Born Rule and Measurement Probabilities

The previous article seated the Born rule among the axioms in a single line: the probability of an outcome is the squared modulus of an amplitude. This article is where that line becomes a working tool — and where you learn why the squaring is the most consequential arithmetic operation in physics. Everything unusual about quantum mechanics funnels through one order of operations. Classical probability adds the chances of alternatives; quantum mechanics adds their amplitudes — complex numbers, phases attached — and squares the sum afterwards. The cross term that squaring produces is interference: the double-slit fringes, the tunnelling resonances, and every quantum algorithm's speedup live inside $2\,\mathrm{Re}(c_1^*c_2)$ . Handle the rule in the wrong order and quantum mechanics collapses into a noisy version of coin-flipping; handle it correctly and the whole subject's strangeness becomes calculable. The wave chapter met the rule's oldest instance — $|\Psi(x)|^2$ as position density, built at Born Statistical Interpretation — and the formalism chapter has been preparing its general home: inner products bounded by Cauchy–Schwarz, Hermitian operators with orthogonal eigenbases, completeness making probabilities sum to one. Here the general rule takes over: any observable, any basis, degenerate outcomes included. You will compute energy statistics in a well, watch probabilities survive a change of sublevel basis because projectors do not care, sweep a relative phase and watch a probability oscillate through interference, and see how close experiment has pinned the exponent to exactly 2.

Key takeaways

The rule: measuring $\hat A$ in normalized state $|\psi\rangle$ returns eigenvalue $a_n$ with probability $P(a_n) = |\langle a_n|\psi\rangle|^2 = |c_n|^2$ .
The complex number $c_n = \langle a_n|\psi\rangle$ is the probability amplitude — phases invisible one at a time, decisive when amplitudes add.
Degenerate outcomes: sum over the eigenspace, $P(a) = \sum_i |\langle a^{(i)}|\psi\rangle|^2 = \langle\psi|\hat P_a|\psi\rangle$ — the projector form is the invariant statement.
Interference: measuring in a second basis mixes amplitudes first, squares second — cross terms $2\,\mathrm{Re}(c_1^*c_2)$ appear that classical probability forbids.
Probabilities sum to one by completeness (proved at The Postulates of Quantum Mechanics); continuous spectra give densities: $|\psi(x)|^2$ , $|\Phi(p)|^2$ .
Unnormalized state? Divide: $P = |\langle a_n|\psi\rangle|^2/\langle\psi|\psi\rangle$ .
Key idea: probability is the squared modulus of the amplitude — amplitudes add, probabilities are what you count.

The core insight

Where this lives today

Every quantum computer's readout stage is a Born-rule sampler: run the circuit, measure in the computational basis, and the output distribution is $|\langle x|\psi\rangle|^2$ over bitstrings — quantum advantage claims literally hinge on how hard that distribution is to sample classically. Error-corrected readout, state tomography, and fidelity benchmarks are all applied Born statistics. On the foundations side, the rule's exact form is under active test: triple-slit interference experiments (Sinha et al., 2010, and successors) bound "higher-order interference" — deviations from pairwise amplitude addition — at the parts-per-thousand level, and precision matter-wave interferometry keeps tightening the exponent's agreement with exactly 2. Meanwhile Gleason's theorem and its modern quantum-information descendants sharpen the sense in which the square is the only self-consistent choice — the least arbitrary axiom in the postulate list, and the most tested.

The physical picture

Picture the state $|\psi\rangle$ as an arrow of unit length, and the eigenstates of the observable as a set of perpendicular axes — one axis per possible reading. Measurement asks the arrow "which axis are you?", and the arrow has no answer prepared: it chooses, randomly, with the chance of each axis given by the squared length of the arrow's shadow on it. Because the axes are perpendicular and the arrow has unit length, the squared shadows automatically total one — Pythagoras is doing the probability bookkeeping.

Figure

Shadows on eigenaxes — the squared projections of a unit state are the outcome probabilities

The picture hides one crucial feature the plane cannot draw: each shadow $c_n$ is a complex number. Its modulus sets the probability; its phase does nothing — until you ask a question in a different basis. Then amplitudes recombine before squaring, phases meet in cross terms, and probabilities oscillate. That two-step — rotate the axes, then square — is the entire mechanism of quantum interference, and Derivation B below runs it explicitly. One more picture for degenerate outcomes: when several perpendicular axes share one eigenvalue (an eigenspace), the outcome's probability is the squared shadow on the whole plane they span — not on any particular axis. The shadow-on-a-plane is a projector's job, which is why the projector form of the rule is the honest general statement.

