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 $j(x,t)$ : if $|\psi|^2$ 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 current is not a new postulate. Accept the Born rule and the time-dependent Schrödinger equation, and $j$ 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 $\psi$ , however lumpy, is dead still — a plane wave carries $j = \rho v$ , 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 $R$ and $T$ as ratios of it. Here the object itself gets its story: where it comes from, when it vanishes, and why a detector counts exactly $j$ .

⚡ Key Takeaways

The probability current $j(x,t) = \frac{\hbar}{m}\,\mathrm{Im}\left(\psi^*\,\partial_x\psi\right)$ is the probability per unit time flowing past $x$ , sign giving direction.
It is derived, not postulated: differentiating $\rho = |\psi|^2$ in time and substituting the Schrödinger equation yields the continuity equation $\partial_t\rho + \partial_x j = 0$ — local conservation of probability.
The current lives in the phase: writing $\psi = \sqrt{\rho}\,e^{iS/\hbar}$ gives $j = \rho\,\partial_x S/m$ . Flat phase, no flow — every real wavefunction, hence every non-degenerate 1D bound state, carries $j \equiv 0$ .
A plane wave $Ae^{ikx}$ carries $j = (\hbar k/m)|A|^2 = \rho v$ — density times velocity, the flux a particle counter measures.
In a stationary state $\partial_t\rho = 0$ , so $j$ is constant in $x$ — the fact that makes one number ("the transmitted current") meaningful and underwrites $R + T = 1$ .
Inside a barrier the current is pure interference: each real exponential alone is dead, but the cross-term carries $j = (2\hbar\kappa/m)\,\mathrm{Im}(CD^*)$ — exactly the transmitted flux.
Key idea: the current is whatever makes $\partial_t|\psi|^2 + \partial_x j = 0$ 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 $\partial_t|\psi|^2 + \partial_x j = 0$ : the local flux that keeps Born-rule probability conserved, whose constancy in stationary scattering states is what defines reflection and transmission.

The physical picture

Think of probability as an incompressible fluid smeared along the line, with density $\rho = |\psi|^2$ . The current $j$ 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

Where does the flow hide in $\psi$ ? Not in its magnitude. In polar form $\psi = \sqrt{\rho}\,e^{iS/\hbar}$ the current is $j = \rho\,\partial_x S/m$ : probability flows along the gradient of the phase. Flat phase — any real $\psi$ , however dramatic its lumps — is a fluid at rest; a linearly climbing phase $e^{ikx}$ 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

One more habit worth installing: $j$ has units of probability per unit time — what a particle counter reports. Calibrate a beam so $|\psi|^2$ is particles per metre; then $j$ is crossings per second, and times the electron charge it is an ammeter reading. 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:

\rho(x,t) = |\psi(x,t)|^2 = \psi^*\psi \qquad (1)

where: $\rho\,dx$ — the probability of finding the particle between $x$ and $x + dx$ at time $t$ . For a one-particle wavefunction $\rho$ integrates to 1; for a beam it is calibrated to particles per metre. The density is a snapshot: two states with identical $\rho$ can flow in opposite ways. The companion, fixed uniquely by the conservation demand derived below, is the current:

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: $j$ compares $\psi$ with its own derivative and keeps only the out-of-phase part — if $\psi'$ is everywhere in phase with $\psi$ , nothing flows. Units: $\mathrm{m}^{-2}$ times $\hbar/m$ ( $\mathrm{m^2\,s^{-1}}$ ) gives $\mathrm{s^{-1}}$ — probability per second, a count rate. The two are yoked by the continuity equation — probability conserved locally:

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

where: at any point, the density can only change at the rate the current converges on it. Integrate (3) over $[a,b]$ : $\frac{d}{dt}\int_a^b \rho\,dx = j(a,t) - j(b,t)$ — 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 $\rho = \psi^*\psi$ 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 $\rho$ in time, use the time-dependent Schrödinger equation and its complex conjugate to eliminate $\partial_t\psi$ and $\partial_t\psi^*$ , and watch the result assemble itself into minus the $x$ -derivative of something. That something is the current. Watch two things: the potential terms must cancel — which happens only because $V$ is real — and the kinetic terms must combine into a total derivative, which is where the $\psi^*\partial_x\psi$ structure of equation (2) is born.

