Parity in a symmetric well: one wall instead of two
Symmetric well plus non-degeneracy: every bound state even or odd.
Before matching anything at the walls of a finite well, notice that it is symmetric. That one observation, backed by a theorem about one-dimensional bound states, turns the interior into a pure cosine or a pure sine and leaves one wall to match. Here the shortcut is proved, used, and priced by what unequal walls take away.
Build the model
A symmetric well and a non-degenerate level force ψ(−x) = ±ψ(x).
1The big idea
A potential with V(−x) = V(x) is unchanged by reflection through the origin, so the reflected function ψ(−x) solves the Schrödinger equation at the same energy as ψ(x). In one dimension a bound state has no independent partner at its energy: the Wronskian of two such solutions is constant and vanishes in the decaying tails, which makes one a multiple of the other. So ψ(−x) = cψ(x), and reflecting twice gives c² = 1. For the finite well centred on the origin that collapses the interior's A sin kx + B cos kx to a single term, ties the left tail to the right one, and leaves one wall to match: k tan(ka) = κ for the even states and −k cot(ka) = κ for the odd. Those are precisely the two factors of the condition that matching both walls the long way produces, so parity adds no physics; it sorts the roots and halves the algebra. The saving has conditions. The origin must sit at the centre of symmetry, and the states must be bound: the free particle's e(ikx) and e(−ikx) share an energy and neither has a parity. Make the walls unequal and ψ(−x) solves a different problem. The interior becomes cos(kx + φ) with a phase that has to be found along with E, both walls must be matched, the state leans towards the lower wall, and a bound state is no longer guaranteed.
2In one sentence
A bound state of a potential with V(−x) = V(x) has definite parity, ψ(−x) = +ψ(x) (even) or ψ(−x) = −ψ(x) (odd), because the symmetry and the non-degeneracy of one-dimensional bound states leave no other option.
Example. In a symmetric well of half-width a = 0.500 nm, an even electron state with k = 2.00 nm⁻¹ is B cos kx inside. The wall at x = +a alone gives κ = k tan(ka) = 3.11 nm⁻¹, so V₀ = 0.522 eV; the wall at x = −a then matches by reflection.
3Reflect the equation, not the wavefunction
Put the origin at the centre of the well, so the walls stand at x = ±a and V(−x) = V(x). Take any solution ψ at energy E and define χ(x) = ψ(−x). The chain rule gives χ′(x) = −ψ′(−x) and χ″(x) = +ψ″(−x): the two sign flips cancel. Now write the Schrödinger equation at the point −x, −(ħ²/2m)ψ″(−x) + V(−x)ψ(−x) = Eψ(−x), and use V(−x) = V(x). It reads −(ħ²/2m)χ″(x) + V(x)χ(x) = Eχ(x), so the mirror image solves the same equation at the same energy, and it is normalisable whenever ψ is. Two cautions. The symmetry has to hold about the origin you chose: the infinite well of UQMI-06, with walls at 0 and L, is symmetric about L/2, and there the reflection x → L − x turns sin(nπx/L) into (−1)(n+1) sin(nπx/L), even about the centre for n = 1, 3, … and odd for n = 2, 4, …. And so far nothing says χ is the same state as ψ, only that it is a state at the same energy.
4One dimension allows one bound state per energy
Let ψ₁ and ψ₂ solve the equation at the same E and form their Wronskian, W = ψ₁ψ₂′ − ψ₂ψ₁′. Differentiate: W′ = ψ₁ψ₂″ − ψ₂ψ₁″, and since each obeys ψ″ = (2m/ħ²)(V − E)ψ, the two terms are equal and W′ = 0. W is built from ψ and ψ′, both continuous at a finite wall (UQMI-07.04), so it is one constant across all three regions. For bound states read it off far out: beyond x = a both are multiples of e(−κx), and W = 0 there, hence everywhere. Then (ψ₂/ψ₁)′ = W/ψ₁² = 0 between the nodes of ψ₁, and at a node W = 0 reads ψ₂ψ₁′ = 0 with ψ₁′ ≠ 0, so ψ₂ vanishes there too and the ratio carries across. So ψ₂ = cψ₁: no one-dimensional bound level is degenerate. The proof leans on the tails. For the free particle, e(ikx) and e(−ikx) share an energy with W = −2ik ≠ 0, and neither has a parity; cos kx and sin kx do, but only because we chose those combinations.
