Normalisation, the ground state, and zero-point energy
A pinned wave must bend, so E₁ is never zero.
Two things finish the ground state of the well: the size of ψ, which one integral fixes, and an energy that will not go to zero. This lesson earns both, traces the lowest energy to the shape the walls force on ψ rather than to heat, and names the one property of ψ that neither normalisation nor any measurement can fix.
Build the model
Pinned at both walls, ψ has to bend, and bending costs energy.
1The big idea
The quantisation of UQMI-06.02 leaves ψₙ = A sin(nπx/L) with A undetermined, and it has to: the Schrödinger equation is linear, so any multiple of a solution is a solution at the same energy. The Born rule supplies the missing condition. The particle is certainly somewhere and ψ is zero beyond the walls, so ∫₀L |ψ|² dx = 1; sin² averages to ½ over any whole number of half-wavelengths, which gives |A|²L/2 = 1 and |A| = √(2/L) for every n, in m(−1/2) so that |ψ|² is a probability per metre. Only the modulus is fixed. Multiplying ψ by a constant e(iα) changes no probability, no expectation value and no energy, so the overall phase is not a property of the state, and A is taken real and positive by convention. The lowest level, E₁ = π²ħ²/(2mL²), is 0.376 eV for an electron in a 1.00 nm well, and no shape can bring it to zero. Inside the well V = 0, so the energy is all kinetic, ⟨T⟩ = (ħ²/2m)∫|ψ′|² dx, and a normalised function pinned to zero at both walls cannot be flat: it must rise and come back, and every slope costs energy. The sine spreads that bending most economically, which is what makes it the ground state. Nothing thermal is involved; E₁ holds no temperature and survives at absolute zero. The model fails where its walls or its kinematics do: finite walls let ψ leak and lower E₁, and near L = 1 pm an electron's E₁ approaches mc².
2In one sentence
The zero-point energy of the infinite square well is its lowest allowed energy, E₁ = π²ħ²/(2mL²), held by the normalised ground state ψ₁ = √(2/L) sin(πx/L) and kept by the particle even at absolute zero.
Example. An electron in a 1.00 nm well has ψ₁ peaking at √(2/1.00 nm) = 1.41 nm(−1/2) and E₁ = 0.376 eV: 14.5 times kB T at 300 K (0.0259 eV), and just as large at 0 K.
3The equation leaves the size of ψ open
UQMI-06.02 ended with ψₙ = A sin(nπx/L) and Eₙ = n²π²ħ²/(2mL²), and said nothing about A. That is not an oversight. The Schrödinger equation is linear and homogeneous in ψ, so if ψ solves it at energy E, so do 3ψ, −ψ and 10⁶ψ, at the same E; the boundary conditions are homogeneous too, since ψ(0) = ψ(L) = 0 holds for 3ψ exactly when it holds for ψ. The size of ψ must come from outside the equation, and it comes from the Born rule: |ψ(x)|² dx is the probability of finding the particle between x and x + dx, and the particle is certainly somewhere. Because ψ vanishes beyond the walls, somewhere means somewhere in [0, L], and the condition is ∫₀L |ψ|² dx = 1. The same statement fixes the units. A probability is a pure number and dx is a length, so in one dimension |ψ|² is a probability per metre and ψ carries m(−1/2).
4One integral fixes |A| = √(2/L) for every n
Write sin²θ = (1 − cos 2θ)/2. Then ∫₀L sin²(nπx/L) dx = L/2 − (L/(4nπ))[sin(2nπx/L)]₀L = L/2, because the cosine runs through n whole periods and integrates to zero. So |A|²L/2 = 1 and |A| = √(2/L), the same for every n: sin² averages to ½ over any whole number of half-wavelengths, however many there are. For an electron in a 1.00 nm well the ground state peaks at √2 = 1.41 nm(−1/2), and |ψ₁|² at the centre is 2/L = 2.00 nm⁻¹, twice the uniform 1/L of a classical particle bouncing between the walls. The antiderivative x/L − sin(2πx/L)/(2π) shows how much gathers in the middle: between L/3 and 2L/3 the probability is 1/3 + √3/(2π) = 0.609, against 1/3 classically. Check the result two ways. Dimensions: |A|²L/2 = 1 is m⁻¹ times m on the left, a pure number on both sides. Scaling: halving L doubles the peak 2/L while halving the width, so the area under |ψ₁|² stays 1.
