Observables, Hermitian operators, and real eigenvalues
Real readings demand Hermitian operators.
Any linear operator can be written down; only some may stand for what a detector reports. Start from the one fact every apparatus guarantees, that its readings are real numbers, derive the condition an observable's operator must meet, then see what breaks when an operator fails it: the averages, the spectrum, and probability itself.
Build the model
A detector prints real numbers, and that fixes which operators may stand for it.
1The big idea
An observable is whatever a measuring device reports, and the formalism has to be built around one plain fact about devices: they print real numbers. Every run of readings therefore has a real mean, so the operator Q̂ that represents the quantity must give a real ⟨Q⟩ = ∫ ψ*(Q̂ψ) dx in every state an experimenter could be handed. Applied to the test states f + g and f + ig, that single demand becomes ∫ f*(Q̂g) dx = ∫ (Q̂f)* g dx for every allowed pair: Hermiticity is not an extra axiom but the reality of averages, stated for pairs. What the measurement postulate needs then follows. The eigenvalues, the only values a reading can take, are real; eigenfunctions of different eigenvalues are orthogonal, so distinct readings exclude one another; every mean lies inside the range of the spectrum; and when the operator is Ĥ, the total probability ∫|Ψ|² dx is conserved. The requirement has sharp edges. It is about every state, so an operator whose eigenvalues happen to be real can still fail it, with complex averages in some superposition. It is about allowed states, so it cannot be judged without the boundary conditions that define them. And it describes closed systems: an effective Hamiltonian for a decaying level is non-Hermitian on purpose, because the probability it loses is carried off by something the model leaves out.
2In one sentence
An observable is represented by a Hermitian operator Q̂: one for which ∫ ψ*(Q̂ψ) dx is real for every allowed state ψ, which is the same as ∫ f*(Q̂g) dx = ∫ (Q̂f)* g dx for every allowed pair f and g.
Example. In a packet ψ = A(x)e(ikx) with A real and normalised and k = 5.0 nm⁻¹, ∫ ψ*(dψ/dx) dx = ik = 5.0i nm⁻¹, a mean no meter can print; −iħ d/dx turns the same integral into the real momentum ħk = 5.3 × 10⁻²⁵ kg m s⁻¹.
3Readings are real, so every mean must be real
A detector ends in a number: a position read off a screen in metres, an energy from an analyser in electronvolts, a count of clicks. Whatever happens between preparation and detection, the record of a run is a list of real numbers, and the mean of real numbers is real. Quantum mechanics predicts that mean for a state ψ as ⟨Q⟩ = ∫ ψ*(Q̂ψ) dx (UQMI-03.08), so the operator standing for Q must make that integral real. The demand holds for every state, not a convenient few, because the operator belongs to the apparatus while the state is whatever the source delivers. An energy analyser looking at an electron in a 1.0 nm box must return real means for the n = 1 state (0.376 eV), for n = 2 (1.504 eV), and for every mixture of the two with every relative phase. That last clause, every relative phase, is where the argument gets its force.
4Real for every state means Hermitian
Call B(f, g) = ∫ f*(Q̂g) dx for any two allowed functions. By linearity, the mean in the unnormalised state f + g is B(f, f) + B(g, g) + B(f, g) + B(g, f). The first two terms are already real, so reality demands that B(f, g) + B(g, f) be real: the imaginary parts of the cross terms must cancel. In the state f + ig the cross terms enter as i[B(f, g) − B(g, f)], and reality now demands that their real parts be equal. Together, B(f, g) = B(g, f)*, which written out is ∫ f*(Q̂g) dx = ∫ (Q̂f)* g dx for every allowed pair. That is Hermiticity, obtained from nothing but real means; normalising the test states divides by a positive number and changes nothing. Each test state catches a different fault. The mix f + g exposes complex couplings that are symmetric but not conjugate; the mix f + ig, a relative phase of 90°, exposes couplings that run one way only.
