The continuity equation and conservation of probability
Probability is conserved locally: it must flow to move
For the rest of this course you will normalise Ψ once, at t = 0, and never check it again. This lesson earns that habit: a few lines of algebra show that the Schrödinger equation keeps the norm fixed, and prove the stronger local fact that probability leaves a region only by flowing out through its edges.
Build the model
Probability is not only conserved. It has to flow to get anywhere.
1The big idea
Quantum mechanics rests on two postulates written down separately. The Born rule reads |Ψ(x, t)|² as a probability density, and so demands ∫|Ψ|² dx = 1 at every instant; the Schrödinger equation decides how Ψ changes. Nothing guarantees that the second respects the first. The heat equation, which has the shape of the free-particle equation without its i, steadily drains ∫u² dx away, so the norm has to be checked, not assumed. The check is short. Differentiate ρ = Ψ*Ψ, replace ∂Ψ/∂t by the equation and ∂Ψ*/∂t by its complex conjugate, and two things happen: the potential terms cancel point by point because V is real, and the kinetic terms collapse into an x-derivative. The result, ∂ρ/∂t + ∂j/∂x = 0, is a local law and stronger than conservation, because probability inside any window can change only by the current through its two ends. Integrate over the whole line, where a normalisable Ψ drives j to zero, and the norm is constant: a theorem, which is why Ψ is normalised once and never again. The proof also marks its own limits. Give V an imaginary part and a source term (2/ħ)(Im V)ρ appears, so the norm decays or grows. Let Ψ fail to vanish at the limits, as a plane wave does, and the global statement says nothing while the local balance still holds. And the whole argument counts one nonrelativistic particle that can never be created or destroyed.
2In one sentence
The continuity equation ∂ρ/∂t + ∂j/∂x = 0, derived from the Schrödinger equation with a real potential, says the probability density at a point changes only as fast as the probability current carries probability in or out.
Example. If the current is +3.0 × 10¹³ s⁻¹ at x = a and +1.0 × 10¹³ s⁻¹ at x = b > a, the probability of finding the particle between a and b is rising at 2.0 × 10¹³ s⁻¹, which is 0.020 per femtosecond.
3Two postulates that could have disagreed
The Born rule reads ρ(x, t) = |Ψ(x, t)|² as a probability density, so ∫|Ψ|² dx must equal 1 at every instant. The Schrödinger equation, iħ ∂Ψ/∂t = −(ħ²/2m) ∂²Ψ/∂x² + VΨ, is a separate postulate about how Ψ changes, and nothing obliges it to keep that integral fixed. A near relative shows it could fail. The heat equation ∂u/∂t = D ∂²u/∂x² has the same shape without the i, and for it d/dt ∫u² dx = 2D ∫u ∂²u/∂x² dx = −2D ∫(∂u/∂x)² dx after one integration by parts, with the boundary term dropped because u → 0. That is strictly negative whenever u varies, so a density built from u² would leak away and never come back. Whatever holds ∫|Ψ|² fixed must therefore come from the i and the complex conjugate. The honest route is to compute the time derivative of ρ itself and see what the equation forces it to be.
4Differentiate ρ, then feed in the equation twice
By the product rule ∂ρ/∂t = Ψ* ∂Ψ/∂t + Ψ ∂Ψ*/∂t, so both time derivatives are needed. The first comes straight from the equation: ∂Ψ/∂t = (iħ/2m) ∂²Ψ/∂x² − (i/ħ)VΨ. The second comes from its complex conjugate, in which every i becomes −i and V becomes V*: ∂Ψ*/∂t = −(iħ/2m) ∂²Ψ*/∂x² + (i/ħ)V*Ψ*. Multiply the first by Ψ*, the second by Ψ, and add. The potential contributes −(i/ħ)V|Ψ|² from one product and +(i/ħ)V*|Ψ|² from the other, a total of (i/ħ)(V* − V)|Ψ|². For a real V that is zero, and zero at each x separately, so the cancellation does not care whether V is a square well, a step, a harmonic bowl, or a field switched on halfway through. What survives, (iħ/2m)(Ψ* ∂²Ψ/∂x² − Ψ ∂²Ψ*/∂x²), is built from the kinetic term alone.
