Transmission and Reflection Coefficients

Send a million identical electrons, one at a time, at the same obstacle — a step, a bump, a spike — and count. Some fraction arrives at a detector beyond the obstacle; the rest comes back. Those two numbers, the transmission coefficient $T$ and the reflection coefficient $R$ , are what a one-dimensional scattering experiment actually measures, and this article is the chapter's official bookkeeping department: it defines them honestly, proves they always add to one, and collects every result the chapter has produced — step, barrier, delta spike — into a single table you can trust. The honesty matters because the obvious definition is wrong. It is tempting to square the transmitted amplitude and call it a probability, but amplitudes measure how much wave is present, not how much is flowing — and a slow wave piles up, tall and lazy, like traffic at the point where a highway narrows. You will see a concrete case below where the transmitted amplitude-squared is 1.50, a number that would be nonsense as a probability. The fix is the probability current $j$ , whose derivation you met back in Probability Current and the Continuity Equation: define $R$ and $T$ as ratios of currents, and a velocity factor $k'/k$ appears in exactly the right place to make probability balance to the last decimal. That one idea — count flux, not amplitude — is the seed of scattering theory. Promoted to three dimensions it becomes the S-matrix; applied to electrons in a wire it becomes the Landauer formula, where an honest-to-goodness electrical conductance is literally $T$ in disguise. Master the ledger here and you own the pattern for both.

Key takeaways

$R$ and $T$ are defined through the probability current: $R=|j_{\text{refl}}|/j_{\text{inc}}=|r|^2$ and $T=j_{\text{trans}}/j_{\text{inc}}=\dfrac{k'}{k}\,|t|^2$ — flux ratios, not amplitude ratios.
For a stationary state the continuity equation collapses to $\mathrm{d}j/\mathrm{d}x=0$ : one constant current threads the entire axis, so what flows in must flow out — $R+T=1$ for any localized potential.
The velocity factor $k'/k$ is mandatory whenever the asymptotic potentials differ: the transmitted wave moves at a different speed, and flux is amplitude-squared times speed.
$|t|^2$ can exceed 1 (amplitude piles up on the slow side); $T$ never can.
Quantum scattering is doubly non-classical: $T>0$ below a barrier (tunnelling) and $R>0$ above one (over-barrier reflection), including reflection from a potential drop.
Key idea: define $R$ and $T$ as ratios of probability currents; current conservation in a stationary state then forces $R+T=1$ , with the velocity factor in $T$ doing exactly the bookkeeping needed when the two sides move at different speeds.

The core insight

Define $R$ and $T$ as ratios of probability currents, not amplitudes; current conservation in a stationary state then forces $R+T=1$ , and the velocity-ratio factor in $T$ is exactly what makes this work when the asymptotic wavenumbers differ.

Where this lives today

The purest modern incarnation of this article is the Landauer picture of electrical conduction: a nanoscale conductor is a scattering obstacle for electron waves arriving from the contacts, and its conductance is $G=(2e^{2}/h)\sum_n T_n$ — transmission coefficients, dressed in ohms. In a quantum point contact each transverse mode's $T_n$ switches from 0 to 1 as the constriction widens, and the conductance climbs a staircase in exact steps of $2e^2/h=(12.9\,\mathrm{k}\Omega)^{-1}$ : an ammeter reading out the ledger of this page. The same current-ratio discipline underwrites tunnel-junction engineering (where $R+T=1$ audits every model of an STM gap or a qubit junction) and, promoted to three dimensions and many channels, becomes the unitarity of the S-matrix — probability conservation as the first axiom of scattering theory, from electrons in wires to protons at the LHC.

The physical picture

Think of a steady stream — water, traffic, probability — flowing toward an obstacle. In steady state nothing accumulates anywhere, so the flow rate arriving must equal the flow rate leaving, split between a reflected stream and a transmitted one. That conservation statement is the entire content of $R+T=1$ , and it holds before you compute a single amplitude, for any potential that levels off at the two ends.

