The Hamiltonian operator
Swap momentum for a derivative; energy becomes an operator
The Schrödinger equation is only as good as its right-hand side, and this topic builds it. Take the classical energy, swap momentum for a derivative, then learn to read what the resulting operator does to a wavefunction, and which physical systems it cannot describe.
Build the model
Trade momentum for a derivative, and the classical energy becomes an operator.
1The big idea
The time-dependent Schrödinger equation iħ ∂Ψ/∂t = ĤΨ is empty until Ĥ is named, and the name is borrowed from classical mechanics. Write the energy as a function of position and momentum, H = p²/2m + V(x), then let x act by multiplication and replace p by −iħ ∂/∂x. The substitution is not arbitrary: a de Broglie wave e(i(kx − ωt)) carries p = ħk, and −iħ ∂/∂x is the derivative instruction that returns exactly ħk times such a wave. Applied twice it gives p̂² = −ħ² ∂²/∂x², so Ĥ = −(ħ²/2m) ∂²/∂x² + V(x). What comes out is an instruction, not a number. Ĥψ is a new function whose kinetic part measures how sharply ψ bends and whose potential part reweights ψ point by point, and only special ψ come back as a constant times themselves. The recipe is unambiguous because p²/2m + V(x) never multiplies x by p, whose operators do not commute, and it is safe because V is a real, local function of position alone. It stops there. A complex V fails to conserve probability; a velocity-dependent force such as qv × B needs p replaced by p − qA before quantising; spin and several particles need a wavefunction richer than one function ψ(x) of one coordinate; and p²/2m itself fails once p approaches mc.
2In one sentence
The Hamiltonian operator is the classical energy p²/2m + V(x) with momentum replaced by −iħ ∂/∂x, giving Ĥ = −(ħ²/2m) ∂²/∂x² + V(x): the instruction that turns a wavefunction into the function driving its time evolution.
Example. An electron (ħ²/2mₑ = 0.0381 eV nm²) with ψ = cos(kx), k = 10.0 nm⁻¹ and V = 0: ψ″ = −k²ψ, so Ĥψ = (0.0381 eV nm²)(100 nm⁻²)ψ = 3.81 eV × ψ, the classical p²/2m with p = ħk.
3Start from the energy written in x and p
The previous topic's equation, iħ ∂Ψ/∂t = ĤΨ, names Ĥ without building it. The construction starts classically. Write the particle's energy as a function of position and momentum, not velocity: H(x, p) = p²/2m + V(x), the classical Hamiltonian. For a 0.20 kg mass on a 5.0 N m⁻¹ spring at x = 0.10 m with p = 0.12 kg m s⁻¹ it evaluates to 0.036 J + 0.025 J = 0.061 J, a number. Quantum mechanics keeps the shape of this function and changes what its letters mean: x becomes the instruction 'multiply by x', p becomes an instruction p̂, and the energy function becomes an energy operator. Momentum rather than velocity is the variable to keep, because the de Broglie relation p = ħk ties momentum directly to the wave, so it is the quantity with a wave-side instruction ready to replace it.
4Why momentum becomes −iħ ∂/∂x
Take the de Broglie wave Ψ = e(i(kx − ωt)) = cos(kx − ωt) + i sin(kx − ωt), a particle of definite momentum p = ħk moving toward +x. Differentiating in x brings down ik: ∂Ψ/∂x = ikΨ. Multiply by −iħ and the i's cancel, (−iħ)(ik) = ħk, so −iħ ∂Ψ/∂x = ħkΨ = pΨ. That is the argument for p̂ = −iħ ∂/∂x: it is the derivative instruction that returns each plane wave multiplied by its own momentum, and because it is linear it then acts correctly on any superposition of them. The i is essential, since differentiating a wave returns i times a real rate. The sign is tied to the time convention: e(−iωt) gives iħ ∂Ψ/∂t = ħωΨ, a positive energy, and with +iħ ∂/∂x instead the same wave would report −ħk, a right-moving particle with leftward momentum.
