The Schrödinger equation with V = 0
Two plane waves at every energy, until a wall appears.
Set V to zero and the Schrödinger equation turns into the oscillator equation from your mechanics course, so the solving takes one line. The real work comes after it: which two solutions to call the basis, what each one says about momentum and current, and how much a wall left out of the model can matter.
Build the model
The simplest Schrödinger equation, solved exactly, for a world with no walls in it.
1The big idea
Where V is zero the time-independent equation loses its potential term and reads ψ″ = −k²ψ, with k = √(2mE)/ħ: the equation of simple harmonic motion with x standing in for t. Being second order, it has two independent solutions at every E > 0, and cos kx, sin kx, e(ikx) and e(−ikx) are different names for one two-dimensional family. Choosing among them is physics, not taste. Only the exponentials are eigenfunctions of p̂ = −iħ d/dx, with momenta +ħk and −ħk; with the time factor e(−iωt) restored they become waves travelling in a definite direction, carrying probability current ±ħk|A|²/m, a density times the classical velocity p/m. A cosine is their equal mixture: a standing wave with fixed nodes, zero current and no definite momentum, though its energy is just as sharp. The energy, E = ħ²k²/2m, is the classical p²/2m with p = ħk, so the equation reproduces the classical kinetic energy and de Broglie's λ = h/p together. The model stops at its own premise. V = 0 for every x is what leaves both exponentials free at every k; one hard wall, at any distance, forces ψ = 0 there, locks e(ikx) and e(−ikx) into a single sine and removes the travelling solutions, and a second wall breaks the continuum into levels. Separately, p²/2m is only the low-speed kinetic energy, and the relation fails once E is no longer small compared with mc².
2In one sentence
The free-particle Schrödinger equation is the time-independent equation with V = 0 for every x, −(ħ²/2m)ψ″ = Eψ, whose solutions at each energy E > 0 are the plane waves e(ikx) and e(−ikx), with k = √(2mE)/ħ.
Example. An electron with E = 1.00 eV has k = √(2mE)/ħ = 5.12 × 10⁹ m⁻¹, so both of its solutions e(±ikx) repeat every 2π/k = 1.23 nm, and they carry the momenta ±ħk = ±5.40 × 10⁻²⁵ kg m s⁻¹.
3Set V to zero and the equation becomes an oscillator in x
Separation of variables, Ψ(x, t) = ψ(x) e(−iEt/ħ), turned the Schrödinger equation into −(ħ²/2m)ψ″ + V(x)ψ = Eψ. With V = 0 for every x the potential term simply goes, and dividing by −ħ²/2m leaves ψ″ = −(2mE/ħ²)ψ. For E > 0 the bracket is positive, so call it k², with k = √(2mE)/ħ: then ψ″ = −k²ψ, the equation of simple harmonic motion with x where t usually stands and k where ω usually stands. The units check at once: 2mE/ħ² is kg J ÷ (J s)² = m⁻², so k is an inverse length, the number of radians of phase per metre. For an electron with E = 1.00 eV, k = 5.12 × 10⁹ m⁻¹, or 5.12 nm⁻¹. Negative E would give ψ″ = +κ²ψ instead, whose solutions e(±κx) grow without limit on one side or the other; that is why k is taken real here, and UQMI-05.04 returns to what this means for the spectrum as a whole.
4Two solutions at every energy, and several ways to name them
A second-order linear equation has exactly two independent solutions, so fixing E fixes k and leaves a two-dimensional family. Written with exponentials it is ψ = A e(ikx) + B e(−ikx); written with trigonometric functions it is ψ = C cos kx + D sin kx; and Euler's identity e(±ikx) = cos kx ± i sin kx converts one into the other, with C = A + B and D = i(A − B). Direct substitution confirms the claim: e(ikx), e(−ikx), cos kx and sin kx each return −k² times themselves when differentiated twice, so every combination with the same k has the same E. Both descriptions are complete, and neither is more correct than the other. What the equation cannot do is choose. The energy fixes k and says nothing at all about the ratio B/A, so something else, either another observable or a boundary, has to settle it.
