Probability Current in One Dimension

Point a beam of electrons at an obstacle and put a detector on the far side. The detector does not report a wavefunction — it clicks, so many times per second. Quantum mechanics owes us a number with those units, and that number is the probability current j(x,t)j(x,t): if ∣ψ∣2|\psi|^2 is where the particle is likely to be, the current is how that likelihood moves — how much probability per second flows past each point, and which way. The probability current is not an independent postulate. Accept the Born rule and the time-dependent Schrödinger equation, and jj is forced on you: probability cannot vanish here and reappear there without streaming through the space between, and that demand hands you one specific formula. That derivation takes five lines, and everything else follows from reading the formula well: the current lives entirely in the phase of the wavefunction — a real-valued ψ\psi, however lumpy, is dead still — a plane wave carries j=ρvj = \rho v, and inside a tunnelling barrier a steady current threads through on an interference cross-term this article computes explicitly. You have already used the current without meeting it — Transmission and Reflection Coefficients defined RR and TT as ratios of it. Here the object itself gets its story: where it comes from, when it vanishes, and why a detector counts exactly jj.

Key takeaways
  • The probability current j(x,t)=ℏm Im(ψ∗ ∂xψ)j(x,t) = \frac{\hbar}{m}\,\mathrm{Im}\left(\psi^*\,\partial_x\psi\right) is the probability per unit time flowing past xx, sign giving direction.
  • It is derived, not postulated: differentiating ρ=∣ψ∣2\rho = |\psi|^2 in time and substituting the Schrödinger equation yields the continuity equation ∂tρ+∂xj=0\partial_t\rho + \partial_x j = 0 — local conservation of probability.
  • Writing the wavefunction in polar form, ψ=ρ eiS/ℏ\psi=\sqrt{\rho}\,e^{iS/\hbar}, gives j=ρ ∂xS/mj=\rho\,\partial_xS/m: probability flows only where the wavefunction has a spatially varying phase. Flat phase, no flow — every real wavefunction, hence every non-degenerate 1D bound state, carries j≡0j \equiv 0.
  • A plane wave AeikxAe^{ikx} carries j=(ℏk/m)∣A∣2=ρvj = (\hbar k/m)|A|^2 = \rho v — density times velocity, the flux a particle counter measures.
  • In a stationary state ∂tρ=0\partial_t\rho = 0, so jj is constant in xx — the fact that makes one number ("the transmitted current") meaningful and underwrites R+T=1R + T = 1.
  • Inside a barrier, neither exponential solution carries current by itself. The entire transmitted flux arises from their interference, j=(2ℏκ/m) Im(CD∗)j = (2\hbar\kappa/m)\,\mathrm{Im}(CD^*) — exactly the transmitted flux.
  • Key idea: the current is whatever makes ∂t∣ψ∣2+∂xj=0\partial_t|\psi|^2 + \partial_x j = 0 hold identically — the flux that keeps Born-rule probability locally conserved.
The core insight

Probability current is the quantity forced by the Schrödinger equation to satisfy ∂t∣ψ∣2+∂xj=0\partial_t|\psi|^2 + \partial_x j = 0: the local flux that keeps Born-rule probability conserved, whose constancy in stationary scattering states is what defines reflection and transmission.

Where this lives today

Every quantum transport calculation ships probability current at its core. The Landauer picture of nanoscale conduction expresses a device's conductance as 2e2h∑Tn\frac{2e^2}{h}\sum T_n — charge times transmitted probability currents summed over channels, the direct descendant of this article's flux bookkeeping. Scanning tunnelling microscopes read a current that is equation (7) wearing laboratory units; time-dependent simulation codes monitor jj on grids and add complex absorbing potentials near the boundaries to remove outgoing waves. Because the potential is then no longer real, the continuity equation acquires a sink term, allowing probability to leave the computational domain without artificial reflections. And weak-measurement experiments now reconstruct averaged trajectories tracing the streamlines of j/ρj/\rho in real interferometers — evidence that the probability current is far more than a bookkeeping device, being measured indirectly in transport experiments, computed routinely in simulations, and, as here, reconstructed experimentally through weak measurements.