The rule, in its three costumes

Costume 1 — amplitudes (nondegenerate spectra). For a normalized state and an observable whose eigenvalue $a_n$ has one eigenstate $|a_n\rangle$ :

P(a_n) = |\langle a_n|\psi\rangle|^2 = |c_n|^2, \qquad c_n = \langle a_n|\psi\rangle \qquad (1)

where: $c_n$ is the expansion coefficient of $|\psi\rangle = \sum_n c_n|a_n\rangle$ over the eigenbasis (Complete Orthonormal Bases built the expansion; Inner Product and Norm of States bounded the amplitude inside the unit disk so that (1) can be a probability at all). Every $|c_n|^2$ the track has quoted — the $|c_n|^2$ energy weights of Stationary States and Separation of Variables, the outcome statistics of every well problem — was this equation on credit; the credit is now paid.

Costume 2 — projectors (the general, degenerate-safe form). When eigenvalue $a$ owns a $d$ -dimensional eigenspace with orthonormal sublevels $|a^{(i)}\rangle$ :

P(a) = \sum_{i=1}^{d} |\langle a^{(i)}|\psi\rangle|^2 = \langle\psi|\hat P_a|\psi\rangle, \qquad \hat P_a = \sum_{i=1}^{d} |a^{(i)}\rangle\langle a^{(i)}| \qquad (2)

where: $\hat P_a$ is the projector onto the eigenspace (Projection Operators owns projector theory, including the lemma this formula depends on: $\hat P_a$ does not care which sublevel basis you chose). Derivation A shows why the sum is forced and why the projector packaging is the invariant statement — the one that survives every legal rewriting of the degenerate basis.

Costume 3 — densities (continuous spectra). For position and momentum the sum becomes an integral and the squared amplitude becomes a density:

\mathrm{d}P = |\psi(x)|^2\,\mathrm{d}x, \qquad \mathrm{d}P = |\Phi(p)|^2\,\mathrm{d}p \qquad (3)

where: $|\psi(x)|^2$ is the wave chapter's original Born density (Born Statistical Interpretation owns that story) and $|\Phi(p)|^2$ is its momentum twin (Momentum-Space Wavefunction owns the transform pair). Only integrals over ranges are probabilities — a density's point values can exceed 1 harmlessly. The delta-normalized machinery that makes costume 3 an honest instance of costume 1 is deferred by name to Continuous Spectra and Dirac Delta Normalization. A quick dictionary for when each costume works:

situation	use	watch for
discrete, nondegenerate	$P = \|c_n\|^2$	normalize the state first
discrete, degenerate	$P = \langle\psi\|\hat P_a\|\psi\rangle$	sum the whole eigenspace
continuous	$\int_{\text{range}} \|\psi\|^2$	densities are not probabilities
unnormalized state	divide by $\langle\psi\|\psi\rangle$	forgetting the divisor inflates everything

Derivation AThe degenerate case — why you sum the eigenspace and why projectors say it best

We derive the projector form (2) from the amplitude form (1) and prove it is well defined. The issue is real: a degenerate eigenvalue owns infinitely many legal sublevel bases (any unitary remix of the eigenspace is as good as any other — Hermitian Operators and Observables flagged exactly this freedom), so "sum the squared amplitudes over the sublevels" is only a formula if the answer ignores the choice. The strategy: treat the outcome $a$ as the event "the reading was $a$ , whichever sublevel", apply ordinary probability addition to the mutually exclusive sublevel outcomes, then show the result is a sandwich of the basis-independent projector.

Step 1. Add exclusive outcomes. The sublevels $|a^{(i)}\rangle$ are orthonormal, so "landed on sublevel $i$ " and "landed on sublevel $j$ " are perfectly distinguishable, mutually exclusive events; classical probability addition applies to them (no interference between orthogonal outcomes — the cross terms below will confirm this is consistent):

P(a) = \sum_i |\langle a^{(i)}|\psi\rangle|^2

Step 2. Rewrite each term as a sandwich and regroup, using linearity of the inner product:

\sum_i \langle\psi|a^{(i)}\rangle\langle a^{(i)}|\psi\rangle = \Big\langle\psi\Big|\Big(\sum_i |a^{(i)}\rangle\langle a^{(i)}|\Big)\Big|\psi\Big\rangle = \langle\psi|\hat P_a|\psi\rangle

Why the regrouping matters: the sum in the middle is an operator — the eigenspace projector — and the state now touches only that operator, not any particular sublevel basis.

