Hermitian operators, and why their eigenvalues are real
Hermitian on a domain: the boundary term decides.
For a derivative operator, every proof that ends “so the eigenvalues are real” runs through an integration by parts and a boundary term somebody chose to drop. Here you do the integration, write the term down, and name the boundary condition that kills it, because one operator can pass the test on one set of functions and fail on the next.
Build the model
An operator is Hermitian on a set of functions, never on its own.
1The big idea
Hermiticity is a claim about moving an operator across an overlap integral: ∫ f*(Âg) dx = ∫ (Âf)* g dx for every pair of allowed functions. For x̂ and a real potential V(x) the move costs nothing, because multiplying by a real function commutes with complex conjugation. For p̂ and the kinetic part of Ĥ the move is an integration by parts, and every integration by parts leaves a term evaluated at the ends of the region: −iħ[f* g] for momentum, −(ħ²/2m)[f* g′ − (f*)′ g] for the Hamiltonian. Whether those terms vanish depends on which functions you let in (decaying ones on the whole line, ones that vanish at the walls of a box, periodic ones on a ring), so Hermiticity belongs to an operator together with its domain, never to a formula alone. Once the test holds, setting f = g = ψ for an eigenfunction gives (a − a*)∫|ψ|² dx = 0, and the eigenvalue is real. Take the domain away and the very same −iħ d/dx has the eigenfunction e(x/b) on a finite interval, with the imaginary eigenvalue −iħ/b. The argument also has an edge: the plane waves e(ikx) have real eigenvalues ħk but cannot be normalised on the whole line, so they sit outside every domain used here, and their reality is argued later through the wave packets built from them.
2In one sentence
An operator  is Hermitian on a set D of allowed wavefunctions when ∫ f*(Âg) dx = ∫ (Âf)* g dx for every f and g in D: it can act on either function in an overlap integral, at the price of a complex conjugate.
Example. On a ring of circumference L = 2.0 nm, where every allowed ψ has ψ(L) = ψ(0), the boundary term −iħ[f* g] cancels and p̂ passes; its eigenfunctions e(2πinx/L) have the real eigenvalues 2πnħ/L = n × 3.3 × 10⁻²⁵ kg m s⁻¹.
3The test needs a list of allowed functions
Call  Hermitian on a domain D when ∫ f*(Âg) dx = ∫ (Âf)* g dx for every f and g in D. The operator may act on the right-hand function, or be moved onto the left-hand one and conjugated, and the integral cannot tell the difference. Read the definition slowly, because “for every f and g in D” is half of it. D is the set of allowed functions: square-integrable, smooth enough for  to act, and obeying the boundary conditions of the problem, such as decay as |x| → ∞ on the whole line, ψ(0) = ψ(L) = 0 in a box of width L, or ψ(L) = ψ(0) on a ring. Change D and you have changed the claim, even with the formula for  untouched. The operator that does the moving in general, ∫ f*(Âg) dx = ∫ (†f)* g dx, is called the adjoint †, so Hermitian means † =  on D. Why a measured quantity must be represented by such an operator is the business of UQMI-04.01; this lesson runs the test.
4Multiplication operators pass without a boundary term
x̂ multiplies by x, and x is real, so ∫ f* x g dx = ∫ (x f)* g dx at once: nothing is differentiated, nothing moves across, and no term appears at the ends. The only demand on D is that x g stays normalisable. Any real potential V(x) passes the same way. Make V complex and the failure sits in the interior instead: ∫ f*(Vg) dx − ∫ (Vf)* g dx = 2i ∫ f* (Im V) g dx. It has a physical signature. With V = V₀ − iΓ/2 the Schrödinger equation gives d/dt ∫|Ψ|² dx = −(Γ/ħ)∫|Ψ|² dx, so the total probability decays as e(−Γt/ħ); for Γ = 1.0 meV the lifetime ħ/Γ is 6.6 × 10⁻¹³ s. Nuclear physicists use such complex potentials on purpose, to model particles removed from a beam, the quantum cousin of a classical damping term. For a closed system, a Hamiltonian that fails the test is one that leaks probability.
