Kinetic energy, potential energy and the Hamiltonian
Differentiate twice for kinetic, multiply by V for potential.
Energy is where operator algebra earns its keep. Build T̂ by letting p̂ act twice, add V(x) as a plain multiplication, and learn the two conditions the construction quietly needs: a potential that depends on x alone, so the translation is unambiguous, and a real one, so the sum is Hermitian.
Build the model
Energy as an instruction: differentiate twice for kinetic, multiply by V for potential.
1The big idea
Classically the energy of a particle on a line is H = p²/2m + V(x), and the quantum rule replaces p by p̂ = −iħ d/dx. For this H the replacement leaves nothing to choose. Kinetic energy is p̂ applied twice and divided by 2m, and because (−i)² = −1 the i's cancel: T̂ = −(ħ²/2m) d²/dx², a real second-derivative operator that responds to the curvature of ψ. Potential energy involves x alone, so V̂ is simply multiplication by V(x) at each point. Their sum Ĥ is linear, and the time-independent Schrödinger equation becomes Ĥψ = Eψ: acting with Ĥ returns the same function scaled by one constant. Two facts make Ĥ more than plausible. Moving T̂ from one function to the other inside an integral takes two integrations by parts; the two sign flips cancel, and each step leaves a boundary term that vanishes only for functions in the right domain. The potential term moves across only if V* = V. Together these make Ĥ Hermitian, which is what conserves probability and what will force real energies and orthogonal stationary states. The construction stops where its hidden assumptions fail. A classical term mixing x and p, as with a position-dependent mass or a magnetic force, has several inequivalent operator orderings and physics must pick one. A complex V, used on purpose to model absorption, breaks Hermiticity and drains probability. Spin and relativity are not in this Ĥ at all.
2In one sentence
The Hamiltonian operator Ĥ = −(ħ²/2m) d²/dx² + V(x) is the energy written as an instruction on ψ: differentiate twice and scale by −ħ²/2m for the kinetic part, multiply by V(x) for the potential part, and add the two results.
Example. For an electron ħ²/2m = 0.0381 eV nm². On ψ = sin(kx) with k = 2.0 nm⁻¹, where V = 0.50 eV, T̂ψ = (0.0381 eV nm²)(2.0 nm⁻¹)² ψ = 0.15 eV × ψ and V̂ψ = 0.50 eV × ψ, so Ĥψ = 0.65 eV × ψ.
3Squaring p̂ means applying it twice
p²/2m becomes an operator by replacing p with p̂ = −iħ d/dx, and p̂²ψ means p̂ acting on the function p̂ψ, not the square of that function. The first p̂ gives −iħψ′; the second gives (−iħ)²ψ″ = −ħ²ψ″, because (−i)² = −1. So T̂ψ = −(ħ²/2m)ψ″: the i's are gone and the minus sign is what they leave behind. Squaring the output instead, (p̂ψ)²/2m = −ħ²(ψ′)²/2m, grows fourfold when ψ doubles, so it is not linear and cannot be an operator on wavefunctions. Check the dimensions: ħ²/2m is in J² s² kg⁻¹ = J m², and d²/dx² supplies m⁻², so T̂ψ carries joules times the units of ψ. For an electron ħ²/2m = 6.10 × 10⁻³⁹ J m², or 0.0381 eV nm². T̂ also asks more of ψ than p̂ does: it needs a second derivative, so its natural domain is functions smooth enough for ψ″ to exist and be square-integrable.
4V̂ multiplies, so each classical term has one translation
V̂ does no differentiating: it multiplies ψ by V(x) point by point, V̂ψ(x) = V(x)ψ(x), exactly as x̂ multiplies by x. That is why H = p²/2m + V(x) becomes an operator with nothing left to choose. Each term is a function of p alone or of x alone, and a function of one operator is unambiguous: p̂² means p̂p̂, and V(x̂) acts by multiplication. A term that mixes x and p is different. Classically xp = px, but x̂(p̂ψ) = −iħxψ′ while p̂(x̂ψ) = −iħ(ψ + xψ′), and the two differ by iħψ; the commutator topic makes this general. Such terms arise when the mass depends on position, as in a graded semiconductor, or when a magnetic force depends on velocity, and each then needs an ordering chosen by physics rather than algebra. A force that dissipates energy, like friction, cannot be written as any V(x), so no Ĥ of this form can describe it.
