Parity Symmetry of One-Dimensional Eigenstates

Fold a sheet of paper down the middle of a symmetric well — the infinite box centred at the origin, the harmonic oscillator's parabola, the finite well — and the potential lands exactly on itself. Quantum mechanics takes that innocent geometric fact and converts it into hard information about every single bound state, before you solve anything: each one must be an even function or an odd function of $x$ . Not "roughly symmetric," not "symmetric on average" — exactly even or exactly odd, with the two types strictly alternating up the energy ladder. This is your first real encounter with the deepest working principle of the subject: symmetry organizes spectra. The mirror operation gets promoted to an operator, the parity operator $\hat\Pi$ ; the symmetry of $V$ becomes the algebraic statement $[\hat H, \hat\Pi] = 0$ ; and that commutator, plus one structural theorem about one-dimensional wells, forces the even/odd classification. The same three-step logic — find the symmetry, build its operator, harvest the labels — will later hand you angular momentum quantum numbers, selection rules for atomic transitions, and conservation laws in field theory. Parity in one dimension is the smallest complete specimen. The payoff is intensely practical too. Knowing a state is odd tells you it vanishes at the origin, that $\langle x\rangle = 0$ forever, and that half of all matrix elements you might ever compute are zero by inspection. It halves the algebra of the finite square well, and it turns a wall-plus-well problem you have never seen into one you already solved. This article builds the operator, proves the classification theorem honestly — including the degeneracy fine print most first courses mumble past — and puts the tool to work.

Key takeaways

The parity operator $\hat\Pi$ reflects a wavefunction through the origin: $\hat\Pi\psi(x) = \psi(-x)$ . It is Hermitian, unitary, and squares to the identity, so its eigenvalues are exactly $\pm 1$ — even and odd.
If the potential is symmetric, $V(-x) = V(x)$ , then $[\hat H, \hat\Pi] = 0$ : parity is conserved and energy eigenstates can be organized by it.
Commuting alone is not enough: a non-degenerate eigenstate of $\hat H$ must be a parity eigenstate. One-dimensional bound states are non-degenerate, so every bound state of a symmetric well is exactly even or exactly odd.
Parity alternates up the ladder: ground state even, first excited odd, $n$ th state $(-1)^n$ — in step with the node count.
Instant consequences: definite-parity states have $\langle x\rangle = \langle p\rangle = 0$ ; matrix elements like $\langle\text{even}|\,\hat x\,|\text{even}\rangle$ vanish by symmetry; odd states solve the "wall at the origin" problem for free.
Key idea: a symmetric well cannot tell left from right, and a bound state — being unique at its energy — must respect that: it is forced to be even or odd, no third option.

The core insight

For an even potential, $[\hat H, \hat\Pi] = 0$ . A non-degenerate energy eigenstate must then be an eigenstate of $\hat\Pi$ — even or odd — and in one dimension bound states are non-degenerate, so parity alternates as $(-1)^n$ with the node count $n$ .

Where this lives today

Parity bookkeeping is daily infrastructure. Spectroscopists label every molecular level by parity (the $g$ / $u$ subscripts of molecular orbitals are the $\pm 1$ of equation (2)) and read allowed transitions off Laporte's rule — the 3D descendant of equation (6). Semiconductor engineers designing quantum-well lasers use the even/odd alternation to know which intersubband transitions couple to light — in a symmetric well $1\to2$ is bright and $1\to3$ is dark — and deliberately growing an asymmetric well is the standard trick for switching forbidden transitions on. In nuclear and particle physics every state carries a $J^P$ assignment, and parity-violating observables — descendants of Wu's asymmetry — remain precision probes of the weak sector, from caesium atomic parity violation to polarized electron scattering. The one-dimensional theorem you now own is the smallest working model of all of it.

The physical picture

Place a mirror at $x = 0$ . If the potential landscape looks identical in the mirror — every hill and valley of $V$ has a twin at the reflected position — then physics itself cannot distinguish left from right. Now ask what a bound state can look like. A bound state is a unique standing wave at its energy (a fact this chapter proves separately, and which we lean on below). Reflect that standing wave: the reflected function solves the same Schrödinger equation at the same energy, so by uniqueness it can only be the original back again, up to a constant. Reflecting twice restores everything, so that constant squares to one: the wave is either identical in the mirror ( $+1$ , even) or exactly inverted ( $-1$ , odd). There is no third option for a unique wave living in a mirror-symmetric world.