5Reflecting twice fixes the sign
Put the two results together. χ(x) = ψ(−x) solves the equation at the energy of the bound state ψ, so χ = cψ for some constant c. Reflect again: ψ(x) = χ(−x) = cψ(−x) = c²ψ(x), so c² = 1 and c = +1 or −1, although c might a priori have been complex. Every bound state of a symmetric potential is even or odd, not by choice but by force. Each parity leaves a mark at the centre: an odd, continuous function has ψ(0) = 0, and the slope of an even function is odd, so an even state has ψ′(0) = 0. The ground state has no node, so it cannot be odd, and since each level up adds one node, parity alternates: even, odd, even, …. What must be even is |ψ|², for both parities, so in every bound state of a symmetric well the probability of x > 0 is ½ and ⟨x⟩ = 0. The classical particle agrees: the time it spends near x goes as 1/|v(x)|, and its speed at −x equals its speed at x.
6Half the unknowns, and one wall to match
Inside the well the general solution is A sin kx + B cos kx, with k = √(2mE)/ħ; outside, normalisability has already left F e(−κx) for x > a and G e(κx) for x < −a, with κ = √(2m(V₀ − E))/ħ. That is four amplitudes and E, pinned by four matching conditions and normalisation. Parity removes half of them. An even state has A = 0 and a mirrored tail, G = F; an odd state has B = 0 and G = −F. Match value and slope at x = +a only. Even: B cos ka = F e(−κa) and −kB sin ka = −κF e(−κa); dividing cancels both amplitudes and leaves k tan(ka) = κ. Odd: A sin ka = F e(−κa) and kA cos ka = −κF e(−κa), giving −k cot(ka) = κ. Both sides are in m⁻¹, and each is one real equation for E. The wall at x = −a adds nothing: ψ′ has the opposite parity to ψ, so ψ′/ψ at −a is minus its value at +a, which is exactly the +κ that the mirrored tail requires.
7Do it the long way and parity falls out
Skip the symmetry argument and match both walls. Write the interior as C cos(kx + φ), with the phase φ unknown; its log derivative is −k tan(kx + φ). At x = +a it must equal −κ and at x = −a it must equal +κ, so tan(ka + φ) = κ/k and tan(ka − φ) = κ/k. Two angles with equal tangents differ by a multiple of π, so 2φ is a multiple of π: φ is 0 or π/2 up to sign, and the interior is ±cos kx or ±sin kx. Parity was not assumed; the walls imposed it. Adding the two conditions instead gives 2ka = 2 arctan(κ/k) + nπ, or tan(2ka) = 2kκ/(k² − κ²), and with t = tan(ka) that factorises as (kt − κ)(k + κt) = 0: the even condition and the odd one, multiplied. They never share a root, since together they would demand −k² = κ². For z₀ = a√(2mV₀)/ħ = 4.00 the roots ka = 1.252, 2.475 and 3.595 each zero one factor only. As V₀ → ∞ the conditions give ka = π/2, π, 3π/2, …, so k(2a) = nπ: the infinite well of width 2a, with cosines for odd n.