5Ĥψ = Eψ is a claim about a whole function
Ĥ turns ψ into a new function, Ĥψ = T̂ψ + V(x)ψ, and the time-independent Schrödinger equation says that new function is E times the old one at every x. Dividing by ψ makes this a test: Ĥψ/ψ must be the same constant everywhere. Try a Gaussian, ψ = e(−x²/2b²), in the spring potential V = ½mω²x². Then T̂ψ/ψ = (ħ²/(2mb²))(1 − x²/b²), which falls as x² grows while V rises as x², so Ĥψ/ψ = ħ²/(2mb²) + ½x²(mω² − ħ²/(mb⁴)). The x² term cancels only when b² = ħ/(mω), and then Ĥψ = ½ħω ψ. For an electron with ħω = 1.0 eV that width is 0.276 nm and E = 0.50 eV. Any other width leaves Ĥψ/ψ varying with x, so that Gaussian is not an energy eigenfunction. How to solve such equations in general is the next topic; the oscillator itself is Unit 9.
6Two integrations by parts make T̂ Hermitian, on a domain
To test T̂, move d²/dx² from ψ to φ* inside ∫φ*ψ″ dx. One integration by parts gives [φ*ψ′] − ∫φ*′ψ′ dx; a second gives [φ*ψ′ − φ*′ψ] + ∫φ*″ψ dx. Each step flips the sign and two flips restore it, so ∫φ*(T̂ψ)dx = ∫(T̂φ)*ψ dx provided the bracket vanishes at both ends; the factor −ħ²/2m is real, so conjugating it changes nothing. On the whole line the bracket dies because normalisable states and their slopes fall to zero. At hard walls it dies because φ and ψ are both zero there, whatever their slopes. Leave the domain and the result fails: in a box 0 ≤ x ≤ a, take φ = sin(πx/a) and the constant ψ = 1, which is not zero at the walls. Then ∫φ(T̂ψ)dx = 0, because ψ″ = 0, but ∫(T̂φ)ψ dx = (ħ²/2m)(2π/a). One derivative alone, d/dx, flips the sign once and is anti-Hermitian, which is why p̂ needs its i and T̂ does not.
7The potential moves across only if it is real
The potential term needs no calculus. ∫φ*(Vψ)dx equals ∫(V*φ)*ψ dx for any V, so it equals ∫(Vφ)*ψ dx for every pair of functions only when V* = V. Nothing else about V matters: it may be negative, discontinuous, or infinite outside a box, and Ĥ stays Hermitian as long as V is real. The condition is not decoration. The rate of change of the norm is dN/dt = (1/iħ)[∫Ψ*(ĤΨ)dx − ∫(ĤΨ)*Ψ dx], which is the Hermiticity test with φ = ψ = Ψ, so a Hermitian Ĥ conserves probability, the result Unit 2 reached through the continuity equation. Give V an imaginary part and the bracket survives: V = V₀ − iW makes N fall as e(−2Wt/ħ). Nuclear and atomic physics use exactly such an optical potential on purpose, to model particles removed from a beam. There Hermiticity fails because probability is meant to leave.
8One integration by parts: kinetic energy is never negative
Integrate ∫ψ*(T̂ψ)dx by parts only once and the boundary term ψ*ψ′ vanishes for a normalisable state, leaving ⟨T⟩ = (ħ²/2m)∫|ψ′|² dx. That is a sum of squares, so ⟨T⟩ ≥ 0, with equality only if ψ′ = 0 everywhere, which on the whole line means ψ = 0. Every normalisable state therefore has positive kinetic energy, and ⟨H⟩ = ⟨T⟩ + ⟨V⟩ lies strictly above the minimum of V: no state can sit motionless at the bottom of a well. The same form prices confinement. Squeeze a normalised ψ to 1/s of its width, ψ(x) → √s ψ(sx), and ∫|ψ′|² dx grows by s², so halving a state's width quadruples ⟨T⟩. The pointwise ratio T̂ψ/ψ can be negative where ψ curves away from the axis; only the mean is bound to be positive.