5Square the momentum by acting twice
p̂² means apply p̂, then apply p̂ to the result: p̂²ψ = (−iħ)(−iħ)ψ″ = −ħ²ψ″, because (−i)² = −1. So the kinetic term is T̂ = −(ħ²/2m) ∂²/∂x², and in three dimensions the same step gives −(ħ²/2m)∇². The minus sign is load-bearing. For ψ = cos(kx), ψ″ = −k²ψ and T̂ψ = +(ħ²k²/2m)ψ, positive as a kinetic energy of a plane wave must be; lose the sign and every plane wave carries negative kinetic energy. Two shortcuts are wrong. Squaring the result, (p̂ψ)², gives −ħ²k² sin²(kx) for that same cosine: it is quadratic in ψ, so doubling ψ would quadruple the term and the Schrödinger equation would stop being linear. And (−iħ)² is −ħ², not −iħ² or +ħ².
6Read Ĥψ as curvature plus a pointwise product
Ĥψ(x) = −(ħ²/2m)ψ″(x) + V(x)ψ(x) is a new function, and its two terms work differently. The potential term is local: its value at x₀ needs only V(x₀) and ψ(x₀). The kinetic term needs ψ's neighbourhood, because ψ″ measures how sharply ψ bends. Where ψ curves back toward the axis, ψ″ has the opposite sign to ψ and −(ħ²/2m)ψ″/ψ is positive; beyond an inflection point, where ψ curves away from the axis, it is negative. For an electron in ψ = A e(−x²/2a²) with a = 0.20 nm and V = 0, normalised on the whole line by A = (πa²)(−1/4), which cancels from any ratio, T̂ψ/ψ = (ħ²/2ma²)(1 − x²/a²) is +0.952 eV at the centre, zero at x = ±a and −2.86 eV at x = 2a. Ĥψ has a different shape from ψ, so this Gaussian has no single energy. The functions for which Ĥψ is a constant times ψ are special, and finding them is the job of the next topics.
7Check dimensions, scale and classical correspondence
Every Ĥ you write should pass three checks. Dimensions: [ħ²/2m] = (J s)²/kg = J m², and ∂²/∂x² adds m⁻², so the kinetic term returns J × [ψ], just as V(x)ψ does; iħ ∂Ψ/∂t carries J s × s⁻¹ = J as well, so the equation balances. Scale: ħ²/2m is 6.10 × 10⁻³⁹ J m² = 0.0381 eV nm² for an electron, which is why electrons bent over 0.1 nm carry a few eV; for a proton it is 1836 times smaller, 20.7 MeV fm², which is why nucleons squeezed into femtometre nuclei carry MeV. Correspondence: where V = V₀ is constant, Ĥ e(ikx) = (ħ²k²/2m + V₀) e(ikx), the classical p²/2m + V₀ with p = ħk. And adding a constant to V adds that constant to every energy without changing any force, exactly as it does classically.
8Where the recipe stops working
The substitution has one answer only because p²/2m + V(x) never multiplies x by p. Classically xp = px, but x̂p̂ψ = −iħxψ′ while p̂x̂ψ = −iħ(ψ + xψ′), so a classical energy containing xp has no unique quantum version. V must be real, or probability leaks, as the continuity equation will show; local, so that Vψ at x needs only ψ at x; and a function of position alone. A magnetic force qv × B depends on velocity and comes from no V(x). There the classical Hamiltonian is (p − qA)²/2m + qφ, with p the canonical momentum rather than mv, and quantising that, called minimal coupling, replaces p̂²/2m + V. The form also assumes one slow, spinless particle: spin needs its own operators, N particles need ψ(x₁, …, xN) with one kinetic term each, and p²/2m fails once p approaches mc.
Key equations
What each relation says, and the units it is written in.