5The phase of A is left free, and nothing can detect it
Normalisation constrains |A|², so it fixes the modulus of A and nothing else. A = √(2/L), −√(2/L) and i√(2/L) all satisfy it, as does e(iα)√(2/L) for any real constant α. These are one physical state. |e(iα)ψ|² = |ψ|², so every position probability agrees; ψ′ picks up the same constant factor, so ∫|ψ′|² dx and the energy agree; and every probability or average the course will compute is an integral of ψ* times something times ψ, where e(−iα) meets e(iα) and cancels. The convention is to take A real and positive, so ψ₁ > 0 between the walls. Two limits keep this freedom in its place. It is one constant for the whole wavefunction: a phase that varies with position, e(iqx) multiplying ψ₁, leaves |ψ|² untouched yet raises the kinetic energy by ħ²q²/2m, so it makes a different state, as the Medium example shows. And it is overall: once two eigenstates are added, their relative phase is physical, it shapes |Ψ|², and its steady advance sets |Ψ|² moving, the subject of UQMI-06.08.
6Why the lowest energy cannot be zero
Inside the well V = 0, so the energy of any state is all kinetic: ⟨T⟩ = −(ħ²/2m)∫₀L ψ*ψ″ dx. Integrate by parts: −∫ψ*ψ″ dx = ∫|ψ′|² dx − [ψ*ψ′]₀L, and the boundary term vanishes because ψ(0) = ψ(L) = 0. So ⟨T⟩ = (ħ²/2m)∫|ψ′|² dx, a sum of squared slopes that is never negative and is zero only if ψ′ = 0 everywhere. A flat ψ that is zero at x = 0 is zero everywhere and cannot be normalised. Every state of the well, stationary or not, therefore has positive energy, and so does every level Eₙ; this is reached without solving anything, where UQMI-06.02 reached it by trying E ≤ 0 and finding only ψ = 0. For the ground state the integral is (2/L)(π/L)²(L/2) = π²/L², which returns E₁ = π²ħ²/(2mL²) exactly. The curvature picture says the same thing point by point. ψ₁″ = −(2mE₁/ħ²)ψ₁, so wherever ψ₁ > 0 it bends back toward the axis at a rate set by E₁. A function that leaves zero at one wall and must return to zero at the other has to bend; E = 0 would make it a straight line, and the only straight line through both zeros is ψ = 0.
7The sine is the cheapest shape that fits
The slope form turns “must bend” into a price that can be compared across shapes. A tent rising straight to the centre and back, normalised, has height √(3/L) and slope ±2√(3/L)/L, so |ψ′|² = 12/L³ everywhere and ⟨T⟩ = 12 ħ²/(2mL²), 21.6% above E₁. Flattening its top does not help much: the flat part costs nothing, but normalisation holds the top up, so the ramps must steepen, and the best trapezoid, with ramps 0.375 L long, still costs 10.67 ħ²/(2mL²). A parabola x(L − x) costs 10 ħ²/(2mL²), only 1.3% high. The sine wins at π² = 9.87 because it bends in proportion to its height everywhere: its local kinetic energy −(ħ²/2m)ψ₁″/ψ₁ equals E₁ at every x, the mark of an eigenstate, while the parabola's is lowest at the centre and grows without limit at the walls. The figure runs the trapezoid through its whole range. No pinned shape beats E₁, and UQMI-06.06 proves it: expanded on the sine basis, any state has ⟨H⟩ = Σ|cₙ|²Eₙ ≥ E₁.