Key equations
What each relation says, and the units it is written in.
Kinetic-energy operator
T̂ = p̂²/2m = −(ħ²/2m) d²/dx²
p̂ acts twice and (−i)² = −1 fixes the sign, so T̂ is a real operator that responds to curvature. Squaring the function p̂ψ instead is not even linear.
ħ = 1.055 × 10⁻³⁴ J s, m in kg; ħ²/2m in J m², 6.10 × 10⁻³⁹ J m² = 0.0381 eV nm² for an electron
Potential-energy operator
V̂ψ(x) = V(x) ψ(x)
A multiplication, well defined because V depends on x alone. A classical term that mixes x and p has more than one operator ordering.
V(x) in J, evaluated at the same x as ψ; no derivative acts
The Schrödinger equation as Ĥψ = Eψ
Ĥψ = −(ħ²/2m) d²ψ/dx² + V(x) ψ = E ψ
Ĥψ/ψ must equal one constant everywhere. A trial ψ for which that ratio varies with x is not an energy eigenfunction.
E in J and the same at every x; ψ in m(−1/2) for a normalised state on a line
Hermiticity of Ĥ
∫ φ* (Ĥψ) dx = ∫ (Ĥφ)* ψ dx
Two integrations by parts move T̂ across, each flipping a sign; the potential moves across only because it is real.
needs [φ*ψ′ − φ*′ψ] = 0 at both ends and V(x)* = V(x); each side in J for normalised φ and ψ
Kinetic energy as a sum of squares
⟨T⟩ = ∫ ψ* (T̂ψ) dx = (ħ²/2m) ∫ |dψ/dx|² dx
Never negative, so ⟨H⟩ ≥ min V. Safe on a kinked ψ, where ψ″ hides a delta function that the first form misses.
one integration by parts with ψ*ψ′ → 0 at the ends; ∫|dψ/dx|² dx in m⁻², so ⟨T⟩ in J
Norm under a complex potential
dN/dt = (2/ħ) ∫ Im V(x) |Ψ|² dx
Zero for every Ψ only if Im V = 0. An absorbing V = V₀ − iW gives N(t) = e(−2Wt/ħ).
N = ∫|Ψ|² dx is dimensionless; Im V in J and ħ in J s, so dN/dt in s⁻¹
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. At ω = 1.00 ω₀, drag b to 1.00 ℓ₀: the dashed curvature term falls exactly as fast as V(x) rises, the solid line lies flat on ⟨H⟩ = 0.50 ħω₀, and Ĥψ = Eψ holds. Then raise ω to 1.50 ω₀ and the same ψ fails until b shrinks to about 0.82 ℓ₀.
Live results
- Ĥψ/ψ AT x = 0
- 0.30 ħω₀
- Ĥψ/ψ AT x = ±b
- 0.85 ħω₀
- MEAN ⟨H⟩ = ⟨T⟩ + ⟨V⟩
- 0.57 ħω₀
- EIGENVALUE ½ħω
- 0.50 ħω₀
Catch the common trap
The mistake this lesson is built to prevent.
A stretch where ψ is a straight line has ψ″ = 0, so it is tempting to say it carries no kinetic energy, and that a wavefunction built from straight segments has ⟨T⟩ = 0. Both halves are wrong. T̂ψ/ψ at a point is not the kinetic energy of anything; only the integral ⟨T⟩ is. And where two segments meet at a kink, ψ′ jumps, so ψ″ contains a delta function holding all the curvature the straight parts lack. Take the normalised tent ψ = √(3/(2a³)) (a − |x|) on |x| ≤ a. Its ψ″ vanishes everywhere except x = 0 and x = ±a, yet ⟨T⟩ = (ħ²/2m)∫|ψ′|² dx = 3ħ²/(2ma²), which is 0.11 eV for an electron with a = 1.0 nm. Compute kinetic energy from ∫|ψ′|² dx, which needs only first derivatives, and the kinks take care of themselves.