The physical picture

Think of probability as an incompressible fluid smeared along the line, with density ρ=∣ψ∣2\rho = |\psi|^2. The current jj is how fast, and which way, that fluid streams past each point. Fence off any interval: if more flows in at the left than leaves at the right, the amount inside rises by exactly the difference. That bookkeeping is the continuity equation, and it is local — probability cannot teleport between regions; it must flow through every point in between, interval by interval, instant by instant.

Figure
Probability as a fluid — the change inside any fenced interval equals inflow minus outflow
x ρ(x,t) = |ψ|² a b j(a) in j(b) out d/dt ∫ρ dx = j(a) − j(b) thicker inflow than outflow → probability piles up between the fences

Where does the flow hide in ψ\psi? Not in its magnitude. In polar form ψ=ρ eiS/ℏ\psi = \sqrt{\rho}\,e^{iS/\hbar} the current is j=ρ ∂xS/mj = \rho\,\partial_x S/m: probability flows along the gradient of the phase. Flat phase — any real ψ\psi, however dramatic its lumps — is a fluid at rest; a linearly climbing phase eikxe^{ikx} is uniform flow. This one idea organises everything: standing waves are still, travelling waves flow, and two real exponentials in a barrier carry current only because their coefficients hold a relative phase.

Figure
The current lives in the phase — a standing wave is still, a travelling wave flows
real ψ (standing) phase flat → j = 0 e^(ikx) (travelling) phase ramps → j = ρv S(x) = const S(x) = ħkx climbs → flow magnitude does not know about motion

One more habit worth installing: jj has units of probability per unit time — what a particle counter reports. If ∣ψ∣2|\psi|^2 is normalized as particles per unit length, then jj gives the number of particles crossing a point each second. Multiplying by the particle charge converts this directly into the measurable electric current. The current is the interface between wave mechanics and laboratory count rates.

The current and what it measures

Start from the two Born-rule objects. The density:

ρ(x,t)=∣ψ(x,t)∣2=ψ∗ψ(1)\rho(x,t) = |\psi(x,t)|^2 = \psi^*\psi \qquad (1)

where: ρ dx\rho\,dx — the probability of finding the particle between xx and x+dxx + dx at time tt. For a one-particle wavefunction ρ\rho integrates to 1; for a beam it is calibrated to particles per metre. The density is only a snapshot of where the particle is likely to be. Different wavefunctions can have the same ρ\rho while carrying completely different probability currents. The companion, fixed uniquely by the conservation demand derived below, is the current:

j(x,t)=ℏ2mi(ψ∗ ∂ψ∂x−ψ ∂ψ∗∂x)=ℏm Im ⁣(ψ∗ ∂ψ∂x)(2)j(x,t) = \frac{\hbar}{2mi}\left(\psi^*\,\frac{\partial\psi}{\partial x} - \psi\,\frac{\partial\psi^*}{\partial x}\right) = \frac{\hbar}{m}\,\mathrm{Im}\!\left(\psi^*\,\frac{\partial\psi}{\partial x}\right) \qquad (2)

where: the two forms are the same number. Read the structure: jj compares ψ\psi with its own derivative and keeps only the out-of-phase part — if the wavefunction has a spatially constant phase (so that ψ\psi and ∂xψ\partial_x\psi are locally in phase), then j=0j = 0. Units: m−2\mathrm{m}^{-2} times ℏ/m\hbar/m (m2 s−1\mathrm{m^2\,s^{-1}}) gives s−1\mathrm{s^{-1}} — probability per second, a count rate. The two are yoked by the continuity equation — probability conserved locally:

∂ρ∂t+∂j∂x=0(3)\frac{\partial\rho}{\partial t} + \frac{\partial j}{\partial x} = 0 \qquad (3)

where: At every point, the probability density changes only because probability flows into or out of an infinitesimal neighbourhood. Integrate (3) over [a,b][a,b]: ddt∫abρ dx=j(a,t)−j(b,t)\frac{d}{dt}\int_a^b \rho\,dx = j(a,t) - j(b,t) — inflow minus outflow, the figure above. Send the fences to infinity for a normalizable state and both boundary currents die, recovering Conservation of Normalization: equation (3) is that theorem's local, sharper parent. The object is first introduced in Probability Current and the Continuity Equation, where it guarantees normalization survives time evolution. This article owns the one-dimensional craft: a clean derivation, the vanishing theorems, its value in every state this chapter uses, and the connection to what detectors count.