8Zero-point energy is not heat
E₁ is there at absolute zero because it is the lowest energy the well allows, not a share of higher levels that cooling could empty. Cooling removes the excited levels' population; it cannot take the particle below E₁. For an electron in a 1.00 nm well E₁ = 0.376 eV, 14.5 times kB T at 300 K and about 1100 times kB T at 4 K. What the energy buys is a spread of momentum, not a motion anyone could follow: ⟨p²⟩ = 2mE₁ = (πħ/L)², so the rms momentum is πħ/L = 3.31 × 10⁻²⁵ kg m s⁻¹ and the rms speed 3.64 × 10⁵ m s⁻¹, while ⟨p⟩ = 0 and |ψ₁|² never changes in time. The 1/(mL²) scaling carries the classical correspondence: a 1.00 g bead in a 1.00 cm box has E₁ = 5.5 × 10⁻⁶¹ J and a zero-point speed of 3.3 × 10⁻²⁹ m s⁻¹, which is why nothing macroscopic shows it. The same scaling marks where the model ends. Finite walls let ψ leak and lower E₁, by 28% for 5.0 eV walls around a 1.00 nm well (UQMI-06.01), and an electron squeezed to L ≈ 0.86 pm would have E₁ = mc², far outside non-relativistic mechanics.
Key equations
What each relation says, and the units it is written in.
Normalisation in the well
∫₀L |ψ(x)|² dx = 1
The Born rule's demand that the particle be somewhere. It fixes the size of ψ, which the linear equation left open, and touches neither E nor the shape.
ψ in m(−1/2), x and L in m, so |ψ|² dx is a pure number. The range stops at the walls because ψ = 0 outside.
Normalised eigenfunctions
ψₙ(x) = √(2/L) sin(nπx/L), n = 1, 2, 3, …
sin² averages to ½ over any whole number of half-wavelengths, so ∫₀L sin²(nπx/L) dx = L/2 and |A|²L/2 = 1.
√(2/L) in m(−1/2): 1.41 nm(−1/2) for L = 1.00 nm. One constant serves every n.
Ground-state (zero-point) energy
E₁ = π²ħ²/(2mL²)
No temperature appears: it is the floor of the spectrum, present at 0 K. It falls as 1/(mL²) and vanishes only as mL² → ∞ or ħ → 0.
m in kg, L in m, E₁ in J. Electron: ħ²/2m = 0.0381 eV nm², so E₁ = 0.376 eV at L = 1.00 nm.
Kinetic energy as summed slope
⟨T⟩ = (ħ²/2m) ∫₀L |ψ′(x)|² dx
Zero only for a flat ψ, and a flat ψ pinned at the walls is zero everywhere. So no state of the well, stationary or not, has zero or negative energy.
ψ′ in m(−3/2). It is −(ħ²/2m)∫ψ*ψ″ dx after integrating by parts; the term ψ*ψ′ vanishes at both walls.
Every pinned shape costs at least E₁
⟨T⟩ ≥ π²ħ²/(2mL²) for any normalised ψ with ψ(0) = ψ(L) = 0
Why the sine is the ground state: it spreads the bending evenly. The proof, ⟨H⟩ = Σ|cₙ|²Eₙ ≥ E₁, arrives with the expansion of UQMI-06.06.
In units of ħ²/(2mL²): tent 12, parabola x(L − x) 10, sine π² = 9.87. Equality holds only for ψ₁.
Freedom of the overall phase
ψ → e(iα)ψ leaves |ψ|², ⟨T⟩ and every probability unchanged
Normalisation fixes |A| only. A phase that varies with x, e(iqx) multiplying a real ψ such as ψₙ, is not overall: it raises ⟨T⟩ by ħ²q²/2m and makes a different state.
α is a real constant angle in rad. Convention: A real and positive, so ψ₁ > 0 inside the well.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Start at the tent, d = 0.500 L: ⟨T⟩ = 12.00, 21.6% above E₁. Shorten the ramps to 0.375 L, leaving a flat top 0.25 L wide, and it falls to 10.67, the best trapezoid; shorten them further and the steeper ramps push it to 42.86 at 0.050 L. Nothing reaches π² = 9.87. Then halve L: both peaks rise by √2 and every energy in eV by 4, while the values in ħ²/(2mL²) stay put.
Live results
- TRAPEZOID ⟨T⟩
- 12.00 ħ²/(2mL²)
- EXCESS OVER E₁
- 21.6 %
- TRAPEZOID ⟨T⟩, ELECTRON
- 0.457 eV
- GROUND STATE E₁, ELECTRON
- 0.376 eV
Catch the common trap
The mistake this lesson is built to prevent.
