Probability density and probability current
Density says where; the phase gradient says which way
Unit 1 taught you to read |Ψ|² as a snapshot of where the particle might be found. This lesson adds the arrows: from the same Ψ you can compute how much probability crosses any point each second, and which way, which is what separates a travelling wave from a standing one.
Build the model
Where the particle is likely to be, and which way that likelihood is flowing.
1The big idea
ρ = |Ψ|² is a density, a probability per metre, so it has dimension L⁻¹ and only its integral over an interval is a probability. It is also blind to the phase of Ψ: e(ikx) and e(−ikx) have the same flat ρ, yet one describes motion to the right and the other to the left. The phase is where the motion lives. Write Ψ = √ρ e(iS/ħ) and the combination j = (ħ/m) Im(Ψ* ∂Ψ/∂x) becomes ρ (1/m) ∂S/∂x: a density times a velocity set by how fast the phase winds in x, which is the classical flux ρv, with v = ħk/m for a plane wave. The definition is not a guess. It is the local expression, built from Ψ and ∂Ψ/∂x at a single instant, whose flow through the ends of any interval accounts for every change in the probability inside, as the next lesson proves from the Schrödinger equation. Two consequences carry the rest of the course. A real wavefunction, or a real one times any overall phase, has no phase gradient and so no current, and every bound stationary state on a line is of that kind: a standing wave in which equal and opposite flows cancel. And opposed waves interfere in ρ but not in j. The picture has limits: j/ρ is a velocity field, not the measured speed of any single particle; for a plane wave ρ is a beam density, because the state cannot be normalised; and in a magnetic field the formula gains a term in the vector potential.
2In one sentence
The probability density ρ = |Ψ|² is the probability per unit length of finding the particle near x, and the probability current j = (ħ/m) Im(Ψ* ∂Ψ/∂x) is the probability per unit time crossing x, counted positive towards +x.
Example. An electron beam Ψ = A e(ikx − iωt) with |A|² = 1.0 nm⁻¹ and k = 5.0 nm⁻¹ has ρ = 1.0 × 10⁹ m⁻¹ everywhere and j = ρħk/m = 5.8 × 10¹⁴ s⁻¹: that many electrons pass any point each second, a current of 93 μA.
3A density, so only its integrals are probabilities
Normalisation, ∫|Ψ|² dx = 1 over the whole line, is a dimensionless statement, so Ψ carries m⁻¹⁄² and ρ = |Ψ|² carries m⁻¹: a probability per metre. Its numerical value depends on the unit of length, so comparing it with 1 means nothing, and it has no upper bound: the ground state of an infinite well of width L, ψ₁ = √(2/L) sin(πx/L), peaks at ρ = 2/L, which grows without limit as the well narrows. For L = 0.50 nm the peak is 4.0 nm⁻¹, or 4.0 × 10⁹ m⁻¹. What is a probability is ρ integrated over an interval: between 0.24 nm and 0.26 nm the exact integral is 0.0799, and ρΔx = 4.0 nm⁻¹ × 0.020 nm = 0.080 is the narrow-window estimate. ρ = Ψ*Ψ is also untouched by an overall phase, since e(iα) and e(−iα) cancel. That blindness is the gap this lesson fills: e(ikx) and e(−ikx) have the same flat density, yet they describe motion in opposite directions.
4A current built from one snapshot of Ψ
Ask how the probability Pab inside a ≤ x ≤ b can change. If probability can only move, never appear or vanish, then dPab/dt = j(a, t) − j(b, t), where j(x, t) is the net probability per second crossing x towards +x. The Schrödinger equation delivers exactly this, as the next lesson shows, with j = (ħ/2mi)(Ψ* ∂Ψ/∂x − Ψ ∂Ψ*/∂x). The bracket is z − z* for z = Ψ* ∂Ψ/∂x, which equals 2i Im z, so j = (ħ/m) Im(Ψ* ∂Ψ/∂x): real by construction. Check the dimensions: ħ/m is in J s kg⁻¹ = m² s⁻¹, about 1.16 × 10⁻⁴ m² s⁻¹ for an electron, and Ψ* ∂Ψ/∂x is in m⁻², so j is in s⁻¹. Notice what is missing: neither V nor ∂Ψ/∂t appears. Hand over Ψ at one instant and the flow at that instant is fixed.
