Orthogonality of eigenfunctions with different eigenvalues
Distinct eigenvalues force zero overlap; degeneracy leaves a choice.
Every expansion this course writes, from box states to spin, leans on one fact: eigenfunctions of a Hermitian operator with different eigenvalues have zero overlap. Here you prove it in two lines, watch each premise do its work, and learn what to do when two eigenfunctions share an eigenvalue and the proof falls silent.
Build the model
Different eigenvalues force a zero overlap; equal ones leave the choice to you.
1The big idea
Take one Hermitian operator Â, one domain D on which the Hermiticity test holds, and two eigenfunctions in D: Âψₘ = aₘψₘ and Âψₙ = aₙψₙ. Evaluate the single integral ∫ ψₘ*(Âψₙ) dx twice. Letting  act to the right gives aₙ times the overlap ∫ ψₘ*ψₙ dx; moving it onto ψₘ, which Hermiticity allows, gives aₘ* times the same overlap, and aₘ* = aₘ because the eigenvalues are already known to be real. The two answers must agree, so (aₙ − aₘ)∫ ψₘ*ψₙ dx = 0, and whenever the eigenvalues differ the overlap is zero with no integral evaluated. It is the argument that made eigenvalues real, run with two different functions instead of one. Each premise can fail on its own, and the failure shows: a non-Hermitian operator, or two functions obeying different boundary conditions, yields pairs with distinct eigenvalues that overlap by 0.97 and 0.85 in the cases worked below. The theorem also stops at degeneracy. When aₘ = aₙ the identity reads 0 = 0, every combination of the two functions is again an eigenfunction, their overlap can take any value up to 1 in modulus, and an orthogonal pair has to be built, by Gram–Schmidt or by a second operator, rather than inherited. Plane waves on the whole line, which cannot be normalised, sit outside it until the continuous spectrum is treated.
2In one sentence
Two eigenfunctions of the same Hermitian operator, taken from one domain and belonging to different eigenvalues, are orthogonal: their overlap integral ∫ ψₘ* ψₙ dx is exactly zero.
Example. In a well of width a = 0.80 nm, an electron's ψ₁ = √(2/a) sin(πx/a) and ψ₃ = √(2/a) sin(3πx/a) have energies 0.588 eV and 5.29 eV. The energies differ, so ∫ ψ₁ψ₃ dx = 0 over 0 < x < a with no integral evaluated.
3One integral, evaluated two ways
Let Âψₘ = aₘψₘ and Âψₙ = aₙψₙ, with  Hermitian on a domain D and both functions in D. Take the single number ∫ ψₘ*(Âψₙ) dx. Let  act where it stands and it returns aₙψₙ, so the integral is aₙ∫ ψₘ*ψₙ dx. Or move  onto the left-hand function, which is exactly what Hermiticity permits: ∫ (Âψₘ)*ψₙ dx = ∫ (aₘψₘ)*ψₙ dx = aₘ*∫ ψₘ*ψₙ dx, and aₘ* = aₘ because the eigenvalues of a Hermitian operator are real. Subtracting, (aₙ − aₘ)∫ ψₘ*ψₙ dx = 0. Three premises carried the load: one operator Hermitian on one domain, both functions inside that domain, and aₘ ≠ aₙ, which lets you divide. The conclusion arrives without evaluating an integral. Units check: with ψ in m⁻¹⁄² the overlap is dimensionless, and each side of the identity carries the units of Â. Put ψₙ = ψₘ and stop before using reality, and the same line reads (a − a*)∫ |ψ|² dx = 0, the proof of UQMI-03.05; take two different functions and it gives orthogonality.
