 10^{-24} \ \mathrm{kg \cdot m/s}\).
- Compute the numerical value: \(0.5275 \times 10^{-24} \ \mathrm{kg \cdot m/s}\).
- Express in scientific notation: \(5.275 \times 10^{-25} \ \mathrm{kg \cdot m/s}\).
Units check. \(\hbar\) has units of \(\mathrm{J \cdot s} = \mathrm{kg \cdot m^2/s^2 \cdot s} = \mathrm{kg \cdot m^2/s}\). Dividing by \(\Delta x\) in \(\mathrm{m}\) gives \(\mathrm{kg \cdot m/s}\), which is the unit of momentum. The calculation is dimensionally consistent.
Three quick tests
- If two operators commute, what is the lower bound on the product of their uncertainties?
Answer
The lower bound is zero. They can both be zero simultaneously. - Why is the ground state of the harmonic oscillator a minimum uncertainty state?
Answer
It satisfies the saturation condition of the Cauchy-Schwarz inequality, leading to \(\Delta x \Delta p = \hbar/2\). - Does the uncertainty principle forbid a particle from having a definite position?
Answer
No, it forbids a definite position and a definite momentum simultaneously. A position eigenstate has \(\Delta x = 0\) but \(\Delta p \to \infty\).
Problem set for PU-202 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.