Step 3. Invoke invariance. The projector onto a subspace is unique — remixing the sublevel basis by any unitary leaves $\hat P_a$ unchanged (the well-definedness lemma of Projection Operators; re-verified numerically for this article). Hence $P(a)$ is a property of the eigenspace, as physics demands: the sublevel labels were bookkeeping, and the answer forgot them.

Step 4. Read the special cases. One-dimensional eigenspace: $\hat P_a = |a\rangle\langle a|$ and (2) collapses to (1). Full space ( $\hat P = \hat 1$ ): $P = \langle\psi|\psi\rangle = 1$ — "some outcome happens" is the completeness statement of the postulates article. Verified: numerically — for a 3-dimensional observable with a two-fold eigenvalue and a random state, the sublevel sum equals the projector sandwich to $10^{-14}$ ; remixing the sublevel pair by a unitary ( $\theta = 0.7$ , $\varphi = 1.9$ ) leaves the projector and the probability unchanged; degenerate + nondegenerate probabilities total 1 (verify script, 28 checks green).

Figure

Two orders of operations — squaring first loses the cross term that squaring last creates

Go deeper · Derivation B · 1 minAmplitudes add, probabilities interfere — measurement in a second basis

Derivation BAmplitudes add, probabilities interfere — measurement in a second basis

We derive the interference structure of the Born rule: what happens when a state prepared over one basis is interrogated in another. This is the formalism-register engine behind every two-path experiment — Double-Slit Experiment told the story with screens and slits; here it is three lines of algebra with the cross term isolated and a phase dial attached. The strategy: expand the state over basis one, overlap with a vector of basis two, and square after the amplitudes have been combined — then compare against the classical bookkeeping that squares first.

\langle b_+|\psi\rangle = \frac{c_1 + c_2}{\sqrt2}

Why "routes": the system can reach the outcome $b_+$ via its $|a_1\rangle$ component or its $|a_2\rangle$ component — indistinguishable alternatives, so their amplitudes add before any squaring.

Step 2. Square honestly:

P(b_+) = \frac{|c_1 + c_2|^2}{2} = \frac{|c_1|^2 + |c_2|^2}{2} + \mathrm{Re}(c_1^*c_2)

where: the first piece is what classical probability would predict (halve each option's weight); the second — the cross term — is interference, present because complex amplitudes met before the modulus. For $c_1 = 3/5$ and $c_2 = (4/5)e^{i\varphi}$ : $P(b_+) = \tfrac12 + \tfrac{12}{25}\cos\varphi$ , sweeping from $0.98$ (in phase) through $0.50$ ( $\varphi = \pi/2$ ) to $0.02$ (anti-phase) while the single-basis probabilities $|c_1|^2 = 0.36$ , $|c_2|^2 = 0.64$ never move.

Step 3. Contrast the classical order of operations. Squaring first and adding halves gives $\tfrac12(|c_1|^2 + |c_2|^2) = \tfrac12$ — flat, phase-blind, wrong. The relative phase $\varphi$ is invisible to the $a$ -basis statistics and decisive for the $b$ -basis: exactly the global-versus-relative phase dichotomy of Hilbert Space and State Vectors, now wearing measurement clothes.

Step 4. Check the bookkeeping still closes: $P(b_+) + P(b_-) = \tfrac12|c_1 + c_2|^2 + \tfrac12|c_1 - c_2|^2 = |c_1|^2 + |c_2|^2 = 1$ — the cross terms cancel between the two outcomes. Interference redistributes probability; it never creates or destroys it. Verified: symbolically — $P(b_+) = \tfrac12 + \tfrac{12}{25}\cos\varphi$ exactly (sympy); numerically — the sweep $0.98 / 0.50 / 0.02$ at $\varphi = 0, \pi/2, \pi$ , with $P(b_+) + P(b_-) = 1$ at every phase, and the squared-first classical value pinned flat at $0.50$ (verify script).

