Ladder operators and the quantisation of j and m

The input is three Hermitian operators with [Jᵢ, Jⱼ] = iħ εᵢⱼₖ Jₖ and nothing else — no r, no p, no wavefunction. Form J² = Jₓ² + J_y² + J_z² and check [J², Jᵢ] = 0 for each i: the commutator of Jₓ² with J_z, for instance, is Jₓ[Jₓ, J_z] + [Jₓ, J_z]Jₓ = −iħ(JₓJ_y + J_yJₓ), which the J_y² term cancels exactly. So J² is compatible with any one component, but the components are not compatible with each other: [Jₓ, J_z] ≠ 0, so Jₓ and J_z share no eigenbasis. The largest commuting set is therefore J² and one component, and convention picks J_z. Write the simultaneous eigenvectors as |λ, m⟩ with J²|λ, m⟩ = ħ²λ|λ, m⟩ and J_z|λ, m⟩ = mħ|λ, m⟩, where λ and m are for now just real numbers — real because both operators are Hermitian. Everything that follows is about which pairs (λ, m) the algebra permits.

Define J± = Jₓ ± iJ_y. They are adjoints of each other and not Hermitian, so they are not observables; they are tools. Two commutators do the work. From [J_z, Jₓ] = iħJ_y and [J_z, J_y] = −iħJₓ, [J_z, J₊] = iħJ_y + ħJₓ = ħJ₊, and likewise [J_z, J₋] = −ħJ₋. From [J², Jᵢ] = 0 follows [J², J±] = 0. Now act on a rung: J_z(J₊|λ, m⟩) = (J₊J_z + ħJ₊)|λ, m⟩ = (m + 1)ħ (J₊|λ, m⟩), and J²(J₊|λ, m⟩) = ħ²λ (J₊|λ, m⟩). So J₊|λ, m⟩ is either the zero vector or an eigenvector with the same λ and with m raised by exactly one; J₋ lowers by exactly one. Applying J₊ repeatedly generates rungs m, m+1, m+2, … all inside one J² eigenspace. Nothing yet stops the climb — that needs the norm.

Compute the squared length of the raised vector. Since (J₊)† = J₋, ‖J₊|λ, m⟩‖² = ⟨λ, m|J₋J₊|λ, m⟩, and multiplying out, J₋J₊ = (Jₓ − iJ_y)(Jₓ + iJ_y) = Jₓ² + J_y² + i[Jₓ, J_y] = J² − J_z² − ħJ_z. So ‖J₊|λ, m⟩‖² = ħ²[λ − m(m+1)], and by the same route ‖J₋|λ, m⟩‖² = ħ²[λ − m(m−1)]. A squared length is never negative, so λ ≥ m(m+1) and λ ≥ m(m−1): for fixed λ, m is bounded above and below. But J₊ raises m by one each time at the same λ, so an unbounded climb would eventually break the bound. The only escape is a rung mₘₐₓ with J₊|λ, mₘₐₓ⟩ = 0, that is λ = mₘₐₓ(mₘₐₓ + 1); call mₘₐₓ = j, so λ = j(j+1). Descending, there is a rung mₘᵢₙ with λ = mₘᵢₙ(mₘᵢₙ − 1), and the two conditions give mₘᵢₙ = −j (the other root, j + 1, lies above the top). The rungs run from −j to j in unit steps, so 2j is a non-negative integer: j = 0, ½, 1, 3⁄2, …, with 2j + 1 rungs. Take λ = 3.75 = (3⁄2)(5⁄2): four rungs at m = ±½ and ±3⁄2, and nothing else.

The norm calculation also fixes the coefficients. With the Condon–Shortley convention of real, non-negative phases, J±|j, m⟩ = ħ√(j(j+1) − m(m±1)) |j, m±1⟩ = ħ√((j∓m)(j±m+1)) |j, m±1⟩. Run it for j = 1: J₊|1, −1⟩ = ħ√((1+1)(1−1+1)) |1, 0⟩ = ħ√2 |1, 0⟩, J₊|1, 0⟩ = ħ√((1−0)(1+0+1)) |1, 1⟩ = ħ√2 |1, 1⟩, and J₊|1, 1⟩ = ħ√(0⋅3) |1, 2⟩ = 0. The factor is not ħ and it is not constant along the ladder: J± are not unitary shifts, and a raised vector must be renormalised before it is a state. Because Jₓ = (J₊ + J₋)/2 and J_y = (J₊ − J₋)/2i, these numbers are the only input to the matrices of topic 9.4: in the basis (|1,1⟩, |1,0⟩, |1,−1⟩) the j = 1 matrix Jₓ is (ħ/√2) times [[0,1,0],[1,0,1],[0,1,0]], whose eigenvalues are 0 and ±ħ — the same set as J_z, because no direction is special.

The derivation is short enough to test in NumPy in a dozen lines, and the test catches the ordering and phase slips that hand calculation hides. Order the basis m = j, j−1, …, −j, so m = np.arange(j, −j-1, −1). Then Jp = ℏ*np.diag(np.√(j*(j+1) - m[1:]*(m[1:]+1)), k=1) places ⟨m+1|J₊|m⟩ on the superdiagonal, Jm = Jp.conj().T, Jz = ℏ*np.diag(m), Jx = (Jp + Jm)/2 and Jy = (Jp - Jm)/(2*1j), where 1j is Python's imaginary unit and not the angular momentum j. Three checks close the loop: np.allclose(Jx@Jy - Jy@Jx, 1j*ℏ*Jz) should return True with residuals near 10⁻¹⁶; Jx@Jx + Jy@Jy + Jz@Jz should equal ħ²j(j+1) times the identity; and np.linalg.eigvalsh of n⋅J for any unit vector n should return the same m values as J_z. A residual of order one instead of 10⁻¹⁶ almost always means the superdiagonal was filled from the wrong end of m — the same slip as raising with the lowering coefficient.

The result is a catalogue of representations, one for each j = 0, ½, 1, …, each of dimension 2j + 1. The commutators cannot say which entries a given system uses; that depends on what the operators J actually are. For orbital motion L = r × p acts on functions of θ and φ, L_z = −iħ ∂/∂φ has eigenfunctions e^(imφ), and only the integer ladders close on the sphere — the half-integer tower fails because lowering below its would-be bottom rung does not give zero, so it never terminates (topic 9.6 finishes that argument). Hence L² = ħ²l(l+1) and L_z = mₗħ with l = 0, 1, 2, …. Spin is the opposite case: S on C² is the j = ½ entry, with no coordinate representation at all. Two consequences carry forward. For every j > 0, m ≤ j < √(j(j+1)), so ⟨Jₓ²⟩ + ⟨J_y²⟩ = ħ²[j(j+1) − m²] ≥ jħ² > 0 on every rung — the transverse components never vanish, which is the origin of the cone picture in topic 9.4. And a physical state need not be a single rung: it can be a superposition over several j. Which j appear is a property of the state, not of the algebra; a Hamiltonian that commutes with J² only guarantees that the weight on each j stays fixed as the state evolves.