Figure

The steady-flow ledger — one current in, two currents out, and the books must balance

The subtlety worth a whole figure is that amplitude is not flow. When the transmitted region sits at higher potential, the particle moves more slowly there ( $k'<k$ ), and a slower wave must stand taller to carry the same current — the quantum version of a river running deep and slow through a narrow gorge. Squaring the transmitted amplitude therefore overcounts transmission on a slow side and undercounts it on a fast side; multiplying by the speed ratio $k'/k$ restores honest books.

Figure

Amplitude piles up where the wave is slow — the k′/k factor turns amplitudes into honest flux

Keep both pictures and the two classic surprises of this chapter stop being surprising. Tunnelling (Rectangular Barrier and Quantum Tunnelling) is a nonzero current threading a region where the wave merely decays. Over-barrier and even downhill reflection (Potential Step and Reflection) is an impedance mismatch: any abrupt change in wave speed scatters a wave, whether the change is up or down — the same reason an open pipe end reflects sound.

The current, and why amplitudes lie

Everything rests on the probability current, whose derivation from the Schrödinger equation belongs to Probability Current and the Continuity Equation. Here we only need its one-dimensional form and one specialization. For a state $\psi(x)$ ,

j(x)=\frac{\hbar}{m}\,\mathrm{Im}\!\left(\psi^{*}\,\frac{\mathrm{d}\psi}{\mathrm{d}x}\right) \qquad (1)

where: $j$ — probability per unit time crossing the point $x$ (units s $^{-1}$ ); $m$ — the particle's mass. Feed it a plane wave $Ae^{ikx}$ and it returns $j=(\hbar k/m)|A|^2$ — density $|A|^2$ times velocity $\hbar k/m$ , exactly the flux formula of any steady stream. A real wavefunction, by contrast, gives $j=0$ identically: standing waves carry no net flow. A stationary scattering state at energy $E$ approaches plane-wave form far from the obstacle, where the potential has levelled off:

\psi(x)\to\begin{cases} e^{ikx}+r\,e^{-ikx}, & x\to-\infty\\[2pt] t\,e^{ik'x}, & x\to+\infty \end{cases} \qquad (2)

where: $k=\sqrt{2mE}/\hbar$ — incident wavenumber; $k'=\sqrt{2m(E-V_{+\infty})}/\hbar$ — transmitted wavenumber, different from $k$ whenever the far side sits at a different potential level $V_{+\infty}$ ; $r,t$ — complex reflection and transmission amplitudes, the outputs of whatever matching problem the obstacle poses. The incident amplitude is normalised to 1; there is no left-mover on the right because nothing is fired from $+\infty$ .

Derivation ACurrents of the three asymptotic waves

We compute the current carried by each asymptotic piece of the scattering state, because the definitions of $R$ and $T$ are built from these three numbers. The only genuinely delicate point is the left region, where incident and reflected waves coexist: the current of a sum is not the sum of the currents in general, so we must check what the interference cross-terms do. The happy outcome — they cancel identically — is what makes "incident flux" and "reflected flux" separately meaningful.

Step 1. Incident piece $e^{ikx}$ : from equation (1), $\psi^*\psi'=ik$ , whose imaginary part is $k$ , so $j_{\text{inc}}=\hbar k/m$ — unit density moving at speed $\hbar k/m$ .

Step 2. Reflected piece $r\,e^{-ikx}$ : now $\psi^*\psi'=-ik|r|^2$ , so $j_{\text{refl}}=-(\hbar k/m)|r|^2$ — the minus sign says the flow is leftward, with magnitude $|r|^2$ of the incident flux.

Step 3. Transmitted piece $t\,e^{ik'x}$ : the same computation with $k'$ gives $j_{\text{trans}}=(\hbar k'/m)|t|^2$ — amplitude-squared times the transmitted speed. This is where the velocity factor is born.

