University Quantum Mechanics II · Deriving the Spherical Harmonics · 1.6

Normalising Yₗᵐ, and the Condon–Shortley phase

The polar equation hands you Pₗᵐ(cos θ) e^(imφ) with its constant and its sign still open. Here the constant comes from one integration by parts, and the sign from a choice that keeps every ladder coefficient positive. Get the constant wrong and probabilities fail; get the sign wrong and only superpositions do, which is harder to catch.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Topic 1.5 left each angular solution as Pₗᵐ(cos θ) e^(imφ) times an undetermined constant, because a linear equation cannot set the size of its own solutions. Two further conditions fix that constant, and they are of different kinds. Its magnitude is physics: |Yₗᵐ|² dΩ is the probability of finding the particle's direction inside the solid angle dΩ, so ∫|Yₗᵐ|² dΩ must be 1, and after x = cos θ that reduces to a single Legendre integral Nₘ, which one integration by parts, repeated, climbs from N₀ = 2/(2l + 1) to any m with no new integral.

Its phase is not physics at all: every e^(iα) normalises equally well, and no measurement on a single state can see it. The Condon–Shortley choice, (−1)ᵐ for positive m and nothing otherwise, makes every ⟨l, m ± 1|L±|l, m⟩ real and positive, which Quantum Mechanics I assumed when it wrote the ladder coefficients down. Positivity fixes only the signs relative to Yₗ⁰, whose own sign is set by keeping it positive on the +z axis, and Yₗ⁻ᵐ = (−1)ᵐ[Yₗᵐ]* then follows.

The construction is exact for every l and contains no constant of nature, so there is no order in α to check and no approximation to break: the Yₗᵐ are the same for hydrogen, a nucleus or a rigid rotor. Where it fails is in bookkeeping. A convention is a relation among many functions, so it holds only if every table, library routine and Clebsch–Gordan coefficient in a calculation shares it; mix two and the error surfaces in the signs of superpositions, never in a single probability.

Simple definition: Normalising Yₗᵐ means fixing the constant in front of Pₗᵐ(cos θ) e^(imφ) so that ∫|Yₗᵐ|² dΩ = 1 over the sphere; the Condon–Shortley phase is the further sign, (−1)ᵐ for positive m, that makes the matrix elements of L₊ and L₋ real and positive.
Example: For l = 1, m = 1: P₁¹ = sin θ, its Legendre norm is ∫₋₁¹ (1 − x²) dx = 4/3, and 1/√(2π × 4/3) = √(3/8π) = 0.3455, so Y₁¹ = −0.3455 sin θ e^(iφ) sr⁻¹⁄², the minus sign being the Condon–Shortley phase.

Normalisation on the sphere∫|Yₗᵐ(θ, φ)|² dΩ = 1, with dΩ = sin θ dθ dφ

Put x = cos θ: dx = −sin θ dθ absorbs the Jacobian, the φ-integral gives 2π, and all that is left is one Legendre integral over −1 ≤ x ≤ 1.

dΩ in sr, 4π sr in all; |Yₗᵐ|² is a probability per steradian, so Yₗᵐ carries sr⁻¹⁄²

The Legendre norm recursionNₘ₊₁ = (l + m + 1)(l − m) Nₘ, where Nₘ = ∫₋₁¹ [Pₗᵐ(x)]² dx

Derived here by one integration by parts, with no boundary term because (1 − x²)ᵐ⁺¹ vanishes at both poles. At m = l the factor l − m is zero, so the ladder stops.

Integers 0 ≤ m ≤ l; x = cos θ, so Nₘ is dimensionless. It starts from N₀ = 2/(2l + 1), derived in topic 1.3

The norm in closed formNₘ = [2/(2l + 1)] (l + m)!/(l − m)!

The raw Pₗᵐ differ wildly in size, a factor 720 in norm between m = 0 and m = 3 at l = 3, which is why no harmonic can skip its constant.

The recursion multiplied out. For l = 2 it gives 0.4, 2.4 and 9.6 at m = 0, 1, 2; for l = 3, m = 3 it gives 205.7

Normalised harmonic, Condon–Shortley conventionYₗᵐ = (−1)ᵐ √[(2l + 1)(l − m)!/(4π (l + m)!)] Pₗᵐ(cos θ) e^(imφ), m ≥ 0

Y₀⁰ = 1/√(4π) = 0.2821 sr⁻¹⁄², and Y₂² = √(15/32π) sin²θ e^(2iφ) = 0.3863 sin²θ e^(2iφ) sr⁻¹⁄², both as the Quantum Mechanics I table printed them.