Derivation AContinuity from the Schrödinger equation

We want to show that the Born-rule density ρ=ψ∗ψ\rho = \psi^*\psi obeys a local conservation law, and to let that demand tell us what the current is — no guessing, no extra postulate. The strategy is direct: differentiate ρ\rho in time, use the time-dependent Schrödinger equation and its complex conjugate to eliminate ∂tψ\partial_t\psi and ∂tψ∗\partial_t\psi^*, and watch the result assemble itself into minus the xx-derivative of something. That something is the current. Watch two things: the potential terms must cancel — which happens only because VV is real — and the kinetic terms must combine into a total derivative, which is where the ψ∗∂xψ\psi^*\partial_x\psi structure of equation (2) is born.

Step 1. Write the equations of motion for ψ\psi and ψ∗\psi^*. The time-dependent Schrödinger equation and its complex conjugate are

iℏ ∂tψ=−ℏ22m ∂x2ψ+Vψ,−iℏ ∂tψ∗=−ℏ22m ∂x2ψ∗+Vψ∗i\hbar\,\partial_t\psi = -\frac{\hbar^2}{2m}\,\partial_x^2\psi + V\psi, \qquad -i\hbar\,\partial_t\psi^* = -\frac{\hbar^2}{2m}\,\partial_x^2\psi^* + V\psi^*

Why conjugate? ρ\rho needs both time derivatives; conjugation flips the ii and — crucially — leaves VV alone only if VV is real.

Step 2. Differentiate the density using the product rule: ∂tρ=ψ∗ ∂tψ+ψ ∂tψ∗\partial_t\rho = \psi^*\,\partial_t\psi + \psi\,\partial_t\psi^*.

Step 3. Substitute Step 1 into Step 2:

∂tρ=1iℏ[ψ∗(−ℏ22m∂x2ψ+Vψ)−ψ(−ℏ22m∂x2ψ∗+Vψ∗)]\partial_t\rho = \frac{1}{i\hbar}\left[\psi^*\Big(-\tfrac{\hbar^2}{2m}\partial_x^2\psi + V\psi\Big) - \psi\Big(-\tfrac{\hbar^2}{2m}\partial_x^2\psi^* + V\psi^*\Big)\right]

Step 4. The potential terms cancel: ψ∗Vψ−ψVψ∗=0\psi^* V\psi - \psi V\psi^* = 0. What survives is pure kinetic:

∂tρ=iℏ2m(ψ∗ ∂x2ψ−ψ ∂x2ψ∗)\partial_t\rho = \frac{i\hbar}{2m}\left(\psi^*\,\partial_x^2\psi - \psi\,\partial_x^2\psi^*\right)

Why this cancellation matters. It used V=V∗V = V^*. Give the potential an imaginary part, V+iWV + iW, and a source +2Wρ/ℏ+2W\rho/\hbar survives: probability created or destroyed in place. This is a tool, not a defect — scattering codes bolt absorbing complex potentials onto grid edges to drain outgoing flux.

Step 5. Recognise a total derivative. Since ∂x(ψ∗∂xψ−ψ ∂xψ∗)=ψ∗∂x2ψ−ψ ∂x2ψ∗\partial_x(\psi^*\partial_x\psi - \psi\,\partial_x\psi^*) = \psi^*\partial_x^2\psi - \psi\,\partial_x^2\psi^* (the first-derivative cross terms cancel), Step 4 reads

∂tρ=− ∂x ⁣[ℏ2mi(ψ∗ ∂xψ−ψ ∂xψ∗)]\partial_t\rho = -\,\partial_x\!\left[\frac{\hbar}{2mi}\left(\psi^*\,\partial_x\psi - \psi\,\partial_x\psi^*\right)\right]

Step 6. Name the bracket jj: continuity (3) holds identically. The definition (2) is thereby derived — unique up to an xx-independent constant that j→0j \to 0 at infinity kills. The combination ∂tρ+∂xj\partial_t\rho + \partial_x j simplifies to zero identically for arbitrary ψ\psi and real V(x)V(x); with V→V+iWV \to V + iW it simplifies to +2Wρ/ℏ+2W\rho/\hbar, confirming both the theorem and the sharpness of the real-VV hypothesis.