5The spectrum is the list of possible readings
The measurement postulate, built in UQMI-04.02 and UQMI-04.03, says a single measurement of Q returns an eigenvalue of Q̂, so the spectrum is the complete list of numbers the dial can show. Hermiticity puts that list on the real line: for an allowed eigenfunction, f = g = ψ gives (q − q*)∫|ψ|² dx = 0, the one-line proof of UQMI-03.05. It also makes distinct readings exclusive. Eigenfunctions of different eigenvalues are orthogonal (UQMI-03.06), so a state prepared with a definite reading q₁ has no overlap with, and never yields, a different reading q₂. And a mean is an average of possible readings, so it lies between the least and greatest eigenvalue without having to equal any of them. For an equal mix of the n = 1 and n = 2 states of that 1.0 nm box, each run prints 0.376 eV or 1.504 eV, and never the mean, 0.940 eV.
6Real eigenvalues are not enough
Let an operator act inside the span of two orthonormal functions ψ₁ and ψ₂, and record it as a grid of overlaps Qₘₙ = ∫ ψₘ*(Q̂ψₙ) dx. The pair condition becomes Qₘₙ = (Qₙₘ)*: the grid equals its own conjugate transpose. Take Q₁₁ = 1, Q₂₂ = 3, Q₁₂ = 1 and Q₂₁ = 0, in units of some q₀. The grid is triangular, so its eigenvalues are its diagonal entries, 1 and 3, both real. Yet in ψ = (ψ₁ + e(iφ)ψ₂)/√2 the mean is 2 + ½(Q₁₂e(iφ) + Q₂₁e(−iφ)), which at φ = 90° is 2 + 0.5i, a value no run of real readings can average to. The eigenfunctions fail as well. The one for 3 is (ψ₁ + 2ψ₂)/√5, and its overlap with ψ₁, the eigenfunction for 1, is 1/√5, so a state certain to read 3 has a squared overlap of 0.20 with the state certain to read 1. Give the reverse coupling the opposite sign, Q₁₂ = 2 and Q₂₁ = −2, and the eigenvalues 2 ± √(1 + Q₁₂Q₂₁) become 2 ± 1.73i: the spectrum itself leaves the real line.
7A non-Hermitian Hamiltonian does not conserve probability
For the Hamiltonian the failure is worse than a complex average: it is a leak. From iħ ∂Ψ/∂t = ĤΨ and its complex conjugate, d/dt ∫|Ψ|² dx = (1/iħ)[∫ Ψ*(ĤΨ) dx − ∫ (ĤΨ)*Ψ dx], and the bracket is 2i Im ∫ Ψ*(ĤΨ) dx, so the rate is (2/ħ) Im ∫ Ψ*(ĤΨ) dx. If Ĥ is Hermitian that vanishes for every Ψ, and a normalised state stays normalised forever, which the Born rule needs. If not, some state has an imaginary part and probability is created or destroyed. Sometimes that is the intended physics. An effective Ĥ with imaginary part −Γ/2 describes a level that decays as e(−Γt/ħ); for the hydrogen 2p level, lifetime 1.6 ns, Γ = ħ/τ = 4.1 × 10⁻⁷ eV. But the lost probability has gone into the emitted photon, which that model leaves out. A closed system has a Hermitian Hamiltonian; a non-Hermitian one marks a model of part of a system. UQMI-03.05 shows a complex potential producing exactly this term.
8Turning a classical product into an observable
Classically xp is a real number, but the operator x̂p̂ is not an observable. For any real normalised ψ, ∫ ψ x(−iħψ′) dx = −(iħ/2)∫ x(ψ²)′ dx = +(iħ/2)∫ ψ² dx = iħ/2, after one integration by parts with xψ² → 0 at the ends: a purely imaginary mean of xp, i × 5.27 × 10⁻³⁵ J s. The general reason is that moving a product of Hermitian operators across an overlap reverses its order, so x̂p̂ turns into p̂x̂, and the two differ by [x̂, p̂] = iħ (UQMI-03.09). Hence Im ∫ ψ*(x̂p̂ψ) dx = ħ/2 in every normalised state, real or not. A product of two Hermitian operators is Hermitian only when they commute, which is why p̂²/2m is safe and x̂p̂ is not. The repair is to symmetrise. (x̂p̂ + p̂x̂)/2 is Hermitian, and its mean is (m/2) d⟨x²⟩/dt, just as the classical xp = (m/2) d(x²)/dt. In a stationary state ⟨x²⟩ is constant and the mean is zero, as the classical xp averages to zero over any periodic orbit.
Key equations
What each relation says, and the units it is written in.
