What is the canonical commutation relation between position and momentum?
[x,p]=iℏ
What operator generates spatial translations?
T(d)=e^(−idp/ℏ)
How does a translation by
d affect the wavefunction?
T(d)ψ(x)=ψ(x−d)
What happens to expectation values under translation by
d?
- ⟨ x ⟩→⟨ x ⟩ + d
- ⟨ p ⟩ and dispersions stay the same.
What does the operator
O(w) = e^(iwx/ℏ) do?
It generates momentum translations:
O(w)∣ p ⟩ = ∣ p + w ⟩
In 3D, how many independent components of angular momentum exist?
Three — corresponding to rotations in the xy, yz, and zx planes.
What is the infinitesimal rotation operator in quantum mechanics?
D(Rij(ϵ))=1−(i/ℏ)J_{ij}ϵ−(1/2ℏ^2)J_{ij}^2ϵ^2+…
Why must rotations preserve vector norms?
Because they are orthogonal transformations satisfying
R^TR = I.
What physical operators generate rotations in quantum mechanics?
The angular momentum operators J_{ij}, since they appear in the infinitesimal rotation operator’s exponent.
Why must [x,p]=Iℏ hold for any consistent quantum theory of continuous variables?
Because it encodes the Fourier duality between position and momentum representations. It ensures that translations in position correspond to phase shifts in momentum space, preserving the uncertainty principle.
What does it physically mean that position and momentum operators do not commute?
It means that measuring position disturbs momentum (and vice versa), reflecting the fact that a wave cannot be perfectly localized in both real and reciprocal space.
Why are angular momentum operators associated with rotations?
Because the generators of infinitesimal rotations in space form the same algebra as the angular momentum operators — both satisfy the [J_{i},J_{j}] = iℏϵ_{ijk}J_{k} commutation relations.
Why are rotations represented by exponentials of angular momentum operators?
Because any continuous symmetry transformation can be expressed as
e^−(iθG/ℏ), where G is the generator. For rotations, that generator is angular momentum.
What does it mean for angular momentum components not to commute?
It means you can’t know all components of orientation simultaneously — knowing
J_{z} precisely destroys knowledge of J_{x} or J_{y}. This reflects the geometry of rotational symmetry, not measurement error.
Why do we define J^2 = J_{x}^2 + J_{y}^2 + J_{z}^2?
Because it’s invariant under rotation — the “magnitude” of angular momentum, representing total rotational symmetry independent of orientation.
What is the geometric meaning of the commutator of two rotation generators?
It gives another rotation — performing one small rotation and then another (in a different plane) does not return you to the same orientation; their difference is another rotation.
What’s the physical insight behind defining infinitesimal rotations?
They represent how an observable changes under an arbitrarily small symmetry operation. Infinitesimal transformations directly expose the generator structure (the algebra) of the symmetry.
Why are infinitesimal rotations sufficient to describe all rotations?
Because any finite rotation can be built as a continuous product (exponential) of infinitesimal ones — the Lie group is generated by exponentiating its algebra.
Why is the total angular momentum operator J^2 a “Casimir operator”?
It commutes with all generators of rotation, meaning it labels irreducible representations of the rotation group — it’s a conserved, group-invariant quantity.
How is the uncertainty principle reflected in rotational measurements?
Since [J_{i},J_{j}] = iℏϵ_{ijk}J_{k}, you cannot measure two perpendicular angular momentum components exactly — the uncertainty in one is bounded by the expectation of the third.