Reading the current

The polar decomposition makes the phase-flow doctrine exact. Write ψ=ρ eiS/ℏ\psi = \sqrt{\rho}\,e^{iS/\hbar} with ρ\rho, SS real; substitute into (2) — the modulus factors are real and drop — leaving

j=ρm ∂S∂x(4)j = \frac{\rho}{m}\,\frac{\partial S}{\partial x} \qquad (4)

where: ∂xS\partial_x S plays the role of a local momentum, making j=ρ×(local velocity)j = \rho \times (\text{local velocity}) — the exact analogue of the hydrodynamic flux ρv\rho v. This is the Madelung form (1926): probability flows down the phase gradient, at a rate set by how steeply the phase climbs. Flat SS means still fluid, whatever ρ\rho looks like. Now evaluate it in the chapter's working states. For a plane wave ψ=Aeikx\psi = Ae^{ikx}, S=ℏkxS = \hbar kx and (4) gives

jplane=ℏkm ∣A∣2=v ρ(5)j_{\text{plane}} = \frac{\hbar k}{m}\,|A|^2 = v\,\rho \qquad (5)

where: v=ℏk/mv = \hbar k/m is the classical velocity at momentum ℏk\hbar k (the group velocity of a packet built around kk — Group Velocity versus Phase Velocity makes that identification honestly). A beam of ∣A∣2|A|^2 particles per metre at speed vv delivers v∣A∣2v|A|^2 per second — the reason this chapter's scattering coefficients carry velocity factors. Superpose both directions, ψ=Aeikx+Be−ikx\psi = Ae^{ikx} + Be^{-ikx}, and a small miracle cleans the bookkeeping:

j=ℏkm(∣A∣2−∣B∣2)(6)j = \frac{\hbar k}{m}\left(|A|^2 - |B|^2\right) \qquad (6)

where: the AA–BB cross-terms — which ripple the density — cancel identically in the current: net flow is rightward flux minus leftward flux, as if the streams passed through each other. This cancellation is what lets scattering theory treat "incident" and "reflected" currents as separate numbers in the same region. (It is special to equal ∣k∣|k|: superpose two different wavenumbers and jj genuinely oscillates.) With (5) and (6), the definitions in Transmission and Reflection Coefficients — RR, TT as current ratios, the velocity-factor law T=(k2/k1)∣t∣2T = (k_2/k_1)|t|^2, the Wronskian proof of R+T=1R + T = 1 — are fully underwritten. That article owns the ratio doctrine; here we add the object-level fact that makes a single transmitted number meaningful: constancy.

Go deeper · Derivation B · 1 minZero-current theorems — and why stationary currents are flat
Derivation BZero-current theorems — and why stationary currents are flat

Three short results organise every current you will compute in this chapter: real wavefunctions carry none, bound states carry none, stationary states carry the same jj at every point. We prove them in order, each leaning on the last, using only definition (2) and continuity (3). The payoff: in a stationary scattering problem "the current" is one number per solution — which is what lets a transmission coefficient exist at all.

Step 1. Real ψ\psi (up to a global phase) has j=0j = 0. If ψ=eiθu(x)\psi = e^{i\theta}u(x) with uu real and θ\theta constant, then ψ∗∂xψ=u u′\psi^*\partial_x\psi = u\,u' — real — so its imaginary part, and with it jj, vanishes everywhere. In the polar representation the phase is spatially constant, so ∂xS=0\partial_xS=0 and therefore j=0j=0.

Step 2. Every non-degenerate 1D bound state carries zero current. With real VV, ψ\psi and ψ∗\psi^* solve the same eigenvalue problem; non-degeneracy (the Wronskian argument previewed in Non-Degeneracy of One-Dimensional Bound States) forces ψ∗=c ψ\psi^* = c\,\psi — real up to a global phase. Step 1 finishes it: confined, kinetically energetic, yet transporting nothing; a steady one-way flow has nowhere to go in a normalizable state. Why non-degeneracy does real work. In 3D degenerate levels exist (hydrogen's m≠0m \neq 0 states), complex combinations are unavoidable, and bound currents circulate — orbital magnetism is exactly that. Bound-state stillness is a genuinely one-dimensional theorem.