Every mean is real, in every state
⟨Q⟩ = ∫ ψ* (Q̂ψ) dx = ⟨Q⟩*, for every allowed normalised ψ
The experimental input. Readings are real, so their mean is; the demand covers every state because the source, not the experimenter, decides which state arrives.
ψ in m⁻¹⁄² in one dimension, so ⟨Q⟩ carries the SI unit of Q: m for x̂, kg m s⁻¹ for p̂, J for Ĥ.
From single states to every pair
Real means in ψ = f + g and in ψ = f + ig ⇒ ∫ f* (Q̂g) dx = ∫ (Q̂f)* g dx
Hermiticity is the reality of averages stated for pairs. The f + g state catches complex couplings that are symmetric but not conjugate; f + ig catches couplings that run one way.
f and g are any two allowed functions, each in m⁻¹⁄²; f* g dx is dimensionless, so both sides carry the unit of Q.
Hermitian on a grid of overlaps
Qₘₙ = (Qₙₘ)*, where Qₘₙ = ∫ ψₘ* (Q̂ψₙ) dx
Symmetric is not enough when entries are complex: Q₁₂ = Q₂₁ = 1 + i fails. A triangular grid fails too, although its eigenvalues are its real diagonal entries.
ψₘ, ψₙ orthonormal, so every Qₘₙ carries the unit of Q. Diagonal entries must be real; off-diagonal pairs, conjugates.
The mean stays inside the spectrum
qₘᵢₙ ≤ ⟨Q⟩ ≤ qₘₐₓ, for any normalised state and a Hermitian Q̂
A mean is a weighted average of possible readings, so it lies within their range without having to be one. A mean outside that range, or off the real line, flags a wrong operator or a slip.
qₘᵢₙ and qₘₐₓ are the least and greatest eigenvalues, in the unit of Q; for an unbounded spectrum that side is open.
Probability is conserved only if Ĥ is Hermitian
d/dt ∫ |Ψ|² dx = (2/ħ) Im ∫ Ψ* (ĤΨ) dx
Zero for every Ψ exactly when Ĥ is Hermitian. An imaginary part −Γ/2 gives decay as e(−Γt/ħ): an open system, whose lost probability is carried by something outside the model.
ħ = 1.055 × 10⁻³⁴ J s; Im ∫ Ψ*(ĤΨ) dx is in J, so the rate is in s⁻¹.
A product of observables must be symmetrised
Im ∫ ψ* (x̂p̂ψ) dx = ħ/2, for every normalised allowed ψ
x̂ and p̂ do not commute, so x̂p̂ is not Hermitian. (x̂p̂ + p̂x̂)/2 is, and its mean is (m/2) d⟨x²⟩/dt, matching the classical xp = (m/2) d(x²)/dt.
x̂p̂ carries m × kg m s⁻¹ = J s, and ħ/2 = 5.27 × 10⁻³⁵ J s.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Start at Q₁₂ = 1, Q₂₁ = 0: both eigenvalues sit on the real axis at 1 and 3, yet the curve bulges off it. Set Q₂₁ equal to Q₁₂ and it flattens onto the axis for every φ; set Q₁₂ = 2, Q₂₁ = −2 and the eigenvalues leave the axis too.
Live results
- RE ⟨Q⟩
- 2.00 q₀
- IM ⟨Q⟩
- 0.50 q₀
- LARGEST |IM ⟨Q⟩| OVER φ
- 0.50 q₀
- IM PART OF EIGENVALUES, ±
- 0.00 q₀
Catch the common trap
The mistake this lesson is built to prevent.
When an expectation integral comes out complex, a common repair is to keep its real part, or to drop a stray i, and carry on. That hides the diagnosis. For a Hermitian operator on allowed states the integral is real identically, so an imaginary part means one of three things: an algebra slip, a wavefunction outside the allowed domain, or an operator that cannot represent an observable. Keeping the real part is not neutral either. Re ∫ ψ*(Q̂ψ) dx is the mean of (Q̂ + Q̂†)/2, a different operator with different eigenvalues and eigenfunctions, quietly substituted for the one written down. For x̂p̂ that substitute happens to be the right symmetrised operator; for a proposed Hamiltonian the discarded imaginary part is ħ/2 times the rate at which total probability is being created or destroyed.