Figure

The mirror test — a symmetric well forces each unique bound state to be even or odd

Why do the two types alternate as you climb the energy ladder? Watch the nodes. Each higher bound state carries exactly one more zero-crossing than the last (the counting law that Node Theorem and Energy Ordering, this chapter's next article, makes precise). A node sitting exactly at the centre forces the function to be odd there; no node at the centre — a hump straddling it — is the even option. Adding nodes one at a time toggles whether the centre is a crossing or a crest, so the parities read even, odd, even, odd, … all the way up: $(-1)^n$ for the $n$ th state, counting the ground state as $n = 0$ . You have already seen the pattern without naming it: the Infinite Square Well's ground state is a single centred hump, its first excited state a full sine that dives through zero at the middle; the Finite Square Well's quantization condition split by itself into an even branch and an odd branch; the harmonic oscillator's Hermite polynomials flip sign under $x \to -x$ exactly when $n$ is odd.

Figure

Parity alternates up the ladder — node at the centre means odd, hump at the centre means even

One warning shot before the algebra, courtesy of the free particle. The plane waves $e^{\pm ikx}$ share the energy $\hbar^2k^2/2m$ in a perfectly symmetric potential — yet neither is even or odd: reflection maps one into the other. The mirror argument silently used uniqueness, and the free particle's doubly degenerate energies are exactly where uniqueness fails. Degeneracy is the fine print of every symmetry-classification theorem, and this article keeps it in view.

The operator and its algebra

To make the mirror an object of the theory, define its action on any wavefunction. This single line is the entire definition; everything else is worked out from it:

\hat\Pi\,\psi(x) = \psi(-x) \qquad (1)

where: $\hat\Pi$ — the parity operator, which evaluates the function at the reflected point. Note what it is not: it does not multiply by $-1$ , and it does nothing at all to a function already symmetric about the origin. Applying it twice evaluates $\psi$ at $-(-x) = x$ : reflection is its own inverse. That double-reflection identity is the engine of the whole classification. From $\hat\Pi^2 = \hat 1$ , any eigenvalue $\lambda$ of $\hat\Pi$ obeys $\lambda^2 = 1$ , and since $\hat\Pi$ is Hermitian (flip $x \to -x$ inside the inner-product integral; the measure is symmetric), $\lambda$ is real:

\hat\Pi^2 = \hat 1 \quad\Longrightarrow\quad \lambda = \pm 1 \qquad (2)

where: $\lambda = +1$ states are the even functions, $\psi(-x) = \psi(x)$ , and $\lambda = -1$ states are the odd functions, $\psi(-x) = -\psi(x)$ . Parity is a discrete symmetry — a two-outcome question, not a continuous dial like translation or rotation — which is why its conserved quantity is a sign rather than a number like momentum. Any function whatsoever splits cleanly into one piece of each type. Write $\psi_{\rm e} = \tfrac12(\psi(x) + \psi(-x))$ and $\psi_{\rm o} = \tfrac12(\psi(x) - \psi(-x))$ ; their sum reconstructs $\psi$ , and each piece is a parity eigenstate. The two eigenspaces of $\hat\Pi$ between them exhaust the whole space of states — parity asks a question every wavefunction must answer, even if the answer is "a superposition of both." Now bring in the dynamics. The kinetic term $-\tfrac{\hbar^2}{2m}\partial_x^2$ is automatically mirror-symmetric (two sign flips from the two derivatives), so the Hamiltonian commutes with parity exactly when the potential does:

V(-x) = V(x) \quad\Longleftrightarrow\quad [\hat H, \hat\Pi] = 0 \qquad (3)

where: the commutator $[\hat H,\hat\Pi] = \hat H\hat\Pi - \hat\Pi\hat H$ measures whether "reflect, then evolve" differs from "evolve, then reflect." Zero means parity is conserved: an even initial state stays even forever, an odd one stays odd. For an asymmetric potential — a tilted well, an applied electric field — equation (3) fails and the even/odd language simply does not apply. (The general operator machinery behind statements like this — commutators, compatible observables, simultaneous eigenbases — is built properly in the Formalism chapter; here we use only what we can verify by direct substitution.)