8Unequal walls take the saving away
Let the left wall stand at V₁ and the right at V₂ ≠ V₁. Now χ(x) = ψ(−x) solves the equation for the mirrored well, a different problem, and nothing forces ψ to be even or odd. The same two conditions become tan(ka − φ) = κ₁/k and tan(ka + φ) = κ₂/k, so 2ka = arctan(κ₁/k) + arctan(κ₂/k) + nπ and φ = ½[arctan(κ₂/k) − arctan(κ₁/k)] + nπ/2: the phase is an unknown that moves with E, and both walls are needed to find it. Inside, ψ² + (ψ′/k)² = C² is constant, and at a wall ψ′ = ∓κψ, so ψ² = C²E/V there: |ψ(−a)/ψ(a)| = √(V₂/V₁), and the state leans towards the lower wall. A guarantee goes too. In a symmetric well k tan(ka) climbs from 0 to ∞ before ka reaches π/2 while κ falls, so an even state exists however shallow the well. An electron in a 1.00 nm well with a 1.00 eV left wall is bound only if V₂ > 0.0656 eV, yet a symmetric 0.050 eV well of that width holds one state.
Key equations
What each relation says, and the units it is written in.
Reflection leaves the equation unchanged
V(−x) = V(x) ⇒ ψ(−x) solves −(ħ²/2m)ψ″ + Vψ = Eψ at the same E
A second solution for free. On its own it does not make ψ(−x) equal to ±ψ(x); that takes non-degeneracy.
x measured from the centre of symmetry, in m; V and E in J. The chain rule gives d²ψ(−x)/dx² = ψ″(−x).
Wronskian of two solutions at one energy
W = ψ₁ψ₂′ − ψ₂ψ₁′, with dW/dx = 0
Bound states vanish at ±∞, so W = 0 everywhere and ψ₂ = cψ₁: a one-dimensional bound level is never degenerate.
ψ in m(−1/2) and ψ′ in m(−3/2), so W is in m⁻²; continuous at a finite wall because ψ and ψ′ both are.
Parity of a bound state
ψ(−x) = cψ(x) with c² = 1, so c = +1 or c = −1
Reflecting twice returns ψ, which fixes c. |ψ|² is even in every bound state, so ⟨x⟩ = 0 and the probability of x > 0 is ½.
c is dimensionless. Even, c = +1: ψ′(0) = 0. Odd, c = −1: ψ(0) = 0.
Even states, matched at x = +a
k tan(ka) = κ
Value and slope at one wall, divided so that both amplitudes cancel: one equation for E. As V₀ → ∞ it gives ka = π/2, 3π/2, …
Interior B cos kx. k = √(2mE)/ħ and κ = √(2m(V₀ − E))/ħ in m⁻¹; a, the half-width, in m; ka in rad.
Odd states, matched at x = +a
−k cot(ka) = κ
As V₀ → ∞ it gives ka = π, 2π, …; with the even roots, k(2a) = nπ, the infinite well of width 2a.
Interior A sin kx, with the same k, κ and a. A root needs ka > π/2, so an odd state needs a√(2mV₀)/ħ > π/2.
Unequal walls: both must be matched
2ka = arctan(κ₁/k) + arctan(κ₂/k) + nπ, n = 0, 1, 2, …
κ₁ = κ₂ splits it into the even (n even) and odd (n odd) conditions; κ₁ ≠ κ₂ leaves a phase φ in cos(kx + φ) to find with E.
κ₁ = √(2m(V₁ − E))/ħ at the left wall and κ₂ = √(2m(V₂ − E))/ħ at the right, in m⁻¹; n counts the nodes.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. With both walls at 4.0 E the dashed mirror hides under ψ for n = 0, 1 and 2. Drop V₂ to 1.6 E: the curves part, ψ(−a)/ψ(a) falls to √(V₂/V₁) = 0.632, P(x > 0) rises from 50% to 65%, and at n = 1 the node leaves the centre for the lower wall's side.
Live results
- LEVEL ENERGY E
- 0.167 eV
- LEFT WALL V₁
- 0.669 eV
- RIGHT WALL V₂
- 0.669 eV
- PROBABILITY x > 0
- 50.0 %
Catch the common trap
The mistake this lesson is built to prevent.