Step 3. For a stationary state ∂tρ=0\partial_t\rho = 0, so continuity immediately makes the current spatially constant, ∂xj=0\partial_x j = 0: Ψ=ψ(x)e−iEt/ℏ\Psi = \psi(x)e^{-iEt/\hbar} has time-independent ρ\rho, so (3) collapses to ∂xj=0\partial_x j = 0 — one constant across the whole line, barrier interiors included. For a bound state that constant is the value at infinity: zero, agreeing with Step 2. For a scattering state it is nonzero and shared by every region: evaluated asymptotically via (6), jL=jRj_L = j_R reads k1(∣A∣2−∣B∣2)=k2∣F∣2k_1(|A|^2 - |B|^2) = k_2|F|^2 — the flux balance Transmission and Reflection Coefficients packages as R+T=1R + T = 1. As a consistency check: jj evaluates to zero for real test functions and for infinite-square-well eigenstates, and for the exact rectangular-barrier scattering solution j(x)j(x) is constant across all three regions to 10−1010^{-10} relative precision.

Go deeper · Derivation C · 1 minThe current inside a barrier — flux through a forbidden region
Derivation CThe current inside a barrier — flux through a forbidden region

Here is the case that breaks intuition the right way. Inside a rectangular barrier with E<V0E < V_0 the solution is ψ=Ceκx+De−κx\psi = Ce^{\kappa x} + De^{-\kappa x} — two real exponentials, which by the last derivation should carry nothing; yet the constancy theorem says the interior current must equal the (nonzero) transmitted current. The resolution is that the boundary conditions generally make CC and DD complex. Neither exponential carries current individually; the transmitted flux arises entirely from their interference. We compute it explicitly — short, exact, and the cleanest working example of "current lives in relative phase" in quantum mechanics.

Step 1. Differentiate: ψ′=κCeκx−κDe−κx\psi' = \kappa Ce^{\kappa x} - \kappa De^{-\kappa x}, so

ψ∗ψ′=κ∣C∣2e2κx−κ∣D∣2e−2κx+κ(D∗C e0−C∗D e0)\psi^*\psi' = \kappa|C|^2e^{2\kappa x} - \kappa|D|^2e^{-2\kappa x} + \kappa\left(D^*C\,e^{0} - C^*D\,e^{0}\right)

Why the cross terms lose their xx. The exponentials multiply to e+κxe−κx=1e^{+\kappa x}e^{-\kappa x} = 1: interference is automatically xx-independent, as constancy demands.

Step 2. Take the imaginary part. The ∣C∣2|C|^2 and ∣D∣2|D|^2 terms are real — each exponential alone is dead still. Using Im(D∗C−C∗D)=2 Im(CD∗)\mathrm{Im}(D^*C - C^*D) = 2\,\mathrm{Im}(CD^*),

jinside=2ℏκm Im ⁣(CD∗)(7)j_{\text{inside}} = \frac{2\hbar\kappa}{m}\,\mathrm{Im}\!\left(C D^*\right) \qquad (7)

where: the forbidden-region current is the out-of-phase part of the pair (C,D)(C, D) — if either coefficient vanishes, or if CC and DD have no relative phase, then Im(CD∗)=0\mathrm{Im}(CD^*) = 0 and the barrier carries no probability current. This explains why both exponential solutions must be retained inside a finite barrier: omitting either destroys the interference term responsible for the transmitted current.

Step 3. Check against the transmitted flux: region III is teikxte^{ikx} with current (ℏk/m)∣t∣2(\hbar k/m)|t|^2, and constancy demands equality with (7). For the worked barrier of Rectangular Barrier and Quantum Tunnelling — electron, V0=10V_0 = 10 eV, a=0.5a = 0.5 nm, E=5E = 5 eV — the 4×44\times4 solve gives T=4.2×10−5T = 4.2\times10^{-5} and the two sides agree to eleven significant figures. Nothing "jumps" across the barrier. A single continuous probability current threads every region, carried inside the barrier entirely by interference between the two exponential solutions. As a consistency check: jj evaluated at seven interior points is constant and equals both (2ℏκ/m)Im(CD∗)(2\hbar\kappa/m)\mathrm{Im}(CD^*) and the transmitted (ℏk/m)∣t∣2(\hbar k/m)|t|^2 to 10−1010^{-10} relative precision. Note the order: Im(CD∗)\mathrm{Im}(CD^*), with CC on the growing exponential — writing Im(C∗D)\mathrm{Im}(C^*D) flips the sign of the flow.