Derivation AWhy a non-degenerate eigenstate must be even or odd

We prove the classification theorem: if $V(-x) = V(x)$ and $\psi$ is a bound state of $\hat H$ with energy $E$ that is non-degenerate — the only normalizable state at that energy — then $\hat\Pi\psi = \pm\psi$ . The strategy is the mirror argument made precise: show the reflected function is again an eigenstate at the same energy, invoke uniqueness to conclude it is proportional to the original, then use $\hat\Pi^2 = \hat 1$ to pin the constant. Watch where non-degeneracy enters — it is the hinge of the whole proof, and the free-particle counterexample below shows the theorem genuinely fails without it.

Step 1. Reflect the eigenvalue equation. Start from $\hat H\psi = E\psi$ and apply $\hat\Pi$ to both sides: $\hat\Pi\hat H\psi = E\,\hat\Pi\psi$ . Because $[\hat H,\hat\Pi] = 0$ , the left side rearranges to $\hat H(\hat\Pi\psi)$ , giving

\hat H\,(\hat\Pi\psi) = E\,(\hat\Pi\psi)

Why this matters: the reflected function $\psi(-x)$ is itself a solution of the time-independent Schrödinger equation at the very same energy $E$ . Symmetry has manufactured a second candidate eigenstate out of the first.

Step 2. Invoke non-degeneracy. By hypothesis there is only one independent normalizable solution at energy $E$ (this chapter proves the theorem separately — Non-Degeneracy of One-Dimensional Bound States — via the Wronskian; here we take it as the certified input). Two solutions at the same energy must therefore be proportional:

\hat\Pi\psi = c\,\psi \quad \text{for some constant } c

Why we may not skip this step: without uniqueness, Step 1 only says $\hat\Pi\psi$ lives somewhere in the degenerate energy subspace — it could be a genuinely different state, as $\hat\Pi e^{ikx} = e^{-ikx}$ is.

Step 3. Pin the constant with $\hat\Pi^2 = \hat 1$ . Apply $\hat\Pi$ once more: $\psi = \hat\Pi^2\psi = c\,\hat\Pi\psi = c^2\psi$ , so $c^2 = 1$ and

c = \pm 1 \quad\Longrightarrow\quad \psi(-x) = \pm\,\psi(x)

Why the answer is a sign and not a phase: $c$ could a priori be any complex number of modulus 1, but reflecting twice must restore the original function exactly, and $c^2 = 1$ has only the two real roots. Even or odd — the third option is eliminated by algebra, not by taste. *Verified: the operator identities were checked by direct substitution (sympy): $f(-(-x)) = f(x)$ , Hermite parity $H_n(-x) = (-1)^n H_n(x)$ for $n = 0\ldots5$ , and the free-particle counterexample $\hat\Pi e^{ikx} = e^{-ikx}$ (a different state, confirming the theorem's degeneracy fine print).*

The theorem's two hypotheses — symmetry and non-degeneracy — are both doing real work, and it is worth seeing the pattern in the standard wells. For the harmonic oscillator the explicit eigenfunctions make the alternation visible in closed form:

\psi_n(-x) = (-1)^n\,\psi_n(x), \qquad \psi_n(x) \propto H_n\!\left(\sqrt{\tfrac{m\omega}{\hbar}}\,x\right)e^{-m\omega x^2/2\hbar} \qquad (4)

where: the Gaussian factor is even, so the parity of $\psi_n$ is exactly the parity of the Hermite polynomial $H_n$ — and $H_n$ contains only even powers of $x$ when $n$ is even, only odd powers when $n$ is odd (Hermite Polynomials and Oscillator Eigenfunctions owns that family's anatomy; we recall the parity fact). The centred infinite well tells the same story with sines: relabel the box to run from $-L/2$ to $L/2$ and the familiar $\psi_n$ become alternately cosines (even) and sines (odd) of the half-width — the ground state even, exactly as the staircase picture demands.

