University Physics V · The Quantum Wavefunction · 3.5
Superposition & Eigenbasis Expansion
Superposition is not a haze between two possibilities — it is linear algebra. Choose an observable, take its eigenvectors as axes, and every state becomes a column of complex components. This lesson is about extracting those components correctly, and about what the expansion quietly assumes.
Build the model
Connect the measurement to the mechanism.
The superposition principle is a statement about the state space, not about the particle. The Schrödinger equation is linear, so states form a complex vector space and any combination of states is again a state. Fix a Hermitian observable Â: the spectral theorem hands you an orthonormal eigenbasis (|n⟩), and the resolution of the identity Σₙ|n⟩⟨n| = 1̂ does the rest.
Insert it into |ψ⟩ = 1̂|ψ⟩ and the expansion |ψ⟩ = Σₙ cₙ|n⟩ with cₙ = ⟨n|ψ⟩ falls out in one line; orthonormality is what makes each coefficient a single projection rather than a coupled linear system. Born's rule then reads the squared moduli as the probabilities of the eigenvalues, and normalisation makes them sum to one — Parseval's identity, not an extra postulate. The costs are three.
Completeness is assumed, never proved by the algebra, and Bessel's inequality guarantees only Σ|cₙ|² ≤ 1, so a short basis loses probability in silence. The coefficients belong to a basis, so "the components of the state" means nothing until the observable is named. And where an eigenvalue is degenerate the expansion fixes a whole eigenspace: individual coefficients inside it can be rotated at will while the physical probability ⟨ψ|P̂ₐ|ψ⟩ stands still.
- Simple definition
- Expanding a state in an eigenbasis means writing |ψ⟩ as a sum Σₙ cₙ|n⟩ over the orthonormal eigenvectors of a chosen observable, with each complex coefficient read off as the projection cₙ = ⟨n|ψ⟩.
- Example
- In the infinite well, |ψ⟩ = (√3/2)|1⟩ + (1/2)e(iφ)|2⟩ has |c₁|² = 0.75 and |c₂|² = 0.25: an energy measurement returns E₁ three times in four, and no value of φ changes that.
Insert it anywhere: |ψ⟩ = 1̂|ψ⟩ = Σₙ|n⟩⟨n|ψ⟩ delivers the whole expansion in a single line.
The sum runs over a complete orthonormal set; each |n⟩⟨n| is a dimensionless projector with P̂² = P̂.
Orthonormality collapses ⟨m|Σₙcₙ|n⟩ to cₘ, so each coefficient is one inner product, not a coupled system.
cₙ is complex and dimensionless in a discrete basis; the bra ⟨n| supplies the conjugation.
Your cheapest audit: a sum short of one means an unnormalised state or a basis that does not close.
Dimensionless. For an incomplete set Bessel gives only Σ|cₙ|² ≤ 1, the deficit being lost probability.
Everything diagonal in this basis is phase-blind, which is why probabilities alone never separate a superposition from a mixture.
aₙ carries the units of Â; the modulus discards arg cₙ, so no phase can enter here.
Only an observable off-diagonal in this basis sees the relative phase: in the box, ⟨x⟩ = L/2 + 2|c₁||c₂| cos φ ⟨1|x̂|2⟩.
Bₘₙ = ⟨m|B̂|n⟩ carries the units of B̂; the m = n terms alone are the classical weighted average.
Rotating the basis inside the eigenspace scrambles every c_{a, k} but leaves the measured probability untouched.
k runs over the g-fold degeneracy; P̂ₐ is dimensionless and basis-independent inside the block.