5Momentum: the i earns its place, the ends decide
For p̂ = −iħ d/dx on x₁ < x < x₂, integrate by parts once: ∫ f*(−iħ g′) dx = −iħ[f* g] from x₁ to x₂ + iħ ∫ (f*)′ g dx. The last integral is ∫ (−iħ f′)* g dx = ∫ (p̂f)* g dx, because the conjugate of −i is +i. So p̂ passes exactly when f*(x₂) g(x₂) = f*(x₁) g(x₁) for every pair in D, and three domains manage it: normalisable functions on the whole line, whose product dies at ±∞; functions vanishing at both walls of a box; and functions with ψ(L) = ψ(0) on a ring, or more generally ψ(L) = e(iθ)ψ(0) with one fixed θ. Now drop the i. The same step gives ∫ f* g′ dx = −∫ (f′)* g dx on those domains, an extra minus sign: d/dx is anti-Hermitian, (d/dx)† = −d/dx, and multiplying by −iħ is precisely what makes it Hermitian. Units check: ħ in J s and [f* g] in m⁻¹ make the boundary term a momentum, in kg m s⁻¹.
6The Hamiltonian: two integrations, one Wronskian
The kinetic term needs integration by parts twice: ∫ f* g″ dx = [f* g′ − (f*)′ g] from x₁ to x₂ + ∫ (f*)″ g dx. With V real, the potential term moves across for free, so ∫ f*(Ĥg) dx − ∫ (Ĥf)* g dx = −(ħ²/2m)[f* g′ − (f*)′ g] from x₁ to x₂. The bracket is the Wronskian of f* and g, and its end values cancel when every allowed function is zero at an end (a hard wall), when every one has zero slope there, when the functions decay, or when ψ and ψ′ both repeat around a ring. In the infinite well every allowed function vanishes at 0 and L, as sin(πx/L) and sin(2πx/L) do, so both halves of the bracket vanish at each end for every pair and Ĥ passes. The word “every” matters again: if some allowed functions vanish at an end while others merely have zero slope there, f* g′ − (f*)′ g need not be zero, and the Hard example below measures the cost. Units: ħ²/2m in J m² times a bracket in m⁻² gives joules.
7Real eigenvalues in one line, for allowed eigenfunctions
Suppose Âψ = aψ, with ψ in a domain on which  passes the test, and put f = g = ψ. The left side is ∫ ψ*(aψ) dx = a∫|ψ|² dx. On the right the constant comes out of (aψ)* conjugated, so ∫ (aψ)* ψ dx = a*∫|ψ|² dx. Subtracting gives (a − a*)∫|ψ|² dx = 0, and since ∫|ψ|² dx > 0 for any ψ that is not zero everywhere, a = a*. Check it on the ring: ψ = e(2πinx/L)/√L satisfies ψ(L) = ψ(0) for every integer n, and p̂ψ = (2πnħ/L)ψ, real as promised. Two premises were used, and each can fail on its own:  must be Hermitian on D, and ψ must belong to D. Run the same algebra with two eigenfunctions of different eigenvalues and it yields orthogonality, which is UQMI-03.06. The converse fails: with no boundary condition, d/dx has the real eigenvalue 1/b on e(x/b), yet on that same interval it also has e(ikx) with the imaginary eigenvalue ik, so it is not Hermitian there, and one real eigenvalue proves nothing about Hermiticity.