5Pick the pair that the momentum operator only rescales
Apply p̂ = −iħ d/dx. p̂ e(ikx) = −iħ(ik) e(ikx) = +ħk e(ikx), and p̂ e(−ikx) = −ħk e(−ikx): each exponential is an eigenfunction with one definite momentum. The cosine fails the test, because p̂ cos kx = iħk sin kx is a different function, so no single momentum belongs to it. Applying p̂ twice does restore it, p̂² cos kx = ħ²k² cos kx, because with V = 0 the Hamiltonian is p̂²/2m and the energy knows only p², not the sign of p. So cos kx = ½e(ikx) + ½e(−ikx) is an equal mixture of +ħk and −ħk, and a momentum measurement returns each value half the time. That is the reason to prefer the exponentials: they are eigenfunctions of Ĥ and p̂ at once, which is possible because the two operators commute when V = 0. It matches the classical picture too, in which a free particle of kinetic energy E moves one way or the other with |p| = √(2mE).
6Travelling and standing waves differ in time and in current
Restore the time factor e(−iωt), with ω = E/ħ. The exponential becomes Ψ = A e(i(kx − ωt)), whose phase is constant along x = (ω/k)t + constant, so its crests move toward +x, while e(−ikx) gives crests moving toward −x; the crest speed ω/k is not the particle's speed, as UQMI-05.07 shows. Yet |Ψ|² = |A|² at every x and every t, as a stationary state's density must be: only the phase travels. The probability current tells the cases apart. For ψ = A e(ikx) + B e(−ikx) the cross terms cancel and j = (ħk/m)(|A|² − |B|²), each term a density times the classical speed p/m. For a 1.00 eV electron ħk/m = 5.93 × 10⁵ m s⁻¹, exactly √(2E/m). The cosine, with B = A, carries no current at all: its density 4|A|² cos² kx has nodes fixed a distance π/k apart, and its real part 2A cos kx cos ωt, for real A, swells and shrinks in place. The figure lets you watch both.
7Checking the solution: three passes, one failure, one edge
Run the course's standing checks on E = ħ²k²/2m and on the plane wave itself. Dimensions: (J s)² m⁻² ÷ kg = J. Limiting case: as k → 0 the energy falls to zero and ψ tends to a constant, a function with no curvature and so no kinetic energy. Classical correspondence: with p = ħk the relation is E = p²/2m, the kinetic energy of a free particle, and de Broglie's p = h/λ comes along with it, carried by p̂ = −iħ d/dx; for the 1.00 eV electron, λ = 1.226 nm gives h/λ = 5.40 × 10⁻²⁵ kg m s⁻¹, and p²/2m returns 1.00 eV. Normalisation is the check that fails: |e(ikx)|² = 1 at every x, so ∫|ψ|² dx over the whole line diverges, and UQMI-05.02 takes that failure seriously. The relation also has an edge of its own. It is the non-relativistic kinetic energy, and at a given p it overstates the exact √((pc)² + (mc²)²) − mc² by a fraction of about E/(2mc²): for an electron that is 1 part in 10⁶ at 1 eV, 1% at 10 keV and 10% at 100 keV, where the Schrödinger equation itself has stopped applying.
8One wall, however far away, changes the spectrum
The strongest assumption is the first one: V = 0 for every x. Put a hard wall at x = 0, with V infinite for x < 0, and ψ must vanish there: A + B = 0, so ψ = A(e(ikx) − e(−ikx)) = 2iA sin kx. Every E > 0 still has a solution, but now only one, and it is a standing wave with |B| = |A| and j = 0: the wall turns each right-moving wave into a left-moving one of equal size. Moving the wall out to x = −D changes only a phase, B = −A e(−2ikD), so no distance brings the travelling states back. A second wall a distance L away adds sin kL = 0, so k = nπ/L and the continuum becomes a ladder of levels. Near energy E adjacent levels sit ΔE ≈ πħv/L apart, with v = ħk/m, which is h divided by the classical round-trip time 2L/v. For a 1.00 eV electron between walls 1.00 cm apart, ΔE = 1.2 × 10⁻⁷ eV, and resolving it takes longer than one round trip, 34 ns. That is the working test for V = 0: it describes a particle that has not yet had time to find the walls.