Test it: what changes?
- Im ⟨Q⟩ = 0 at every phase φ
- Eigenvalues 2 ± √(1 + Q₁₂²), both real
- Mean stays between the two eigenvalues
- Eigenfunctions of the two eigenvalues become orthogonal
- Im ⟨Q⟩ = ½ sin φ, in units of q₀
- Mean is real only at φ = 0° and 180°
- Largest |Im ⟨Q⟩|, 0.5 q₀, at 90° and 270°
- Eigenvalues stay real, at 1 and 3 q₀
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An operator acts within the span of two orthonormal functions ψ₁ and ψ₂. Three of its overlaps are Q₁₁ = 2.0 eV, Q₂₂ = 5.0 eV and Q₁₂ = (1.0 + 1.0i) eV. Find the Q₂₁ that lets it represent an observable, evaluate ⟨Q⟩ in ψ = (ψ₁ + iψ₂)/√2, and check the result against its eigenvalues.
View solution →
- The grid must equal its own conjugate transpose, so Q₂₁ = (Q₁₂)* = (1.0 − 1.0i) eV. The diagonal entries 2.0 eV and 5.0 eV are already real, as they must be.
- Normalisation: |1/√2|² + |i/√2|² = ½ + ½ = 1. Write ⟨Q⟩ = |c₁|²Q₁₁ + |c₂|²Q₂₂ + c₁*c₂Q₁₂ + c₂*c₁Q₂₁ with c₁ = 1/√2 and c₂ = i/√2.
- Cross terms: (i/2)(1.0 + 1.0i) + (−i/2)(1.0 − 1.0i) = (−0.5 + 0.5i) + (−0.5 − 0.5i) = −1.0 eV; the imaginary parts cancel, as the conjugate pairing guarantees. So ⟨Q⟩ = 3.5 − 1.0 = 2.5 eV, real.
- Spectrum check: the eigenvalues are 3.5 ± √(1.5² + |Q₁₂|²) = 3.5 ± √4.25 = 1.44 eV and 5.56 eV, and 2.5 eV lies between them, as a mean of readings must. Limiting case: with Q₁₂ → 0 the mean returns to 3.5 eV and the eigenvalues to 2.0 and 5.0 eV.
- The tempting symmetric choice Q₂₁ = Q₁₂ = (1.0 + 1.0i) eV fails. It gives 3.5 eV in this state, but in (ψ₁ + ψ₂)/√2 its mean is 3.5 + ½(Q₁₂ + Q₂₁) = (4.5 + 1.0i) eV: it is the f + g test state that catches it.
Answer. Q₂₁ = (1.0 − 1.0i) eV. Then ⟨Q⟩ = 2.5 eV in (ψ₁ + iψ₂)/√2: real, and between the eigenvalues 1.44 eV and 5.56 eV.
MediumExample 2
An electron packet ψ(x) = (2/(πa²))¹⁄⁴ e(−(x − x₀)²/a²) e(ikx) has a = 0.200 nm, x₀ = 0.500 nm and k = 5.00 nm⁻¹. A student proposes x̂p̂ as the operator for the classical product xp. Evaluate ∫ ψ*(x̂p̂ψ) dx and ∫ ψ*(p̂x̂ψ) dx, then the mean of (x̂p̂ + p̂x̂)/2, and compare with the classical value.
View solution →
- Write ψ = G e(ikx) with G real, ∫ G² dx = 1 (the Gaussian is normalised) and ∫ x G² dx = x₀. Then p̂ψ = −iħψ′ = (ħkG − iħG′)e(ikx).
- ∫ ψ*(x̂p̂ψ) dx = ħk∫ xG² dx − iħ∫ xGG′ dx. The first integral is x₀. For the second, xGG′ = ½x(G²)′, and integrating by parts with xG² → 0 at ±∞ gives −½. So ∫ ψ*(x̂p̂ψ) dx = ħkx₀ + iħ/2.
- p̂(xψ) = −iħψ + x(p̂ψ), so ∫ ψ*(p̂x̂ψ) dx = ħkx₀ + iħ/2 − iħ = ħkx₀ − iħ/2, the complex conjugate of the first. The symmetrised mean is their average, ħkx₀, and it is real.