8When the boundary term survives, the spectrum turns complex
Take 0 < x < L and impose nothing at the ends. Then ψ = N e(x/b) is an eigenfunction of p̂ with a = −iħ/b, purely imaginary, though the operator's formula has not changed; only the domain has gone. The proof fails exactly where it should: with f = g = ψ, normalised, the test leaves a − a* = −iħ(|ψ(L)|² − |ψ(0)|²). For L = 1.0 nm and b = 0.50 nm, N² = 2/(b(e(2L/b) − 1)) = 0.0746 nm⁻¹, the end gap |ψ(L)|² − |ψ(0)|² is 2/b = 4.0 nm⁻¹, and −iħ × 4.0 nm⁻¹ = −i × 4.2 × 10⁻²⁵ kg m s⁻¹, which is a − a* = −2iħ/b to the digit. Impose a domain and the whole family disappears: ψ(L) = ψ(0) would need e(L/b) = 1, which no finite b allows, and hard walls exclude every exponential, so a particle in a box has no state of definite momentum at all. The same leak reaches expectation integrals: a real trial function that ignores a wall returns an imaginary ∫ ψ*(p̂ψ) dx, as the Medium example shows.
Key equations
What each relation says, and the units it is written in.
The Hermiticity test
∫ f* (Â g) dx = ∫ (Â f)* g dx, for every f and g in the domain D
 may act on either function inside the overlap, at the price of a conjugate. With f = g it makes ∫ ψ*(Âψ) dx real.
f, g in m⁻¹⁄² in one dimension, so each side has the units of Â. D lists the allowed functions and their boundary conditions.
Multiplication operators
∫ f* (V g) dx − ∫ (V f)* g dx = 2i ∫ f* (Im V) g dx
x̂ and every real V pass on any domain. A complex V = V₀ − iΓ/2 fails in the interior, and ∫|Ψ|² dx then decays as e(−Γt/ħ).
V(x) in J; x̂ multiplies by x, in m. Nothing is differentiated, so no boundary term appears.
Momentum: one integration by parts
∫ f* (p̂ g) dx − ∫ (p̂ f)* g dx = −iħ [f* g] from x₁ to x₂
p̂ passes exactly when f* g takes one value at both ends for every pair: decay at ±∞, ψ = 0 at walls, or ψ(L) = ψ(0) on a ring.
p̂ = −iħ d/dx, ħ = 1.055 × 10⁻³⁴ J s; [f* g] in m⁻¹, so the defect is a momentum in kg m s⁻¹.
Hamiltonian: two integrations by parts
∫ f* (Ĥ g) dx − ∫ (Ĥ f)* g dx = −(ħ²/2m) [f* g′ − (f*)′ g] from x₁ to x₂
The Wronskian of f* and g must cancel between the ends: every allowed function zero there, every slope zero there, or ψ and ψ′ both repeating around a ring.
Real V(x) assumed. ħ²/2m in J m² and the bracket in m⁻², so the defect is in J.
Why the eigenvalues come out real
(a − a*) ∫ |ψ|² dx = 0, so a = a*, when Âψ = aψ with ψ in D
Set f = g = ψ in the test and divide by ∫ |ψ|² dx. An eigenfunction outside D escapes the proof, and its eigenvalue can be complex.
a carries the units of Â; ∫ |ψ|² dx > 0 for any ψ that is not zero everywhere.
The leak in a trial function
Im ∫ ψ* (p̂ψ) dx = −(ħ/2) (|ψ(x₂)|² − |ψ(x₁)|²)
An imaginary part in a momentum integral is a diagnosis, not a result: the function is not in a domain on which p̂ is Hermitian.
Normalised ψ, |ψ|² in m⁻¹, result in kg m s⁻¹. It holds for any ψ, allowed or not.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Drag κL to 0 and the imaginary part and the end gap vanish together. Then step kL/2π through whole numbers: the ring mismatch also reaches zero, and only those functions, with real eigenvalues 2πnħ/L, belong to the ring's domain.
Live results
- RE EIGENVALUE ħk
- 6.28 ħ/L
- IM EIGENVALUE −ħκ
- -0.50 ħ/L
- END GAP |ψ(L)|² − |ψ(0)|²
- 1.00 L⁻¹
- RING MISMATCH |ψ(L) − ψ(0)|
- 0.49 L⁻¹⁄²
Catch the common trap
The mistake this lesson is built to prevent.