Momentum becomes a derivative
p̂ = −iħ ∂/∂x
Fixed by the de Broglie wave: −iħ ∂/∂x acting on e(ikx) returns ħk e(ikx). The opposite sign would report a right-moving wave as moving left.
ħ = 1.055 × 10⁻³⁴ J s; ∂/∂x carries m⁻¹, so p̂ψ has units kg m s⁻¹ × [ψ]
Kinetic energy operator
p̂²/2m = −(ħ²/2m) ∂²/∂x²
Kinetic energy becomes curvature: T̂ψ/ψ is positive where ψ bends back toward the axis, negative where it bends away, zero where ψ is straight.
p̂² means p̂ applied twice, and (−i)² = −1 supplies the minus sign; m in kg
The Hamiltonian in one dimension
Ĥ = −(ħ²/2m) ∂²/∂x² + V(x)
ĤΨ is the right-hand side of iħ ∂Ψ/∂t = ĤΨ, so this operator decides how any wavefunction changes in the next instant.
V(x) in J, real and a function of x alone; both terms return J × [ψ]
Kinetic prefactor for an electron
ħ²/2mₑ = 6.10 × 10⁻³⁹ J m² = 0.0381 eV nm²
Divide by the squared length over which ψ bends to get the kinetic scale: 0.0381 eV at 1 nm, 3.81 eV at 0.1 nm.
mₑ = 9.109 × 10⁻³¹ kg; the proton value is 1836 times smaller, 20.7 MeV fm²
Correspondence check with a plane wave
Ĥ e(ikx) = (ħ²k²/2m + V₀) e(ikx)
With p = ħk the multiplier is p²/2m + V₀, the classical energy. Any proposed Ĥ must pass this test in a uniform region.
valid where V = V₀ is constant; k in m⁻¹, V₀ in J
Minimal coupling for magnetic forces
Ĥ = (−iħ∇ − qA)²/2m + qφ
The magnetic force qv × B depends on velocity, so no V(x) can supply it. Shift the momentum by qA before quantising; this Ĥ replaces p̂²/2m + V.
q in C; A the vector potential in T m; φ the scalar potential in V
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Start at s = 0.70 ℓ, κ = 1.0: the solid Ĥψ dips below zero where ψ never does. Slide s to 1.00 ℓ and it lands exactly on the dashed curve, because T̂ψ and Vψ now add to ½ψ. Then set κ = 0: with no potential, no width makes the two curves agree.
Live results
- Ĥψ/ψ AT x = 0
- 1.020 ε
- Ĥψ/ψ AT x = ℓ
- -0.562 ε
- MISMATCH κ − 1/s⁴
- -3.165 ε/ℓ²
- T̂ψ NEGATIVE FOR |x| BEYOND
- 0.70 ℓ
Catch the common trap
The mistake this lesson is built to prevent.
The commonest slip is to read Ĥ as 'the energy' and write Ĥψ = Eψ for whatever ψ is on the page, as if applying Ĥ multiplied by a number. It does not. Ĥ is an instruction: differentiate twice, scale by −ħ²/2m, add V(x) times ψ, and the output is generally a different function. In a box of width L, sin(πx/L) and sin(2πx/L) each satisfy Ĥψ = Eψ on their own, with energies E₁ and 4E₁. Their sum does not: Ĥ returns E₁ sin(πx/L) + 4E₁ sin(2πx/L), which is no single number times sin(πx/L) + sin(2πx/L). Ĥψ = Eψ is an equation that picks out special functions and has to be solved, not assumed. The time-dependent equation iħ ∂Ψ/∂t = ĤΨ holds for every state; the eigenvalue equation holds only for the special ones.
Test it: what changes?