5The kinetic terms are a derivative in disguise
Differentiate Ψ* ∂Ψ/∂x − Ψ ∂Ψ*/∂x with respect to x. The product rule gives four terms; the two cross terms (∂Ψ*/∂x)(∂Ψ/∂x) arrive with opposite signs and cancel, leaving exactly Ψ* ∂²Ψ/∂x² − Ψ ∂²Ψ*/∂x². So ∂ρ/∂t = −∂j/∂x with j = (ħ/2mi)(Ψ* ∂Ψ/∂x − Ψ ∂Ψ*/∂x), the probability current of the previous lesson, and the derivation is what justifies that definition: it is the one combination whose slope accounts for every change in ρ. Check the dimensions. Ψ carries m(−1/2), so ρ is in m⁻¹ and ∂ρ/∂t in m⁻¹ s⁻¹; ħ/m is in m² s⁻¹, so j is in s⁻¹ and ∂j/∂x is in m⁻¹ s⁻¹ again. Check the classical correspondence. For a plane wave A e(i(kx − ωt)), j = |A|² ħk/m = ρv with v = p/m, and the equation becomes ∂ρ/∂t + ∂(ρv)/∂x = 0, the continuity equation of a fluid or of electric charge.
6Integrate over a window, then over the whole line
Integrate the continuity equation from a fixed a to a fixed b. The first term becomes the rate of change of Pab = ∫ₐᵇ |Ψ|² dx, and ∫ₐᵇ ∂j/∂x dx = j(b) − j(a), so dPab/dt = j(a, t) − j(b, t). Read the signs with care: j > 0 is flow toward +x, so a positive current at a feeds the window and a positive current at b drains it. This is local conservation, and it is stronger than the global kind, because probability cannot vanish at one place and reappear at another without crossing every point between. Now let a → −∞ and b → +∞. A physical, normalisable Ψ goes to zero there with bounded slope, so j vanishes at both ends and dN/dt = 0. In a box the walls do the same job, since Ψ = 0 there makes j = 0. A plane wave is not normalisable and its global statement is empty, but the window version still holds, and in scattering it becomes the flux bookkeeping.
7Watch it balance in a state that sloshes
Put an electron in a 1.0 nm infinite well in Ψ = (ψ₁e(−iE₁t/ħ) + ψ₂e(−iE₂t/ħ))/√2, with ψₙ = √(2/L) sin(nπx/L) and u = πx/L. The density is ρ = (1/L)[sin²u + sin²2u + 2 sin u sin 2u cos ωt], where ω = (E₂ − E₁)/ħ = 3π²ħ/(2mL²) = 1.71 × 10¹⁵ s⁻¹. The current works out to j = (2πħ/(mL²)) sin³u sin ωt, zero at both walls as it must be. Both ∂ρ/∂t and ∂j/∂x are proportional to sin²u cos u sin ωt, with coefficients −4ω/L and +6π²ħ/(mL³), equal and opposite once ω is substituted, so the equation holds at every x and t. The probability in the left half is ½ + (4/(3π)) cos ωt, swinging from 0.924 to 0.076 and back every 3.67 fs, and its rate of change is exactly −j(L/2), peaking at 7.27 × 10¹⁴ s⁻¹. A single stationary state is the opposite case: ρ is frozen, so ∂j/∂x = 0, and with j = 0 at the walls j = 0 throughout.