The usual slip is to treat Hermiticity as a property of the formula: −iħ d/dx is Hermitian, so the boundary term is dropped by reflex, with “wavefunctions vanish at infinity” quoted even on a finite interval where nothing is at infinity. The integration by parts says otherwise. p̂ passes only when f* g takes the same value at both ends for every allowed pair, and that is a statement about the functions, not about the operator. On 0 < x < L with no boundary condition, the unchanged formula has the eigenfunction e(x/b) with the imaginary eigenvalue −iħ/b, and a real trial function gives an imaginary ∫ ψ*(p̂ψ) dx. “p̂ is Hermitian” is an unfinished sentence until the domain is named.
Test it: what changes?
- Eigenvalue gains imaginary part −ħκ
- End gap |ψ(L)|² − |ψ(0)|² opens to 2κ
- Real part ħk unchanged
- ψ leaves the ring's domain
- Eigenvalue ħk stays real
- End gap stays zero
- Ring mismatch vanishes only at k = 2πn/L
- Ring eigenvalues are 2πnħ/L
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An infinite well of width L = 0.50 nm allows only functions that vanish at x = 0 and x = L. For f = √(2/L) sin(πx/L) and g = √(2/L) sin(2πx/L), evaluate both sides of the Hermiticity test for p̂ = −iħ d/dx and confirm that they agree.
View solution →
- Read off the boundary term before integrating: −iħ[f* g] from 0 to L. Both functions vanish at both walls, so it is zero and the two sides must come out equal.
- Left side: p̂g = −iħ(2π/L)√(2/L) cos(2πx/L), so ∫ f*(p̂g) dx = −iħ(4π/L²)∫ sin(πx/L) cos(2πx/L) dx from 0 to L. With u = πx/L the integral is (L/π)∫ sin u cos 2u du from 0 to π = (L/π)(−2/3), giving +8iħ/(3L).
- Right side: p̂f = −iħ(π/L)√(2/L) cos(πx/L), whose conjugate is +iħ(π/L)√(2/L) cos(πx/L). Then ∫ (p̂f)* g dx = iħ(2π/L²)(L/π)∫ cos u sin 2u du from 0 to π = iħ(2/L)(4/3) = 8iħ/(3L). The sides agree.
- Numbers: ħ/L = 1.055 × 10⁻³⁴ J s ÷ 5.0 × 10⁻¹⁰ m = 2.11 × 10⁻²⁵ kg m s⁻¹, so each side is i × 5.6 × 10⁻²⁵ kg m s⁻¹, a momentum as the units demand.
- The value is imaginary, and that is allowed: Hermiticity ties ∫ f*(p̂g) dx to the conjugate of ∫ g*(p̂f) dx; here ∫ g*(p̂f) dx = −8iħ/(3L), whose conjugate is the +8iħ/(3L) found above. Only the diagonal integrals, with f = g, are forced to be real.
Answer. Both sides equal 8iħ/(3L) = i × 5.6 × 10⁻²⁵ kg m s⁻¹; the boundary term vanishes because f and g are zero at both walls.
MediumExample 2
A student models an electron between hard walls at x = 0 and x = L = 1.0 nm with the normalised trial function ψ = √(3/L³) x, which vanishes at x = 0 but not at x = L. Compute ∫ ψ*(p̂ψ) dx, account for the result with the boundary term, then repeat with ψ = √(30/L⁵) x(L − x).
View solution →
- Normalisation first: ∫ (3/L³) x² dx from 0 to L = (3/L³)(L³/3) = 1, so ψ is a properly normalised function, just not an allowed one.
- p̂ψ = −iħ√(3/L³), so ∫ ψ*(p̂ψ) dx = −iħ(3/L³)∫ x dx from 0 to L = −3iħ/(2L). It is purely imaginary: −i × 1.6 × 10⁻²⁵ kg m s⁻¹ for L = 1.0 nm.
- Account for it: with f = g = ψ the test leaves 2i Im ∫ ψ*(p̂ψ) dx = −iħ(|ψ(L)|² − |ψ(0)|²). Here |ψ(L)|² = 3/L = 3.0 nm⁻¹ and |ψ(0)|² = 0, so the imaginary part is −(ħ/2)(3/L) = −3ħ/(2L). The whole answer is leakage through the wall the function ignored.
