How many bound states does the one-dimensional attractive potential V(x) = −αδ(x) (α > 0) have?

None, because the well has zero width

For a one-dimensional finite square well of depth V₀, which statements are correct?

There is at least one bound state however shallow the well

The number of bound states grows with V₀a²

Bound-state wavefunctions extend into the classically forbidden region

There is at least one bound state however shallow the well

The number of bound states grows with V₀a²

Bound-state wavefunctions extend into the classically forbidden region

The bound-state energies equal those of an infinite well of the same width

For the one-dimensional harmonic oscillator, which statements are correct?

The energy levels are equally spaced

The eigenstates alternate in parity

The ground state is a minimum-uncertainty state

The energy levels are equally spaced

The eigenstates alternate in parity

The ground state is a minimum-uncertainty state

Each level is two-fold degenerate

Quantum Mechanics I: Foundations, Operators in Hilbert Space, One-Dimensional Potentials, Boxes, Delta Potentials and the Harmonic Oscillator

The first of two chapters for Section 7 of the GATE Physics paper. The section is the largest in the syllabus and the most heavily examined, and it divides naturally: this chapter is the formalism and every exactly solvable one-dimensional (and separable box) problem; the second takes central potentials, hydrogen, angular momentum and the approximation methods. Here: the basic ideas and the uncertainty principle; linear vectors and operators in Hilbert space, expectation values and measurement; the time-independent Schrödinger equation; the step potential, the finite rectangular well and tunnelling through a barrier; the particle in one-, two- and three-dimensional boxes; single and double delta-function potentials; and the one-, two- and three-dimensional harmonic oscillator, with the degeneracies that make the counting questions. Constants: h = 6.626 × 10⁻³⁴ J s, ħ = h/2π, m_e = 9.109 × 10⁻³¹ kg, e = 1.602 × 10⁻¹⁹ C.

1. Basic ideas and the uncertainty principle

Matter has a wavelength λ = h/p (de Broglie), light comes in quanta E = hν, and a particle is described by a wavefunction ψ(x, t) whose squared modulus is a probability density (Born): ∫|ψ|² dx = 1. Because a localised wave packet needs a spread of wavenumbers, position and momentum cannot both be sharp: Δx Δp ≥ ħ/2. In general, for any two observables σ_A σ_B ≥ ½|⟨[A, B]⟩|, and [x, p] = iħ gives the Heisenberg form.

The uncertainty principle estimates ground-state energies. For the oscillator, E ≈ (Δp)²/2m + ½mω²(Δx)² with ΔxΔp = ħ/2 is minimised at E = ħω/2, the exact answer (the Gaussian ground state is a minimum-uncertainty state). A particle confined to a box of width L has Δp ≳ ħ/L and so a kinetic energy of order ħ²/(mL²) — the origin of zero-point energy.

2. Hilbert space, operators, expectation values and measurement

States are vectors |ψ⟩ in a Hilbert space with ⟨φ|ψ⟩ = ∫φ*ψ dx; observables are Hermitian operators, so their eigenvalues are real and their eigenvectors complete and orthogonal.
Measurement: expand |ψ⟩ = Σcₙ|n⟩ in the eigenstates of the observable. A measurement returns eigenvalue aₙ with probability |cₙ|², and leaves the system in |n⟩. The expectation value is ⟨A⟩ = ⟨ψ|A|ψ⟩ = Σ|cₙ|²aₙ.
Commutators: [x, p] = iħ, [x, p²] = 2iħp, [H, x] = −iħp/m. Commuting observables share eigenstates and can be measured together.
Time evolution: iħ∂ψ/∂t = Hψ. A stationary state ψₙe^−iEₙt/ħ has time-independent probabilities; a superposition of different energies does not. Ehrenfest’s theorem, d⟨A⟩/dt = (i/ħ)⟨[H, A]⟩ + ⟨∂A/∂t⟩, gives d⟨x⟩/dt = ⟨p⟩/m and d⟨p⟩/dt = −⟨dV/dx⟩.

