Angular momentum algebra and spin: operators for CSIR NET
Learn angular momentum commutators, eigenvalues, ladder operators and spin for CSIR NET, GATE and IIT-JAM, with two worked practice questions and a study plan
This post gives you the commutator rules and eigenvalue formulas for angular momentum in one place.
You will also see how spin fits in, solve two practice questions, and get a short plan to revise it.
Operators and commutators you must know
In quantum mechanics, a physical quantity becomes an operator. An operator acts on a wavefunction. If the result is the same function times a number, that function is an eigenfunction and the number is the eigenvalue. The eigenvalue is the value you can measure.
Orbital angular momentum is defined like the classical one: L = r × p. Here r is the position operator and p is the momentum operator. L is a vector operator, meaning it has three components, written Lx, Ly and Lz. Each component is itself an operator.
Three kinds of angular momentum operator appear in quantum mechanics. They are the orbital one (L), the spin one (S) and the total one (J). Be careful when a question says only "angular momentum operator". It can mean either the total or the orbital one, so read the context.
A commutator is [A, B] = AB − BA. If it is zero, the two quantities can be measured together, with sharp values. If not, they cannot. Learn these results by heart:
- [Lx, Ly] = iħLz. The other two follow by cyclic change of x, y, z.
- [L2, Lx] = [L2, Ly] = [L2, Lz] = 0.
- Here ħ is the reduced Planck constant, h/2π.
The first line says you cannot fix Lx and Ly together. The second line says you can fix L2 and Lz together. So we label states by two numbers, one for L2 and one for Lz.
Eigenvalues and ladder operators
Solving the algebra gives two quantum numbers. Here l is a non-negative integer for orbital motion, and ml runs from −l to +l in steps of one.
| Operator | Eigenvalue | Remark |
|---|---|---|
| L2 | l(l+1)ħ2 | Magnitude of L is √[l(l+1)] ħ |
| Lz | mlħ | 2l + 1 values for a given l |
| Lz maximum | lħ | Always less than the magnitude of L |
The last row matters. Lz can never equal the full length of L. If it did, Lx and Ly would both be zero, which would break the uncertainty rule above.
To move between states we use the ladder operators:
- L+ = Lx + iLy raises m by 1.
- L− = Lx − iLy lowers m by 1.
- [Lz, L±] = ±ħL±. This is the reason they shift m by one unit.
Their action on a state |l, m⟩ is:
L±|l, m⟩ = ħ √[l(l+1) − m(m±1)] |l, m±1⟩
Check the ends. For m = l, the factor under the root is zero when raising. So L+ gives zero at the top of the ladder, and L− gives zero at the bottom. This is how the limits −l to +l arise.
Spin: same algebra, new source
Spin is a separate kind of angular momentum, written S. It has three components, also operators. Spin is often depicted as a particle literally turning about an axis, but this is only a metaphor. Treat it as an intrinsic property of the particle.
The spin components obey the same commutator rules as L. For example, the first two spin components give back the third, with the same factor iħ. Hence the same formulas hold, with quantum number s and projection ms:
- The square of spin has eigenvalue s(s+1)ħ2.
- The z-component of spin has eigenvalue msħ, from −s to +s.
The big difference is that s can be a half-integer. Spin is the property that separates two types of elementary particles. Particles with half-integer spin are fermions. Particles with integer spin are bosons. The electron has s = ½, so the square of its spin is ¾ħ2 and the z-component is ±ħ/2.
Light is made of photons, which are spin-1 bosons. When a light beam is circularly polarised, each photon carries a spin angular momentum of ±ħ along the beam axis. The sign depends on whether the polarisation is left or right.
For several sources of angular momentum, add them as vectors. The total J has quantum number j from |l − s| up to l + s in steps of one. For an electron with l = 1 and s = ½, j is ½ or 3/2. Total angular momentum of an isolated system is always constant, because no net external torque acts on it.
The exam angle
In CSIR NET, GATE and IIT-JAM, questions on this topic are mostly short and algebraic. You are rarely asked to derive anything. You are asked to apply a rule fast. Expect these kinds of tasks:
- Decide whether two operators commute, and what that means for simultaneous measurement.
- Find the magnitude of L, or the allowed Lz values, for a given l.