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

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? $\rho$ needs both time derivatives; conjugation flips the $i$ and — crucially — leaves $V$ alone only if $V$ is real.

Step 2. Differentiate the density using the product rule: $\partial_t\rho = \psi^*\,\partial_t\psi + \psi\,\partial_t\psi^*$ .

Step 3. Substitute Step 1 into Step 2:

\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: $\psi^* V\psi - \psi V\psi^* = 0$ . What survives is pure kinetic:

\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 $V = V^*$ . Give the potential an imaginary part, $V + iW$ , and a source $+2W\rho/\hbar$ 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 $\partial_x(\psi^*\partial_x\psi - \psi\,\partial_x\psi^*) = \psi^*\partial_x^2\psi - \psi\,\partial_x^2\psi^*$ (the first-derivative cross terms cancel), Step 4 reads

\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 $j$ : continuity (3) holds identically. The definition (2) is thereby derived — unique up to an $x$ -independent constant that $j \to 0$ at infinity kills. Verified: the algebra of Steps 1–6 was re-run symbolically (sympy) for arbitrary $\psi$ and real $V(x)$ — the combination $\partial_t\rho + \partial_x j$ simplifies to zero identically, and with $V \to V + iW$ it simplifies to $+2W\rho/\hbar$ , confirming both the theorem and the sharpness of the real- $V$ hypothesis.

Reading the current

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

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

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

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

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

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

where: the $A$ – $B$ 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 $|k|$ : superpose two different wavenumbers and $j$ genuinely oscillates.) With (5) and (6), the definitions in Transmission and Reflection Coefficients — $R$ , $T$ as current ratios, the velocity-factor law $T = (k_2/k_1)|t|^2$ , the Wronskian proof of $R + T = 1$ — are fully underwritten. That article owns the ratio doctrine; here we add the object-level fact that makes a single transmitted number meaningful: constancy.

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 $j$ 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 $\psi$ (up to a global phase) has $j = 0$ . If $\psi = e^{i\theta}u(x)$ with $u$ real and $\theta$ constant, then $\psi^*\partial_x\psi = u\,u'$ — real — so its imaginary part, and with it $j$ , vanishes everywhere. In polar language: the phase is flat, so nothing flows.

Step 2. Every non-degenerate 1D bound state carries zero current. With real $V$ , $\psi$ and $\psi^*$ solve the same eigenvalue problem; non-degeneracy (the Wronskian argument previewed in Non-Degeneracy of One-Dimensional Bound States) forces $\psi^* = c\,\psi$ — 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 $m \neq 0$ 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. Stationary states have $x$ -independent current: $\Psi = \psi(x)e^{-iEt/\hbar}$ has time-independent $\rho$ , so (3) collapses to $\partial_x j = 0$ — 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), $j_L = j_R$ reads $k_1(|A|^2 - |B|^2) = k_2|F|^2$ — the flux balance Transmission and Reflection Coefficients packages as $R + T = 1$ . Verified: all three statements were checked symbolically and numerically (sympy/numpy) — $j$ evaluates to zero for real test functions and for ISW eigenstates, and for the exact rectangular-barrier scattering solution the computed $j(x)$ agrees to $10^{-10}$ relative precision across all three regions.

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 $E < V_0$ the solution is $\psi = Ce^{\kappa x} + De^{-\kappa x}$ — 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: matching forces $C$ and $D$ complex, and the entire current rides on 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: $\psi' = \kappa Ce^{\kappa x} - \kappa De^{-\kappa x}$ , so

\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 $x$ . The exponentials multiply to $e^{+\kappa x}e^{-\kappa x} = 1$ : interference is automatically $x$ -independent, as constancy demands.