Limiting cases

case	result	meaning
eigenstate input, $\|\psi\rangle = \|a_k\rangle$	$P(a_n) = \delta_{nk}$	certainty — the deterministic corner of the rule
degenerate eigenvalue	$P(a) = \langle\psi\|\hat P_a\|\psi\rangle$	sum the whole eigenspace; projector form is invariant
orthogonal to an eigenstate	$P = 0$	zero overlap, impossible outcome — distinguishability made quantitative
continuous spectrum	$\mathrm{d}P = \|\psi(x)\|^2\mathrm{d}x$	density, not probability; integrate over a range
unnormalized state	divide by $\langle\psi\|\psi\rangle$	the rule's built-in unit check
all outcomes summed	$\sum_n P(a_n) = 1$	completeness at work — proved at The Postulates of Quantum Mechanics
second-basis measurement	cross terms $2\,\mathrm{Re}(c_i^*c_j)$	interference — amplitudes add first, square after

Common trapProbabilities of alternatives add, so I can add the slit-1 and slit-2 probabilities

Only when the alternatives are distinguishable — recorded somewhere, in principle readable. If the routes are indistinguishable, quantum mechanics adds their amplitudes, and the squared sum carries a cross term the probability sum lacks: $|c_1 + c_2|^2 \ne |c_1|^2 + |c_2|^2$ . The order of operations is the entire content of the Born rule: add amplitudes for indistinguishable routes; add probabilities for distinguishable outcomes. Two companion traps ride along. A density is not a probability — $|\psi(x)|^2$ can exceed 1 pointwise; only $\int|\psi|^2\mathrm{d}x$ over a range is countable. And an unnormalized state inflates every prediction — divide by $\langle\psi|\psi\rangle$ or normalize first.

Worked examples

Example 1 — energy statistics in a box (confidence builder). An electron in a 1 nm infinite square well ( $E_1 = 0.376$ eV; spectrum from Infinite Square Well) is prepared in

\Psi = \frac{\psi_1 + 2\psi_2 + 2\psi_3}{3}

Find each outcome's probability, the mean energy, and the spread.

Step 1. Normalization check: $(1^2 + 2^2 + 2^2)/9 = 1$ . Legal state, no divisor needed.

Step 2. Probabilities are squared coefficients: $P(E_1) = \boxed{1/9}$ , $P(E_2) = \boxed{4/9}$ , $P(E_3) = \boxed{4/9}$ . Sum: 1. Nothing but $E_1 = 0.376$ , $E_2 = 1.504$ , or $E_3 = 3.384$ eV is ever read out.

Step 3. Mean: $\langle E\rangle = \tfrac19 E_1 + \tfrac49(4E_1) + \tfrac49(9E_1) = \tfrac{53}{9}E_1 = \boxed{2.21\ \mathrm{eV}}$ . Note it sits between allowed readings — a statistical mean, not a possible single-shot value.

Step 4. Spread: $\langle E^2\rangle = \tfrac{389}{9}E_1^2$ , so $\sigma_E = \sqrt{\tfrac{389}{9} - \tfrac{2809}{81}}\,E_1 = \tfrac{\sqrt{692}}{9}E_1 = \boxed{1.10\ \mathrm{eV}}$ . (Both routes — Born weights and the sandwich $\langle\Psi|\hat H|\Psi\rangle$ — agree by machine check; the general theorem connecting them is owned by Expectation Values and Variance, next in this chapter.)

Example 2 — a momentum window on a Gaussian (exam level). A particle is in the minimum-uncertainty Gaussian state whose momentum-space wavefunction (built at Momentum-Space Wavefunction) is a Gaussian of width $\sigma_p$ . What is the probability that a momentum measurement lands within $\pm\sigma_p$ of the mean?

Step 1. Set up the density. Costume 3: $\mathrm{d}P = |\Phi(p)|^2\mathrm{d}p$ with $|\Phi|^2$ a normalized Gaussian of standard deviation $\sigma_p$ — machine-checked to integrate to 1 over all $p$ .

