The time-independent Schrödinger equation
V − E decides whether ψ bends toward the axis
Separation of variables left you an ordinary differential equation with an unknown number inside it. Here you learn to read that equation before solving it: which way ψ must bend at every x, why a real potential lets you draw real curves, and why the energy it returns is real and above the floor of V, proved rather than assumed.
Build the model
Before you solve it, read it: the sign of V − E tells ψ which way to bend.
1The big idea
For one particle in one dimension in a static, real potential, the equation to solve is −(ħ²/2m)ψ″ + V(x)ψ = Eψ. Written Ĥψ = Eψ it is an eigenvalue problem with two unknowns: which functions ψ does the Hamiltonian merely rescale, and by what factor E? As a second-order linear equation it has two independent solutions at every E, complex values included, so it picks no energy by itself; what a state must do at its edges does the choosing. Before solving anything, rearrange it as ψ″/ψ = (2m/ħ²)(V − E). That is a local curvature rule. Where E > V, ψ bends back toward the axis and oscillates, with wavenumber √(2m(E − V))/ħ, de Broglie's value for the classical momentum. Where E < V it bends away and grows or dies exponentially, and where E = V it has an inflection. Two properties usually taken for granted then become theorems. Multiply by ψ* and integrate once by parts: when ψ vanishes at both ends, E∫|ψ|² dx equals a positive kinetic integral plus ∫V|ψ|² dx, so E is real and lies above the minimum of V. Both proofs need V real and ψ normalisable, and both fail without them: a plane wave with complex k solves the equation at complex E, and an absorbing complex potential gives states that decay. The equation itself stops at relativistic speeds, for a potential that changes in time, and for more than one particle.
2In one sentence
The time-independent Schrödinger equation, −(ħ²/2m)ψ″ + V(x)ψ = Eψ, asks for the pairs of a number E and a function ψ for which the Hamiltonian returns ψ multiplied by E; the admissible pairs are the energy eigenvalues and eigenfunctions.
Example. For an electron, ħ²/2m = 0.0381 eV nm². Where E − V = 1.50 eV, ψ″/ψ = −39.4 nm⁻² and ψ oscillates with wavelength 1.00 nm; where V − E = 1.50 eV, ψ″/ψ = +39.4 nm⁻² and ψ changes by a factor e every 0.159 nm.
3Two unknowns: a number and a function
In −(ħ²/2m)ψ″ + V(x)ψ = Eψ, E is the separation constant and ψ(x) the spatial factor. Written Ĥψ = Eψ, with Ĥ = −(ħ²/2m)d²/dx² + V(x), it asks for functions that Ĥ merely rescales, and most functions fail. With V = 0, Ĥ acting on cos kx with k = 6.00 nm⁻¹ returns (ħ²k²/2m) cos kx = 1.37 eV × cos kx, using ħ²/2m = 0.0381 eV nm² for an electron. Add a second wave with k′ = 3.00 nm⁻¹ and Ĥ returns 1.37 eV × cos kx + 0.343 eV × cos k′x, which is not a multiple of the input: Ĥ rescales each wave by its own factor, so their sum is not an energy eigenfunction. The equation is linear and homogeneous, so any multiple of a solution is a solution and the scale of ψ is left to normalisation. It is also second order: choose E, fix ψ and ψ′ at one point, and the equation marches out a unique solution. So for every E, real or complex, there are two independent solutions. Nothing in the equation refuses an energy. What refuses it is what we demand of ψ at the edges, and that is the work of the last two lessons of this unit.
4Read the equation as a curvature rule
Divide by ψ and the equation becomes ψ″/ψ = (2m/ħ²)(V(x) − E), a statement about the shape of ψ at each point separately. Where E > V the right side is negative, so ψ″ has the opposite sign to ψ: above the axis ψ curves down, below it curves up, and either way it bends back toward the axis and oscillates. Where E < V, ψ″ has the same sign as ψ, so the curve bends away from the axis and grows or dies exponentially. The size of V − E sets how sharply. An electron 4.00 eV above V has ψ″/ψ = −105 nm⁻², so k = 10.2 nm⁻¹ and λ = 0.613 nm; 4.00 eV below V, the same 105 nm⁻² with the opposite sign makes ψ change by a factor e every 0.0976 nm. Where E = V, the classical turning point, ψ″ = 0 and ψ has an inflection. Nodes are inflection points too, and a forbidden region can hold at most one of them: between two nodes ψ would need a hump, and a hump curves toward the axis, which E < V forbids.