“The well is symmetric, so ψ is symmetric.” Students who read the symmetry of V as a property of ψ write B cos kx inside, match at x = a, and stop, finding only the even ladder. For a well with z₀ = a√(2mV₀)/ħ = 2.0 that reports one bound state where there are two, and it loses the first excited state, whose node sits exactly at the centre. The symmetry argument never said ψ(−x) = ψ(x). It said ψ(−x) = cψ(x), and reflecting twice allows c = −1 as readily as c = +1. What the symmetry of V forces to be even is |ψ|², the quantity a position measurement sees, and the odd states have an even |ψ|² as surely as the even states do. The opposite slip is as common: carrying parity to a well whose origin sits at a wall, or whose walls differ, where ψ(−x) is not a solution at all.
Test it: what changes?
- ψ parts from its mirror image ψ(−x)
- ψ(−a)/ψ(a) = √(V₂/V₁) falls below 1
- Probability shifts towards the lower right wall
- Phase φ must be found along with E
- Parity alternates: even, odd, even
- One more node at each step
- Odd states keep a node at x = 0
- |ψ|² stays even, so P(x > 0) = ½
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron's first excited state in a symmetric finite well with walls at x = ±0.300 nm has interior wavenumber k = 7.50 nm⁻¹. Decide which interior function it has, use one wall to find κ and the wall height V₀, and show that the other wall needs no separate matching.
View solution →
- The ground state is even and nodeless and parity alternates, so the first excited state, with one node, is odd: inside, ψ = A sin kx, which vanishes at x = 0 as an odd function must.
- Match at x = +a: ka = 7.50 × 0.300 = 2.25 rad, and −k cot(ka) = κ gives κ = −7.50 nm⁻¹ × cot(2.25) = −7.50 × (−0.8073) = 6.06 nm⁻¹. It is positive, as a decaying tail needs, and both sides are in nm⁻¹.
- With ħ²/2m = 0.0381 eV nm²: E = 0.0381 × 7.50² = 2.14 eV and V₀ − E = 0.0381 × 6.055² = 1.40 eV, so V₀ = 3.54 eV. Checks: ka = 2.25 lies between π/2 and π and would reach π only as V₀ → ∞; z₀ = 0.300 × √(3.54/0.0381) = 2.89 is also below π, so this odd state is the top level.
- Left wall: sin kx is odd and its slope k cos kx is even, so at x = −a, ψ′/ψ = k cos(ka)/(−sin ka) = −k cot(ka) = +6.06 nm⁻¹. That is exactly the log derivative of the tail G e(κx) on the left, so the second wall is satisfied already.
Answer. Odd, ψ = A sin kx inside; κ = 6.06 nm⁻¹ and V₀ = 3.54 eV, with E = 2.14 eV. The wall at x = −a matches by reflection.
MediumExample 2
An electron is held against an impenetrable surface at x = 0 by a step: V = ∞ for x < 0, V = 0 for 0 < x < 0.800 nm, and V = 0.600 eV beyond. Use a symmetric partner well to write the quantisation condition without doing any new matching, count the bound states, and find the ground-state energy.
View solution →
- Reflect the problem: the symmetric well with V = 0.600 eV for |x| > b, b = 0.800 nm, has odd states with ψ(0) = 0. On x > 0 each obeys the half-well's equation and its conditions, ψ(0) = 0 and a decaying tail matched at x = b, so the half-well's states are exactly the partner's odd states. Its even states, with ψ(0) ≠ 0, are excluded.
- The condition is therefore the odd one, −k cot(kb) = κ. In z = kb and z₀ = b√(2mV₀)/ħ = 0.800 × √(0.600/0.0381) = 3.17, it reads −z cot z = √(z₀² − z²).
- Odd roots lie in (π/2, π), (3π/2, 2π), …, each needing z₀ above the start of its interval. z₀ = 3.17 clears π/2 but not 3π/2 = 4.71, so there is exactly one bound state. Any binding needs z₀ > π/2, which is V₀ > 0.147 eV at this width, whereas the partner's even ground state exists at any depth.