Key equations
What each relation says, and the units it is written in.
The free time-independent equation
ψ″ = −k² ψ, with k = √(2mE)/ħ
Set V = 0 in −(ħ²/2m)ψ″ + Vψ = Eψ and divide by −ħ²/2m: the harmonic-oscillator equation, with x where t usually stands.
V = 0 for every x. ħ = 1.055 × 10⁻³⁴ J s, m in kg and E in J, so k is in m⁻¹.
Every solution at one energy
ψ(x) = A e(ikx) + B e(−ikx)
Two constants for a second-order equation. C cos kx + D sin kx is the same family, with C = A + B and D = i(A − B).
A and B are complex constants in m⁻¹⁄²; E fixes k > 0 but not the ratio B/A.
Momentum eigenvalues of the exponentials
−iħ (d/dx) e(±ikx) = ±ħk e(±ikx)
The reason to prefer e(±ikx) to sines and cosines: each is an eigenfunction of Ĥ and p̂ at once, with one definite momentum.
ħk in kg m s⁻¹. By contrast −iħ (d/dx) cos kx = iħk sin kx, a different function, so cos kx has no single momentum.
Energy and wavenumber
E = ħ²k²/2m
The classical p²/2m with p = ħk. E fixes k², not the sign of k, and halving the wavelength quadruples the energy.
E in J, k in m⁻¹. For an electron ħ²/2m = 0.0381 eV nm², so k = 5.12 nm⁻¹ gives E = 1.00 eV.
Probability current of the general solution
j = (ħk/m)(|A|² − |B|²)
Density times velocity for each direction, with the interference terms cancelling. A standing wave, |A| = |B|, carries no current however large ψ is.
j in s⁻¹ in one dimension; |A|² and |B|² in m⁻¹; ħk/m is the classical speed p/m, in m s⁻¹.
Level spacing once two walls exist
ΔE ≈ πħv/L = h/(2L/v)
h times the classical round-trip rate. For a 1.00 eV electron and walls 1.00 cm apart, ΔE = 1.2 × 10⁻⁷ eV: tiny, but never zero.
Hard walls a distance L apart force kL = nπ. v = ħk/m in m s⁻¹, L in m; ΔE in J, between adjacent levels near E.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Set r = 0 and drag t: the crests of Re Ψ march toward +x while |Ψ|² stays flat and still, a stationary state carrying current. Take r to 1 or −1 and drag again: the waves only swell and shrink between fixed nodes, the current reads zero, and at r = −1 the circle sits on zero at x = 0.
Live results
- ENERGY ħ²k²/2m
- 0.953 eV
- WAVELENGTH 2π/k
- 1.257 nm
- CURRENT j/|A|²
- 5.79 × 10⁵ m s⁻¹
- WEIGHT AT p = +ħk
- 100 %
Catch the common trap
The mistake this lesson is built to prevent.
The tempting shortcut treats energy and momentum as the same information: E = p²/2m, so a state of definite energy must have momentum √(2mE), and cos kx, which has that energy, must be a particle moving with momentum ħk. The equation fixes only p², never the sign of p. At each E > 0 the free equation allows every combination A e(ikx) + B e(−ikx), and only the two pure exponentials have a definite momentum. The cosine is their equal mixture, so a momentum measurement gives +ħk or −ħk with probability ½ each, its mean momentum is zero, and it carries no current. Nor is it a particle bouncing back and forth: its |ψ|² never moves. Definite energy and definite momentum coincide only for e(ikx) and e(−ikx).