Pₗᵐ(x) = (1 − x²)^(m/2) dᵐPₗ/dxᵐ carries no sign here; the root equals 1/√(2πNₘ), in sr⁻¹⁄²

Negative mYₗ⁻ᵐ = (−1)ᵐ [Yₗᵐ]*

Conjugating L₊ gives −L₋, so this relation carries the positive L₊ coefficients over to L₋. Writing Yₗ⁻ᵐ = [Yₗᵐ]* instead turns every coefficient on the negative-m side negative.

m ≥ 0 and * is complex conjugation. Written out, Yₗ⁻ᵐ = +√[…] Pₗᵐ(cos θ) e^(−imφ), with no sign in front

Why the sign is neededL₊[Pₗᵐ(cos θ) e^(imφ)] = −ħ Pₗᵐ⁺¹(cos θ) e^{(i[m + 1]φ)}

Every step up in m costs a minus sign, which the (−1)ᵐ cancels, leaving ⟨l, m + 1|L₊|l, m⟩ = +ħ√[(l − m)(l + m + 1)], as Quantum Mechanics I assumed.

m ≥ 0, unphased Pₗᵐ; L₊ = ħ e^(iφ)(∂/∂θ + i cot θ ∂/∂φ), quoted from Quantum Mechanics I; ħ = 1.055 × 10⁻³⁴ J s

The sphere's measure leaves one integral in x

Write the angular solution as A Pₗᵐ(cos θ) e^(imφ), take m ≥ 0 for now, and ask what fixes |A|. The full state ψ = R(r) Yₗᵐ(θ, φ) normalises as ∫R² r² dr × ∫|Yₗᵐ|² dΩ, and the convention sets each factor to 1 so that radial and angular tables can be printed separately. Over the sphere dΩ = sin θ dθ dφ, where sin θ is the radius of the ring of latitude at polar angle θ on the unit sphere, shrinking to zero at the poles, so a band of angular width dθ has area 2π sin θ dθ. The φ-integral of |e^(imφ)|² = 1 gives 2π. Setting x = cos θ, dx = −sin θ dθ swallows the Jacobian and maps θ = 0 → π onto x = 1 → −1, so the condition becomes |A|² × 2π × Nₘ = 1, with Nₘ = ∫₋₁¹ [Pₗᵐ(x)]² dx. Units: dΩ is in steradians and |Y|² is a probability per steradian, so Y carries sr⁻¹⁄²; the uniform distribution has |Y|² = 1/(4π) = 0.0796 sr⁻¹. Leaving out sin θ is the common slip: it weights a thin polar cap as heavily as the broad equator.

One integration by parts climbs the norm ladder

Write Pₗᵐ = (1 − x²)^(m/2) uₘ with uₘ = dᵐPₗ/dxᵐ, so that uₘ₊₁ = uₘ′. Differentiating Legendre's equation m times by the Leibniz rule gives (1 − x²)uₘ″ − 2(m + 1)x uₘ′ + (l − m)(l + m + 1)uₘ = 0, and multiplying by (1 − x²)ᵐ folds the first two terms into one derivative: d/dx[(1 − x²)ᵐ⁺¹uₘ′] = −(l − m)(l + m + 1)(1 − x²)ᵐuₘ. Now Nₘ₊₁ = ∫₋₁¹ (1 − x²)ᵐ⁺¹(uₘ′)² dx. Integrate by parts once, treating one uₘ′ as the derivative of uₘ: the boundary term (1 − x²)ᵐ⁺¹uₘuₘ′ vanishes at x = ±1, and what remains is (l − m)(l + m + 1)∫(1 − x²)ᵐuₘ² dx = (l − m)(l + m + 1)Nₘ. Climb from N₀ = 2/(2l + 1), derived in topic 1.3: for l = 2 the rungs are 0.4, 2.4 and 9.6. Check the top rung directly: P₂² = 3(1 − x²), and ∫9(1 − x²)² dx = 9 × 16/15 = 9.6. At m = l the factor l − m vanishes, so Nₗ₊₁ = 0: the recursion itself knows that Pₗᵐ is identically zero beyond m = l.