Test it: what changes?
- Centre value ħ²/(2mb²) of Ĥψ/ψ falls
- Value at x = ±b, ½mω²b², rises
- ⟨T⟩ = ħ²/(4mb²) falls
- Ĥψ/ψ is flat only at b² = ħ/(mω)
- V(x) = ½mω²x² steepens
- Centre value ħ²/(2mb²) unchanged
- Matching width √(ħ/(mω)) shrinks
- Eigenvalue ½ħω rises
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron has ψ(x) = A e(−x²/2b²) with b = 0.20 nm, in a region where V = 0. Build T̂ψ by letting p̂ act twice, find the value of T̂ψ/ψ at the centre and where it changes sign, and find ⟨T⟩.
View solution →
- First p̂: p̂ψ = −iħ dψ/dx = −iħ(−x/b²)ψ = (iħx/b²)ψ.
- The second p̂ acts on that whole function, product rule included: d(xψ)/dx = ψ + xψ′ = (1 − x²/b²)ψ, so p̂(iħxψ/b²) = −iħ × (iħ/b²)(1 − x²/b²)ψ = (ħ²/b²)(1 − x²/b²)ψ.
- Divide by 2m: T̂ψ = (ħ²/(2mb²))(1 − x²/b²)ψ. Dimensions: ħ²/2m in J m² over b² in m² gives J, times ψ.
- Numbers: ħ²/(2mb²) = 0.0381 eV nm² ÷ 0.040 nm² = 0.95 eV. So T̂ψ/ψ is 0.95 eV at x = 0, zero at x = ±b = ±0.20 nm (the inflection points of ψ), and negative beyond, where ψ curves away from the axis.
- For the mean, integrate by parts once: ⟨T⟩ = (ħ²/2m)∫|dψ/dx|² dx = (ħ²/2m)⟨x²⟩/b⁴, and |ψ|² ∝ e(−x²/b²) has ⟨x²⟩ = b²/2, so ⟨T⟩ = ħ²/(4mb²) = 0.48 eV.
- Checks: ⟨T⟩ is positive although T̂ψ/ψ is negative in the tails. As b → ∞ the Gaussian flattens and ⟨T⟩ → 0, the particle at rest; halving b would quadruple it.
Answer. T̂ψ = (ħ²/(2mb²))(1 − x²/b²)ψ: T̂ψ/ψ = 0.95 eV at the centre, zero at x = ±0.20 nm and negative beyond; ⟨T⟩ = ħ²/(4mb²) = 0.48 eV.
MediumExample 2
An absorbing region is modelled by adding a constant imaginary part to an otherwise real potential, V(x) = V₀(x) − iW with W = 5.0 meV. Show that Ĥ is no longer Hermitian, find how the norm N(t) = ∫|Ψ|² dx evolves, and find the time for N to fall from 1 to 0.50.
View solution →
- T̂ and the real part V₀(x) pass the Hermiticity test as before, so only the new term matters: ∫φ*(−iWψ)dx − ∫(−iWφ)*ψ dx = −iW∫φ*ψ dx − iW∫φ*ψ dx = −2iW∫φ*ψ dx. Conjugating −iW flips its sign, so the two sides no longer match.
- Differentiate the norm and use the Schrödinger equation, ∂Ψ/∂t = ĤΨ/(iħ), with its conjugate: dN/dt = (1/iħ)[∫Ψ*(ĤΨ)dx − ∫(ĤΨ)*Ψ dx]. The bracket is the Hermiticity test with φ = ψ = Ψ.
- Substitute step 1: dN/dt = (1/iħ)(−2iW N) = −(2W/ħ)N, so N(t) = e(−2Wt/ħ). Dimensions: W in J over ħ in J s gives s⁻¹, as a rate must.