5A real potential lets you draw real curves
Fix a real E. Every coefficient in −(ħ²/2m)ψ″ + (V − E)ψ = 0 is then real, so the complex conjugate of a solution is another solution, and so are Re ψ = (ψ + ψ*)/2 and Im ψ = (ψ − ψ*)/2i. Any complex solution is therefore a combination of two real ones: in a flat region with E > V, e(ikx) = cos kx + i sin kx, and the real pair cos kx and sin kx spans the same two-dimensional space. That is why the curvature reading, the sketches in this unit and the figure below may treat ψ as a real curve with a sign, which is what gives “toward the axis” a meaning. Two cautions. A real E does not make every solution real: e(ikx) solves the equation at real E, is complex, and carries the probability current met two lessons on. And the argument needs V real. If V had an imaginary part, conjugating the equation would produce a different equation, and ψ* would no longer solve the original one.
6Why E comes out real
Multiply the equation by ψ* and integrate over the whole domain. Integrating the ψ*ψ″ term by parts once gives E∫|ψ|² dx = (ħ²/2m)∫|ψ′|² dx + ∫V|ψ|² dx − (ħ²/2m)[ψ*ψ′], the bracket taken between the ends. For a normalisable state ψ dies at ±∞, or vanishes at a wall, and the bracket is zero. What remains is real: |ψ′|² and |ψ|² are real and non-negative, and V is real by assumption. Since ∫|ψ|² dx > 0, E is real. Nothing was assumed about E. The result rests on two inputs, and removing either one breaks it. Drop normalisability: with V = 0, e(ikx) with k = (4.00 + 0.30i) nm⁻¹ solves the equation at E = (0.606 + 0.0914i) eV, and its |ψ|² = e(−0.60x/nm) grows without limit to the left, so the bracket never vanishes. Drop real V: an absorbing potential V − iW with W > 0 gives Im E = −∫W|ψ|² dx for a normalised ψ, and the factor e(−iEt/ħ) then shrinks the norm with time. That is how a detector that swallows particles is modelled.
7Curvature puts a floor under the energy
A classical particle can sit at rest at the bottom of a well with E = Vₘᵢₙ. A normalisable ψ cannot. With the bracket gone and ∫|ψ|² dx = 1, the energy identity reads E = (ħ²/2m)∫|ψ′|² dx + ∫V|ψ|² dx. The second term is at least Vₘᵢₙ, because |ψ|² is a probability density. The first is strictly positive, because ψ′ = 0 everywhere would make ψ a constant, and a nonzero constant can neither be normalised on an infinite line nor vanish at a wall. So E > Vₘᵢₙ, with equality excluded. The curvature rule gives the same result as a picture. If E ≤ V at every x, then ψ″/ψ ≥ 0 everywhere, so a ψ that sets off from zero on the left can only swing away from the axis and never returns to zero on the right. The price of confinement is kinetic: squeezing ψ steepens it and raises ∫|ψ′|² dx. In the Hard example below the kinetic integral is 0.122 eV and E sits 0.610 eV above Vₘᵢₙ. Whether E can also be discrete depends on V far away, which the last lesson of this unit settles.