Assembling the constant, and the checks it must pass

Multiplying the rungs gives Nₘ = [2/(2l + 1)](l + m)!/(l − m)!, so |A| = 1/√(2πNₘ) = √[(2l + 1)(l − m)!/(4π(l + m)!)], derived. Four checks. Dimensions: the 4π carries the steradians, and |A|² × 2π × Nₘ is a pure number. The limit l = 0: |A| = 1/√(4π) = 0.2821 sr⁻¹⁄², the uniform distribution. The pole: Pₗ(1) = 1, so Yₗ⁰ on the +z axis equals √[(2l + 1)/4π], which is 0.4886 sr⁻¹⁄² for l = 1 and 0.6308 sr⁻¹⁄² for l = 2, values any table must reproduce. A sum: |Y₁⁰|² + |Y₁¹|² + |Y₁⁻¹|² = (3/4π)cos²θ + 2(3/8π)sin²θ = 3/4π = 0.2387 sr⁻¹ at every θ, so a full p shell is spherical; the general sum Σₘ|Yₗᵐ|² = (2l + 1)/4π is derived from the addition theorem in topic 1.9. The constant depends on |m| only, so Yₗᵐ and Yₗ⁻ᵐ share a modulus. What normalisation cannot do is choose a phase: e^(iα)A normalises exactly as well as A, for any real α.

The sign that keeps the ladder positive

Quantum Mechanics I wrote L₊ = ħ e^(iφ)(∂/∂θ + i cot θ ∂/∂φ) and took ⟨l, m + 1|L₊|l, m⟩ = ħ√[(l − m)(l + m + 1)], real and positive. Test that on the unphased functions, m ≥ 0. On sinᵐθ uₘ(cos θ) e^(imφ), ∂/∂φ brings down im, so the bracket acts as ∂/∂θ − m cot θ. Differentiating sinᵐθ gives m cos θ sinᵐ⁻¹θ uₘ, which the −m cot θ term cancels exactly; what survives comes from uₘ(cos θ) through dx/dθ = −sin θ, namely −sinᵐ⁺¹θ uₘ′ = −Pₗᵐ⁺¹. So each step up in m costs a minus sign, and with plain constants every coefficient from m ≥ 0 would be −ħ√[(l − m)(l + m + 1)]. The (−1)ᵐ in Yₗᵐ cancels one sign per step. At l = 1: L₊Y₁⁰ = ħ e^(iφ) d/dθ[√(3/4π) cos θ] = −ħ√(3/4π) sin θ e^(iφ) = √2 ħ Y₁¹. The sign is chosen, not derived: positive ladder coefficients fix only the signs relative to Yₗ⁰, which is kept positive on the +z axis, and no |Yₗᵐ|² feels either choice. Photoelectrons freed from helium's 1s shell by light polarised along z leave in an l = 1, m = 0 wave (quoted from topic 1.8 and Unit 13), so they emerge as |Y₁⁰|² ∝ cos²θ, an asymmetry parameter β = 2; measured values for helium sit close to 2.

Negative m, and the convention that lives in Pₗᵐ

For positive m this lesson writes Yₗ⁻ᵐ = √[(2l + 1)(l − m)!/(4π(l + m)!)] Pₗᵐ(cos θ) e^(−imφ), with no sign. Conjugating Yₗᵐ gives (−1)ᵐ times that same function, hence Yₗ⁻ᵐ = (−1)ᵐ[Yₗᵐ]*, derived from the definitions. It matters because the complex conjugate of L₊, as a differential operator, is −L₋: conjugating L₊Yₗᵐ = ħ√[(l − m)(l + m + 1)] Yₗᵐ⁺¹ and inserting the relation gives L₋Yₗ⁻ᵐ = ħ√[(l − m)(l + m + 1)] Yₗ⁻ᵐ⁻¹, positive on the negative-m side too. Many texts and numerical libraries instead put the (−1)ᵐ inside the Legendre function, writing Pₗᵐ(x) = (−1)ᵐ(1 − x²)^(m/2) dᵐPₗ/dxᵐ, and extend that phased function to negative order by Pₗ⁻ᵐ = (−1)ᵐ[(l − m)!/(l + m)!]Pₗᵐ, after which one formula with no separate sign serves every m. The harmonics that come out are identical; only the bookkeeping has moved. The danger is using both at once. Before trusting a routine, evaluate P₁¹(0.5): it returns +0.866 in this lesson's convention and −0.866 in the other, and a routine of the second kind must not be multiplied by (−1)ᵐ again.

