Questions: Ladder Operators for the Harmonic Oscillator

Question 1

The ground state |0⟩ satisfies â|0⟩ = 0. Applying the Hamiltonian H = ℏω(â†â + 1/2) to |0⟩, what is the ground state energy, and where does the 1/2 come from?

• A. E₀ = 0, because â|0⟩ = 0 means the ground state has no energy
• B. E₀ = ℏω/2, because â†â|0⟩ = 0 so H|0⟩ = ℏω(0 + 1/2)|0⟩ = (ℏω/2)|0⟩; the 1/2 comes from the commutator [â, â†] = 1 when factoring the Hamiltonian
• C. E₀ = ℏω, because the ground state contains one quantum of energy
• D. E₀ = ℏω/2 only if the wavefunction is Gaussian; for other ground states E₀ differs

Answer: (B)

â†â|0⟩ = â†(â|0⟩) = â†(0) = 0, so the number operator gives eigenvalue 0. Thus H|0⟩ = ℏω(0 + 1/2)|0⟩ = (ℏω/2)|0⟩. The 1/2 arises from the commutator correction when factoring: (â†â + 1/2) comes from working out that p̂²/2m + mω²x̂²/2 = ℏω(â†â + 1/2), where the extra +1/2 is the commutator [â, â†] = 1 contributing ℏω/2. This zero-point energy is fixed by algebra alone — no differential equation is required.

Question 2

How does the commutation relation [â, â†] = 1 guarantee that if |n⟩ is an energy eigenstate with eigenvalue Eₙ, then â†|n⟩ is an eigenstate with eigenvalue Eₙ + ℏω?

• A. It doesn't — you need to solve the Schrödinger equation to verify the energy of â†|n⟩
• B. From [â, â†] = 1 you can derive H↠= â†(H + ℏω), so H(â†|n⟩) = â†H|n⟩ + ℏωâ†|n⟩ = (Eₙ + ℏω)(â†|n⟩)
• C. The commutation relation sets the spacing between energy levels by convention, not by calculation
• D. â†|n⟩ is not an eigenstate; it is a superposition of energy eigenstates

Answer: (B)

From [â, â†] = 1 and H = ℏω(â†â + 1/2), compute the commutator [H, â†] = ℏω[â†â, â†] = ℏω(â†[â, â†]) = ℏωâ†. So H↠= â†H + ℏωâ†. Applying this to |n⟩: H(â†|n⟩) = (â†H + ℏωâ†)|n⟩ = â†(Eₙ|n⟩) + ℏω(â†|n⟩) = (Eₙ + ℏω)(â†|n⟩). The commutation relation alone — not a differential equation — proves â†|n⟩ is an eigenstate with energy exactly ℏω higher.

Question 3

The zero-point energy ℏω/2 of the quantum harmonic oscillator should be derived by solving the Schrödinger differential equation for the Hermite polynomial ground state wavefunction.

• T. True
• F. False

Answer: False

The zero-point energy follows directly from the algebraic requirement that â|0⟩ = 0 (the ladder must have a lowest rung) and the form of the Hamiltonian H = ℏω(â†â + 1/2). Applying H to the ground state gives H|0⟩ = ℏω(0 + 1/2)|0⟩ = (ℏω/2)|0⟩ with no differential equations involved. The Hermite polynomial wavefunction is the coordinate-space representation of |0⟩, but the energy eigenvalue is determined purely by the algebra. This is the key point of the ladder operator method: the spectrum is fixed by operator algebra, not by solving differential equations.

Question 4

The number operator n̂ = â†â has eigenvalues n = 0, 1, 2, ... where n counts the number of energy quanta ℏω above the ground state.

• T. True
• F. False

Answer: True

The ladder structure forces this. Starting from â|0⟩ = 0 (ground state, n = 0), successive application of ↠generates states |n⟩ = (â†)ⁿ|0⟩/√(n!) with n̂|n⟩ = n|n⟩ and energy Eₙ = ℏω(n + 1/2). Each application of ↠adds one quantum ℏω of energy; each application of â removes one. The number operator counts exactly how many times ↠has been applied to the ground state — how many 'quanta' are present. This language directly generalizes to quantum field theory, where ↠creates and â destroys particles.

Question 5

Why must the energy spectrum of the harmonic oscillator be bounded below, and how does this force the existence of a ground state from the algebraic structure alone?

Model answer: The energy is bounded below by zero because the Hamiltonian H = p̂²/2m + mω²x̂²/2 is a sum of squares of Hermitian operators — kinetic energy and potential energy are both non-negative. Algebraically: H = ℏω(â†â + 1/2) ≥ ℏω/2 > 0 for all states (since â†â has non-negative expectation value). Now, applying â repeatedly to any energy eigenstate produces states of energy Eₙ - ℏω, Eₙ - 2ℏω, ... This chain would eventually produce a state with negative energy, contradicting the lower bound, unless it terminates. It must terminate: there exists a state |0⟩ for which â|0⟩ = 0, stopping the ladder. This state is the ground state, and its energy ℏω/2 is fixed by applying H.

The argument has the structure of a reductio: the energy is bounded below (physical argument from non-negativity of kinetic + potential energy), but the lowering operator â would generate an infinite descending chain of energy eigenstates unless the chain terminates. Termination requires â|0⟩ = 0 for some |0⟩. This is not an assumption — it follows from the physics. Once â|0⟩ = 0 is established, the ground state energy is determined by algebra. The elegance of this argument is that it derives the entire spectrum from two ingredients: the commutation relation [â, â†] = 1 and the non-negativity of energy.