What parity buys you

Expectation values vanish by symmetry. For any definite-parity state, the probability density $|\psi|^2$ is even regardless of whether $\psi$ itself is even or odd — the sign squares away. Integrating the odd function $x|\psi(x)|^2$ over a symmetric domain gives exactly zero:

\langle x\rangle = 0, \qquad \langle p\rangle = 0 \qquad \text{(any definite-parity state)} \qquad (5)

where: the momentum average vanishes by the same integrand-parity bookkeeping ( $\psi$ and $\psi'$ have opposite parities, so $\psi^*\psi'$ is odd). This is a structural zero — no integral ever needs computing, and it holds for the 47th excited state as surely as for the ground state. A bound particle in a symmetric well is, on average, dead centre with no drift, forever.

Matrix elements vanish by selection rule. The same parity bookkeeping clears whole tables of integrals at a glance. The operator $\hat x$ is odd; sandwiched between two states it gives an integrand of total parity (parity of bra) × (odd) × (parity of ket):

\langle \psi_m |\, \hat x \,| \psi_n \rangle = 0 \quad \text{whenever } \psi_m,\ \psi_n \text{ have the same parity} \qquad (6)

where: only opposite-parity pairs survive. This is the prototype of every selection rule you will meet: when light drives transitions (the machinery of the time-dependent chapter), the coupling is essentially $\hat x$ , and equation (6) says even↔even and odd↔odd transitions are switched off at this order. In the centred infinite well, $\langle 1|\hat x|2\rangle = 16L/9\pi^2 \approx 0.180\,L$ (even↔odd: allowed and large), while $\langle 1|\hat x|3\rangle = 0$ identically (even↔even: forbidden).

Half the problem, free. An odd state is pinned to zero at the origin — $\psi(0) = 0$ is forced by $\psi(-0) = -\psi(0)$ . So the odd states of a symmetric well are exactly the full solution set of a different problem: the same well with an impenetrable wall inserted at $x = 0$ . Conversely, even states have a flat crest there, $\psi'(0) = 0$ . This "half-well dictionary" is a genuine solving technique, not a curiosity, and Worked example 2 cashes it on a well you would otherwise have to attack from scratch.

Go deeper · Derivation B · 2 minThe half-well dictionary — odd states solve the wall problem

Derivation BThe half-well dictionary — odd states solve the wall problem

We show that the bound states of the half-space problem (impenetrable wall at $x = 0$ , potential $V(x)$ for $x > 0$ ) are in exact one-to-one correspondence with the odd bound states of the symmetric full-space problem with potential $V(|x|)$ , at identical energies. The strategy is to check boundary conditions in both directions: every odd full-well state restricts to a legal half-well state, and every half-well state extends by odd reflection to a legal full-well state. The point to watch is the derivative at the origin — continuity there is what makes the odd extension a genuine solution rather than a paste-up.

Step 1. Restrict an odd state. Let $\psi$ be an odd bound state of the symmetric well at energy $E$ . On $x > 0$ it solves the Schrödinger equation with the same $V$ , and oddness forces $\psi(0) = 0$ — precisely the boundary condition an impenetrable wall imposes. Its restriction to $x \ge 0$ is therefore a legitimate half-well bound state at energy $E$ (renormalize by $\sqrt2$ : the half-line carries half the norm).

Step 2. Extend a half-well state. Let $\phi$ solve the wall problem at energy $E$ , with $\phi(0) = 0$ . Define $\psi(x) = \phi(x)$ for $x \ge 0$ and $\psi(x) = -\phi(-x)$ for $x < 0$ . On each open half-line, $\psi$ solves the symmetric-well equation (the reflected equation is the same equation, by $V(-x) = V(x)$ ).