Rebuilding the table, and where a wrong sign shows

From P₂ = (3x² − 1)/2 come P₂¹ = 3x√(1 − x²) = 3 sin θ cos θ and P₂² = 3(1 − x²) = 3 sin²θ. The constants are √(5/4π) for m = 0, √(5/24π) for |m| = 1 and √(5/96π) for |m| = 2, so Y₂⁰ = √(5/16π)(3cos²θ − 1) = 0.3154(3cos²θ − 1), Y₂¹ = −√(15/8π) sin θ cos θ e^(iφ) = −0.7725 sin θ cos θ e^(iφ), Y₂⁻¹ is the same with +0.7725 and e^(−iφ), and Y₂^(±2) = √(15/32π) sin²θ e^(±2iφ) = 0.3863 sin²θ e^(±2iφ), all in sr⁻¹⁄². With Y₀⁰, Y₁⁰ and Y₁^(±1) from the sections above, that is the nine-entry table Quantum Mechanics I quoted, now derived, signs included. A wrong sign never shows in a |Yₗᵐ|²; it shows in superpositions. The real orbital pₓ is (Y₁⁻¹ − Y₁¹)/√2 = √(3/4π) sin θ cos φ. Run the same recipe on harmonics without the (−1)ᵐ and it returns −i√(3/4π) sin θ sin φ: a p_y orbital, the pₓ lobes turned through 90° about z.

Change one variable at a time

Make the relationship visible.

Interactive model

Degree l2

Order m1

Condon–Shortley phase (1 on, 0 off)1

The default is Y₂¹, where ⟨2, 2|L₊|2, 1⟩ = +2 ħ. Set the switch to 0: the solid curve flips over while the dashed modulus stays, and the coefficient turns to −2 ħ. Back at 1, set m = −1: Y₂⁻¹ shares the modulus of Y₂¹ but not its sign. Then push |m| past l and watch Nₘ fall to zero.

LEGENDRE NORM Nₘ = ∫[Pₗᵐ]² dx2.400

FACTOR cₗₘ = 1/√(2πNₘ)0.2575 sr⁻¹⁄²

⟨l, m+1|L₊|l, m⟩ / ħ2.000

Live interpretationLEGENDRE NORM Nₘ = ∫[Pₗᵐ]² dx: 2.400. FACTOR cₗₘ = 1/√(2πNₘ): 0.2575 sr⁻¹⁄². SIGN IN FRONT ε: −1. ⟨l, m+1|L₊|l, m⟩ / ħ: 2.000

Catch the common trap

Explain before calculating.

Misconception check

The tempting argument is that a sign cannot matter because |Yₗᵐ|² is the same with or without it, so students drop the (−1)ᵐ, or apply it twice when their Legendre routine already carries it. The premise is true and the conclusion is false. A phase convention is not a property of one function but a relation among all of them, and every ladder coefficient, Clebsch–Gordan table and real-orbital recipe you will meet is written assuming a particular one. Drop the sign and ⟨1, 1|L₊|1, 0⟩ comes out as −√2 ħ instead of +√2 ħ, the standard Clebsch–Gordan tables of Unit 5 build the wrong coupled states, and (Y₁⁻¹ − Y₁¹)/√2 turns from pₓ into −i p_y, a different orbital. None of these errors changes a single |Yₗᵐ|², which is exactly why they survive until an interference term exposes them. Choose one convention, test it once with L₊Y₁⁰ = +√2 ħ Y₁¹, and carry it to the last line.

Using P₂¹(x) = 3x√(1 − x²), the normalising factor √[(2l + 1)(l − |m|)!/(4π (l + |m|)!)] and this lesson's Condon–Shortley convention, which is Y₂⁻¹(θ, φ)?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyStarting from P₃(x) = (5x³ − 3x)/2 and N₀ = 2/(2l + 1), use the norm recursion, not the closed form, to find Y₃²(θ, φ) in the Condon–Shortley convention. Then confirm the constant against the closed form and the norm by direct integration.

Build the function: d²P₃/dx² = 15x, so P₃²(x) = (1 − x²) × 15x = 15 sin²θ cos θ.
Climb the ladder with Nₘ₊₁ = (l + m + 1)(l − m)Nₘ at l = 3: N₀ = 2/7 = 0.2857, N₁ = (4)(3)N₀ = 24/7 = 3.429, N₂ = (5)(2)N₁ = 240/7 = 34.29.
Constant: 1/√(2πN₂) = 1/√(480π/7) = √(7/480π) = 0.06813 sr⁻¹⁄². The closed form gives √(7 × 1!/(4π × 5!)) = √(7/480π), the same.
Sign: m = 2 is even, so (−1)² = +1. Then 15 × 0.06813 = 1.022 = √(105/32π).
Direct check of the norm: ∫₋₁¹ 225x²(1 − x²)² dx = 225 × (2/3 − 4/5 + 2/7) = 225 × 16/105 = 34.29, matching N₂.