4Orthogonal is not the same as separated
Orthogonality says nothing about where the functions live. In a well of width a, ψ₁ = √(2/a) sin(πx/a) and ψ₂ = √(2/a) sin(2πx/a) are both nonzero at almost every point, yet their overlap vanishes. The product ψ₁ψ₂ is positive on the left half and negative on the right, and the halves contribute +4/(3π) = +0.424 and −4/(3π) = −0.424, so the signed areas cancel. For any pair, sin A sin B = (1/2)[cos(A − B) − cos(A + B)] turns ∫₀ᵃ sin(mπx/a) sin(nπx/a) dx into integrals of cos((m ∓ n)πx/a), which vanish over 0 to a whenever m ≠ n, while m = n leaves a/2. The theorem predicted every one of those zeros from the energies n²π²ħ²/(2ma²) being different. The classical ancestor is a string clamped at both ends: its standing waves are the same sines, orthogonal for the same reason, which is why a plucked shape can be split into harmonics one at a time.
5Parity is a Hermitian operator too
Define P̂ψ(x) = ψ(−x) on the whole line. Substituting x → −x shows ∫ f*(x) g(−x) dx = ∫ f*(−x) g(x) dx, so P̂ is Hermitian, and since P̂ applied twice returns ψ, its only eigenvalues are +1 for even functions and −1 for odd ones. The theorem then says every even function is orthogonal to every odd one: the rule that an odd integrand integrates to zero is this theorem, applied to parity. But +1 is shared by every even function, so parity is silent about two even ones, and indeed the normalised Gaussians e(−x²/2b²) and e(−x²/b²) overlap by 0.971. That is where a second operator earns its keep. The oscillator's ψ₀ ∝ e(−x²/2b²) and ψ₂ ∝ (2x²/b² − 1)e(−x²/2b²) are both even, so parity cannot separate them, yet they are eigenfunctions of Ĥ with energies ħω/2 and 5ħω/2, and that difference alone forces their overlap to zero, as the Medium example confirms.
6Break a premise and the overlap comes back
Each premise can be removed on its own, and the zero goes with it. Drop Hermiticity: on 0 < x < L = 1.0 nm with no boundary condition, e(x/b₁) and e(x/b₂) are eigenfunctions of −iħ d/dx with different eigenvalues −iħ/b₁ and −iħ/b₂, but both are positive everywhere, so their overlap cannot vanish; for b₁ = 0.50 nm and b₂ = 1.0 nm the normalised overlap is 0.972. Drop the single domain: sin(πx/L), zero at both ends, and sin(πx/2L), with zero slope at x = L, have different energies under one formula for Ĥ and, once normalised, overlap by 8/(3π) = 0.849, as UQMI-03.05 measured. Vectors show the same thing. The symmetric matrix with rows (2, −1) and (−1, 2) has eigenvectors (1, 1) and (1, −1), eigenvalues 1 and 3, dot product 0; times k/M it gives the squared mode frequencies of two masses M joined to each other and to two walls by three springs of stiffness k, whose normal modes are orthogonal for this reason. The non-symmetric matrix with rows (2, 1) and (0, 1) has eigenvectors (1, 0) and (1, −1), and their normalised dot product is 0.707.
7Equal eigenvalues: the identity reads 0 = 0
On a ring of circumference L, where every allowed function repeats after one lap, the free Ĥ = −(ħ²/2m) d²/dx² is Hermitian, and the level with k = 2πn/L, n ≥ 1, has energy ħ²k²/2m = 2π²n²ħ²/(mL²): 0.376 eV for an electron at n = 1 on a 2.0 nm ring. cos(kx), sin(kx), e(ikx) and e(−ikx) all have that energy, and by linearity so does every combination of them, such as cos(kx − φ) = cos φ cos(kx) + sin φ sin(kx). The theorem's last line is now (aₙ − aₘ)∫ ψₘ*ψₙ dx = 0 with aₙ − aₘ = 0, true whatever the overlap. The overlaps confirm the silence: normalised, cos(kx) and cos(kx − φ) overlap by cos φ, anything from −1 to +1, and cos(kx) and e(ikx) overlap by 1/√2 = 0.707. Degenerate eigenfunctions are not forced apart, nor forced together: cos(kx) and sin(kx) happen to be orthogonal. They are simply not covered, and an orthogonal pair has to be chosen.