5The phase gradient sets the flow
Write Ψ = R e(iS/ħ) with R and S real, so ρ = R². Then ∂Ψ/∂x = (R′ + iRS′/ħ) e(iS/ħ), and Ψ* ∂Ψ/∂x = RR′ + iR²S′/ħ. The real part RR′ = ½ ∂ρ/∂x never reaches j; the imaginary part gives j = ρS′/m. The current is the density times a local velocity S′/m, set by how fast the phase winds along x. For a plane wave S = ħkx − Et, so S′ = ħk and j = ρħk/m = ρp/m, the classical flux of particles moving at p/m. For an electron with k = 5.0 nm⁻¹ that speed is 5.79 × 10⁵ m s⁻¹, twice the phase velocity ω/k = ħk/2m = 2.89 × 10⁵ m s⁻¹: the current follows the particles, not the crests. The sign of k sets the direction, so e(−ikx) has the same ρ and the opposite j.
6A real ψ carries no current
If ψ(x) is real, Ψ = ψ e(−iEt/ħ) gives Ψ* ∂Ψ/∂x = ψψ′, which is real, so j = 0 at every x and every t. The same holds for ψ times any constant phase, because a phase that does not depend on x cancels in Ψ* ∂Ψ/∂x. Every bound stationary state on a line is of this kind: with V real the time-independent equation has real coefficients, so the real and imaginary parts of a solution are solutions too, and for a non-degenerate bound state they are proportional. Zero current is not a particle at rest. In a 0.50 nm well, ψ₁ = √(2/L) sin(πx/L) is the sum of e(iπx/L) and e(−iπx/L) halves, each with density 1/(2L) and speed πħ/(mL). Each carries 7.27 × 10¹⁴ s⁻¹, one rightward and one leftward, and they cancel exactly while the kinetic energy stays at E₁ = 1.50 eV.
7Opposed waves interfere in ρ but not in j
Take Ψ = A e(ikx) + B e(−ikx) with A real and B = |B| e(iφ), the form found on the incident side of every step and barrier later in the course. Then ρ = |A|² + |B|² + 2|A||B| cos(2kx − φ): fringes spaced π/k, half a de Broglie wavelength. In Ψ* ∂Ψ/∂x the same cross terms combine into 2k|A||B| sin(φ − 2kx), which is real, so they drop out of j, leaving j = (ħk/m)(|A|² − |B|²), constant in x. For |B| = 0.60|A| the net current is 64% of the incident one, while ρ swings by a factor (1.6/0.4)² = 16 across a fringe. A flat j over a rippling ρ is consistent, because in a steady state no interval gains or loses probability. This is why a reflection coefficient is a ratio of currents, R = |B|²/|A|², and why a transmitted wave with a different wave number k′ picks up the speed factor k′/k.
8Superpositions of energies carry current, and how to check it
Mix two real stationary states: Ψ = c₁ψ₁ e(−iE₁t/ħ) + c₂ψ₂ e(−iE₂t/ħ) with c₁ and c₂ real. The phase now differs from point to point, and the same algebra gives j = (ħ/m) c₁c₂ sin(ωt)(ψ₂ψ₁′ − ψ₁ψ₂′), with ω = (E₂ − E₁)/ħ. The current oscillates at the beat frequency, vanishes at the instants ωt = 0 and π when the density is momentarily still, and vanishes identically if either coefficient is zero. Run four checks on any current you compute. It must be real and in s⁻¹. It must vanish for a real ψ. It must reduce to ρħk/m for a plane wave. And its integral over all x must equal d⟨x⟩/dt, the velocity of the mean position, which follows from the next lesson's continuity equation by one integration by parts.