Test it: what changes?
- Energy ħ²k²/2m rises as k²
- Wavelength 2π/k shortens
- Current j/|A|² rises in proportion to k
- Period T = h/E shortens
- Energy and wavelength unchanged
- Current falls as 1 − r², reaching zero
- Weight at +ħk falls from 100% to 50%
- |Ψ|² ripple deepens; zeros appear at r = 1
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron moves where V = 0 in the stationary state ψ(x) = A e(ikx), with k = 8.00 nm⁻¹. Show that ψ solves the time-independent equation, then find its energy in eV, its momentum, and the classical speed that goes with it, and check that ½mv² returns the same energy.
View solution →
- Differentiate twice: ψ′ = ikψ and ψ″ = (ik)²ψ = −k²ψ. Then −(ħ²/2m)ψ″ = (ħ²k²/2m)ψ, a constant times ψ, so ψ is a solution with E = ħ²k²/2m, whatever the value of A.
- Momentum: p̂ψ = −iħ(ik)ψ = ħkψ, so ψ has the definite momentum p = ħk = 1.055 × 10⁻³⁴ J s × 8.00 × 10⁹ m⁻¹ = 8.44 × 10⁻²⁵ kg m s⁻¹, directed toward +x.
- Energy: E = (ħk)²/2m = (8.44 × 10⁻²⁵ kg m s⁻¹)² ÷ (2 × 9.11 × 10⁻³¹ kg) = 3.91 × 10⁻¹⁹ J = 2.44 eV. Units: (kg m s⁻¹)² ÷ kg = kg m² s⁻² = J.
- Classical check: v = p/m = 9.26 × 10⁵ m s⁻¹, and ½mv² = ½ × 9.11 × 10⁻³¹ kg × (9.26 × 10⁵ m s⁻¹)² = 3.91 × 10⁻¹⁹ J, the same energy. The wavelength 2π/k = 0.785 nm is also h/p, the de Broglie wavelength.
- Normalisation flag: |ψ|² = |A|² at every x, so ∫|ψ|² dx over the whole line diverges for any A other than zero. The function solves the equation without yet being a normalisable state, which is the business of UQMI-05.02.
Answer. E = 2.44 eV, p = +8.44 × 10⁻²⁵ kg m s⁻¹ and v = 9.26 × 10⁵ m s⁻¹; ½mv² = 3.91 × 10⁻¹⁹ J reproduces E.
MediumExample 2
A free electron (V = 0) with energy 0.500 eV is in the stationary state ψ(x) = N(3 cos kx + i sin kx), where |ψ|² is read as the number of electrons per metre along a beam. Rewrite ψ in the travelling basis, find the share of electrons moving each way, the net number crossing a point per second if the mean density is 2.00 × 10⁶ m⁻¹, and the spacing between maxima of |ψ|².
View solution →
- Use cos kx = (e(ikx) + e(−ikx))/2 and sin kx = (e(ikx) − e(−ikx))/(2i). Then 3 cos kx + i sin kx = (3/2 + 1/2) e(ikx) + (3/2 − 1/2) e(−ikx), so ψ = N(2e(ikx) + e(−ikx)), with A = 2N and B = N.
- Weights: |A|² : |B|² = 4 : 1, so 80% of the electrons carry +ħk and 20% carry −ħk. The mean momentum is (0.8 − 0.2)ħk = 0.6ħk: neither ħk nor zero.
- Density: |ψ|² = |N|²(5 + 4 cos 2kx), which runs from |N|² to 9|N|² with mean 5|N|². Setting 5|N|² = 2.00 × 10⁶ m⁻¹ gives |N|² = 4.00 × 10⁵ m⁻¹.
- Speed: k = √(2mE)/ħ = 3.62 × 10⁹ m⁻¹, so ħk/m = 4.19 × 10⁵ m s⁻¹, equal to √(2E/m), the classical speed at 0.500 eV.