🧠 Expectation values from the coefficients

ψ = (φ₁ + 2φ₂)/√5 in energy eigenstates: P(E₂) = 4/5 = 0.8 and ⟨E⟩ = (E₁ + 4E₂)/5. For the oscillator state √(1/3) φ₀ + √(2/3) φ₁: ⟨E⟩ = (1/3)(½ħω) + (2/3)(3/2 ħω) = (7/6)ħω ≈ 1.17ħω. No integral is needed once the state is written in the eigenbasis.

3. The step, the finite well and tunnelling

Solve −(ħ²/2m)ψ″ + Vψ = Eψ piecewise and match ψ and ψ′ at each boundary (ψ′ jumps only at an infinite or delta-function potential). For a step of height V₀ with E > V₀, k₁ = √(2mE)/ħ and k₂ = √(2m(E − V₀))/ħ, and R = ((k₁ − k₂)/(k₁ + k₂))² — non-zero, unlike a classical particle. At E = 2V₀, k₂ = k₁/√2 and R = 0.0294. For E < V₀ the wave is totally reflected, but decays as e^−κx into the step, κ = √(2m(V₀ − E))/ħ.

A finite square well of depth V₀ and width 2a always has at least one bound state in one dimension, and the number grows with the strength parameter z₀ = a√(2mV₀)/ħ (one new state for each π/2). Bound states alternate in parity, their energies lie below the infinite-well values, and the wavefunction penetrates the classically forbidden region. For a rectangular barrier of height V₀ and width a with E < V₀, the transmission for κa ≫ 1 is T ≈ 16(E/V₀)(1 − E/V₀)e^−2κa, dominated by e^−2κa. For an electron with E = 1 eV, V₀ = 5 eV and a = 0.5 nm, κ = 1.02 × 10¹⁰ m⁻¹ and e^−2κa ≈ 3.6 × 10⁻⁵ — the exponential sensitivity that a scanning tunnelling microscope exploits.

4. Boxes in one, two and three dimensions, and delta potentials

In an infinite well 0 < x < L, ψₙ = √(2/L) sin(nπx/L) and Eₙ = n²π²ħ²/(2mL²) = n²h²/(8mL²); for an electron in a 1 nm box E₁ = 0.376 eV. In a box the separable solution multiplies: E = (π²ħ²/2m)(n_x²/L_x² + n_y²/L_y² + n_z²/L_z²). In a cube, E = E₀(n_x² + n_y² + n_z²) and degeneracy is the number of ordered triples with the same sum of squares.

The lowest levels of a cubic box
n_x² + n_y² + n_z²	Triples	Degeneracy
3	(1, 1, 1)	1
6	(1, 1, 2) and permutations	3
9	(1, 2, 2) and permutations	3
11	(1, 1, 3) and permutations	3
14	(1, 2, 3) and permutations	6

An attractive delta well V = −αδ(x) has exactly one bound state, ψ = √κ e^−κ|x| with κ = mα/ħ² and E = −mα²/(2ħ²). ψ is continuous but ψ′ jumps by −(2mα/ħ²)ψ(0). The probability of finding the particle within |x| < 1/κ is 1 − e⁻² = 0.865. A double delta well −α[δ(x − a) + δ(x + a)] has an even bound state always and an odd one only if the wells are strong or far enough apart (maα/ħ² > ½) — the one-dimensional model of a bonding and an antibonding orbital. A delta barrier transmits T = 1/(1 + mα²/(2ħ²E)).

5. The harmonic oscillator in one, two and three dimensions

With a = √(mω/2ħ)(x + ip/(mω)) and a†, H = ħω(a†a + ½), [a, a†] = 1, and Eₙ = (n + ½)ħω: equally spaced, with zero-point energy ħω/2. a†|n⟩ = √(n + 1)|n + 1⟩ and a|n⟩ = √n|n − 1⟩, x = √(ħ/2mω)(a + a†), so ⟨n|x|n⟩ = 0 and ⟨n|x²|n⟩ = (ħ/2mω)(2n + 1). The eigenfunctions are Hermite polynomials times a Gaussian, with parity (−1)ⁿ, and ⟨T⟩ = ⟨V⟩ = Eₙ/2 in every eigenstate.