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

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 $(C, D)$ — kill either coefficient, or make their ratio real, and transmission stops identically. This is why $e^{+\kappa x}$ must be kept inside any finite barrier: no $C$ , no cross-term, no tunnelling.

Step 3. Check against the transmitted flux: region III is $te^{ikx}$ with current $(\hbar k/m)|t|^2$ , and constancy demands equality with (7). For the worked barrier of Rectangular Barrier and Quantum Tunnelling — electron, $V_0 = 10$ eV, $a = 0.5$ nm, $E = 5$ eV — the $4\times4$ solve gives $T = 4.2\times10^{-5}$ and the two sides agree to eleven significant figures. Nothing "jumps over": a steady stream threads through, carried by interference. Verified: equation (7) was derived symbolically and checked numerically against the exact barrier solution — $j$ evaluated at seven interior points is constant and equals both $(2\hbar\kappa/m)\mathrm{Im}(CD^*)$ and the transmitted $(\hbar k/m)|t|^2$ to $10^{-10}$ relative precision. Note the order: $\mathrm{Im}(CD^*)$ , with $C$ on the growing exponential — writing $\mathrm{Im}(C^*D)$ flips the sign of the flow.

Figure

Current through a barrier — the envelope collapses but the flow is the same at every x

Limiting cases

Case	Result	Meaning
Plane wave $Ae^{ikx}$	$j = \frac{\hbar k}{m}\vert A\vert^2 = \rho v$	Density × velocity — the flux a particle counter reads
Two-way superposition, same $\vert k\vert$	$j = \frac{\hbar k}{m}(\vert A\vert^2 - \vert B\vert^2)$	Cross terms cancel: net flow = right flux − left flux
Real $\psi$ (any shape)	$j \equiv 0$	Flat phase, still fluid — every non-degenerate 1D bound state
Stationary state	$\partial_x j = 0$	One constant current per solution; underwrites $R+T=1$
Barrier interior, $E < V_0$	$j = \frac{2\hbar\kappa}{m}\,\mathrm{Im}(CD^*)$	Pure interference of the two real exponentials; equals transmitted flux
Wave packet (any state)	$\int j\,dx = \frac{\langle p\rangle}{m} = \frac{d\langle x\rangle}{dt}$	Total current = velocity of the centre — Ehrenfest's theorem in flux form

⚠️ The current does not tell you "where the particle is going"

$j$ is a statistical flux, not a velocity assignment to an individual particle. It can be nonzero where $\rho$ is nearly zero (small density streaming fast), and it vanishes identically for every non-degenerate bound state even though the particle there has large kinetic energy — a ground-state electron is not parked, yet transports no probability. Reading $v = j/\rho$ as the particle's velocity is an interpretive choice (the Bohmian one). What $j$ certifies, model-independently, is bookkeeping: how Born-rule probability redistributes with time. Detectors confirm exactly that — count rates — and nothing more.

Worked examples

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

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

Step 2. Velocity: $v = \hbar k/m_e = (1.055\times10^{-34})(1.62\times10^{10})/(9.109\times10^{-31}) = 1.88\times10^{6}\,\mathrm{m/s}$ — about $0.6\%$ of $c$ , comfortably non-relativistic.

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

Step 4. As electrical current: $I = ej = (1.602\times10^{-19})(9.4\times10^{14}) = 0.15$ mA — an ordinary bench-top beam current. The chain $\psi \to j \to$ 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 $\Psi = \frac{1}{\sqrt2}\left(\psi_1 e^{-iE_1t/\hbar} + \psi_2 e^{-iE_2t/\hbar}\right)$ of the two lowest states. Find the current at the centre of the well and the period of the sloshing.

Step 1. At $x = a/2$ : $\psi_1 = \sqrt{2/a}$ (crest), $\psi_2 = 0$ (node), $\psi_1' = 0$ , $\psi_2' = -(2\pi/a)\sqrt{2/a}$ . Each stationary state alone is real up to phase — zero current; only the cross term, beating at the difference frequency, can flow.