Figure
Current through a barrier — the envelope collapses but the flow is the same at every x
barrier V₀ |ψ| — standing-wave beats (R ≈ 1) transmitted |t| = 0.0065 (invisible at this scale) j equal everywhere electron, V₀ = 10 eV, a = 0.5 nm, E = 5 eV — the net current is the same in all three regions

Limiting cases

CaseResultMeaning
Plane wave AeikxAe^{ikx}j=ℏkm∣A∣2=ρvj = \frac{\hbar k}{m}\vert A\vert^2 = \rho vDensity × velocity — the flux a particle counter reads
Two-way superposition, same ∣k∣\vert k\vertj=ℏkm(∣A∣2−∣B∣2)j = \frac{\hbar k}{m}(\vert A\vert^2 - \vert B\vert^2)Cross terms cancel: net flow = right flux − left flux
Real ψ\psi (any shape)j≡0j \equiv 0Flat phase, still fluid — every non-degenerate 1D bound state
Stationary state∂xj=0\partial_x j = 0One constant current per solution; underwrites R+T=1R+T=1
Barrier interior, E<V0E < V_0j=2ℏκm Im(CD∗)j = \frac{2\hbar\kappa}{m}\,\mathrm{Im}(CD^*)Pure interference of the two real exponentials; equals transmitted flux
Wave packet (any state)∫j dx=⟨p⟩m=d⟨x⟩dt\int j\,dx = \frac{\langle p\rangle}{m} = \frac{d\langle x\rangle}{dt}Total current = velocity of the centre — Ehrenfest's theorem in flux form

Worked examples

Example 1 — from wavefunction to ammeter (confidence builder). A monoenergetic electron beam of energy E=10E = 10 eV is described by ψ=Aeikx\psi = Ae^{ikx} with ∣A∣2=5×108|A|^2 = 5\times10^{8} electrons per metre of beam. What current does a detector register?

Step 1. Wavenumber: k=2meE/ℏ=2(9.109×10−31)(10)(1.602×10−19)/(1.055×10−34)=1.62×1010 m−1k = \sqrt{2m_eE}/\hbar = \sqrt{2(9.109\times10^{-31})(10)(1.602\times10^{-19})}/(1.055\times10^{-34}) = 1.62\times10^{10}\,\mathrm{m^{-1}}.

Step 2. Velocity: v=ℏk/me=(1.055×10−34)(1.62×1010)/(9.109×10−31)=1.88×106 m/sv = \hbar k/m_e = (1.055\times10^{-34})(1.62\times10^{10})/(9.109\times10^{-31}) = 1.88\times10^{6}\,\mathrm{m/s} — about 0.6%0.6\% of cc, comfortably non-relativistic.

Step 3. Current: j=v∣A∣2=(1.88×106)(5×108)=9.4×1014 electrons per secondj = v|A|^2 = (1.88\times10^{6})(5\times10^{8}) = \boxed{9.4\times10^{14}\ \text{electrons per second}} past any checkpoint.

Step 4. As electrical current: I=ej=(1.602×10−19)(9.4×1014)=0.15I = ej = (1.602\times10^{-19})(9.4\times10^{14}) = 0.15 mA — an ordinary bench-top beam current. The chain ψ→j→\psi \to j \to ammeter is two multiplications long.

Example 2 — the sloshing box (exam level). An electron in a 1 nm infinite square well is prepared in the equal superposition Ψ=12(ψ1e−iE1t/ℏ+ψ2e−iE2t/ℏ)\Psi = \frac{1}{\sqrt2}\left(\psi_1 e^{-iE_1t/\hbar} + \psi_2 e^{-iE_2t/\hbar}\right) of the two lowest states. Find the current at the centre of the well and the period of the sloshing.