- Numbers: ħk = 1.055 × 10⁻³⁴ J s × 5.00 × 10⁹ m⁻¹ = 5.27 × 10⁻²⁵ kg m s⁻¹, so ħkx₀ = 5.27 × 10⁻²⁵ × 5.00 × 10⁻¹⁰ = 2.64 × 10⁻³⁴ J s, while ħ/2 = 0.53 × 10⁻³⁴ J s: the imaginary part is a fifth of the real part, far too large to wave away.
- Checks: every term is in J s, as m × kg m s⁻¹ must be. The width a has dropped out. Setting k = 0 or x₀ = 0 removes the real part but leaves iħ/2, so the defect belongs to the operator, not the packet. The symmetrised mean equals x₀ × ħk = ⟨x⟩⟨p⟩, the classical product.
Answer. ∫ ψ*(x̂p̂ψ) dx = (2.64 + 0.53i) × 10⁻³⁴ J s and ∫ ψ*(p̂x̂ψ) dx = (2.64 − 0.53i) × 10⁻³⁴ J s; the symmetrised operator gives the real 2.64 × 10⁻³⁴ J s = x₀ħk, the classical value.
HardExample 3
A proposed model of two levels in a quantum dot uses orthonormal states ψ₁ and ψ₂ and a Hamiltonian with overlaps H₁₁ = 10.0 meV, H₂₂ = 14.0 meV, H₁₂ = 1.0 meV and H₂₁ = 0. Show that its energy eigenvalues are real, then show that it is still unphysical: find ⟨H⟩ and the initial rate of change of total probability in Ψ = (ψ₁ + iψ₂)/√2, and the overlap of its two energy eigenfunctions. Repair it and give the new eigenvalues.
View solution →
- The grid is triangular, so (10.0 − E)(14.0 − E) − (1.0)(0) = 0 and the eigenvalues are 10.0 meV, with eigenfunction ψ₁, and 14.0 meV. Both are real: the model passes the test most people try first.
- Ψ is normalised, ½ + ½ = 1. With c₁ = 1/√2 and c₂ = i/√2, ⟨H⟩ = ½(H₁₁ + H₂₂) + (i/2)(H₁₂ − H₂₁) = (12.0 + 0.50i) meV. At a relative phase of 0 the imaginary part vanishes, which is why a check with real mixtures alone misses it.
- d/dt ∫|Ψ|² dx = (2/ħ) Im⟨H⟩ = 2 × 0.50 × 10⁻³ eV ÷ 6.58 × 10⁻¹⁶ eV s = 1.5 × 10¹² s⁻¹, in s⁻¹ as it must be. At that initial rate the total probability passes 1.01 within about 6.6 fs, in a model meant to describe a closed dot.
- The eigenfunction for 14.0 meV obeys (10.0 − 14.0)u + 1.0v = 0, so v = 4u and it is (ψ₁ + 4ψ₂)/√17. Its overlap with ψ₁ is 1/√17 = 0.243, a squared overlap of 0.059 between two supposedly exclusive energies.
- Repair: H₂₁ = (H₁₂)* = 1.0 meV. Now Im⟨H⟩ = 0 at every phase, and the eigenvalues are 12.0 ± √(2.0² + 1.0²) = 12.0 ± 2.24 meV, that is 9.76 meV and 14.24 meV, with orthogonal eigenfunctions.
- Checks: as H₁₂ → 0 the repaired eigenvalues return to 10.0 and 14.0 meV, and a real coupling pushes the two levels apart, by 0.24 meV each, as level repulsion requires.
Answer. Real eigenvalues 10.0 and 14.0 meV, yet ⟨H⟩ = (12.0 + 0.50i) meV, total probability initially changing at 1.5 × 10¹² s⁻¹, and eigenfunctions overlapping by 0.243. With H₂₁ = 1.0 meV the eigenvalues are 9.76 and 14.24 meV.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Hermiticity is the reality of every mean, stated for pairs.
Test a proposed operator with ∫ f*(Q̂g) dx = ∫ (Q̂f)* g dx for every allowed pair, or Qₘₙ = (Qₙₘ)* on a grid of overlaps. Real eigenvalues alone prove nothing. The test depends on the boundary conditions, and a non-Hermitian Ĥ fails to conserve probability.