Commutators

Canonical commutator [r̂i, p̂j]

iħδij

[r̂i, r̂j] and [p̂i, p̂j]

Both = 0 (position components commute; momentum components commute)

[ẑ, p̂z] and [p̂z, ẑ]

[ẑ, p̂z] = iħ, [p̂z, ẑ] = −iħ

Commutator product identity [ÂB̂, Ĉ]

[Â, Ĉ]B̂ + Â[B̂, Ĉ]

[L̂x, L̂y]

iħL̂z

[L̂y, L̂z]

iħL̂x

[L̂z, L̂x]

iħL̂y

[L̂z, L̂y]

−iħL̂x (reversing the order flips the sign)

Compact form of the L commutators

[L̂i, L̂j] = iħ εijk L̂k

Levi-Civita symbol εijk

+1 cyclic (xyz, yzx, zxy), −1 anticyclic (zyx, xzy, yxz), 0 if any two indices are equal

Deriving [L̂x, L̂y]: which terms survive after expanding [ŷp̂z − ẑp̂y, ẑp̂x − x̂p̂z]?

[ŷp̂z, ẑp̂x] + [ẑp̂y, x̂p̂z] = ŷ[p̂z, ẑ]p̂x + x̂[ẑ, p̂z]p̂y = iħ(x̂p̂y − ŷp̂x) = iħL̂z

Deriving [L̂x, L̂y]: why is [ŷp̂z, x̂p̂z] = 0?

p̂z commutes with x̂ and ŷ, so it equals [ŷ, x̂]p̂z² = 0

Deriving [L̂x, L̂y]: why is [ẑp̂y, ẑp̂x] = 0?

ẑ commutes with p̂x and p̂y, so it equals ẑ²[p̂y, p̂x] = 0

Generalised uncertainty principle

ΔA² ΔB² ≥ ( (1/2i) ⟨[Â, B̂]⟩ )²

Uncertainty relation for L̂x and L̂y

ΔLx² ΔLy² ≥ (ħ²/4) ⟨L̂z⟩²

Physical consequence of [L̂x, L̂y] ≠ 0

Angular momentum components cannot all have definite values at once; if L̂z is definite, L̂x and L̂y are uncertain

Key difference between position and angular momentum components

x̂ and ŷ commute (simultaneously specifiable); L̂x and L̂y do not

[L̂², L̂x] expanded

[L̂x², L̂x] + [L̂y², L̂x] + [L̂z², L̂x]

[L̂x², L̂x]

0 (an operator commutes with any function of itself)

[L̂², L̂x] after using the identity

L̂y(−iħL̂z) + (−iħL̂z)L̂y + L̂z(iħL̂y) + (iħL̂y)L̂z = 0

[L̂², L̂i]

0 for i = x, y, z (compactly [L̂², L̂] = 0)

Consequence of [L̂², L̂z] = 0

Simultaneous eigenstates of L̂² and L̂z exist: L̂²|ψ⟩ = λ|ψ⟩, L̂z|ψ⟩ = μ|ψ⟩

Why define L̂± = L̂x ± iL̂y?

The Cartesian components don't have neat commutators with L̂z, but L̂± do

[L̂z, L̂±] derivation

[L̂z, L̂x] ± i[L̂z, L̂y] = iħL̂y ± i(−iħL̂x) = ±ħ(L̂x ± iL̂y)

[L̂z, L̂±]

±ħL̂±

[L̂², L̂±]

0, since [L̂², L̂x] ± i[L̂², L̂y] = 0

Using [L̂², L̂±] = 0 on an eigenstate

L̂²(L̂±|ψ⟩) = L̂±(L̂²|ψ⟩) = λ(L̂±|ψ⟩), so λ is unchanged

Using [L̂z, L̂±] = ±ħL̂± on an eigenstate

L̂z(L̂±|ψ⟩) = (L̂±L̂z ± ħL̂±)|ψ⟩ = (μ ± ħ)(L̂±|ψ⟩), so μ shifts by ±ħ

L̂±L̂∓ expanded (where the commutator appears)

L̂x² + L̂y² ∓ i[L̂x, L̂y] = L̂x² + L̂y² ± ħL̂z

L̂² in terms of ladder operators

L̂² = L̂±L̂∓ + L̂z² ∓ ħL̂z