Step 1. At x=a/2x = a/2: ψ1=2/a\psi_1 = \sqrt{2/a} (crest), ψ2=0\psi_2 = 0 (node), ψ1′=0\psi_1' = 0, ψ2′=−(2π/a)2/a\psi_2' = -(2\pi/a)\sqrt{2/a}. Each stationary eigenstate individually carries zero current. The entire oscillating current arises from the interference (cross) term between the two energy eigenstates.

Step 2. Assemble equation (2): Ψ∗∂xΨ∣a/2=12 ψ1(a/2) ψ2′(a/2) ei(E1−E2)t/ℏ=−2πa2 e−iΔE t/ℏ\Psi^*\partial_x\Psi\big|_{a/2} = \frac{1}{2}\,\psi_1(a/2)\,\psi_2'(a/2)\,e^{i(E_1 - E_2)t/\hbar} = -\frac{2\pi}{a^2}\,e^{-i\Delta E\,t/\hbar} with ΔE=E2−E1\Delta E = E_2 - E_1. Taking the imaginary part,

j(a/2,t)=2πℏma2 sin⁡ ⁣(ΔE tℏ)j(a/2, t) = \frac{2\pi\hbar}{ma^2}\,\sin\!\left(\frac{\Delta E\,t}{\hbar}\right)

Step 3. Numbers: E1=π2ℏ2/2mea2=0.376E_1 = \pi^2\hbar^2/2m_ea^2 = 0.376 eV at a=1a = 1 nm, so ΔE=3E1=1.13\Delta E = 3E_1 = 1.13 eV, and jmax⁡=2πℏ/(mea2)=7.3×1014 s−1j_{\max} = 2\pi\hbar/(m_ea^2) = \boxed{7.3\times10^{14}\ \mathrm{s^{-1}}}.

Step 4. Period: T=2πℏ/ΔE=3.7 fsT = 2\pi\hbar/\Delta E = \boxed{3.7\ \text{fs}}. The lump sloshes across the nanometre box at petahertz rates, the centre current passing through zero exactly as it turns around. A femtosecond movie of charge sloshing in a quantum well is, almost literally, a plot of this jj.

The current made ontological

Origins · 1 minDavid Bohm and how this was discovered
DB
David Bohm
1917–1992 · Princeton, São Paulo, Birkbeck College London

Bohm took the velocity field hiding in equation (4) and promoted it to physics: in his 1952 papers, particles have definite positions at all times, guided along the current's streamlines — v=j/ρv = j/\rho read literally as an equation of motion. Every statistical prediction survives intact — the current already carries the theory's full bookkeeping, which is why the reinterpretation is possible. Built at Princeton and refined in exile in Brazil after McCarthyism ended his U.S. career, the pilot-wave picture remains the sharpest demonstration that this article's formalism underdetermines the story beneath it: same jj, same counts, different world.

History
1926
— Max Born attaches the probability interpretation to |ψ|², creating the object whose conservation demands a current
1926
— Erwin Madelung recasts the Schrödinger equation as hydrodynamics: density ρ and velocity field j/ρ
1927
— Louis de Broglie presents a pilot-wave dynamics at the Solvay Congress; particles ride the wave's phase gradient
1952
— David Bohm publishes the complete guidance-equation formulation, v = j/ρ, reviving and sharpening de Broglie's programme
2011
— Kocsis and collaborators reconstruct average photon trajectories in a double-slit interferometer via weak measurements — the streamlines of j made visible

Going further

References. Griffiths, Introduction to Quantum Mechanics, 3rd ed., §1.5 (probability current) and §2.6 (scattering) · Cohen-Tannoudji, Quantum Mechanics, Vol. I, Complement D_III · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 5 · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., §2.4 · Townsend, A Modern Approach to Quantum Mechanics, Ch. 6 Key results to keep in view: the plane-wave current j=ℏk∣A∣2/mj=\hbar k|A|^2/m; the R+T=1R+T=1 identity relying on the k2/k1k_2/k_1 flux factor; the cancellation of VV terms in the continuity derivation requiring real potential; the imaginary-part identity yielding the two equivalent current forms in equation (2); the barrier interior current (7); and the Ehrenfest flux identity ⟨x⟩˙=⟨p⟩/m\dot{\langle x\rangle}=\langle p\rangle/m.