8The classical limit, and where the equation stops
The local wavelength λ(x) = h/√(2m(E − V(x))) is de Broglie's, built on the classical momentum at x, and that is the first correspondence check. At fixed kinetic energy it shrinks as 1/√m: 1.50 eV gives an electron λ = 1.00 nm and a proton 0.0234 nm, 42.9 times shorter. A 1.0 mg grain drifting at 1.0 mm s⁻¹ has λ = 6.6 × 10⁻²⁵ m, so ψ oscillates far too fast for any detector to resolve and only its average survives. When V varies little over one wavelength, the envelope of ψ goes as k(x)(−1/2), so the averaged |ψ|² goes as 1/v(x): the particle is most likely found where a classical one moves slowly and lingers. The penetration length 1/κ also shrinks as 1/√m, so forbidden regions become strictly forbidden. The limits of the equation are set by its inputs. V must not depend on t, or there is no separation constant to be an energy. V must be real, or E is complex. The kinetic term p²/2m is non-relativistic. And it describes one particle, without spin.
Key equations
What each relation says, and the units it is written in.
Time-independent Schrödinger equation
−(ħ²/2m) ψ″(x) + V(x) ψ(x) = E ψ(x)
Ĥψ = Eψ: find the ψ that the Hamiltonian returns multiplied by a number. The equation is linear and homogeneous, so the scale of ψ is left free.
ψ in m(−1/2); ħ = 1.055 × 10⁻³⁴ J s; m in kg; V and E in J, usually quoted in eV
Curvature rule
ψ″/ψ = (2m/ħ²)(V(x) − E)
Negative where E > V: ψ bends toward the axis. Positive where E < V: it bends away. Zero where E = V, the turning point, where ψ has an inflection.
ψ″/ψ in m⁻²; for an electron ħ²/2m = 0.0381 eV nm², so each eV of V − E gives 26.2 nm⁻²
Local wavenumber where E > V
k(x) = √(2m(E − V(x)))/ħ
Where V is constant the solutions are A cos kx + B sin kx. More kinetic energy means tighter bending and a shorter wavelength.
k in m⁻¹; λ = 2π/k = h/p, with p = √(2m(E − V)) the classical momentum at x
Decay constant where E < V
κ(x) = √(2m(V(x) − E))/ħ
Where V is constant the solutions are C e(κx) + D e(−κx). On a side that runs to infinity, only the one that dies there can be normalised.
κ in m⁻¹; ψ changes by a factor e over 1/κ, and |ψ|² over 1/(2κ)
Energy identity
E ∫|ψ|² dx = (ħ²/2m) ∫|ψ′|² dx + ∫V|ψ|² dx − (ħ²/2m)[ψ*ψ′]
Multiply the equation by ψ* and integrate by parts once. When ψ vanishes at both ends the bracket goes, every remaining term is real for real V, and so is E.
Integrals over the whole domain, with [ψ*ψ′] taken between its two ends; each term is an energy when ∫|ψ|² dx = 1
Floor on the energy
E > Vₘᵢₙ
In the identity, (ħ²/2m)∫|ψ′|² dx > 0 because a nonzero constant ψ can neither be normalised on the whole line nor vanish at a wall, and ∫V|ψ|² dx ≥ Vₘᵢₙ for a normalised ψ.
Vₘᵢₙ is the least value of V(x); holds for every normalisable solution, with equality excluded
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. At E = 0.50 eV, φ = 45° gives the one tail that dies away; move φ to 40° or 50° and the tail leaves the panel within a nanometre, since bending away amplifies any error. At E = 1.00 eV the right-hand ψ is straight; at 1.90 eV no phase runs away, and the right-hand wavelength is longer.
Live results
- LEFT ψ″/ψ
- -13.1 nm⁻²
- RIGHT ψ″/ψ
- 13.1 nm⁻²
- LEFT WAVELENGTH
- 1.73 nm
- BOUNDED-TAIL φ
- 45.0 °
Catch the common trap
The mistake this lesson is built to prevent.
The curvature rule is often memorised as “ψ curves down where E > V and up where E < V”. It fixes ψ″/ψ, not ψ″, so the direction depends on the sign of ψ. In an allowed region a negative ψ has ψ″ > 0 and curves up, back toward the axis; drawn curving down, it would run off to −∞, which is forbidden-region behaviour. The error bites hardest in the tails. Beyond a turning point a positive bound-state ψ has ψ″ > 0: it is concave up and approaches the axis only asymptotically. A tail sketched curving down to meet the axis at a finite point has ψ > 0 and ψ″ < 0 there, which needs E > V, so the sketch has quietly moved the turning point. Since ψ″ = (2m/ħ²)(V − E)ψ, ψ″ can vanish only where ψ = 0 or V = E: inflection points belong at nodes and turning points, and nowhere else.