The ground state gets pictured as the particle cooled to rest on the floor of the well, with E₁ as leftover thermal motion that enough cooling, or a gentle enough measurement, would remove. Nothing in E₁ = π²ħ²/(2mL²) refers to a temperature or to a measuring device: it depends only on m, L and ħ, and as the lowest eigenvalue it is exactly what remains once cooling has emptied every excited level. A particle at rest would have ψ′ = 0 everywhere, and the only such ψ that vanishes at both walls is zero, which cannot be normalised. The energy is the price of the shape the walls force on ψ. Nor is anything jiggling: ψ₁ is stationary, |ψ₁|² is frozen in time and ⟨p⟩ = 0. What is nonzero is ⟨p²⟩ = (πħ/L)², a spread of momentum outcomes across identically prepared particles, not a trajectory being shaken.
Test it: what changes?
- Flat top opens to 0.25 L wide
- ⟨T⟩ falls from 12.00 to 10.67 ħ²/(2mL²)
- Still above the sine's π² ħ²/(2mL²)
- Ground-state E₁ unchanged
- Peak of ψ rises by √2
- Peak |ψ|² doubles
- Every energy in eV quadruples
- Energies in ħ²/(2mL²) unchanged
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron is in the ground state ψ₁ = A sin(πx/L) of an infinite well of width L = 0.800 nm. Find A with its units, the probability density at the centre, the probability of finding the electron in the central 0.200 nm, and E₁.
View solution →
- Normalise: ∫₀L A² sin²(πx/L) dx = A²L/2, because sin² averages to ½ over the half-wavelength. Setting A²L/2 = 1 gives A = √(2/L) = √(2/0.800 nm) = 1.58 nm(−1/2), taken real and positive.
- At the centre sin(π/2) = 1, so |ψ₁(L/2)|² = 2/L = 2.50 nm⁻¹. Units: nm⁻¹ times a length in nm is a pure number, as a probability must be.
- The central 0.200 nm runs from 0.300 nm to 0.500 nm. With ∫(2/L) sin²(πx/L) dx = x/L − sin(2πx/L)/(2π), P = 0.250 − [sin(1.25π) − sin(0.75π)]/(2π) = 0.250 + 1.414/(2π) = 0.475.
- Check: peak density times width, 2.50 nm⁻¹ × 0.200 nm = 0.500, is a little above 0.475 because |ψ₁|² falls away from the centre; a classical particle would give 0.200/0.800 = 0.250.
- E₁ = (ħ²/2m)π²/L² = 0.0381 eV nm² × 9.870/0.640 nm² = 0.588 eV.
Answer. A = √(2/L) = 1.58 nm(−1/2); |ψ₁|² = 2.50 nm⁻¹ at the centre; P = 0.475 in the central 0.200 nm, against 0.250 classically; E₁ = 0.588 eV.
MediumExample 2
Four students report the ground state of an electron in an infinite well of width L = 0.600 nm: (i) √(2/L) sin(πx/L); (ii) −√(2/L) sin(πx/L); (iii) ((1 + i)/√L) sin(πx/L); (iv) √(2/L) sin(πx/L) e(iπx/L). Which are normalised, which share |ψ|², and which describe the same state? Find ⟨T⟩ for each.
View solution →
- |ψ|²: (i) and (ii) give (2/L) sin²(πx/L). For (iii), |1 + i|² = 2, so again (2/L) sin²(πx/L). For (iv), |e(iπx/L)|² = 1, so once more (2/L) sin²(πx/L). All four integrate to 1 over the well and predict identical position probabilities.
- (ii) is e(iπ)ψ₁ and (iii) is e(iπ/4)ψ₁, since (1 + i)/√L = √(2/L) e(iπ/4). Each is ψ₁ times one constant, so ψ′ is multiplied by the same constant and |ψ′|² is unchanged.
- For (i) to (iii), ⟨T⟩ = (ħ²/2m)∫|ψ₁′|² dx = (ħ²/2m)π²/L² = 0.0381 eV nm² × 9.870/0.360 nm² = 1.04 eV, which is E₁.
- For (iv) the phase depends on x. With u = πx/L, ψ′ = √(2/L)(π/L)(cos u + i sin u)e(iu) = √(2/L)(π/L)e(2iu), so |ψ′|² = (2/L)(π/L)², constant across the well, and ∫₀L |ψ′|² dx = 2π²/L².