Superposition is closure of the state space under addition
Nothing in the postulates says a particle is "in two places". What they say is that states live in a complex vector space and the Schrödinger equation is linear: if |ψ₁⟩ and |ψ₂⟩ both solve it, so does α|ψ₁⟩ + β|ψ₂⟩ for any complex α and β. A superposition is therefore one vector, not an alternation between two. Two constraints follow. The combination must be renormalised — for orthonormal |1⟩ and |2⟩ that means |α|² + |β|² = 1, while for non-orthogonal kets the cross term 2Re(α*β⟨1|2⟩) joins the sum. And multiplying the whole vector by e(iθ) changes nothing measurable, so a state is a ray; only the relative phase between components survives. Keep those two phases apart and half the confusion in this topic disappears.
Insert the identity and the coefficients fall out
Pick a Hermitian  with a discrete, non-degenerate spectrum. The spectral theorem gives an orthonormal eigenbasis (|n⟩) with ⟨m|n⟩ = δₘₙ and Σₙ|n⟩⟨n| = 1̂. The derivation is then one line: |ψ⟩ = 1̂|ψ⟩ = Σₙ|n⟩⟨n|ψ⟩ = Σₙ cₙ|n⟩ with cₙ = ⟨n|ψ⟩. In the position representation that inner product is the integral cₙ = ∫ψₙ*(x)ψ(x)dx, which is a Fourier coefficient whenever the basis is sinusoidal. The step doing the work is orthonormality: acting with ⟨m| on the sum kills every term but one, so each coefficient is a single projection. Lose orthogonality and you lose that — on a non-orthogonal set the coefficients obey Gc = v with G the Gram matrix ⟨m|n⟩, and you must invert G. Orthonormalise first; it is cheaper.
Moduli are the odds, phases are everything else
Born's rule reads off the eigenbasis directly: measuring  returns aₙ with probability |cₙ|², and normalisation makes Σₙ|cₙ|² = ⟨ψ|ψ⟩ = 1, which is Parseval's identity rather than a new postulate. Because the modulus discards the argument of cₙ, no measurement of  can see any phase, and anything diagonal in this basis inherits that blindness: ⟨Â⟩ = Σₙ|cₙ|²aₙ. An operator with off-diagonal elements does not. Take |ψ⟩ = (√3/2)|1⟩ + (1/2)e(iφ)|2⟩ in the infinite well, where ⟨n|x̂|n⟩ = L/2 and ⟨1|x̂|2⟩ = −16L/(9π²) = −0.1801L. Then ⟨x⟩ = L/2 + 2|c₁||c₂|cos φ ⟨1|x̂|2⟩ = 0.500L − 0.156L cos φ, running from 0.344L at φ = 0 to 0.656L at φ = π while ⟨H⟩ sits at 1.75E₁ throughout. Same probabilities, different states.
Completeness is assumed, and truncation is where you pay
The resolution of the identity is an assumption about the basis, not a theorem the algebra supplies. For any orthonormal set, complete or not, Bessel's inequality gives Σₙ|cₙ|² ≤ ⟨ψ|ψ⟩; completeness is exactly the case of equality. A short basis therefore loses probability silently — the sum simply comes out under one and nothing in the arithmetic complains. Put numbers on it. The nodeless tent state ψ(x) = √(30/L⁵)⋅x(L−x) expands on the box eigenfunctions with cₙ = 4√60/(n³π³) for odd n and zero for even n by parity, giving |c₁|² = 0.99856, |c₃|² = 1.370×10⁻³ and |c₅|² = 6.39×10⁻⁵. One term already carries 99.856% of the probability, but stopping there leaves 1.44×10⁻³ unassigned; keeping n ≤ 3 cuts the deficit to 7.5×10⁻⁵. Auditing Σ|cₙ|² against 1 is the cheapest check you own.