Test it: what changes?
- Left wavelength gets shorter
- Right-hand ψ″/ψ shrinks toward zero
- Decay length 1/κ on the right grows
- Right-hand ψ″/ψ turns negative
- ψ oscillates on both sides of the step
- No phase φ makes the tail run away
- Right-hand wavelength longer than left
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron with E = 2.30 eV crosses from a region where V = 0.50 eV into one where V = 3.20 eV. For each region find ψ″/ψ, say which way ψ bends, and give the length that sets its shape: the wavelength in one, the decay length in the other. Take ħ²/2m = 0.0381 eV nm².
View solution →
- Curvature rule: ψ″/ψ = (V − E)/(ħ²/2m). First region: V − E = 0.50 − 2.30 = −1.80 eV, so ψ″/ψ = −1.80/0.0381 = −47.2 nm⁻². Negative, so ψ bends toward the axis and oscillates.
- k = √(47.2 nm⁻²) = 6.87 nm⁻¹, so λ = 2π/k = 0.914 nm. Correspondence check: de Broglie's λ = h/√(2mK) with K = 1.80 eV = 2.884 × 10⁻¹⁹ J is 6.626 × 10⁻³⁴ J s ÷ √(2 × 9.109 × 10⁻³¹ kg × 2.884 × 10⁻¹⁹ J) = 0.914 nm, the same.
- Second region: V − E = 3.20 − 2.30 = +0.90 eV, so ψ″/ψ = +23.6 nm⁻². Positive, so ψ bends away from the axis, and the solutions are e(±κx) with κ = √(23.6 nm⁻²) = 4.86 nm⁻¹.
- ψ changes by a factor e every 1/κ = 0.206 nm, and |ψ|² every 1/(2κ) = 0.103 nm. Dimension check: eV ÷ (eV nm²) = nm⁻², so both square roots are inverse lengths, as they must be.
Answer. Region 1: ψ″/ψ = −47.2 nm⁻², oscillating with λ = 0.914 nm. Region 2: ψ″/ψ = +23.6 nm⁻², bending away from the axis with decay length 1/κ = 0.206 nm.
MediumExample 2
For an electron in V = 0, a student finds that ψ = e(ikx) with k = (4.00 + 0.30i) nm⁻¹ satisfies the time-independent equation. Find E, show that this ψ cannot be normalised, and check the energy identity on −1.00 nm ≤ x ≤ 1.00 nm to see exactly where the imaginary part of E comes from.
View solution →
- ψ″ = (ik)²ψ = −k²ψ, so −(ħ²/2m)ψ″ = (ħ²k²/2m)ψ and E = (ħ²/2m)k². With k² = 16.00 − 0.09 + 2.40i = (15.91 + 2.40i) nm⁻², E = 0.0381 eV nm² × k² = (0.606 + 0.0914i) eV.
- |ψ|² = e(ikx) e(−ik*x) = e(−2 Im(k) x) = e(−0.60x/nm). It grows without limit as x → −∞: |ψ(−5 nm)|²/|ψ(5 nm)|² = e(6.0) = 403, and ∫|ψ|² dx over the whole line diverges.
- Take the imaginary part of E∫|ψ|² dx = (ħ²/2m)∫|ψ′|² dx + ∫V|ψ|² dx − (ħ²/2m)[ψ*ψ′]. With V = 0 both integrals are real, so Im E ∫|ψ|² dx = −(ħ²/2m) Im[ψ*ψ′] between the ends, and ψ*ψ′ = ik|ψ|² gives Im[ψ*ψ′] = Re(k)|ψ|².