- So ⟨T⟩ = 2 × (ħ²/2m)π²/L² = 2.09 eV: the extra ħ²q²/2m with q = π/L equals E₁ again. Same |ψ|², double the energy, and since 2E₁ is none of the levels n²E₁, (iv) is not an eigenstate at all.
Answer. All four are normalised with the same |ψ|². (i), (ii) and (iii) differ only by an overall phase and are one state, with ⟨T⟩ = E₁ = 1.04 eV; (iv) is a different state with ⟨T⟩ = 2.09 eV.
HardExample 3
A trial state for an electron in an infinite well of width L = 1.20 nm is ψ = B x²(L − x) inside the well, a lopsided hump that peaks at x = 2L/3. Normalise it, find ⟨T⟩ from the slope form and again from −(ħ²/2m)∫ψψ″ dx, compare with E₁, and use the local kinetic energy −(ħ²/2m)ψ″/ψ at x = L/2 and x = 0.2L to decide whether ψ could be a stationary state.
View solution →
- ψ vanishes at x = 0 and x = L, so it is allowed. ∫₀L x⁴(L − x)² dx = L⁷(1/5 − 1/3 + 1/7) = L⁷/105, so B²L⁷/105 = 1 and B = √(105/L⁷) = √(105/3.583 nm⁷) = 5.41 nm(−7/2). Units: nm(−7/2) × nm³ = nm(−1/2), as ψ needs.
- Slope form: ψ′ = B(2Lx − 3x²) and ∫₀L (2Lx − 3x²)² dx = L⁵(4/3 − 3 + 9/5) = 2L⁵/15, so ∫|ψ′|² dx = 2B²L⁵/15 = 14/L². Second form: ψ″ = B(2L − 6x), and −∫ψψ″ dx = −B²∫₀L (2L²x² − 8Lx³ + 6x⁴) dx = −B²L⁵(2/3 − 2 + 6/5) = 2B²L⁵/15 = 14/L² again; the boundary term [ψψ′]₀L separating the two forms vanishes because ψ = 0 at both walls.
- ⟨T⟩ = (ħ²/2m) × 14/L² = 0.0381 eV nm² × 14/1.44 nm² = 0.370 eV. The ground state has E₁ = 0.0381 eV nm² × 9.870/1.44 nm² = 0.261 eV, so the hump sits 14/π² − 1 = 41.8% above E₁, far worse than the symmetric parabola's 1.3%: packing the hump against one wall forces a steep fall there, and every slope costs energy.
- Local kinetic energy: −ψ″/ψ = (6x − 2L)/(x²(L − x)). At x = L/2 this is L/(L³/8) = 8/L², giving 0.0381 × 8/1.44 = 0.212 eV. At x = 0.2L it is −0.8L/(0.032L³) = −25/L², giving 0.0381 × (−25)/1.44 = −0.661 eV: negative, because for x < L/3 the hump curves away from the axis, ψ″ > 0 where ψ > 0.
- In a stationary state ψ″ = −(2mE/ħ²)ψ inside the well, so the local kinetic energy equals E at every x; for ψ₁ it is 0.261 eV everywhere. This ψ swings from −0.661 eV at 0.2L to +0.212 eV at the centre, so no single E balances ψ″ against ψ: it is not an eigenstate, and its |ψ|² will change in time. Weighted by |ψ|², the varying local values still average to the 0.370 eV found above.
Answer. B = √(105/L⁷) = 5.41 nm(−7/2); ⟨T⟩ = 14ħ²/(2mL²) = 0.370 eV by both routes, 41.8% above E₁ = 0.261 eV; the local kinetic energy is 0.212 eV at the centre and −0.661 eV at 0.2L, so ψ is not stationary.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
|A| = √(2/L) for every n; E₁ survives absolute zero.
Normalise with ∫₀L |ψ|² dx = 1, giving ψₙ = √(2/L) sin(nπx/L) in m(−1/2); only the modulus of A is fixed, and an overall phase changes nothing. E₁ = π²ħ²/(2mL²) contains no temperature and holds while the walls are effectively infinite.