Step 3. Check the seam. At $x = 0$ : the function is continuous ( $0 = -0$ ), and the odd extension has $\psi'(0^-) = \phi'(0^+)$ — the derivative of an odd function is even, so the slopes match automatically. Both matching conditions of a finite potential (Matching Conditions and Wavefunction Continuity owns the general rules) are satisfied; $\psi$ is a genuine bound state, and it is odd by construction. Why even states do not sneak in: an even function has $\psi'(0) = 0$ but generically $\psi(0) \ne 0$ — it fails the wall's boundary condition. The wall kills exactly half the symmetric well's spectrum, the even half. *Verified: numerically for the infinite well — the spectrum of a hard-wall box of width $a$ equals the odd- $n$ subset of the centred box of width $2a$ ( $E_m^{\rm half} = m^2\pi^2\hbar^2/2ma^2$ matches $E_{n=2m}^{\rm full}$ , checked for $m = 1\ldots5$ ) — and for the finite well of Worked example 2 (odd-branch roots reproduce the wall problem's spectrum).*

Limiting cases

case	result	meaning
symmetric $V$ , bound state	$\psi(-x) = \pm\psi(x)$ , parity $(-1)^n$	the theorem at full strength: uniqueness + mirror symmetry
symmetric $V$ , degenerate level	eigenstates need not have definite parity	$\hat\Pi$ only maps the energy subspace into itself; definite-parity combinations exist but are a choice
free particle	$e^{\pm ikx}$ swap under $\hat\Pi$	the canonical degenerate counterexample; $\cos kx$ , $\sin kx$ are the optional even/odd repackaging
asymmetric $V$	$[\hat H,\hat\Pi] \ne 0$ , no classification	tilt the well (add $\varepsilon x$ ) and even/odd labels dissolve immediately
superposition of parities	$\langle x\rangle(t)$ oscillates	mixing $n=0$ and $n=1$ makes the density slosh — the seed of radiating dipoles
odd states at the origin	$\psi(0) = 0$ always	solves the wall-at-origin problem free of charge (half-well dictionary)

Common trapSince the Hamiltonian commutes with parity, every energy eigenstate is even or odd

The commutator alone guarantees less than that. $[\hat H,\hat\Pi] = 0$ says a common eigenbasis of the two operators can be chosen — it does not say every eigenstate of $\hat H$ individually already has definite parity. The gap is degeneracy: when two or more states share an energy, parity is free to rotate them into each other, and generic members of that subspace are neither even nor odd ( $e^{ikx}$ is the standing counterexample, at a doubly degenerate free-particle energy). What closes the gap in one dimension is the non-degeneracy of bound states — a separate theorem with its own proof and its own fine print. Keep the logic straight in your notes: commutation + non-degeneracy ⟹ definite parity. In three dimensions, where degeneracy is rampant, exactly this distinction is why a hydrogen level contains states of mixed character until you deliberately diagonalize the symmetry operators.

Worked examples

Example 1 — reading off zeros and sloshing charge (confidence builder). An electron sits in a centred infinite well of width $L = 1.0$ nm. (a) Without integrating, state $\langle x\rangle$ for the ground and first excited states. (b) Compute the one matrix element that parity permits between them, and the frequency at which a 50–50 superposition of the two sloshes.

Step 1. Parity bookkeeping. In the centred well, $\psi_1$ (one hump) is even and $\psi_2$ (one central node) is odd. Both have definite parity, so by equation (5): $\langle x\rangle_1 = \langle x\rangle_2 = \boxed{0}$ — no integral required.

Step 2. The allowed cross element. Even↔odd survives the selection rule; the standard integral gives $\langle 1|\hat x|2\rangle = 16L/9\pi^2 = 0.180$ nm — about a fifth of the box width, a large dipole by atomic standards.

Step 3. Energies and the beat. $E_1 = \pi^2\hbar^2/2mL^2 = 0.376$ eV and $E_2 = 4E_1 = 1.504$ eV, so the superposition's density oscillates at $f = (E_2 - E_1)/2\pi\hbar = \boxed{273\ \text{THz}}$ — near-infrared. The centroid swings as $\langle x\rangle(t) = 0.180\,\text{nm} \times \cos(2\pi f t)$ : an oscillating dipole of sub-nanometre amplitude at optical frequency, which is precisely the kind of object that radiates. Parity told us which pair of states could do this at all.

Example 2 — a well you never solved, solved by parity (exam level). An electron is confined by an impenetrable wall at $x = 0$ and a finite barrier of height $V_0 = 10$ eV beyond $x = a = 0.4$ nm (inside, $V = 0$ ). Find all bound states.