8Gram–Schmidt builds the partner inside the level
Given normalised degenerate φ₁ and φ₂, keep ψ₁ = φ₁ and remove from φ₂ its component along φ₁: χ = φ₂ − Sφ₁ with S = ∫ φ₁*φ₂ dx. Then ∫ φ₁*χ dx = S − S = 0 and ∫ |χ|² dx = 1 − |S|², so ψ₂ = χ/√(1 − |S|²). On the ring, φ₁ = √(2/L) cos(kx) and φ₂ = √(2/L) cos(kx − π/3) give S = 1/2 and χ = √(2/L)(√3/2) sin(kx); dividing by √(3/4) = 0.866 returns ψ₂ = √(2/L) sin(kx). χ stays an eigenfunction only because φ₁ and φ₂ share the eigenvalue a: Âχ = aφ₂ − Saφ₁ = aχ. Applied to functions that are not eigenfunctions, Gram–Schmidt still orthogonalises but yields no eigenfunctions. The result also depends on the starting pair and its order; a second operator can remove that freedom, as p̂ does by picking e(±ikx), a thread for UQMI-03.09. With every level treated, ∫ ψₘ*ψₙ dx = δₘₙ, the property UQMI-03.07 uses to extract expansion coefficients; what they predict for a measurement is Unit 4's business.
Key equations
What each relation says, and the units it is written in.
Orthogonality from one Hermitian operator
(aₙ − aₘ) ∫ ψₘ* ψₙ dx = 0
Different eigenvalues force the overlap to zero without evaluating it. Equal ones reduce the line to 0 = 0, which constrains nothing.
ψₘ, ψₙ in m⁻¹⁄², both in the one domain D on which  is Hermitian; aₘ, aₙ real, in the units of Â.
An orthonormal set
∫ ψₘ* ψₙ dx = δₘₙ
Normalise every eigenfunction and orthogonalise inside every degenerate level, and the whole table of overlaps becomes the identity.
δₘₙ = 1 when m = n and 0 otherwise; dimensionless, as the overlap of two functions in m⁻¹⁄² must be.
Box sines, checked by direct integration
∫₀ᵃ sin(mπx/a) sin(nπx/a) dx = (a/2) δₘₙ
The zero comes out only because both sines vanish at both walls: that shared boundary condition is the one-domain premise, seen in the integral.
m, n positive integers, a in m. Product-to-sum leaves cosines of whole multiples of π, which integrate to zero.
A degenerate pair on a ring
(2/L) ∫₀ᴸ cos(kx) cos(kx − φ) dx = cos φ
Equal eigenvalues leave the overlap anywhere from −1 to +1. Only φ = ±90° is orthogonal, and picking it is a choice, not a theorem.
k = 2πn/L in m⁻¹ with n a positive integer, circumference L in m, φ any phase; both functions have the energy ħ²k²/2m.
One Gram–Schmidt step
χ = φ₂ − S φ₁, with S = ∫ φ₁* φ₂ dx
χ is orthogonal to φ₁ and still an eigenfunction, because it mixes only functions that share one eigenvalue.
φ₁, φ₂ normalised and degenerate, in m⁻¹⁄²; S dimensionless and in general complex.
Norm left after the subtraction
∫ |χ|² dx = 1 − |S|²
|S| = 1 means φ₂ is φ₁ up to a phase and nothing is left over: a genuinely independent eigenfunction has to be found first.
Dimensionless for normalised φ₁, φ₂. Divide χ by √(1 − |S|²) to get the normalised partner ψ₂.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. At the default m = 2, g shares f's energy and the overlap is cos φ: swing φ and only 90° gives zero, a choice rather than a theorem. Set any other m and the mean locks at zero for every φ, because the energies differ.
Live results
- ENERGY RATIO Eg/Ef
- 1.00
- OVERLAP S = ∫ f g dx
- 0.500
- (Eg − Ef) S / Ef
- 0.000
- NORM OF g − S f
- 0.866
Catch the common trap
The mistake this lesson is built to prevent.