- Current: j = (ħk/m)(|A|² − |B|²) = 4.19 × 10⁵ m s⁻¹ × 3 × 4.00 × 10⁵ m⁻¹ = 5.03 × 10¹¹ s⁻¹, net electrons crossing any point per second toward +x. Multiplying by e = 1.602 × 10⁻¹⁹ C gives a charge current of magnitude 80.6 nA. j is the same at every x, as a stationary state requires.
- Maxima of |ψ|² sit where cos 2kx = 1, at x = nπ/k, a spacing of π/k = 0.867 nm, half a wavelength. The ripple is the interference of the two beams, and it stands still because both carry the same ω.
Answer. ψ = N(2e(ikx) + e(−ikx)): 80% of the electrons move toward +x and 20% toward −x, the net flux is 5.03 × 10¹¹ s⁻¹ (80.6 nA), and the density maxima are 0.867 nm apart.
HardExample 3
An electron with energy near 1.00 eV moves along x with V = 0 between hard walls at x = 0 and x = L = 1.00 μm. Show what the wall at x = 0 alone does to the solutions, then add the wall at L, find the level nearest 1.00 eV, find the spacing to the next level exactly and from ΔE ≈ πħv/L, and compare it with h divided by the classical round-trip time.
View solution →
- Wall at x = 0: ψ = A e(ikx) + B e(−ikx) must vanish there, so B = −A and ψ = 2iA sin kx for every k > 0. One solution per energy survives instead of two, and with |B| = |A| the current j = (ħk/m)(|A|² − |B|²) is zero.
- Wall at x = L: sin kL = 0, so k = nπ/L with n = 1, 2, 3, … and Eₙ = n² π²ħ²/(2mL²). With ħ²/2m = 0.03810 eV nm² and L = 1000 nm, π²ħ²/(2mL²) = 3.7603 × 10⁻⁷ eV.
- Nearest level: n = √(1.00 eV ÷ 3.7603 × 10⁻⁷ eV) = 1630.8, so n = 1631, with E₁₆₃₁ = 1631² × 3.7603 × 10⁻⁷ eV = 1.0003 eV.
- Exact spacing: Eₙ₊₁ − Eₙ = (2n + 1) × 3.7603 × 10⁻⁷ eV = 3263 × 3.7603 × 10⁻⁷ eV = 1.2270 × 10⁻³ eV.
- Estimate: at this level k = nπ/L = 5.124 × 10⁹ m⁻¹ and v = ħk/m = 5.932 × 10⁵ m s⁻¹, so πħv/L = 1.2266 × 10⁻³ eV. It keeps 2n where the exact result has 2n + 1, so it is low by 1 part in 3263.
- Round trip: 2L/v = 2.00 × 10⁻⁶ m ÷ 5.932 × 10⁵ m s⁻¹ = 3.372 × 10⁻¹² s, and h ÷ (2L/v) = hv/(2L) = 4.1357 × 10⁻¹⁵ eV s × 5.932 × 10⁵ m s⁻¹ ÷ 2.00 × 10⁻⁶ m = 1.2266 × 10⁻³ eV, the estimate again, since πħv/L = h/(2L/v). As L → ∞ the spacing falls as 1/L toward zero but never reaches it.
Answer. One wall leaves ψ = 2iA sin kx, one state per energy, with j = 0. Two walls give k = nπ/L; n = 1631 sits at 1.0003 eV, and the spacing 1.227 × 10⁻³ eV equals h divided by the 3.37 ps round trip to within 0.03%.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Energy fixes k², not the sign of k.
Write ψ = A e(ikx) + B e(−ikx) with k = √(2mE)/ħ; energy leaves B/A free. Only the pure exponentials have definite momentum ±ħk, and j = (ħk/m)(|A|² − |B|²). This needs V = 0 everywhere and E ≪ mc²: one hard wall forces a standing sine with zero current.