- Left side: ∫|ψ|² dx from −1.00 to 1.00 nm = sinh(0.60)/0.30 nm = 2.122 nm, so Im E ∫|ψ|² dx = 0.0914 eV × 2.122 nm = 0.194 eV nm.
- Right side: −0.0381 eV nm² × 4.00 nm⁻¹ × (e(−0.60) − e(0.60)) = −0.1524 eV nm × (0.549 − 1.822) = 0.194 eV nm. The two agree: the imaginary part of E is paid for entirely by the boundary term, which never vanishes, because |ψ|² refuses to die at the left end.
Answer. E = (0.606 + 0.0914i) eV. Since |ψ|² = e(−0.60x/nm) diverges to the left, ψ is not a state; on the interval both sides of the identity equal 0.194 eV nm, all of it from the boundary term.
HardExample 3
A proposed ground state for an electron is ψ(x) = A(1 + |x|/a)e(−|x|/a), with a = 0.250 nm. Use the equation to read off the potential: find V(x) − E, fix E by requiring V → 0 far away, and give V(x). Then check E > Vₘᵢₙ, locate the turning points, normalise ψ, and find the probability that the electron is in the classically forbidden region.
View solution →
- Write u = |x|/a. For x > 0, ψ′ = −(A/a)u e(−u) and ψ″ = (A/a²)(u − 1)e(−u); ψ′ → 0 at x = 0 from both sides, so ψ has no kink there. Then ψ″/ψ = (u − 1)/[a²(u + 1)], and the curvature rule gives V(x) − E = (ħ²/2ma²)(u − 1)/(u + 1).
- Here ħ²/2ma² = 0.0381 eV nm² ÷ 0.0625 nm² = 0.610 eV. As u → ∞, V − E → +0.610 eV, so V → 0 requires E = −0.610 eV. Then V(x) = E + 0.610 eV × (u − 1)/(u + 1) = −1.22 eV/(1 + |x|/a), a well with a cusp at x = 0.
- Vₘᵢₙ = V(0) = −1.22 eV, and E = −0.610 eV lies above it, as it must. Turning points: V = E where 1 + u = 2, so x = ±a = ±0.250 nm. There ψ″ ∝ (u − 1) = 0, so ψ's inflection points sit exactly at the classical turning points; for |x| > a, ψ″/ψ > 0 and the tail bends away from the axis.
- Normalise: ∫ψ² dx = 2aA² ∫(1 + u)²e(−2u) du from 0 to ∞ = 2aA²(1/2 + 2 × 1/4 + 2/8) = 2.5aA², so A = 1/√(2.5a) = 1/√(0.625 nm) = 1.26 nm(−1/2).
- Forbidden region |x| > a: P = (2a/2.5a) ∫(1 + u)²e(−2u) du from 1 to ∞ = 0.8 × (1/2 + 2 × 3/4 + 5/4)e(−2) = 0.8 × 3.25 × 0.1353 = 0.352.
- Check with the energy identity: (ħ²/2m)∫ψ′² dx = 0.0381 eV nm²/(5a²) = 0.122 eV, and ∫Vψ² dx = −1.22 eV × 0.8 × ∫(1 + u)e(−2u) du from 0 to ∞ = −1.22 eV × 0.8 × 3/4 = −0.732 eV. The sum is −0.610 eV = E. Dimensions: ħ²/(ma²) is J² s² kg⁻¹ m⁻² = J.
Answer. V(x) = −1.22 eV/(1 + |x|/0.250 nm) and E = −0.610 eV, 0.610 eV above Vₘᵢₙ; turning points at ±0.250 nm, exactly where ψ has its inflections; A = 1.26 nm(−1/2); P(forbidden) = 0.352.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
ψ″/ψ = (2m/ħ²)(V − E): read the shape before solving.
Where E > V, ψ bends toward the axis with k = √(2m(E − V))/ħ; where E < V it bends away, with κ = √(2m(V − E))/ħ. For real V and normalisable ψ, the energy identity makes E real and strictly above Vₘᵢₙ. Apply the rule to ψ″/ψ, never to ψ″ alone.