The theorem is routinely quoted as “eigenfunctions of a Hermitian operator are orthogonal”, with the clause about different eigenvalues lost. A student then meets cos(kx) and e(ikx) on a ring, both eigenfunctions of the free Ĥ with energy ħ²k²/2m, finds their normalised overlap is 1/√2, and decides that either the integral is wrong or Ĥ is not Hermitian there. Neither is true. The proof ends at (aₙ − aₘ)∫ ψₘ*ψₙ dx = 0, and with aₙ = aₘ that line is 0 = 0 whatever the overlap. Inside a degenerate level every combination of eigenfunctions is again an eigenfunction, so orthogonal and non-orthogonal pairs both exist there. An orthogonal set is a choice, made by Gram–Schmidt or by a second operator such as p̂, and it has to be made before an eigenbasis can be called orthonormal.
Test it: what changes?
- Overlap S = cos φ, from 1 to −1
- Zero overlap only at φ = 90°
- Energy ratio stays 1
- (Eg − Ef)S stays zero
- Energy ratio m²/4 no longer 1
- Overlap locks at zero for every φ
- Norm of g − Sf stays 1
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron in an infinite well of width a = 0.40 nm is described by Φ = ψ₁ − 3iψ₂, where ψₙ = √(2/a) sin(nπx/a) on 0 < x < a. Find the constant N that normalises NΦ, using orthogonality instead of integrating the cross terms, then confirm the one cross integral directly.
View solution →
- Premise check: ψ₁ and ψ₂ are eigenfunctions of the well's Ĥ, which is Hermitian on functions vanishing at x = 0 and x = a, and both vanish there. Their energies differ, E₁ = π²ħ²/(2ma²) = 2.35 eV and E₂ = 4E₁ = 9.40 eV, so ∫ ψ₁ψ₂ dx = 0.
- Expand, conjugating the coefficient in Φ*: ∫ |Φ|² dx = ∫ (ψ₁ + 3iψ₂)(ψ₁ − 3iψ₂) dx = ∫ ψ₁² dx + 9∫ ψ₂² dx − 3i∫ ψ₁ψ₂ dx + 3i∫ ψ₂ψ₁ dx.
- Each ψₙ is normalised and both cross integrals are zero, so ∫ |Φ|² dx = 1 + 9 = 10 and N = 1/√10 = 0.316, dimensionless because each ψₙ already carries m⁻¹⁄².
- Direct check: ∫₀ᵃ sin(πx/a) sin(2πx/a) dx = (1/2)∫₀ᵃ [cos(πx/a) − cos(3πx/a)] dx = (a/2π)[sin(πx/a) − (1/3) sin(3πx/a)] from 0 to a = 0, since sin π, sin 3π and sin 0 all vanish.
- Limiting case: remove the ψ₂ term and N → 1, as a single normalised eigenfunction requires. The width a set the energies but never entered N.
Answer. N = 1/√10 = 0.316. The cross terms vanish because E₁ = 2.35 eV and E₂ = 9.40 eV differ, and the direct integral of sin(πx/a) sin(2πx/a) over the well is 0.
MediumExample 2
Take as supplied data the oscillator functions ψ₀ = A₀ e(−x²/2b²) and ψ₂ = A₂(2x²/b² − 1)e(−x²/2b²), with b = √(ħ/mω), for an electron with ħω = 0.20 eV. Show that ψ₂ is an eigenfunction of Ĥ = −(ħ²/2m) d²/dx² + ½mω²x² and find its energy, explain why parity cannot make ψ₀ and ψ₂ orthogonal, and verify ∫ ψ₀ψ₂ dx = 0 with Gaussian integrals.
View solution →
- Scale first. With u = x/b, ħ²/(mb²) = ħω and mω²b² = ħω, so Ĥ = (ħω/2)(−d²/du² + u²). Numerically ω = 0.20 eV/ħ = 3.04 × 10¹⁴ s⁻¹ and b = √(ħ/mω) = 0.617 nm.
- Differentiate (2u² − 1)e(−u²/2) twice: the first derivative is (5u − 2u³)e(−u²/2) and the second (2u⁴ − 11u² + 5)e(−u²/2). Then (−d²/du² + u²) returns (10u² − 5)e(−u²/2) = 5(2u² − 1)e(−u²/2), so E₂ = (5/2)ħω = 0.50 eV. The same steps on e(−u²/2) give E₀ = ħω/2 = 0.10 eV.