A degenerate eigenvalue fixes a subspace, not a basis
When an eigenvalue a is g-fold degenerate, the label n no longer names a unique vector: any orthonormal basis (|a, k⟩) of the g-dimensional eigenspace will serve. The individual coefficients c_{a, k} = ⟨a, k|ψ⟩ are then convention, and only the projector P̂ₐ = Σₖ|a, k⟩⟨a, k| is basis-independent. So the physical quantity is P(a) = ⟨ψ|P̂ₐ|ψ⟩ = Σₖ|c_{a, k}|², a sum over the whole block, invariant under any unitary mixing inside it. Concretely, a square two-dimensional box has E ∝ nₓ² + ny², so |1,2⟩ and |2,1⟩ share E = 5E₁, and the pair (|1,2⟩ ± |2,1⟩)/√2 spans the same eigenspace just as well. A state with c₁₂ = c₂₁ = 1/2 gives P(5E₁) = 1/4 + 1/4 = 1/2 in either basis. Lüders projection then returns P̂ₐ|ψ⟩ ⁄ ‖P̂ₐ|ψ⟩‖ — a vector in the subspace, not one basis element of it.
Doing it in NumPy, and what the machine will not tell you
Discretise on a grid: the state becomes a length-N complex vector, the observable a Hermitian N×N matrix, and numpy.linalg.eigh returns eigenvalues w together with a matrix V whose columns are the |n⟩. The whole expansion is then one matrix product, c = V.conj().T @ ψ, with probabilities np.abs(c)**2 and the audit np.sum(np.abs(c)**2) against 1. Three warnings. eigh fixes each eigenvector only up to a phase, and that phase is library- and machine-dependent, so an individual cₙ's argument means nothing until you pin a convention — force the largest component real and positive, for instance. For a degenerate eigenvalue eigh hands back an arbitrary orthonormal pair spanning the block, so only the summed probability over that block is reproducible. And the grid basis is complete on the grid, so Σ|cₙ|² = 1 there even when the physical basis has been badly truncated: the grid hides the very deficit Bessel's inequality was warning you about.
Change one variable at a time
Make the relationship visible.
Sweep the phase from 0° to 180°: the density slides from the left half of the well to the right and ⟨x⟩ crosses L/2, while both bars stand perfectly still. Then set the weight to 1.00, where the phase does nothing at all — one eigenstate has no relative phase to carry.
|c₁|²0.75
|c₂|²0.25
⟨x⟩0.344 L
INTERFERENCE IN ⟨x⟩-0.156 L
Live interpretation|c₁|²: 0.75. |c₂|²: 0.25. ⟨x⟩: 0.344 L. INTERFERENCE IN ⟨x⟩: −0.156 L
Catch the common trap
Explain before calculating.
In the infinite well's energy eigenbasis a state |ψ⟩ = (√3/2)|1⟩ + (1/2)|2⟩ is replaced by |ψ′⟩ = (√3/2)|1⟩ − (1/2)|2⟩. What is different about |ψ′⟩?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn orthonormal eigenbasis of  has eigenvalues a₁ = 1 eV, a₂ = 3 eV and a₃ = 5 eV. A state is handed to you unnormalised as |χ⟩ = 2|1⟩ − i|2⟩ + 2|3⟩. Normalise it, give the three outcome probabilities, and find ⟨Â⟩.
- Orthonormality kills every cross term in the norm: ⟨χ|χ⟩ = |2|² + |−i|² + |2|² = 4 + 1 + 4 = 9, so |ψ⟩ = |χ⟩/3.
- The coefficients are c₁ = 2/3, c₂ = −i/3, c₃ = 2/3. Audit: Σ|cₙ|² = 4/9 + 1/9 + 4/9 = 1 ✓.
- Born's rule: P(1 eV) = 4/9 = 0.444, P(3 eV) = 1/9 = 0.111, P(5 eV) = 4/9 = 0.444. The factor −i is invisible here because |−i/3|² = 1/9.
- Â is diagonal in its own eigenbasis, so no cross terms survive: ⟨Â⟩ = Σ|cₙ|²aₙ = (4⋅1 + 1⋅3 + 4⋅5)/9 = 27/9 = 3.00 eV.