Two dimensions (isotropic): E = (n_x + n_y + 1)ħω = (N + 1)ħω, degeneracy N + 1.
Three dimensions (isotropic): E = (N + 3/2)ħω, degeneracy (N + 1)(N + 2)/2 — 1, 3, 6, 10 for N = 0, 1, 2, 3.
Anisotropic frequencies break the degeneracy; it is a consequence of symmetry, and counting questions always state whether ω_x = ω_y = ω_z.

Key takeaways

σ_Aσ_B ≥ ½|⟨[A, B]⟩|; measurement gives aₙ with probability |cₙ|², and ⟨A⟩ = Σ|cₙ|²aₙ.
Step: R = ((k₁ − k₂)/(k₁ + k₂))² even for E > V₀; barrier: T ≈ e^−2κa with κ = √(2m(V₀ − E))/ħ.
Box: Eₙ = n²h²/(8mL²); a cube’s level 14E₀ is six-fold degenerate. A 1D finite well always binds at least one state.
A delta well binds exactly one state, E = −mα²/(2ħ²); a double delta well binds one or two.
Oscillator: Eₙ = (n + ½)ħω, ⟨x²⟩ = (ħ/2mω)(2n + 1); 3D degeneracy (N + 1)(N + 2)/2, 2D N + 1.