8Where the theorem stops, and what a drifting norm means
Keep V complex and the leftover term is (2/ħ)(Im V)ρ, so ∂ρ/∂t + ∂j/∂x = (2/ħ)(Im V)ρ. With Im V = −Γ/2 wherever the particle can be found, dN/dt = −(Γ/ħ)N and the norm decays as e(−Γt/ħ); Γ = 1.0 μeV gives a lifetime ħ/Γ = 0.66 ns. Such optical potentials are used on purpose, to model absorption or decay into channels the calculation leaves out, and the probability that vanishes is exactly what those channels receive. Everywhere else the theorem turns a drifting norm into a diagnosis. If ∫|Ψ|² changes in your own work, look for a V that picked up an imaginary part, a boundary current that did not vanish, or a time step that is not unitary. Forward Euler multiplies an eigenstate by 1 − iEΔt/ħ, raising N by a factor 1 + (EΔt/ħ)² per step, so E = 1.0 eV and Δt = 0.10 fs inflate N nearly tenfold in 100 steps.
Key equations
What each relation says, and the units it is written in.
The continuity equation
∂ρ/∂t + ∂j/∂x = 0
Density can fall at a point only as fast as current carries it to the neighbouring points. Probability never jumps a gap.
ρ = |Ψ|² in m⁻¹ and j in s⁻¹ in one dimension, so both terms carry m⁻¹ s⁻¹.
What the equation and its conjugate leave
∂ρ/∂t = (iħ/2m)(Ψ* ∂²Ψ/∂x² − Ψ ∂²Ψ*/∂x²) + (i/ħ)(V* − V)|Ψ|²
The whole proof in one line: a real V removes the last term at every x, whatever shape V has and even if it changes with time.
m is the mass in kg; each term is a rate density in m⁻¹ s⁻¹, and (i/ħ)(V* − V) = (2/ħ) Im V.
The kinetic terms as a derivative
Ψ* ∂²Ψ/∂x² − Ψ ∂²Ψ*/∂x² = ∂/∂x (Ψ* ∂Ψ/∂x − Ψ ∂Ψ*/∂x)
Turns what is left of ∂ρ/∂t into −∂j/∂x, with j = (ħ/2mi)(Ψ* ∂Ψ/∂x − Ψ ∂Ψ*/∂x), so it integrates down to end-point values.
Ψ is in m(−1/2), so each side is in m⁻³; the cross terms (∂Ψ*/∂x)(∂Ψ/∂x) cancel.
Probability in a fixed window
d/dt ∫ₐᵇ |Ψ|² dx = j(a, t) − j(b, t)
The window gains what crosses a heading toward +x and loses what crosses b the same way. Nothing that happens inside it enters the balance.
a < b are fixed positions in m; both sides are rates in s⁻¹.
The norm is constant
dN/dt = j(−∞, t) − j(+∞, t) = 0
Normalise once at t = 0 and the Born rule holds at every later time. Any finite N is preserved, not only N = 1.
N = ∫|Ψ|² dx is dimensionless; needs Ψ → 0 at ±∞ with bounded slope, or Ψ = 0 at hard walls.
Stationary states carry a uniform current
∂ρ/∂t = 0 ⟹ ∂j/∂x = 0
A free check on any stationary solution, and later the flux balance at a barrier: incident minus reflected current equals transmitted.
j is then one value in s⁻¹ at every x; a 1D bound state has j → 0 at ±∞, so j = 0 everywhere.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. At Γ = 0 sweep ωt through a full turn: the hump sloshes right and back, j reverses every half turn, and dP/dt matches j(0) − j(b) at every setting. Then raise Γ at any ωt past 0: the two readouts part by exactly the sink term, and the norm falls below 1.
Live results
- dP/dt FROM ρ
- -6.30 × 10¹⁴ s⁻¹
- FLUX IN j(0) − j(b)
- -6.30 × 10¹⁴ s⁻¹
- SINK −(Γ/ħ)P
- 0.00 × 10¹⁴ s⁻¹
- NORM N
- 1.000
Catch the common trap
The mistake this lesson is built to prevent.
Conservation of probability does not mean |Ψ(x, t)|² stops changing. It fixes the integral over all x, while the density itself is free to move, and in any superposition of different energies it does. An electron wavepacket 1 nm wide travelling at 1.0 × 10⁶ m s⁻¹ crosses its own width in 1 fs, and within a few femtoseconds the density where it started has fallen almost to zero, yet N stays exactly 1. Writing ∂ρ/∂t = 0 is the stationary-state special case, not the conservation law. The law is ∂ρ/∂t = −∂j/∂x, so ρ can fall at one point only as fast as the current carries it to the next. The reverse slip is just as common: seeing ρ fall at one point and concluding that probability is being lost, when the current says exactly where it has gone.
