Commutators, Compatible Observables and the General Uncertainty Relation
For finite-dimensional observables, commuting operators share an eigenbasis. The general uncertainty relation bounds the product of standard deviations using the expectation of their commutator. A nonzero operator need not give a positive bound in every state.
What a learner can do afterwards
- Evaluates a commutator and states whether the two observables share an eigenbasis
- Derives the uncertainty bound for a general pair from their commutator
- Applies the relation to a pair other than position and momentum
1 · Read
Define [A,B]=AB-BA. AB applies B first, then A; BA reverses that order. The commutator subtracts the two operator products. These definitions apply to general linear operators, not only Hermitian observables. For vectors, <a|b> conjugates a's components and sums their products with b's. A vector's norm is ||a||=sqrt(<a|a>). Cauchy-Schwarz gives ||a|| ||b||>=|<a|b>|. Also, |<a|b>|>=|Im<a|b>|; Im means imaginary part.
For finite-dimensional Hermitian observables, [A,B]=0 gives a common orthonormal eigenbasis. Commutation makes each eigenspace of one invariant under the other. Diagonalize the other observable within those spaces, including degenerate ones. Each resulting basis state has definite values of both observables. Compatibility does not mean every superposition has zero spread.
To evaluate a commutator on a vector, subtract the two actions: [A,B]v=ABv-BAv. For example, ABw=iw and BAw=-iw give [A,B]w=2iw, where i²=-1. Such actions can occur for Hermitian A and B. A nonzero commutator rules out a full common eigenbasis, but some common eigenstates may remain. A single shared eigenstate and a shared basis are different claims.
For normalized psi, <A> is its expectation and ΔA=sqrt(<A²>-<A>²). Set a=(A-<A>)psi and b=(B-<B>)psi. For Hermitian A,B, <[A,B]>=2i Im<a|b>. Cauchy-Schwarz gives ΔA ΔB>=|<a|b>|>=|Im<a|b>|=|<[A,B]>|/2. Assume the required products and finite variances exist. This Robertson bound uses the expectation in the chosen state. The spreads describe repeated measurements on identically prepared states, not detector errors.
For a free particle, an ideal momentum eigenstate has one wave number and is an extended plane wave. It is not a normalizable localized packet. A localized packet needs a range of momenta. This wave picture explains why exact free-particle momentum does not describe a particle localized at one position.
Position and momentum obey [x,p]=i hbar I on a suitable common domain. I is the identity and hbar is Planck's constant divided by 2 pi. Insert A=x and B=p into ΔA ΔB>=|<[A,B]>|/2. In a normalized state, <I>=1, so |<[x,p]>|=hbar. Thus Δx Δp>=hbar/2: the position-momentum relation is a case of the general bound.
For spin, [Sx,Sy]=i hbar Sz, so ΔSx ΔSy>=hbar |<Sz>|/2. If <Sz>=hbar/2, the lower bound is hbar²/4. If <Sz>=0, this bound is zero even though the commutator operator is nonzero. A zero bound does not prove both spreads vanish. For any stated expectation, take its absolute value, multiply by hbar, then divide by two.
Commutation tests full compatibility. The uncertainty bound uses the commutator's expectation in the chosen state, not just whether the operator is nonzero.
Original lesson, practice questions and flashcards authored for this dataset.
Reading notes and flashcards were authored for this dataset. Video references credit TED-Ed, Chad Orzel and Khan Academy. No reuse license for the videos has been verified.
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