Practice questions (13)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The commutator [x, p²] equals:
1. 2iħp
2. iħ
3. −2iħp
4. 0
Show answer
Answer: A — 2iħp
[x, p²] = [x, p]p + p[x, p] = iħp + piħ = 2iħp. Treating [x, p²] as [x, p]² gives −ħ², and assuming it vanishes forgets that p² contains p.
A particle is in the state ψ = (φ₁ + 2φ₂)/√5, where φ₁ and φ₂ are normalised energy eigenstates. The probability that an energy measurement gives E₂ is ____.
Numerical answer — type the value.
Show answer
Answer: 0.8
c₂ = 2/√5, so P(E₂) = |c₂|² = 4/5 = 0.8. Taking the amplitude 2/√5 = 0.894 as the probability forgets to square it; 2/3 comes from comparing coefficients without normalising.
A harmonic oscillator is in the state √(1/3) φ₀ + √(2/3) φ₁. Its expectation value of energy, in units of ħω, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 1.17
⟨E⟩ = (1/3)(0.5) + (2/3)(1.5) = 0.1667 + 1.0 = 1.1667ħω → 1.17. Weighting by the amplitudes instead of their squares gives 0.577 × 0.5 + 0.816 × 1.5 = 1.51; using Eₙ = nħω gives 0.67.
An electron is confined in a one-dimensional infinite well of width 1.0 nm. Its ground-state energy, in eV, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.38
E₁ = h²/(8mL²) = (6.626 × 10⁻³⁴)²/(8 × 9.109 × 10⁻³¹ × 10⁻¹⁸) = 6.03 × 10⁻²⁰ J = 0.376 eV → 0.38. Using ħ in place of h in h²/(8mL²) gives a value 4π² times too small, 0.0095 eV.
A particle is in a cubical box, with energies E = E₀(n_x² + n_y² + n_z²). The degeneracy of the level E = 14E₀ is ____.
Numerical answer — type the value.
Show answer
Answer: 6
14 = 1 + 4 + 9 only, i.e. (1, 2, 3), and its 3! = 6 orderings are distinct states. Levels with a repeated quantum number, such as 6E₀ from (1, 1, 2), are only three-fold.
For an isotropic three-dimensional harmonic oscillator, the degeneracy of the energy level (9/2)ħω is ____.
Numerical answer — type the value.
Show answer
Answer: 10
E = (N + 3/2)ħω = (9/2)ħω gives N = 3, and the degeneracy is (N + 1)(N + 2)/2 = 4 × 5/2 = 10. Reading 9/2 as n + ½ in one dimension gives n = 4 and degeneracy 1; the 2D formula N + 1 gives 4.
An electron of energy 1.0 eV meets a rectangular barrier of height 5.0 eV and width 0.50 nm. Using T ≈ e^−2κa with κ = √(2m(V₀ − E))/ħ, the value of T × 10⁵, to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 3.6
V₀ − E = 4 eV = 6.408 × 10⁻¹⁹ J; κ = √(2 × 9.109 × 10⁻³¹ × 6.408 × 10⁻¹⁹)/1.0546 × 10⁻³⁴ = 1.025 × 10¹⁰ m⁻¹; 2κa = 10.25, T = e^−10.25 = 3.6 × 10⁻⁵. Using V₀ instead of V₀ − E gives 2κa = 11.45 and T ≈ 1.1 × 10⁻⁵.
A beam of particles with energy E = 2V₀ meets a potential step of height V₀. The reflection coefficient, to three decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.029
k₂/k₁ = √((E − V₀)/E) = 1/√2 = 0.7071, so R = ((1 − 0.7071)/(1 + 0.7071))² = (0.1716)² = 0.029. Classically R = 0; quoting the amplitude 0.172 forgets to square it.
A particle is bound in the attractive delta potential V = −αδ(x), with ground state ψ = √κ e^−κ|x|. The probability of finding it in −1/κ < x < 1/κ, to three decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.865
P = 2∫₀^1/κ κ e^−2κx dx = 1 − e⁻² = 1 − 0.1353 = 0.865. Using |ψ| instead of |ψ|² (so e^−κx) gives 1 − e⁻¹ = 0.632.
How many bound states does the one-dimensional attractive potential V(x) = −αδ(x) (α > 0) have?
1. Exactly one, for any α > 0
2. None, because the well has zero width
3. Infinitely many
4. One or two, depending on α
Show answer
Answer: A — Exactly one, for any α > 0
The only normalisable even solution is √κ e^−κ|x| with κ = mα/ħ², which exists for every α > 0; an odd solution would vanish at x = 0 and never feel the potential. "One or two" describes the double delta well.
For a one-dimensional finite square well of depth V₀, which statements are correct?
1. There is at least one bound state however shallow the well
2. The number of bound states grows with V₀a²
3. Bound-state wavefunctions extend into the classically forbidden region
4. The bound-state energies equal those of an infinite well of the same width
Show answer
Answer: A — There is at least one bound state however shallow the well; B — The number of bound states grows with V₀a²; C — Bound-state wavefunctions extend into the classically forbidden region
In one dimension the even state always exists; the count is set by z₀ = a√(2mV₀)/ħ, so by V₀a²; outside the well ψ ∝ e^−κ|x| is non-zero. Because ψ leaks out, the effective width is larger and each energy lies below the infinite-well value.
For the one-dimensional harmonic oscillator, which statements are correct?
1. The energy levels are equally spaced
2. The eigenstates alternate in parity
3. The ground state is a minimum-uncertainty state
4. Each level is two-fold degenerate
Show answer
Answer: A — The energy levels are equally spaced; B — The eigenstates alternate in parity; C — The ground state is a minimum-uncertainty state
Eₙ = (n + ½)ħω is equally spaced; ψₙ has parity (−1)ⁿ; the Gaussian ground state has ΔxΔp = ħ/2 exactly. Bound states in one dimension are never degenerate, so the last statement is false.
Which property guarantees that every measured value of an observable is a real number?
1. The operator is Hermitian
2. The operator is unitary
3. The operator commutes with H
4. The operator is linear
Show answer
Answer: A — The operator is Hermitian
A Hermitian operator has ⟨ψ|Aψ⟩ = ⟨Aψ|ψ⟩, which forces its eigenvalues — the possible results — to be real. A unitary operator has eigenvalues of modulus 1, generally complex; commuting with H makes a quantity conserved, not real.