Step 2. Integrate over the window:

P(|p - \langle p\rangle| \le \sigma_p) = \int_{-\sigma_p}^{+\sigma_p} \frac{e^{-p^2/2\sigma_p^2}}{\sigma_p\sqrt{2\pi}}\,\mathrm{d}p = \mathrm{erf}\!\left(\tfrac{1}{\sqrt2}\right) = \boxed{0.683}

Step 3. Read it. The familiar "68% within one sigma" of statistics is here a quantum prediction about single-particle momentum readouts — and the number is exact for the Gaussian state, not a rule of thumb. Widening to $\pm 2\sigma_p$ gives $0.954$ ; the tails never vanish, so arbitrarily large momenta remain possible at exponentially small rates.

Step 4. Note what was not legal: quoting $|\Phi(\langle p\rangle)|^2$ as "the probability of the mean momentum." That number is a density with units of 1/momentum — only the integral is an answer.

The man who generalized the square

Origins · 1 minWolfgang Pauli and how this was discovered

Wolfgang Pauli

1900–1958 · Hamburg and Zurich

Born's 1926 collision papers attached probability to $|\psi(x)|^2$ — position only, scattering only (the famous footnote's story is told at Potential Step and Reflection; the reception battle, at Born Statistical Interpretation). The rule as this article states it — any observable, any basis — crystallized in other hands, and the sharpest early push was Pauli's. In an October 1926 letter to Heisenberg, Pauli proposed that $|\Phi(p)|^2$ must give momentum probabilities exactly as $|\psi(x)|^2$ gives position ones, treating the two representations symmetrically before transformation theory existed to justify it — and in a footnote of his own soon after, he stated the general amplitude-squared reading that Dirac's and Jordan's transformation theory then built into the formalism's foundations. It was characteristic: Pauli published cautiously but corresponded fearlessly, and half the era's ideas passed through his letters before anyone's journals. His famous severity — "not even wrong" was his phrase — made him the working conscience of the quantum generation; Ehrenfest called him "the whip of God." The exclusion principle, the neutrino hypothesis, spin matrices: those stories belong to later chapters of this track. Here he is the physicist who first insisted the Born square was not about position, or about scattering — it was about every question you can put to a state.

History

1926

— Born's collision papers: probability enters via the scattered wave, the square fixed in a footnote

1926

— Pauli's letter to Heisenberg extends the square to momentum space: the rule is representation-blind

1927

— Dirac and Jordan's transformation theory makes any-basis amplitudes the formalism's native objects

1932

— von Neumann's Grundlagen gives the rule its projector form and axiomatic seat

1957

— Gleason's theorem: in dimension three or more, the squared amplitude is the only consistent probability assignment on projectors

2010

— Sinha and collaborators' triple-slit experiment bounds genuine three-path interference, testing the Born square directly

Going further

References. Griffiths, Introduction to Quantum Mechanics, 3rd ed., §1.4, §3.4 · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., §1.4 · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 4 · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. 1, Ch. III · Townsend, A Modern Approach to Quantum Mechanics, Ch. 1–2 · Born, Z. Phys. 37, 863 (1926).

Provenance. Equations and derivations are standard results cross-checked against Griffiths (3rd ed., §1.4, §3.4), Sakurai & Napolitano (2nd ed., §1.4), Shankar (2nd ed., Ch. 4), and Cohen-Tannoudji Vol. 1 (Ch. III). A verifier should confirm: the completeness sum $\sum_n|c_n|^2 = \langle\psi|\psi\rangle$ ; the projector form for degenerate spectra and its sublevel-basis invariance; the interference cross term; and both worked examples. For this edition, verification was machine-executed (sympy/numpy, 28 checks): the $(1,2,2)/3$ well statistics ( $P = 1/9, 4/9, 4/9$ ; $\langle E\rangle = \tfrac{53}{9}E_1 = 2.21$ eV and $\sigma_E = \tfrac{\sqrt{692}}{9}E_1 = 1.10$ eV at $L = 1$ nm, sandwich route agreeing); the degenerate projector sandwich vs sublevel sum to $10^{-14}$ with unitary-remix invariance; the symbolic $P(b_+) = \tfrac12 + \tfrac{12}{25}\cos\varphi$ with the $0.98/0.50/0.02$ sweep and $P(b_+) + P(b_-) = 1$ ; the unnormalized $(1,2)$ case; and the Gaussian window integrals $\mathrm{erf}(1/\sqrt2) = 0.6827$ , $\mathrm{erf}(2/\sqrt2) = 0.9545$ . Per the chapter ownership map: the position-density story and its history belong to Born Statistical Interpretation, the momentum transform pair to Momentum-Space Wavefunction, projector theory (including the invariance lemma consumed here) to Projection Operators, the axiom's seat and the probabilities-sum-to-one proof to The Postulates of Quantum Mechanics, collapse to Measurement and Wavefunction Collapse, and the expectation-value theorem $\langle\hat A\rangle = \sum_n a_nP(a_n) = \langle\psi|\hat A|\psi\rangle$ to Expectation Values and Variance — stated here in one line only.