Step 1. Recognize the disguise. This wall-plus-well problem is the half-well of the dictionary: its spectrum is exactly the odd-state spectrum of the symmetric finite square well of full width $2a = 0.8$ nm and depth $10$ eV. No new matching analysis is needed — the Finite Square Well already derived the odd-branch quantization condition $-\cot z = \sqrt{(z_0/z)^2 - 1}$ , with $z = ka$ and $z_0 = a\sqrt{2mV_0}/\hbar$ .

Step 2. Size the well. $z_0 = a\sqrt{2mV_0}/\hbar = 6.48$ . Odd-branch roots live where $-\cot z$ is positive, i.e. in $(\pi/2, \pi)$ , $(3\pi/2, 2\pi)$ , …, capped by $z < z_0$ . With $z_0 = 6.48 < 5\pi/2$ , exactly two windows are open: the wall problem has two bound states.

Step 3. Solve the transcendental (numerically). The roots come out $z_1 = 2.710$ and $z_2 = 5.320$ , giving $E = z^2\hbar^2/2ma^2$ : $E_1 = \boxed{1.75\ \text{eV}}$ and $E_2 = \boxed{6.74\ \text{eV}}$ — bound by $8.25$ eV and $3.26$ eV below the barrier lip respectively.

Step 4. Sanity checks. Both energies sit below $V_0$ ✓; the count matches the window argument ✓. And the physics reads correctly: the wall deleted the even states — the symmetric well of the same dimensions holds roughly twice as many levels, including a ground state below $1.75$ eV. Walls raise ground states: forcing $\psi(0) = 0$ costs curvature, and curvature is kinetic energy.

The physicist who caught the mirror lying

Origins · 1 minChien-Shiung Wu and how this was discovered

Chien-Shiung Wu

1912–1997 · Columbia University

By the 1950s parity symmetry was an article of faith: mirror-image experiments must give mirror-image results. Then the "θ–τ puzzle" in kaon decays led Lee and Yang, in 1956, to notice that parity conservation had never actually been tested for the weak interaction. Wu, the acknowledged master of precision beta-decay experiments, aligned cobalt-60 nuclei at millikelvin temperatures and watched where the decay electrons went: preferentially opposite to the nuclear spin — a screw sense the mirror world would reverse. Parity is violated, maximally, by the weak force. The result stunned physics (Pauli had bet against it in writing); Lee and Yang received the 1957 Nobel Prize within a year, while Wu's exclusion remains a canonical grievance. The theorem in this article is untouched — electromagnetic and strong forces, and every $V(|x|)$ built from them, respect the mirror — but Wu's experiment marks its jurisdiction: parity is a symmetry of some of nature's laws, and knowing which is experimental knowledge, not logic.

History

1836

— Sturm's oscillation theorems organize the zeros of second-order ODEs, the mathematics beneath the even–odd–even staircase

1925

— Heisenberg, Born, and Jordan's matrix mechanics makes symmetry operators actors in the theory rather than commentary

1927

— Wigner introduces parity as a quantum number in atomic spectra and derives the Laporte selection rule from it

1928

— Wigner's group-theoretic program: symmetries classify spectra — parity is the simplest worked case

1932

— von Neumann's spectral theory puts the "commuting operators share an eigenbasis" toolkit on rigorous footing

1956

— Lee and Yang question parity conservation in weak interactions; Wu's cobalt-60 experiment finds it maximally violated

1957

— Lee and Yang share the Nobel Prize; parity becomes the textbook example of a symmetry with a jurisdiction

Going further

References. Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed., §2.1–2.6 · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 5 & 11 · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. I, Complement F_II · Landau & Lifshitz, Quantum Mechanics, §21.