Key equations
What each relation says, and the units it is written in.
Probability density
ρ(x, t) = |Ψ(x, t)|² = Ψ*Ψ
Only ∫ρ dx over an interval is a probability. ρ itself has no upper bound, and it cannot see an overall phase of Ψ.
ρ in m⁻¹ in one dimension (m⁻³ in three), because ∫|Ψ|² dx = 1 gives Ψ units of m⁻¹⁄²
Probability current
j(x, t) = (ħ/m) Im(Ψ* ∂Ψ/∂x)
Probability per second crossing x, read off one snapshot of Ψ: neither V nor ∂Ψ/∂t enters it.
ħ/m in m² s⁻¹ and Ψ* ∂Ψ/∂x in m⁻², so j is in s⁻¹. Positive j is net flow towards +x.
Current from the phase gradient
j = (ρ/m) ∂S/∂x, for Ψ = √ρ e(iS/ħ)
Only a phase that varies with x carries current. A real ψ, or a real ψ times an overall phase, has j = 0 everywhere.
S(x, t) real, in J s, so (1/m) ∂S/∂x is a local velocity in m s⁻¹
One plane wave
Ψ = A e(ikx − iωt) gives j = |A|² ħk/m
Density times particle speed, the classical flux ρv. The phase velocity ω/k = ħk/2m would give only half.
|A|² in m⁻¹, read as particles per metre of beam. ħk/m = p/m in m s⁻¹.
Two opposed plane waves
Ψ = A e(ikx) + B e(−ikx) gives j = (ħk/m)(|A|² − |B|²)
Currents subtract, amplitudes do not. |B| = |A| is a standing wave with no current, and |B|²/|A|² is the reflected fraction.
Same k both ways. The cross terms put fringes of spacing π/k into ρ and cancel out of j.
Total current
∫ j dx = d⟨x⟩/dt
The classical check on any current you compute: the summed flow equals the velocity of the mean position.
Integral over all x for a normalisable Ψ, in m s⁻¹. Unit 3 writes the same quantity as ⟨p⟩/m.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Raise r from 0 to 1: the fringes in ρ deepen until they touch zero while the solid current line sinks onto the axis, because a standing wave carries no current. Then sweep φ: the fringes slide along x and j does not move.
Live results
- INCIDENT CURRENT |A|²ħk/m
- 5.79 × 10¹⁴ s⁻¹
- REFLECTED CURRENT |B|²ħk/m
- 1.45 × 10¹⁴ s⁻¹
- NET CURRENT j
- 4.34 × 10¹⁴ s⁻¹
- FRINGE SPACING π/k
- 0.628 nm
Catch the common trap
The mistake this lesson is built to prevent.
“The particle in a box bounces between the walls, so its ground state must carry a current, and j = 0 must mean it sits still.” Both halves are wrong. j is a net flow: probability moving right counts positive and probability moving left counts negative. The ground state ψ₁ = √(2/L) sin(πx/L) is an equal mix of e(iπx/L) and e(−iπx/L), and their currents cancel at every point and every instant, so j = 0 exactly; that is what a standing wave is. None of this makes the particle still: its kinetic energy is π²ħ²/(2mL²), not zero, and position measurements on many identical copies find it spread across the box. What j = 0 does say is that ρ never changes, which is why the state is called stationary. Current appears only when the phase varies with x, as in a travelling wave or a superposition of different energies.
Test it: what changes?
- Fringes in ρ deepen toward true nodes
- Reflected current grows as r²
- Net current j falls as 1 − r²
- Fringe spacing π/k unchanged
- Fringes slide along x
- Net current j unchanged
- Fringe depth and spacing unchanged
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
A beam of slow neutrons (m = 1.675 × 10⁻²⁷ kg) is modelled by Ψ = A e(ikx − iωt) with k = 3.5 × 10¹⁰ m⁻¹ and |A|² = 4.0 × 10³ m⁻¹. Work out ρ and j from their definitions, and find how many neutrons cross a detector plane in one minute.