- Parity: both functions are even, so both have parity eigenvalue +1 and parity says nothing about their overlap. The integrand ψ₀ψ₂ is even, not odd, and nothing cancels by symmetry.
- Ĥ is Hermitian on normalisable functions on the whole line, with V real, and E₀ ≠ E₂, so the overlap must vanish. Check with ∫ e(−x²/b²) dx = √π b and ∫ x² e(−x²/b²) dx = √π b³/2: ∫ (2x²/b² − 1)e(−x²/b²) dx = √π b − √π b = 0.
- Normalisation, for completeness: A₀² = 1/(√π b) and A₂² = 1/(2√π b), so A₀ = 0.956 nm⁻¹⁄² and A₂ = 0.676 nm⁻¹⁄². The overlap is dimensionless, as it must be, and its zero holds for every b.
Answer. E₀ = ħω/2 = 0.10 eV and E₂ = 5ħω/2 = 0.50 eV, with b = 0.617 nm. ∫ ψ₀ψ₂ dx = 0 exactly, forced by the different energies, since parity is +1 for both.
HardExample 3
On a ring of circumference L = 1.5 nm, an electron's level n = 1 (k = 2π/L) is twofold degenerate. Starting from the normalised eigenfunctions φ₁ = e(ikx)/√L and φ₂ = √(2/L) sin(kx + π/6), build an orthonormal pair by Gram–Schmidt, identify the partner it produces, and check it against the momentum operator and against level n = 2.
View solution →
- Both are periodic eigenfunctions of Ĥ = −(ħ²/2m) d²/dx² with E₁ = ħ²k²/2m = 2π²ħ²/(mL²) = 0.668 eV. Equal energies, so the theorem is silent and the overlap has to be computed.
- Write sin(kx + π/6) = [e(i(kx + π/6)) − e(−i(kx + π/6))]/(2i). Against e(−ikx) only the first term survives a full lap, so S = ∫ φ₁*φ₂ dx = (√2/L)(L e(iπ/6)/(2i)) = e(−iπ/3)/√2 = 0.354 − 0.612i, with |S| = 0.707.
- Subtract: χ = φ₂ − Sφ₁. The e(ikx) part of φ₂ equals Sφ₁ and cancels, leaving χ = (i/√(2L)) e(−iπ/6) e(−ikx). The conjugate belongs on φ₁ inside S; placing it on φ₂ would give S* and leave an e(ikx) remainder.
- Norm: ∫ |χ|² dx = 1/2 = 1 − |S|², as it must be. So ψ₂ = √2 χ = e(iπ/3) e(−ikx)/√L, and ∫ φ₁*ψ₂ dx = (e(iπ/3)/L)∫₀ᴸ e(−2ikx) dx = 0.
- Physics: p̂φ₁ = +ħkφ₁ and p̂ψ₂ = −ħkψ₂, with ħk = 2πħ/L = 4.42 × 10⁻²⁵ kg m s⁻¹: the pair circulates in opposite senses with one kinetic energy, like a bead sent either way round a hoop. Level n = 2 has E₂ = 4E₁ = 2.67 eV, so ψ₂ is orthogonal to it with no work.
Answer. S = 0.707 e(−iπ/3); ψ₂ = e(iπ/3) e(−2πix/L)/√L, the counter-rotating state with p = −4.42 × 10⁻²⁵ kg m s⁻¹. Both members have E₁ = 0.668 eV; level 2 sits at 2.67 eV.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Check all three premises before claiming a zero overlap.
(aₙ − aₘ)∫ψₘ*ψₙ dx = 0 forces orthogonality only when one operator is Hermitian on one domain holding both functions and aₙ ≠ aₘ. Inside a degenerate level the identity reads 0 = 0: build an orthogonal pair by Gram–Schmidt, χ = φ₂ − Sφ₁ with S = ∫φ₁*φ₂ dx, or let a second operator choose.