AnswerY₃² = +√(105/32π) sin²θ cos θ e^(2iφ) ≈ 1.022 sin²θ cos θ e^(2iφ) sr⁻¹⁄², with N₂ = 240/7 ≈ 34.3.

MediumTake Y₂¹ = −√(15/8π) sin θ cos θ e^(iφ) and apply L₊ = ħ e^(iφ)(∂/∂θ + i cot θ ∂/∂φ) to it directly. Show that the result is a positive multiple of Y₂², compare the multiple with ħ√[(l − m)(l + m + 1)], and find what a student whose Y₂¹ lacks the Condon–Shortley sign obtains.

∂/∂φ acting on e^(iφ) gives i, so i cot θ ∂/∂φ becomes −cot θ, and the bracket acts on f(θ) = −√(15/8π) sin θ cos θ as f′ − cot θ f.
f′ = −√(15/8π)(cos²θ − sin²θ) and cot θ f = −√(15/8π) cos²θ, so f′ − cot θ f = +√(15/8π) sin²θ.
Hence L₊Y₂¹ = ħ√(15/8π) sin²θ e^(2iφ). Since Y₂² = √(15/32π) sin²θ e^(2iφ), the ratio is √(32/8) = 2, so L₊Y₂¹ = 2ħ Y₂².
The formula gives ħ√[(2 − 1)(2 + 1 + 1)] = ħ√4 = 2ħ, the same, and positive. In SI units 2ħ = 2 × 1.0546 × 10⁻³⁴ J s = 2.109 × 10⁻³⁴ J s.
Without the sign, the student's Y₂¹ is the negative of ours while Y₂² is unchanged, since m = 2 is even. So L₊ returns −2ħ times the student's Y₂², though every |Y|² agrees with ours.

AnswerL₊Y₂¹ = +2ħ Y₂², with 2ħ = 2.109 × 10⁻³⁴ J s; without the Condon–Shortley sign the coefficient comes out as −2ħ.

HardAn angular state is ψ(θ, φ) = A sin θ cos φ (1 + cos θ). Find A, expand ψ on the Condon–Shortley Yₗᵐ, give the probability of each (l, m) and ⟨L²⟩, and say which of these results would change if the harmonics carried no (−1)ᵐ.

Normalise: the φ-integral of cos²φ over a full turn is π, and x = cos θ turns the θ-integral into ∫₋₁¹ (1 − x²)(1 + x)² dx = 2 − 2/5 = 8/5. So ∫|ψ|² dΩ = (8π/5)A² = 1, and A = √(5/8π) = 0.4460 sr⁻¹⁄².
Write cos φ = (e^(iφ) + e^(−iφ))/2 and use Y₁^(±1) = ∓√(3/8π) sin θ e^(±iφ): sin θ cos φ = √(2π/3)(Y₁⁻¹ − Y₁¹). Likewise Y₂^(±1) = ∓√(15/8π) sin θ cos θ e^(±iφ) gives sin θ cos θ cos φ = √(2π/15)(Y₂⁻¹ − Y₂¹).
Multiply by A: ψ = √(5/12)(Y₁⁻¹ − Y₁¹) + √(1/12)(Y₂⁻¹ − Y₂¹), coefficients of modulus 0.6455 and 0.2887.
Probabilities: 5/12 = 0.417 for each of (1, −1) and (1, 1), and 1/12 = 0.083 for each of (2, −1) and (2, 1); they sum to 1, confirming A, so P(l = 1) = 5/6 and P(l = 2) = 1/6.
⟨L²⟩ = ħ²[1(2) × 5/6 + 2(3) × 1/6] = (8/3)ħ² = 2.667 × (1.0546 × 10⁻³⁴ J s)² = 2.97 × 10⁻⁶⁸ J² s².
Without the (−1)ᵐ, Y₁¹ and Y₂¹ each change sign, so ψ = √(5/12)(Y₁⁻¹ + Y₁¹) + √(1/12)(Y₂⁻¹ + Y₂¹) in those harmonics. The coefficients' signs change; ψ itself, the four probabilities and ⟨L²⟩ do not.

AnswerA = √(5/8π) = 0.446 sr⁻¹⁄²; ψ = √(5/12)(Y₁⁻¹ − Y₁¹) + √(1/12)(Y₂⁻¹ − Y₂¹); probabilities 0.417, 0.417, 0.083, 0.083; ⟨L²⟩ = (8/3)ħ² = 2.97 × 10⁻⁶⁸ J² s². Only the coefficient signs depend on the convention.