- Numbers: ħ = 6.582 × 10⁻¹⁶ eV s, so ħ/(2W) = 6.582 × 10⁻¹⁶ eV s ÷ 1.0 × 10⁻² eV = 6.6 × 10⁻¹⁴ s. N = 0.50 when t = (ħ/2W) ln 2 = 4.6 × 10⁻¹⁴ s.
- Checks: W → 0 gives an infinite decay time and Hermiticity returns. With W positive, probability leaves, as an absorber should; the opposite sign would create probability from nothing.
Answer. ∫φ*(Ĥψ)dx − ∫(Ĥφ)*ψ dx = −2iW∫φ*ψ dx ≠ 0; N(t) = e(−2Wt/ħ) with ħ/2W = 66 fs, so N halves after 46 fs.
HardExample 3
In a graded semiconductor layer an electron's effective mass varies near x = 0 as 1/m(x) = (1/m₀)(1 + αx/b), with m₀ = 0.067 mₑ, α = 0.20 and b = 5.0 nm. Two candidate kinetic operators are T̂₁ = −(ħ²/2m(x)) d²/dx² and T̂₂ = p̂ (1/2m(x)) p̂. Test each for Hermiticity with the normalised pair φ ∝ e(−x²/2b²) and ψ ∝ (x/b) e(−x²/2b²).
View solution →
- Expand the second: T̂₂ψ = −(ħ²/2) d/dx[(1/m) dψ/dx] = T̂₁ψ − (ħ²/2)(d(1/m)/dx)(dψ/dx). The extra term is nonzero wherever the mass varies and ψ has a slope, so p²/2m(x) names no single operator: the order of p̂ and m(x) is a physical choice.
- Use u = x/b and ε = ħ²/(2m₀b²) = (0.0381 eV nm² ÷ 0.067) ÷ (5.0 nm)² = 22.7 meV. Normalised, φ = (b√π)(−1/2) e(−u²/2) and ψ = (b√π/2)(−1/2) u e(−u²/2), so that b²φ″ = (u² − 1)φ and b²ψ″ = (u² − 3)ψ.
- One way: ∫φ(T̂₁ψ)dx = −(√2/√π)αε∫(u⁴ − 3u²)e(−u²)du, since the terms without α are odd and vanish. With ∫u²e(−u²)du = √π/2 and ∫u⁴e(−u²)du = 3√π/4, this is +(3√2/4)αε = 4.8 meV.
- The other way: ∫(T̂₁φ)ψ dx = −(√2/√π)αε∫(u⁴ − u²)e(−u²)du = −(√2/4)αε = −1.6 meV. The two differ by √2 αε = 6.4 meV, so T̂₁ is not Hermitian: moving its derivatives across 1/m(x) creates terms in d(1/m)/dx that T̂₁ does not contain.
- For T̂₂ integrate by parts once; the Gaussian tails kill the boundary term: ∫φ(T̂₂ψ)dx = (ħ²/2)∫(1/m)(dφ/dx)(dψ/dx)dx. That is symmetric in φ and ψ, so it equals ∫(T̂₂φ)ψ dx for every pair; here it is (√2/√π)αε∫(u⁴ − u²)e(−u²)du = (√2/4)αε = 1.6 meV.
- Checks: at α = 0 both operators reduce to −(ħ²/2m₀) d²/dx² and all three numbers vanish, because φ is even and ψ odd. Every result is ε times a pure number, so each is an energy.
Answer. T̂₁ gives 4.8 meV one way and −1.6 meV the other, so it is not Hermitian; T̂₂ = p̂(1/2m(x))p̂ gives 1.6 meV both ways and is the ordering to use.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Kinetic energy is p̂ applied twice; potential energy just multiplies.
Build T̂ by applying p̂ twice, not by squaring p̂ψ, and multiply by V(x) for the potential. Ĥ is Hermitian only if V is real and the boundary term [φ*ψ′ − φ*′ψ] vanishes. Compute ⟨T⟩ as (ħ²/2m)∫|dψ/dx|² dx, which is never negative.