AnswerP = 4/9, 1/9, 4/9 and ⟨Â⟩ = 3.00 eV. The −i changes nothing at all here; it would matter only for an observable with non-zero off-diagonal elements in this basis.
MediumAn electron occupies a 1.00 nm infinite well in the state |ψ⟩ = (√3/2)|1⟩ + (1/2)e(iπ/3)|2⟩. Find ⟨H⟩ in eV and ⟨x⟩ in nm, given ⟨n|x̂|n⟩ = L/2 and ⟨1|x̂|2⟩ = −16L/(9π²).
- Check the state: |c₁|² + |c₂|² = 3/4 + 1/4 = 1. The phase e(iπ/3) drops out of every modulus, so normalisation cannot see it.
- Ĥ is diagonal in this basis, so ⟨H⟩ = Σ|cₙ|²Eₙ = 0.75E₁ + 0.25(4E₁) = 1.75E₁, independent of φ. For an electron with L = 1.00 nm, E₁ = h²/(8mL²) = 6.02×10⁻²⁰ J = 0.376 eV, giving ⟨H⟩ = 0.658 eV.
- x̂ is not diagonal, so the m ≠ n terms survive: ⟨x⟩ = (|c₁|² + |c₂|²)(L/2) + 2Re(c₁*c₂)⟨1|x̂|2⟩, with ⟨1|x̂|2⟩ = −16L/(9π²) = −0.1801L.
- 2Re(c₁*c₂) = 2(√3/2)(1/2)cos 60° = 0.4330, so ⟨x⟩ = 0.500L − 0.4330(0.1801L) = 0.500L − 0.0780L = 0.4220L = 0.422 nm.
Answer⟨H⟩ = 1.75E₁ = 0.658 eV, untouched by φ; ⟨x⟩ = 0.422L = 0.422 nm, and it sweeps from 0.344L at φ = 0 to 0.656L at φ = 180° while both |cₙ|² stand still.
HardExpand the normalised tent state ψ(x) = √(30/L⁵)⋅x(L − x) on the infinite-well eigenbasis ψₙ(x) = √(2/L) sin(nπx/L). Find cₙ, evaluate |c₁|², state how much probability a one-term truncation loses, and compute ⟨H⟩ in units of E₁.
- First confirm the prefactor: ∫₀ᴸ x²(L − x)²dx = L⁵/30, so √(30/L⁵) does normalise ψ.
- cₙ = ⟨n|ψ⟩ = √(2/L)⋅√(30/L⁵)∫₀ᴸ x(L − x) sin(nπx/L)dx, and the integral evaluates to 2L³[1 − (−1)ⁿ]/(nπ)³. Even n vanish by parity: ψ is symmetric about L/2 while ψ₂, ψ₄, … are antisymmetric.
- For odd n the integral is 4L³/(n³π³), so cₙ = √(60/L⁶)⋅4L³/(n³π³) = 4√60/(n³π³) = 0.99928/n³ — every power of L cancels, as a dimensionless coefficient must.
- Hence |c₁|² = 0.99856, |c₃|² = 1.370×10⁻³, |c₅|² = 6.39×10⁻⁵. Truncating after n = 1 leaves 1.44×10⁻³ of the probability unassigned; keeping n ≤ 3 cuts that to 7.5×10⁻⁵.
- ⟨H⟩ = Σ|cₙ|²n²E₁ = (960/π⁶)Σ_{odd}n⁻⁴ = (960/π⁶)(π⁴/96)E₁ = 10E₁/π² = 1.0132E₁. The same coefficients give Σ|cₙ|² = (960/π⁶)(π⁶/960) = 1 exactly, confirming the sine basis is complete.
Answercₙ = 4√60/(n³π³) for odd n and zero for even n. |c₁|² = 0.9986, so a one-term truncation loses only 1.44×10⁻³, and ⟨H⟩ = 10E₁/π² = 1.013E₁ — a smooth nodeless trial state sits barely above the ground level.