Provenance. All operator identities ( $\hat P^2=\mathbb1$ , Hermiticity, $[\hat H,\hat P]=0$ for even $V$ ) are elementary and checkable by direct substitution $x\to-x$ . The non-degeneracy theorem uses the constancy of the Wronskian and normalisability — verify the Wronskian derivative vanishes and that bound states decay at infinity (Griffiths §2.6, Shankar Ch. 5). Hermite parity $H_n(-x)=(-1)^n H_n(x)$ and the listed low-order polynomials $H_0,H_1,H_2$ are standard (Griffiths §2.3; Abramowitz & Stegun). Selection-rule logic and the finite-well transcendental conditions are reproduced in Cohen-Tannoudji Complement F_II and Griffiths §2.6. No empirical numbers are quoted; all results are analytic and recomputable. For this edition, every claim was re-verified symbolically and numerically (sympy/numpy): the parity algebra, Hermite parity $n=0\ldots5$ , centred-well parities and selection rules with the exact $16L/9\pi^2$ element, the half-well spectrum identity, and both worked examples ( $E_1 = 0.376$ eV, $f = 273$ THz; $z_0 = 6.48$ , $E = 1.75$ and $6.74$ eV). The non-degeneracy proof itself is owned by Non-Degeneracy of One-Dimensional Bound States, this chapter's closing article.

Builds on: Time-Independent Schrödinger Equation in One Dimension · Bound States versus Scattering States

Leads to: Infinite Square Well · Finite Square Well · Quantum Harmonic Oscillator: Analytic Solution · Hermite Polynomials and Oscillator Eigenfunctions

Why does the ground state always come out even and not odd?

An odd ground state would be forced to vanish at the origin, and that extra zero-crossing costs curvature — kinetic energy — without any compensating benefit in potential energy. The variational argument (made precise in Node Theorem and Energy Ordering) shows the lowest-energy normalizable function never changes sign at all; a sign-definite function cannot be odd, so the nodeless, even option wins the ground floor, and the alternation $(-1)^n$ follows as nodes are added one per level.

Does an odd wavefunction mean the particle avoids the centre?

It means the probability density vanishes exactly at $x=0$ — you will never find the particle precisely there — but "avoids the centre" oversells it: the density typically peaks quite close to the origin on both sides. And remember the density $|\psi|^2$ is symmetric for both parities, so the particle shows no left/right preference either way. The node is a genuine interference zero, not a repulsion.

What happens to the even/odd classification if the well is slightly asymmetric?

It dissolves — but continuously. Add a small tilt $\varepsilon x$ to a symmetric well and $[\hat H,\hat\Pi] \ne 0$ : exact parity labels are gone, each eigenstate becomes mostly-even or mostly-odd with a small admixture of the other type, and "forbidden" matrix elements like $\langle 0|\hat x|2\rangle$ become small-but-nonzero (proportional to $\varepsilon$ at leading order). Perturbation theory, in the approximation-methods chapter, quantifies exactly this leakage — and engineered asymmetry is how designers switch on transitions that symmetry had turned off.

If parity commutes with H, is parity conserved in time the way energy is?

Yes, in the precise sense that expectation values and probabilities of parity outcomes never change: an even state evolves into an even state, and a superposition keeps its even/odd amplitudes' magnitudes fixed. What can change is the relative phase between the even and odd components — that phase evolution is exactly the sloshing of Worked example 1. Parity conservation forbids even↔odd conversion, not interference between the sectors.

Why do we say scattering states "restore" degeneracy — where do the two states come from physically?

At a given positive energy a particle can travel left-to-right or right-to-left, and no boundary condition kills either option. Two distinct histories at one energy = twofold degeneracy. Bound states have no directional freedom — a decaying tail is a decaying tail — which is the physical root of their uniqueness, and hence of forced parity.

Is the (−1)^n alternation a separate law from the node theorem?

Two faces of one fact. Parity forces each bound state to be even or odd; the node count decides which — an odd number of nodes puts a crossing at the centre (odd), an even count leaves a crest (even). The node theorem supplies "exactly $n$ nodes for state $n$ ," and $(-1)^n$ drops out. This article owns the parity half; the node half is the next article's theorem.

Does parity exist in three dimensions, and is it still just ±1?

Yes — $\hat\Pi\psi(\mathbf r) = \psi(-\mathbf r)$ , still Hermitian and involutive, eigenvalues $\pm1$ . What changes: $-\mathbf r$ is a point inversion (a mirror is inversion times a rotation), and central-potential levels are highly degenerate, so definite parity comes from choosing the angular-momentum basis, where spherical harmonics carry parity $(-1)^\ell$ . That story belongs to the central-potentials chapter; the algebra transfers verbatim.