core35 min studyLesson 377 in Physics

Harmonic-oscillator ladder algebra

Obtain energies and matrix elements from positive norms and operator commutators rather than solving each eigenfunction separately.

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What you’ll understand

  • Factor the oscillator Hamiltonian with dimensionless ladder operators.
  • Use norms to derive allowed occupation numbers and ladder factors.
  • Compute energies and position moments without treating annihilation as a deterministic measurement.

The harmonic oscillator has infinitely many normalizable bound states, yet one commutator organizes all of them. Ladder operators turn the spectral problem into an algebra of positive norms. Their square-root coefficients are essential: they preserve the normalization bookkeeping that links abstract operator actions to probabilities and expectation values. The method also reveals why zero-point energy is unavoidable without assigning a classical trajectory to an energy eigenstate.

Factor a positive Hamiltonian

For mass m, positive angular frequency ω\omega, and the usual full-line position and momentum operators, define a dimensionless annihilation operator from a scaled combination of x and p. Its adjoint reverses the imaginary momentum sign. The canonical commutator gives [a,a†]=1[a,a^\dagger]=1. Multiplication then factors the oscillator Hamiltonian into the number operator plus one half. The algebra is valid on a common invariant dense domain, such as finite linear combinations of oscillator eigenstates; unbounded operators still require domains.

The number operator is nonnegative because ⟨ψ∣a†a∣ψ⟩=∥aψ∥2\langle\psi|a^\dagger a|\psi\rangle=\|a\psi\|^2. Lowering a number eigenstate reduces its eigenvalue by one until a zero-norm lowering action terminates the ladder. A normalizable lowest state therefore satisfies a∣0⟩=0a|0\rangle=0, and repeated raising gives integer labels beginning at zero. Allowing negative occupation would contradict positivity rather than create another branch of bound states. This is an algebraic spectral argument supported by the normalizable ground solution.

a=mω2ℏ x+ip2mℏω,[a,a†]=1,N=a†a,H=ℏω(N+12)a=\sqrt{\frac{m\omega}{2\hbar}}\,x+\frac{i p}{\sqrt{2m\hbar\omega}},\qquad [a,a^\dagger]=1,\qquad N=a^\dagger a,\qquad H=\hbar\omega\left(N+\frac12\right)
x is length, p is momentum, m is mass, ω is rad/s, and ℏω is energy. The ladder operators and N are dimensionless. This factorization assumes the standard positive-stiffness oscillator on L² of the full real line with its physical self-adjoint Hamiltonian.↔ Scroll wide equations horizontally.

Norm factors make the ladder quantitative

For a normalized number state, the squared norm of its lowered vector equals n. The squared norm of its raised vector equals n + 1 because aa†=N+1aa^\dagger=N+1. Dividing by these norms gives normalized adjacent states and fixes all ladder matrix elements. A lowered superposition generally remains unnormalized, so its norm must be calculated before it is used as a state for a new expectation value. Dropping the square roots changes both spectra and matrix elements.

Position is proportional to a+a†a+a^\dagger, and momentum to i(a†−a)i(a^\dagger-a). Orthogonality of number states immediately makes their mean position and mean momentum zero. Applying position twice leaves only number-preserving terms in a diagonal expectation. The resulting nonzero position variance and nonzero ground energy are compatible with stationary probability density. They do not imply that the ground state follows a hidden classical sinusoid whose phase has merely been forgotten.

a∣n⟩=n ∣n−1⟩,a†∣n⟩=n+1 ∣n+1⟩,En=ℏω(n+12),⟨x2⟩n=ℏmω(n+12)a|n\rangle=\sqrt n\,|n-1\rangle,\qquad a^\dagger|n\rangle=\sqrt{n+1}\,|n+1\rangle,\qquad E_n=\hbar\omega\left(n+\frac12\right),\qquad \langle x^2\rangle_n=\frac{\hbar}{m\omega}\left(n+\frac12\right)
n is a nonnegative integer and |n⟩ is normalized. The lowered vacuum is the zero vector, not a state with negative label. The last expression is a squared length and equals position variance because the mean position is zero in a number eigenstate.↔ Scroll wide equations horizontally.

Operator action is not automatically an experiment

Applying a lowering operator is a mathematical operation. A physical photon-subtraction or transition experiment needs a coupling, an outcome model, and normalization conditioned on the outcome. A closed oscillator Hamiltonian does not lower its own number state with time: each energy eigenstate acquires only a phase. An expectation of a noninteger occupation can occur in a superposition or mixture even though a number measurement yields an integer. Mean and possible outcome are distinct concepts.

