Questions: Ladder Operators for the Harmonic Oscillator

Question 1

The lowering operator â is applied repeatedly to some eigenstate |n⟩. What determines when this process terminates?

• A. When n equals the zero-point energy ½ℏω, the operator returns a state with zero energy
• B. When n = 0, because â|0⟩ = 0 — the ground state is annihilated rather than mapped to a negative-n state
• C. When the eigenvalue becomes negative, signaling the state has no physical meaning
• D. When n = 1, because no quantum number below 1 can have positive energy

Answer: (B)

The number operator N̂ = â†â is positive semidefinite — its eigenvalues cannot be negative. Repeatedly applying â would lower the eigenvalue by 1 each time, eventually producing a negative eigenvalue unless the sequence terminates. It terminates at n = 0 because â|0⟩ = 0 (the zero vector, not a new state), which is consistent with N̂|0⟩ = 0. This termination condition defines the ground state and establishes the zero-point energy ½ℏω algebraically, without solving any differential equation.

Question 2

Using the ladder operator expression for x̂, which matrix element ⟨m|x̂|n⟩ is nonzero?

• A. ⟨3|x̂|3⟩ — the diagonal element dominates in the position basis
• B. ⟨3|x̂|4⟩ — because x̂ connects states differing by exactly one quantum number
• C. ⟨2|x̂|4⟩ — because the states are separated by two quanta of excitation
• D. ⟨0|x̂|5⟩ — because the ground state has the widest spatial distribution

Answer: (B)

Writing x̂ = √(ℏ/2mω)(â + â†), we see that x̂ maps |n⟩ to a linear combination of |n−1⟩ and |n+1⟩. So ⟨m|x̂|n⟩ is nonzero only when |m − n| = 1. This selection rule is a direct algebraic consequence of the ladder structure: ⟨3|x̂|4⟩ has |3−4| = 1 and is nonzero; the other options have |m−n| = 0, 2, or 5 and all vanish.

Question 3

The energy spectrum of the quantum harmonic oscillator can be derived entirely from the commutation relation [â, â†] = 1, without solving any differential equation.

• T. True
• F. False

Answer: True

This is the central point of the ladder operator approach. From [â, â†] = 1 alone, one can prove that if |n⟩ is an eigenstate of N̂ = â†â with eigenvalue n, then â†|n⟩ has eigenvalue n+1 and â|n⟩ has eigenvalue n−1. Positive-semidefiniteness of N̂ forces termination at n = 0, establishing the ground state. The entire spectrum En = ℏω(n + ½) for n = 0, 1, 2, ... follows from algebra — Hermite polynomials never appear.

Question 4

Applying the lowering operator â to the ground state |0⟩ produces a new quantum state with energy −½ℏω.

• T. True
• F. False

Answer: False

â|0⟩ = 0 — the zero vector, not a normalizable quantum state. There is no state below the ground state because N̂ cannot have negative eigenvalues. The ground state is defined precisely as the state annihilated by â. This is why the energy spectrum has a lowest rung (E₀ = ½ℏω) but no highest rung — there is no upper termination condition because the raising operator ↠never annihilates any state.

Question 5

Why must the ladder of energy levels have a lowest rung (ground state) but no highest rung?

Model answer: The ladder terminates at the bottom because the number operator N̂ = â†â is positive semidefinite — its eigenvalues cannot be negative. Repeatedly applying â would subtract 1 from the eigenvalue at each step, eventually producing a negative eigenvalue, which is impossible. The sequence terminates at n = 0 because â|0⟩ = 0. There is no corresponding upper termination because applying ↠to any eigenstate always produces a valid normalized state with eigenvalue n+1; no physical law prevents arbitrarily large n.

This argument is purely algebraic — it does not require knowing anything about wavefunctions or Hermite polynomials. The key steps are: (1) N̂ is positive semidefinite because N̂ = â†â and ⟨ψ|â†â|ψ⟩ = ‖â|ψ⟩‖² ≥ 0; (2) â lowers the eigenvalue by 1; (3) the sequence must terminate, and the termination condition â|0⟩ = 0 defines the ground state uniquely.