View solution →
- ρ = Ψ*Ψ = |A|² e(−ikx + iωt) e(ikx − iωt) = |A|² = 4.0 × 10³ m⁻¹, the same at every x: on average four neutrons per millimetre of beam. The plane wave cannot be normalised, so ρ is read as a beam density rather than one particle's probability per metre.
- ∂Ψ/∂x = ikΨ, so Ψ* ∂Ψ/∂x = ik|A|², whose imaginary part is k|A|². Then j = (ħ/m) k|A|² = ρħk/m.
- ħk/m = (1.055 × 10⁻³⁴ J s)(3.5 × 10¹⁰ m⁻¹) ÷ (1.675 × 10⁻²⁷ kg) = 2.20 × 10³ m s⁻¹, so j = (4.0 × 10³ m⁻¹)(2.20 × 10³ m s⁻¹) = 8.8 × 10⁶ s⁻¹.
- Checks: m⁻¹ × m s⁻¹ = s⁻¹, a rate. j is positive, matching the positive k and motion towards +x. It is the classical ρv with v = p/m, the standard thermal-neutron speed; the phase velocity, 1.10 × 10³ m s⁻¹, would have given half.
- In 60 s the number crossing is N = jt = 8.8 × 10⁶ s⁻¹ × 60 s = 5.3 × 10⁸.
Answer. ρ = 4.0 × 10³ m⁻¹ and j = 8.8 × 10⁶ s⁻¹ (speed 2.2 × 10³ m s⁻¹), so about 5.3 × 10⁸ neutrons cross the plane in one minute.
MediumExample 2
A 2.00 nA beam of 3.00 eV electrons meets a barrier. On the incident side Ψ = [A e(ikx) + B e(−ikx)] e(−iEt/ħ), with |B| = 0.400|A|. Find |A|², the reflected and net currents, and the largest and smallest values of ρ and the spacing of its fringes.
View solution →
- E = 3.00 eV = 4.81 × 10⁻¹⁹ J, so k = √(2mE)/ħ = √(2 × 9.109 × 10⁻³¹ kg × 4.81 × 10⁻¹⁹ J) ÷ (1.055 × 10⁻³⁴ J s) = 8.87 × 10⁹ m⁻¹, and the particle speed is ħk/m = 1.03 × 10⁶ m s⁻¹.
- The incident current is the beam current over e: 2.00 × 10⁻⁹ A ÷ 1.602 × 10⁻¹⁹ C = 1.248 × 10¹⁰ s⁻¹ = |A|²ħk/m. So |A|² = 1.248 × 10¹⁰ s⁻¹ ÷ 1.027 × 10⁶ m s⁻¹ = 1.22 × 10⁴ m⁻¹, about one electron per 82.3 μm of beam.
- j = (ħk/m)(|A|² − |B|²). The reflected part is 0.400² = 0.160 of the incident current, 2.00 × 10⁹ s⁻¹ (0.320 nA). The net current is 0.840 × 1.248 × 10¹⁰ s⁻¹ = 1.05 × 10¹⁰ s⁻¹ (1.68 nA), the same at every x.
- With φ the phase of B relative to A, ρ = |A|²[1.16 + 0.800 cos(2kx − φ)] runs from (0.600)²|A|² = 4.37 × 10³ m⁻¹ to (1.400)²|A|² = 2.38 × 10⁴ m⁻¹, a ratio of 5.44, with maxima every π/k = 0.354 nm, half the de Broglie wavelength of 0.708 nm.
- Checks: every current is in s⁻¹ and every density in m⁻¹. At |B| = 0 the net current equals the incident one and ρ is flat; at |B| = |A| the net current is zero and ρ has true nodes, the standing-wave limit.