The algebra is exact for a quadratic potential with positive curvature. An anharmonic term changes both level spacing and the simple ladder relation with energy, even if oscillator functions remain a useful expansion basis. A free particle, inverted oscillator, or finite well cannot inherit this bound spectrum by replacing a parameter name. Recognizing the physical Hamiltonian and its domain is therefore as important as recognizing the commutator identity in a formal calculation.

Lower and measure a superposition

Let ∣ψ⟩=(∣0⟩+3∣2⟩)/2|\psi\rangle=(|0\rangle+\sqrt3|2\rangle)/2 and ℏω=2 meV\hbar\omega=2\,\mathrm{meV}. Check normalization, find mean energy and energy uncertainty, and identify the normalized state proportional to a∣ψ⟩a|\psi\rangle.

  1. Orthogonality gives probabilities P0=1/4P_0=1/4 and P2=3/4P_2=3/4, which sum to one. The occupation moments are ⟨N⟩=0(1/4)+2(3/4)=3/2\langle N\rangle=0(1/4)+2(3/4)=3/2 and ⟨N2⟩=4(3/4)=3\langle N^2\rangle=4(3/4)=3.

  2. The mean energy is (3/2+1/2)ℏω=4 meV(3/2+1/2)\hbar\omega=4\,\mathrm{meV}. The occupation variance is 3−(3/2)2=3/43-(3/2)^2=3/4, so ΔE=ℏω3/4=3 meV≃1.73 meV\Delta E=\hbar\omega\sqrt{3/4}=\sqrt3\,\mathrm{meV}\simeq1.73\,\mathrm{meV}. The mean energy need not equal one eigenvalue.

  3. Lowering kills the vacuum part and gives a∣ψ⟩=(6/2)∣1⟩a|\psi\rangle=(\sqrt6/2)|1\rangle. Its squared norm is 6/4=3/2=⟨N⟩6/4=3/2=\langle N\rangle, providing an independent check. Dividing by its norm leaves the normalized number state ∣1⟩|1\rangle, up to overall phase.

The result

The original state is normalized, with mean energy 4 meV and uncertainty √3 meV. The normalized lowered vector is |1⟩; the unnormalized lowering action is not itself a probability-preserving time evolution.

Explore the model

Interactive lab

Interactive lab · quantum

Represent two selected oscillator amplitudes

At a fixed instant interpret the lab basis as |0⟩ and |2⟩. Reset and set both controls to θ = 120° and φ = 0°; predict probabilities 1/4 and 3/4, matching the example. Reset and set both controls to θ = 90° and φ = 0°; predict equal probabilities. The pure two-state instrument neither supplies the infinite oscillator ladder nor applies a lowering operator; playback is not calibrated to the oscillator gap.

Represent two selected oscillator amplitudesA normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Current readings: Probability |0⟩: 71.1%; Probability |1⟩: 28.9%; Bloch (x, y, z): 0.78, 0.45, 0.42. All parameters are adjustable with the labeled controls below.xyz|0⟩|1⟩Basis probabilitiesP(0)71.1%P(1)28.9%
Bloch vector / P(0)P(1)Phase latitude
Probability |0⟩
71.1%
Probability |1⟩
28.9%
Bloch (x, y, z)
0.78, 0.45, 0.42
t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA two-component population picture can represent a truncated preparation, but oscillator matrix elements and its infinite spectrum require the ladder calculation.

Check your understanding

Why does a normalized oscillator number state acquire a factor √n when acted on by its annihilation operator?

Try it yourself

For the normalized number state ∣3⟩|3\rangle, find the mean energy, energy uncertainty, and position variance in terms of ℏω\hbar\omega and oscillator length ℓ0=ℏ/(mω)\ell_0=\sqrt{\hbar/(m\omega)}. Give the vector a†∣3⟩a^\dagger|3\rangle and its normalized direction.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

A number eigenstate is also an energy eigenstate; use the raised-state norm separately from its final normalization.

Compare with the solution

The energy is (3+1/2)ℏω=(7/2)ℏω(3+1/2)\hbar\omega=(7/2)\hbar\omega, with zero energy uncertainty because this is one exact eigenstate. Its mean position is zero and position variance is (3+1/2)ℓ02=(7/2)ℓ02(3+1/2)\ell_0^2=(7/2)\ell_0^2. Raising gives a†∣3⟩=4∣4⟩=2∣4⟩a^\dagger|3\rangle=\sqrt4|4\rangle=2|4\rangle, whose squared norm is four. After dividing by its norm, the direction is the normalized state ∣4⟩|4\rangle. Zero energy uncertainty does not imply zero position uncertainty, because the two observables have different eigenstates.

Take it with you

  • Positive ladder norms enforce a nonnegative integer number spectrum.
  • Square-root ladder factors control normalization and matrix elements.
  • Stationary number states can have zero energy uncertainty and nonzero position uncertainty.

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