- Repair: ψ = √(30/L⁵) x(L − x) vanishes at both walls, and ∫ x²(L − x)² dx = L⁵/30 confirms the normalisation. Because ψ is real, ∫ ψ(−iħψ′) dx = −iħ[ψ²/2] from 0 to L = 0.
- Checks: ħ/L carries kg m s⁻¹, a momentum. As the end value |ψ(L)|² is pulled to zero, the imaginary part goes with it. And zero is the classical answer: a particle bouncing between two walls spends as long moving left as moving right.
Answer. The linear trial gives ∫ ψ*(p̂ψ) dx = −3iħ/(2L) = −i × 1.6 × 10⁻²⁵ kg m s⁻¹, purely imaginary, and its imaginary part is exactly the boundary term −(ħ/2)|ψ(L)|²; the allowed function x(L − x) gives exactly 0.
HardExample 3
An electron (m = 9.11 × 10⁻³¹ kg) moves freely in 0 < x < L with L = 1.0 nm. f = √(2/L) sin(πx/L) has f(0) = f(L) = 0; g = √(2/L) sin(πx/(2L)) has g(0) = 0 and g′(L) = 0. Each is an eigenfunction of Ĥ = −(ħ²/2m) d²/dx² with a real eigenvalue. Find ∫ f*(Ĥg) dx − ∫ (Ĥf)* g dx from the boundary term and directly, and say what it means for a domain holding both.
View solution →
- Eigenvalues: Ef = π²ħ²/(2mL²) and Eg = π²ħ²/(8mL²). With ħ²/(mL²) = 1.222 × 10⁻²⁰ J = 0.0763 eV, Ef = 0.376 eV and Eg = 0.0941 eV.
- Boundary term: −(ħ²/2m)[f g′ − f′ g] from 0 to L. At x = 0 both functions vanish. At x = L, f = 0 and g′ = 0, but f′(L) = −(π/L)√(2/L) and g(L) = √(2/L), so the bracket is +2π/L² and the defect is −πħ²/(mL²).
- Directly: Ĥg = Eg g and (Ĥf)* = Ef f, so the defect is (Eg − Ef)∫ f g dx. The overlap is (2/π)∫ sin u sin(u/2) du from 0 to π = (2/π)(4/3) = 8/(3π) = 0.849, and Eg − Ef = −(3π²/8)ħ²/(mL²). Their product is −πħ²/(mL²), matching.
- Numbers: −π × 1.222 × 10⁻²⁰ J = −3.8 × 10⁻²⁰ J, or −0.24 eV. Units: J m² × m⁻² = J, an energy, as a test on Ĥ must give.
- Control check: swap g for √(2/L) sin(2πx/L), which also vanishes at x = L. The bracket at L becomes 0 and the overlap ∫ f g dx is 0, so the defect vanishes, as it must when both functions share one domain.
- Meaning: the bracket vanishes on the well's domain (ψ = 0 at both ends) and on the domain with ψ(0) = 0 and ψ′(L) = 0, so Ĥ is Hermitian on each. On a domain holding both it is not, and two eigenfunctions with different real eigenvalues then overlap by 0.85 instead of 0; the orthogonality of UQMI-03.06 needs one domain.
Answer. The defect is −πħ²/(mL²) = −3.8 × 10⁻²⁰ J (−0.24 eV) both ways. Ĥ is Hermitian on each boundary condition separately but not on a domain containing both, where eigenfunctions of different energy overlap by 8/(3π) = 0.85.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Name the domain before calling an operator Hermitian.
Check ∫f*(Âg) dx = ∫(Âf)* g dx for every allowed pair. x̂ and real V pass with no boundary term; p̂ needs f*g equal at both ends, Ĥ a cancelling Wronskian bracket. Then (a − a*)∫|ψ|² dx = 0 makes eigenvalues real, but only for eigenfunctions inside the domain.
Next: Orthogonality of eigenfunctions with different eigenvalues →