Answer. |A|² = 1.22 × 10⁴ m⁻¹; reflected current 2.00 × 10⁹ s⁻¹ (0.320 nA); net current 1.05 × 10¹⁰ s⁻¹ (1.68 nA) at every x; ρ from 4.37 × 10³ to 2.38 × 10⁴ m⁻¹, with fringes every 0.354 nm.
HardExample 3
An electron in an infinite well of width L = 0.800 nm is prepared in Ψ = [ψ₁ e(−iE₁t/ħ) + ψ₂ e(−iE₂t/ħ)]/√2, with ψₙ = √(2/L) sin(nπx/L). Find j(x, t) and its largest value, then check it against the left half's probability Pleft = ½ + (4/(3π)) cos ωt and against ⟨x⟩ = L/2 − (16L/(9π²)) cos ωt, where ω = (E₂ − E₁)/ħ.
View solution →
- Ψ* ∂Ψ/∂x = ½[ψ₁ψ₁′ + ψ₂ψ₂′ + ψ₁ψ₂′ e(−iωt) + ψ₂ψ₁′ e(iωt)]. The first two terms are real, and the imaginary part of the rest is ½ sin(ωt)(ψ₂ψ₁′ − ψ₁ψ₂′). So j = (ħ/2m) sin(ωt)(ψ₂ψ₁′ − ψ₁ψ₂′).
- With u = πx/L, ψ₂ψ₁′ − ψ₁ψ₂′ = (2π/L²)(sin 2u cos u − 2 sin u cos 2u). Using sin 2u = 2 sin u cos u and cos 2u = 2cos²u − 1, the bracket is 2 sin u − 2 sin u cos²u = 2 sin³u. So j = (2πħ/(mL²)) sin³(πx/L) sin ωt.
- ω = 3π²ħ/(2mL²) = 2.68 × 10¹⁵ rad s⁻¹, a period of 2.35 fs. The current peaks at the midpoint when ωt = π/2: jₘₐₓ = 2πħ/(mL²) = 2π(1.055 × 10⁻³⁴ J s) ÷ (9.109 × 10⁻³¹ kg × 6.40 × 10⁻¹⁹ m²) = 1.14 × 10¹⁵ s⁻¹.
- Basic checks: ħ/(mL²) is in (m² s⁻¹)/m² = s⁻¹. j = 0 at both walls, where sin³ vanishes, and at ωt = 0 and π. Dropping ψ₂ leaves a single real stationary state and j = 0, as it must.
- Midpoint check: dPleft/dt = −(4ω/(3π)) sin ωt = −(4/(3π))(3π²ħ/(2mL²)) sin ωt = −(2πħ/(mL²)) sin ωt = −j(L/2, t). The left half loses probability exactly as fast as it crosses the midpoint, and the wall at x = 0 passes none.
- Mean-position check: ∫₀ᴸ sin³(πx/L) dx = 4L/(3π), so ∫ j dx = (8ħ/(3mL)) sin ωt, amplitude 3.86 × 10⁵ m s⁻¹. d⟨x⟩/dt = (16L/(9π²)) ω sin ωt, amplitude (0.144 nm)(2.68 × 10¹⁵ s⁻¹) = 3.86 × 10⁵ m s⁻¹. They agree.
Answer. j = (2πħ/(mL²)) sin³(πx/L) sin ωt, with ω = 2.68 × 10¹⁵ rad s⁻¹ and a peak of 1.14 × 10¹⁵ s⁻¹ through the midpoint; it matches −dPleft/dt, and ∫ j dx = d⟨x⟩/dt with amplitude 3.86 × 10⁵ m s⁻¹.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
j = (ħ/m) Im(Ψ* ∂Ψ/∂x): current needs a phase gradient.
Compute j from one snapshot of Ψ; neither V nor ∂Ψ/∂t enters. A real ψ, even times a constant phase, carries none. For opposed waves currents subtract, j = (ħk/m)(|A|² − |B|²), while fringes appear only in ρ. A plane wave's ρ is a beam density, not a normalised probability.
Next: The continuity equation and conservation of probability →