- Ĥψ/ψ at x = 0, 1/(2s²) ε, falls
- T̂ψ sign change at x = ±s moves outward
- Mismatch κ − 1/s⁴ increases
- Ĥψ/ψ at x = 0 unchanged
- Ĥψ/ψ at x = ℓ rises
- Mismatch κ − 1/s⁴ increases
- T̂ψ still changes sign at x = ±s
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
Inside a nucleus a neutron (mₙc² = 939.6 MeV) is modelled by a constant potential V = −45.0 MeV, and there its wavefunction is ψ(x) = A sin(kx) with k = 0.800 fm⁻¹. Apply Ĥ, show that the result is a constant times ψ, find the constant, and check it against the classical energy with p = ħk. Use ħc = 197.3 MeV fm.
View solution →
- Write the operator: Ĥ = −(ħ²/2mₙ) ∂²/∂x² + V, where the constant V = −45.0 MeV simply multiplies ψ.
- Differentiate twice: ψ″ = −k²A sin(kx) = −k²ψ, so the kinetic term gives −(ħ²/2mₙ)(−k²ψ) = +(ħ²k²/2mₙ)ψ.
- Evaluate the prefactor in nuclear units: ħ²/2mₙ = (ħc)²/(2mₙc²) = (197.3 MeV fm)²/(2 × 939.6 MeV) = 20.71 MeV fm². Multiplying by k² = 0.640 fm⁻² gives 13.26 MeV; MeV fm² × fm⁻² = MeV, the same unit as V.
- So Ĥψ = (13.26 − 45.0) MeV × ψ = −31.7 MeV × ψ, a constant times ψ everywhere in the region.
- Classical check: pc = ħck = 0.800 × 197.3 = 157.8 MeV, and p²/2m = (pc)²/(2mₙc²) = 13.26 MeV, so the constant is exactly p²/2m + V. Since pc is only 0.17 mₙc², p²/2m is adequate; the relativistic kinetic energy is about 0.7% lower.
Answer. Ĥψ = −31.7 MeV × ψ: a kinetic 13.3 MeV plus the potential −45.0 MeV, identical to the classical p²/2m + V with p = ħk.
MediumExample 2
An electron is held in 0 ≤ x ≤ L, L = 0.500 nm, by infinite walls, with V = 0 inside. A student proposes ψ(x) = x(L − x), which vanishes at both walls like the true ground state. Apply Ĥ, evaluate Ĥψ/ψ at x = L/2 and at x = L/10, and decide whether ψ has a definite energy. Compare with sin(πx/L). Use ħ²/2mₑ = 0.0381 eV nm².
View solution →
- Inside the box Ĥ = −(ħ²/2m) d²/dx². Here ψ″ = −2, a constant, so Ĥψ = ħ²/m = 0.0762 eV nm² at every x. Units: ψ carries nm², so eV nm² is energy × [ψ], as it must be. The normalising factor √(30/L⁵), from ∫₀ᴸ x²(L − x)² dx = L⁵/30, would multiply ψ and Ĥψ alike and cancels from Ĥψ/ψ, so it is left out.
- At x = L/2: ψ = 0.250 nm × 0.250 nm = 0.0625 nm², so Ĥψ/ψ = 0.0762/0.0625 = 1.22 eV.
- At x = L/10 = 0.0500 nm: ψ = 0.0500 nm × 0.450 nm = 0.0225 nm², so Ĥψ/ψ = 0.0762/0.0225 = 3.39 eV. Toward either wall ψ → 0 while Ĥψ stays at 0.0762 eV nm², so the ratio grows without bound.
- A ratio that changes with x means Ĥψ is not a constant times ψ, so the parabola has no single energy. The curvature shows why: a parabola bends by the same amount everywhere, while ψ itself shrinks toward the walls.
- Contrast sin(πx/L): its second derivative is −(π/L)² times itself, so Ĥψ/ψ = π²(ħ²/2m)/L² = 9.870 × 0.0381/0.250 = 1.50 eV at every x. The sine's curvature shrinks in step with the function, which is exactly what the parabola lacks.