Test it: what changes?
- Norm N falls below 1
- dP/dt and j(0) − j(b) part
- Gap equals the sink term −(Γ/ħ)P
- Density hump sloshes across and back
- j reverses sign every half turn
- dP/dt matches j(0) − j(b) throughout
- Norm stays exactly 1
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
At one instant an electron wavepacket has probability current j = +2.4 × 10¹³ s⁻¹ at x = −2.0 nm and j = −0.6 × 10¹³ s⁻¹ at x = +2.0 nm. Find the rate at which the probability of finding the electron between these points is changing, how long that rate would take to add 0.015 to it, and check that the total probability is conserved.
View solution →
- Integrate the continuity equation over the fixed window a = −2.0 nm to b = +2.0 nm: dPab/dt = j(a, t) − j(b, t). In one dimension j is in s⁻¹, and so is the rate of change of a pure probability.
- Signs first. j(a) > 0 is flow toward +x across the left edge, into the window. j(b) < 0 is flow toward −x across the right edge, also into the window. Both ends feed it.
- dPab/dt = (+2.4 × 10¹³) − (−0.6 × 10¹³) = +3.0 × 10¹³ s⁻¹. Subtracting magnitudes to get 1.8 × 10¹³ s⁻¹ is the classic slip: it treats the inflow at b as an outflow.
- At a constant rate, Δt = ΔP ÷ (dPab/dt) = 0.015 ÷ (3.0 × 10¹³ s⁻¹) = 5.0 × 10⁻¹⁶ s = 0.50 fs.
- Conservation check. The same balance on (−∞, a) gives j(−∞) − j(a) = 0 − 2.4 × 10¹³ s⁻¹, and on (b, +∞) gives j(b) − j(+∞) = −0.6 × 10¹³ s⁻¹ − 0. The three rates sum to −2.4 + 3.0 − 0.6 = 0 in units of 10¹³ s⁻¹. Limiting case: equal currents at a and b would leave Pab fixed however large they were.
Answer. dPab/dt = +3.0 × 10¹³ s⁻¹, so 0.015 is added in 0.50 fs. The two outer regions lose 2.4 × 10¹³ s⁻¹ and 0.6 × 10¹³ s⁻¹, and the total rate is zero.
MediumExample 2
To model an electron that can tunnel out of a quantum dot into a contact, its potential is given an imaginary part: V(x) = VR(x) − iΓ/2, with VR real and Γ = 0.25 meV constant over the region the electron occupies. Rederive the continuity equation for this V, find N(t) given N(0) = 1, and find the lifetime and the time at which N has fallen to 0.10.
View solution →
- The kinetic part of the derivation is unchanged and still gives −∂j/∂x. The potential leaves (i/ħ)(V* − V)|Ψ|², and V* − V = (VR + iΓ/2) − (VR − iΓ/2) = iΓ, so the leftover term is (i/ħ)(iΓ)ρ = −(Γ/ħ)ρ.
- So ∂ρ/∂t + ∂j/∂x = −(Γ/ħ)ρ. Dimensions: Γ/ħ is J ÷ (J s) = s⁻¹ and ρ is in m⁻¹, so the new term carries m⁻¹ s⁻¹ like the other two.
- Integrate over all x. The current still vanishes at ±∞, so dN/dt = −(Γ/ħ)N and N(t) = e(−Γt/ħ). The sign matters: Im V < 0 is a sink, and a positive imaginary part would make N grow, which no loss model can allow.
- Lifetime: τ = ħ/Γ = (6.582 × 10⁻¹⁶ eV s) ÷ (2.5 × 10⁻⁴ eV) = 2.63 × 10⁻¹² s = 2.63 ps.