- Apply a ladder operator to a given state.
- Count states, such as 2l + 1 or the j values from vector addition.
- Pick the correct statement about spin, fermions and bosons.
Common traps: writing l² instead of l(l+1), forgetting the ħ, and taking the maximum Lz as the magnitude of L. Also remember that [Lx, Ly] is not zero.
A good habit is to check every answer against two limits. First, Lz must be smaller than the magnitude of L. Second, a ladder operator must give zero at the end of its ladder. A wrong answer usually breaks one of these checks.
Two practice questions
These are practice questions written for this post. They are not past-paper questions.
Practice question 1. An electron is in a state with orbital quantum number l = 2. Find (a) the magnitude of its orbital angular momentum, (b) the allowed Lz values, and (c) the smallest angle between L and the z-axis.
Solution.
- L2 = l(l+1)ħ2 = 6ħ2. So |L| = √6 ħ, about 2.45ħ.
- ml = −2, −1, 0, +1, +2. So Lz = −2ħ, −ħ, 0, +ħ, +2ħ. That is 2l + 1 = 5 values.
- The smallest angle comes from the largest Lz. cos θ = 2ħ / (√6 ħ) = 2/√6 = 0.816. So θ is about 35°.
Check: Lz = 2ħ is less than 2.45ħ, as the rule demands. The angle can never be zero.
Practice question 2. A system is in the state |l = 1, m = 0⟩. (a) Find L+|1, 0⟩. (b) In the state |1, 1⟩, what is the lower limit on the product of the uncertainties ΔLx and ΔLy? Use ΔLx ΔLy ≥ ½ |⟨[Lx, Ly]⟩|.
Solution.
- Use the ladder formula with l = 1, m = 0 and the upper sign: l(l+1) − m(m+1) = 2 − 0 = 2.
- So L+|1, 0⟩ = √2 ħ |1, 1⟩.
- For part (b), [Lx, Ly] = iħLz. In |1, 1⟩, ⟨Lz⟩ = ħ.
- So ⟨[Lx, Ly]⟩ = iħ², and its modulus is ħ².
- Therefore ΔLx ΔLy ≥ ħ²/2.
Check: The product cannot be zero. This is the commutator showing up as a real limit on measurement. In the state |1, 0⟩, ⟨Lz⟩ = 0 and the bound would be zero.
A short study plan
Spend a few focused sittings on this topic. Do not just read; write each result by hand.
- Sitting 1: Write the commutators from memory. Then explain in one line why L2 and Lz can be sharp together.
- Sitting 2: Fill the eigenvalue table for l = 1, 2, 3 yourself. Compute |L| and the allowed angles each time.
- Sitting 3: Practise ladder operators on several states, including the end points where the answer is zero.
- Sitting 4: Do spin and vector addition. List j values for a few (l, s) pairs, and count the states.
- Sitting 5: Solve timed chemistry questions from the ABC Chemistry question bank on quantum chemistry. Redo every error the next day.
Keep a one-page sheet of formulas. Read it before each practice set.
Key takeaways
- [Lx, Ly] = iħLz (cyclic), while L2 commutes with every component.
- L2 = l(l+1)ħ2 and Lz = mlħ, with ml from −l to +l.
- L± change m by one unit and give zero at the ends of the ladder.
- Spin follows the same algebra but allows half-integer values; electrons have s = ½.
- Always use l(l+1), not l², and keep the ħ in your answer.
Frequently asked questions
Why can't Lx, Ly and Lz all be known together?
Their commutators are not zero. For example, [Lx, Ly] = iħLz. Operators that do not commute have no common set of sharp values, in general.
Is spin a real rotation of the particle?
No. Spin is often depicted as a particle turning on an axis, but this is only a metaphor. Spin is an intrinsic angular momentum with its own operators.
What is the difference between fermions and bosons?
Fermions have half-integer spin, such as the electron with ½. Bosons have integer spin, such as the photon with 1.
Is angular momentum always conserved?
Total angular momentum of a system is constant when no net external torque acts on it. In quantum mechanics, the total angular momentum J is always conserved.
Do I need to derive the ladder operator formula for the exam?
Usually not. Learn the formula, practise using it, and know where it gives zero. Do the derivation once for understanding.