Builds on: Hilbert Space and State Vectors · Inner Product and Norm of States · Hermitian Operators and Observables · Eigenvalues and Eigenstates · Completeness Relation and Resolution of Identity

Leads to: Measurement and Wavefunction Collapse · Expectation Values and Variance · Projection Operators · The Postulates of Quantum Mechanics

Why the square? Why not |c| or |c|⁴ or something else?

Three converging answers. Bookkeeping: only the square makes probabilities sum to one in every basis, because Pythagoras ( $\sum|c_n|^2 = \|\psi\|^2$ ) is a squared-length statement. Theorem: Gleason proved (dimension ≥ 3) that any consistent, non-contextual probability assignment on projectors must have the form $\langle\psi|\hat P|\psi\rangle$ — the square is forced, not chosen. Experiment: triple-slit tests bound deviations from pairwise amplitude addition at the $10^{-3}$ level and below. The square is the most over-determined convention in the theory.

Is the amplitude itself ever measurable, or only its square?

No single measurement reads an amplitude — but amplitudes are not private bookkeeping either. Relative phases between amplitudes are measurable through interference (Derivation B's sweep is a measurement protocol for $\varphi$ ), and modern tomography reconstructs the full set of amplitudes, up to the global phase, from many measurements in complementary bases. The honest summary: one basis's statistics see only moduli; the state — moduli and relative phases — is recoverable from statistics across several bases.

When do I add amplitudes and when do I add probabilities?

Ask: could I, even in principle, know which alternative occurred? If the alternatives leave no record anywhere — two slits with no detector, two routes to the same final state — they are indistinguishable: add amplitudes, then square. If a record exists (even an unread one — a stray photon, a detector's memory), the alternatives are distinguishable: add probabilities. The double-slit article's visibility formula interpolates smoothly between the two regimes as the record's quality varies; the Born rule contains both as limiting cases of one bookkeeping.

Does the Born rule tell me what the state collapses to after the measurement?

No — that is the collapse postulate's job, stated at The Postulates of Quantum Mechanics and developed at Measurement and Wavefunction Collapse: the post-measurement state is $\hat P_a|\psi\rangle$ renormalized. The Born rule only prices the outcomes. The two work as a pair: Born gives the odds of each branch, collapse says which description to carry forward once the outcome is known.

What does a probability mean for a single system measured once?

Operationally, the Born rule is verified on ensembles: prepare identically, measure many times, compare histograms — the reading built at Born Statistical Interpretation. For a single shot the rule still constrains rational expectation (a 0.98 outcome should not surprise you; a 0.02 one should), but "the probability of this one event" is philosophically loaded, and interpretations differ on what, if anything, it attaches to. The calculations of this chapter are interpretation-independent; the ensemble reading is the safe operational floor.

For a degenerate eigenvalue, why can't I just pick my favourite eigenstate and use |⟨a|ψ⟩|²?

Because the answer would depend on the pick — and physics can't. Different orthonormal bases of one eigenspace assign different squared amplitudes to individual sublevels; only the sum over a full sublevel basis is invariant (Derivation A, Step 3). The projector form $\langle\psi|\hat P_a|\psi\rangle$ bakes the invariance in: it references the eigenspace as a whole, never a basis of it. Single-sublevel probabilities become meaningful only if a finer measurement actually distinguishes the sublevels — a different observable, with its own Born statistics.

Do probabilities change if I write the state in a different basis before applying the rule?

No — and this is a theorem, not luck. The amplitude $\langle a_n|\psi\rangle$ is a basis-independent inner product; expanding $|\psi\rangle$ over any intermediate basis and resumming (the insertion craft of Completeness Relation and Resolution of Identity) reproduces the same number. The postulates article's representation-independence derivation makes it general: unitary relabelings leave every prediction fixed. What does change probabilities is measuring a different observable — rotating the question, not the bookkeeping.