- N = 0.10 when Γt/ħ = ln 10 = 2.303, which is at t = 2.303 × 2.63 ps = 6.06 ps.
- Limiting case: Γ → 0 sends τ to infinity and restores dN/dt = 0. The probability that disappears has not been destroyed; it has gone into the contact, which this model does not describe.
Answer. ∂ρ/∂t + ∂j/∂x = −(Γ/ħ)ρ, so N(t) = e(−Γt/ħ); τ = ħ/Γ = 2.63 ps, and N = 0.10 at t = 6.06 ps.
HardExample 3
An electron in an infinite well of width L = 0.80 nm starts in Ψ(x, 0) = (√3 ψ₁ + ψ₂)/2, with ψₙ = √(2/L) sin(nπx/L). Find ρ(x, t) and j(x, t), verify ∂ρ/∂t + ∂j/∂x = 0 at every point, and find how the probability in the left half of the well evolves and the largest current through the midpoint.
View solution →
- Normalisation: (√3/2)² + (1/2)² = 3/4 + 1/4 = 1. With Eₙ = n²π²ħ²/(2mL²) and ħ²/2mₑ = 0.0381 eV nm², E₁ = π² × 0.0381/0.64 = 0.588 eV and E₂ = 4E₁ = 2.35 eV, so ω = (E₂ − E₁)/ħ = 1.763 eV ÷ (6.582 × 10⁻¹⁶ eV s) = 2.68 × 10¹⁵ s⁻¹, a period of 2.35 fs.
- Density, with u = πx/L: ρ = (2/L)[¾ sin²u + ¼ sin²2u + (√3/2) sin u sin 2u cos ωt]. Only the cross term moves, and it moves at ω.
- Current: j = (ħ/m) Im(Ψ* ∂Ψ/∂x). The diagonal terms are real and drop out, leaving j = (ħ/m)(√3/4) sin ωt (ψ₂ψ₁′ − ψ₁ψ₂′). Since sin 2u cos u − 2 sin u cos 2u = 2 sin³u, ψ₂ψ₁′ − ψ₁ψ₂′ = (4π/L²) sin³u, so j = (√3πħ/(mL²)) sin³u sin ωt, which vanishes at both walls.
- Point by point: ∂ρ/∂t = −(2√3ω/L) sin²u cos u sin ωt and ∂j/∂x = (3√3π²ħ/(mL³)) sin²u cos u sin ωt. Substituting ω = 3π²ħ/(2mL²) makes 2√3ω/L = 3√3π²ħ/(mL³), so the two cancel at every x and t.
- Left half: ∫₀(L/2) ψ₁ψ₂ dx = 4/(3π), so Pleft = ½ + (√3/2)(4/(3π)) cos ωt = 0.500 + 0.368 cos ωt, swinging between 0.868 and 0.132. Its rate dPleft/dt = −0.3676 ω sin ωt peaks at 0.3676 × 2.678 × 10¹⁵ s⁻¹ = 9.84 × 10¹⁴ s⁻¹.
- Cross-check with the current: dPleft/dt = j(0) − j(L/2) = −(√3πħ/(mL²)) sin ωt, and √3πħ/(mL²) = 9.84 × 10¹⁴ s⁻¹ for L = 0.80 nm, in s⁻¹ as a 1D current must be. Two routes, one number. Limiting case: remove the ψ₂ admixture and the state is stationary, with j = 0 and Pleft = ½ at all times.
Answer. ω = 2.68 × 10¹⁵ s⁻¹ (period 2.35 fs); j = (√3πħ/(mL²)) sin³(πx/L) sin ωt satisfies the continuity equation identically; Pleft = 0.500 + 0.368 cos ωt, and the peak midpoint current is 9.84 × 10¹⁴ s⁻¹.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
∂ρ/∂t + ∂j/∂x = 0 holds whenever V is real.
Use dP/dt = j(a) − j(b) for a fixed window, counting j positive toward +x. Over the whole line the norm stays constant if V is real and j vanishes at the ends; time dependence of V does not matter. An imaginary part of V adds a sink or source.