- Solve on (π/2, π): at z = 2.30, −z cot z = 2.06 against √(z₀² − z²) = 2.19; at z = 2.35 the two are 2.32 and 2.13, so the root lies between. Bisection closes on z = 2.321, where both sides equal 2.166.
- E = 0.0381 × (2.321/0.800)² = 0.321 eV, and κ = 2.166/0.800 = 2.71 nm⁻¹, a tail of 0.369 nm. Limiting check: the infinite well of width 0.800 nm has π² × 0.0381/0.800² = 0.588 eV; the finite step lets the state spread and lowers it to 55% of that, and as V₀ → ∞, z → π recovers it.
Answer. One bound state, at E = 0.321 eV (z = kb = 2.321), with κ = 2.71 nm⁻¹; binding at this width needs V₀ > 0.147 eV.
HardExample 3
An electron sits in a flat-bottomed well 1.00 nm wide, walls at x = ±0.500 nm, whose left wall stands at V₁ = 1.00 eV and right wall at V₂ = 0.400 eV. Find the ground-state energy, the phase φ in the interior ψ = C cos(kx + φ), and ψ(−a)/ψ(a). Then show what the parity shortcut, B cos kx matched at the right wall, gets wrong, and decide whether a first excited state exists.
View solution →
- No parity: V(−x) ≠ V(x). Setting ψ′/ψ = −k tan(kx + φ) equal to −κ₂ at x = +a and +κ₁ at x = −a gives tan(ka + φ) = κ₂/k and tan(ka − φ) = κ₁/k. Adding, the ground state satisfies f(E) = 2ka − arctan(κ₁/k) − arctan(κ₂/k) = 0 with 0 < E < 0.400 eV.
- With ħ²/2m = 0.0381 eV nm², f = −0.101 rad at E = 0.150 eV and +0.106 rad at 0.170 eV. Bisection gives E = 0.160 eV, where k = 2.047 nm⁻¹, κ₁ = 4.697 nm⁻¹ and κ₂ = 2.512 nm⁻¹: 2ka = 2.047 rad = 1.160 + 0.887 rad.
- Phase: φ = ½(0.887 − 1.160) = −0.136 rad, so cos(kx + φ) peaks at x = −φ/k = +0.067 nm, towards the lower wall. At each wall ψ² = C²E/V, so ψ(−a)/ψ(a) = √(V₂/V₁) = √0.400 = 0.632; directly, cos(−1.160)/cos(0.887) = 0.3995/0.6317 = 0.632.
- Bracket check: symmetric wells of this width have ground states at 0.137 eV for V₀ = 0.400 eV and 0.191 eV for V₀ = 1.00 eV. Raising one wall of the lower well lifts the level, and 0.160 eV sits between the two, as it must.
- The shortcut: B cos kx matched at x = +a is the symmetric 0.400 eV well, E = 0.137 eV. At x = −a its log derivative is k tan(ka) = 2.63 nm⁻¹, but at that energy the 1.00 eV wall needs κ₁ = 4.76 nm⁻¹. The left wall is violated, and beyond it the solution grows.
- First excited state: n = 1 needs 2ka = θ₁ + θ₂ + π below the lower wall, with θᵢ = arctan(κᵢ/k). At E = 0.400 eV, 2ka = 3.240 while θ₁ + π = 0.886 + 3.142 = 4.028, and f only rises with E, so there is none, although the symmetric 0.400 eV well, with z₀ = 1.62 > π/2, holds two states.
Answer. E = 0.160 eV, φ = −0.136 rad with the peak at x = +0.067 nm, and ψ(−a)/ψ(a) = 0.632; the shortcut gives 0.137 eV and breaks the left wall; no first excited state exists.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Parity halves the work: match one wall, not two.
With the origin at the centre of symmetry, each bound state is B cos kx inside, with k tan(ka) = κ, or A sin kx, with −k cot(ka) = κ; the other wall then matches by reflection. Unequal walls void this: the phase must be found with E, matching both walls.
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