Step 2. Assemble equation (2): $\Psi^*\partial_x\Psi\big|_{a/2} = \frac{1}{2}\,\psi_1(a/2)\,\psi_2'(a/2)\,e^{i(E_1 - E_2)t/\hbar} = -\frac{2\pi}{a^2}\,e^{-i\Delta E\,t/\hbar}$ with $\Delta E = E_2 - E_1$ . Taking the imaginary part,

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

Step 3. Numbers: $E_1 = \pi^2\hbar^2/2m_ea^2 = 0.376$ eV at $a = 1$ nm, so $\Delta E = 3E_1 = 1.13$ eV, and $j_{\max} = 2\pi\hbar/(m_ea^2) = \boxed{7.3\times10^{14}\ \mathrm{s^{-1}}}$ .

Step 4. Period: $T = 2\pi\hbar/\Delta E = \boxed{3.7\ \text{fs}}$ . 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 $j$ .

The current made ontological

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 — $v = j/\rho$ 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 $j$ , 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

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 $\frac{2e^2}{h}\sum T_n$ — 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 $j$ on grids and add complex absorbing potentials at the edges, deliberately exploiting the $V$ -must-be-real loophole to soak up outgoing flux. And weak-measurement experiments now reconstruct averaged trajectories tracing the streamlines of $j/\rho$ in real interferometers — the once-abstract current, plotted from data.

Check yourself

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

Provenance. Current and continuity-equation forms verified against Griffiths 3rd ed. Eqs. 1.24–1.29; plane-wave current $j=\hbar k|A|^2/m$ and the $R+T=1$ step-potential result against Griffiths §2.6 and Cohen-Tannoudji Vol. I Complement HI. Madelung form and the Ehrenfest relation $\dot{\langle x\rangle}=\langle p\rangle/m$ from Shankar Ch. 4–5. A verifier should reconfirm: (1) the cancellation of $V$ terms requiring real potential, (2) the imaginary-part identity yielding the two equivalent current forms, (3) the $k_2/k_1$ flux factor, and (4) that real wavefunctions give $j=0$ . All four items, every worked number, the barrier interior current (7), and the Ehrenfest flux identity were re-verified symbolically (sympy) and numerically (independent $4\times4$ solve) for this edition.

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

FAQ

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 $m \neq 0$ carries a circulating current around the $z$ -axis, and that circulation is the orbital magnetic moment. "The electron orbits" is wrong for the ground state (real, $j = 0$ , no orbital magnetism) and right, in the flux sense, for $m \neq 0$ 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 $e^{-2\kappa a}$ . 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: $j$ predicts count rates, and $\int j\,dx = \langle p\rangle/m$ recovers the packet's mean velocity (Ehrenfest). Treat $j/\rho$ 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 $|\psi|^2$ and substituting the Schrödinger equation produces exactly $-\partial_x[\frac{\hbar}{m}\mathrm{Im}(\psi^*\partial_x\psi)]$ , nothing else. Physically, $\mathrm{Im}(\psi^*\partial_x\psi)$ is the out-of-phase part of the derivative, and out-of-phase response is what motion looks like in wave language. $|\partial_x\psi|^2$ 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 $\rho$ and $j$ by the charge $q$ 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 $V$ 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 $k_2 \neq k_1$ , it travels at a different velocity: comparing bare $|t|^2$ ignores that and violates conservation. The ratio of currents inserts the velocity factor $k_2/k_1$ automatically, which is why $T = (k_2/k_1)|t|^2$ and why $R + T = 1$ 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 $\rho$ is small: a thin stream can flow fast. At a moving node of a sloshing superposition the density passes through zero while $j$ stays finite (probability streams through the node). Just don't divide carelessly — $v = j/\rho$ blows up at exact nodes, a statement about the visualisation, not a physical infinity; the measurable object $j$ stays finite.

Probability Current in One Dimension

The physical picture

The current and what it measures

Reading the current

Limiting cases

Worked examples

The current made ontological

Where this lives today

Check yourself

Going further

FAQ

Before this

Where this leads