Answer. Ĥψ = ħ²/m = 0.0762 eV nm² everywhere, so Ĥψ/ψ is 1.22 eV at the centre and 3.39 eV at x = L/10: no definite energy. The sine gives 1.50 eV at every point.
HardExample 3
A rubidium-87 atom (m = 1.4432 × 10⁻²⁵ kg) is released from an optical trap at t = 0, when its wavefunction is the real Gaussian Ψ(x, 0) = A e(−x²/2s²) with s = 1.00 μm and A = (πs²)(−1/4); from then on V = 0. Use Ĥ and iħ ∂Ψ/∂t = ĤΨ to find ∂Ψ/∂t at x = 0, s and 2s. Show that |Ψ|² does not change at first, and say what does change during the first 0.100 ms. Use ħ = 1.0546 × 10⁻³⁴ J s.
View solution →
- With V = 0, Ĥ = −(ħ²/2m) ∂²/∂x². Differentiating twice, Ψ′ = −(x/s²)Ψ and Ψ″ = (x²/s⁴ − 1/s²)Ψ, so at t = 0 ĤΨ = (ħ²/2ms²)(1 − x²/s²)Ψ. A cancels from every ratio; its value makes A²∫e(−x²/s²) dx = A²√π s = 1 over the whole line.
- Divide by iħ: ∂Ψ/∂t = −iω₀(1 − x²/s²)Ψ with ω₀ = ħ/(2ms²) = (1.0546 × 10⁻³⁴ J s)/(2 × 1.4432 × 10⁻²⁵ kg × 1.00 × 10⁻¹² m²) = 365 s⁻¹. Units: J s/(kg m²) = s⁻¹.
- At x = 0, ∂Ψ/∂t = −i(365 s⁻¹)Ψ; at x = s the bracket vanishes and ∂Ψ/∂t = 0; at x = 2s the bracket is −3, so ∂Ψ/∂t = +i(1.10 × 10³ s⁻¹)Ψ.
- Each rate is i times a real number times Ψ, so ∂|Ψ|²/∂t = Ψ*(∂Ψ/∂t) + Ψ(∂Ψ*/∂t) = −iω(x)|Ψ|² + iω(x)|Ψ|² = 0 at every x, where ω(x) = ω₀(1 − x²/s²). At t = 0 neither the density nor the normalisation is changing. What changes is the phase: to first order in t, Ψ ≈ Ψ(x, 0) e(−iω(x)t).
- After 0.100 ms the phase at the centre has turned by −ω₀t = −0.0365 rad, at x = s not at all, and at x = 2s by +3ω₀t = +0.110 rad. A phase rising with x looks locally like e(ikx) with k > 0; here the phase rises with |x| on both sides, so Ψ has acquired momentum pointing away from the centre, and |Ψ|² is about to widen, as a released gas cloud expands classically. Checks: ħω₀ = 3.85 × 10⁻³² J is an energy, as ĤΨ/Ψ must be; as s → ∞, ω₀ → 0 and a very wide Gaussian, nearly a zero-momentum plane wave, barely evolves; and only if ω(x) were the same at every x would Ψ merely rotate as a whole, which it does not once the trap is off.
Answer. ∂Ψ/∂t = −iω₀(1 − x²/s²)Ψ with ω₀ = ħ/2ms² = 365 s⁻¹: −i(365 s⁻¹)Ψ at the centre, 0 at x = s and +i(1.10 × 10³ s⁻¹)Ψ at x = 2s. |Ψ|² is momentarily unchanged, but after 0.100 ms the phase at 2s exceeds the centre's by 0.146 rad: momentum pointing outward, the start of spreading.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Replace p by −iħ ∂/∂x; Ĥψ is a new function.
Build Ĥ = −(ħ²/2m)∂²/∂x² + V(x) by applying p̂ twice, never by squaring p̂ψ. Read the kinetic term as curvature and Vψ as a pointwise product. The recipe needs V real, local and a function of x alone; magnetic forces need minimal coupling instead.