Builds on: Time-Independent Schrödinger Equation in One Dimension · Free Particle and Plane Waves · Bound States versus Scattering States

Leads to: Finite Square Well · Wave Packets and Their Construction · Group Velocity versus Phase Velocity

If j is zero for a bound state, how does the electron in a hydrogen atom "orbit"?

In 1D nothing orbits: a non-degenerate bound state is still — kinetic energy stored in curvature, zero transport. In three dimensions degeneracy allows intrinsically complex eigenstates: a hydrogen state with m≠0m \neq 0 carries a circulating current around the zz-axis, and that circulation is the orbital magnetic moment. "The electron orbits" is wrong for the ground state (real, j=0j = 0, no orbital magnetism) and right, in the flux sense, for m≠0m \neq 0 states.

How can there be a current inside the barrier if the wavefunction there is "exponentially decaying"?

The decay describes the envelope, not the flow. The two interior pieces have complex coefficients (matching forces it); each alone is still, but the pair, holding a relative phase, carries the constant interference current (7). The current cannot decay across the barrier — constancy forbids it — even though the density collapses by e−2κae^{-2\kappa a}. Small density moving effectively fast, if you want a fluid slogan.

Is v = j/ρ the actual velocity of the particle?

It is a consistent velocity field — the one Bohmian mechanics adopts as literal guidance, and the one weak measurements reconstruct as averaged trajectories. Standard quantum mechanics does not commit: interpretations without trajectories reproduce identical statistics. Interpretation-free content: jj predicts count rates, and ∫j dx=⟨p⟩/m\int j\,dx = \langle p\rangle/m recovers the packet's mean velocity (Ehrenfest). Treat j/ρj/\rho as an honest visualisation that becomes ontology only if you buy an interpretation.

Why does the current use Im(ψ*∂ψ/∂x) rather than something symmetric like |∂ψ/∂x|²?

The continuity equation dictates the form — there is no freedom. Differentiating ∣ψ∣2|\psi|^2 and substituting the Schrödinger equation produces exactly −∂x[ℏmIm(ψ∗∂xψ)]-\partial_x[\frac{\hbar}{m}\mathrm{Im}(\psi^*\partial_x\psi)], nothing else. Physically, Im(ψ∗∂xψ)\mathrm{Im}(\psi^*\partial_x\psi) is the out-of-phase part of the derivative, and out-of-phase response is what motion looks like in wave language. ∣∂xψ∣2|\partial_x\psi|^2 is real, positive, directionless — kinetic-energy density, a different object.

The continuity equation looks exactly like charge conservation in electromagnetism. Is that an analogy or an identity?

A structural identity with different conserved stuff. Multiply ρ\rho and jj by the charge qq and you get expected charge and current densities. The deeper connection: any locally conserved quantity obeys a continuity equation, and probability conservation follows from unitarity of time evolution — in the Formalism chapter this reappears as "Hermitian Hamiltonians generate norm-preserving evolution." The derivation's failure for complex VV is that statement read backwards: non-Hermitian generators leak probability.

Why do scattering coefficients need currents at all — why not just compare |amplitude|² of the waves?

Because probability per second is what must balance, and a wave's flux is density times speed. When the transmitted wave moves in a region with a different wavenumber k2≠k1k_2 \neq k_1, it travels at a different velocity: comparing bare ∣t∣2|t|^2 ignores that and violates conservation. The ratio of currents inserts the velocity factor k2/k1k_2/k_1 automatically, which is why T=(k2/k1)∣t∣2T = (k_2/k_1)|t|^2 and why R+T=1R + T = 1 comes out exact — the full doctrine, with the Wronskian proof, lives in Transmission and Reflection Coefficients. The current is the reason that article's definitions are the right ones.

Can the current be nonzero where the density is essentially zero?

Yes — nothing caps the phase gradient where ρ\rho is small: a thin stream can flow fast. At a moving node of a sloshing superposition the density passes through zero while jj stays finite (probability streams through the node). Just don't divide carelessly — v=j/ρv = j/\rho blows up at exact nodes, a statement about the visualisation, not a physical infinity; the measurable object jj stays finite.

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