Step 4. The full left-region state $\psi=e^{ikx}+re^{-ikx}$ . Write $r=|r|e^{i\varphi}$ and grind out equation (1):

j_{\mathrm{left}}=\frac{\hbar k}{m}\left(1-|r|^2\right)

Why the cross terms vanish: the interference terms in $\psi^*\psi'$ are proportional to $r\,e^{-2ikx}$ minus its complex conjugate taken with an extra sign — the two contributions to the imaginary part cancel exactly, position by position, not merely on average. Counter-propagating waves of equal wavenumber exchange no net flux.

Step 5. Read off the physics: the left region carries the incident flux minus the reflected flux, the right region carries the transmitted flux, and all three pieces have honest, separate meanings. Verified: all four currents were computed symbolically (sympy) with complex $r$ ; the left-region current is exactly $(\hbar k/m)(1-|r|^2)$ with no $x$ -dependent remainder.

Defining the coefficients

With the currents in hand, the definitions write themselves — and they are definitions about flow:

R\equiv\frac{|j_{\text{refl}}|}{j_{\text{inc}}}=|r|^{2}, \qquad T\equiv\frac{j_{\text{trans}}}{j_{\text{inc}}}=\frac{k'}{k}\,|t|^{2} \qquad (3)

where: $R$ — the fraction of incident flux that returns; $T$ — the fraction that gets through; $k'/k$ — the velocity ratio, equal to 1 only when the two asymptotic regions sit at the same potential (as for the rectangular barrier, which is why Rectangular Barrier and Quantum Tunnelling could write $T=|t|^2$ without sin). In a beam experiment these are precisely the counting fractions: detectors upstream and downstream measure rates, and rates are currents. The centrepiece of this article is that these two numbers are not independent.

Go deeper · Derivation B · 1 minR plus T equals one from current conservation

Derivation BR plus T equals one from current conservation

We prove $R+T=1$ for any localized potential — no formula for $r$ and $t$ required. The strategy is to show that a stationary state's current is one global constant, then simply evaluate that constant at the two ends of the axis, where the asymptotic forms of equation (2) hold. This is the payoff of defining the coefficients through currents: conservation is structural, not an algebraic accident to be checked case by case. Watch the single hypothesis doing the work — the potential is real.

Step 1. Differentiate the current:

\frac{\mathrm{d}j}{\mathrm{d}x}=\frac{\hbar}{m}\,\mathrm{Im}\!\left(\frac{\mathrm{d}\psi^*}{\mathrm{d}x}\frac{\mathrm{d}\psi}{\mathrm{d}x}+\psi^*\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2}\right)

Why the first term dies: $|\psi'|^2$ is real, and real numbers have no imaginary part.

Step 2. Eliminate $\psi''$ with the time-independent Schrödinger equation, $\psi''=(2m/\hbar^2)\,(V(x)-E)\,\psi$ :

\psi^*\psi''=\frac{2m}{\hbar^{2}}\,\big(V(x)-E\big)\,|\psi|^{2}

Why this is real: $V(x)$ and $E$ are real and $|\psi|^2$ is real — this is the hermiticity of the Hamiltonian showing up as a conservation law. (A complex "optical" potential, used to model absorption, breaks exactly this step and deliberately lets $R+T<1$ .)

Step 3. Conclude $\mathrm{d}j/\mathrm{d}x=0$ : the current of an energy eigenstate is a single constant along the entire axis — through the obstacle, inside any forbidden region, everywhere. This is the Wronskian of $\psi^*$ and $\psi$ in disguise, and it is the stationary-state remnant of the continuity equation (density static, divergence of flux zero).

Step 4. Evaluate the constant at both ends using Steps 1–4 of the previous derivation:

\frac{\hbar k}{m}\left(1-|r|^{2}\right)=j(-\infty)=j(+\infty)=\frac{\hbar k'}{m}\,|t|^{2}

Step 5. Divide by the incident flux $\hbar k/m$ :

1-R=\frac{k'}{k}\,|t|^{2}=T \qquad\Longrightarrow\qquad R+T=1

Why the velocity factor was inevitable: without $k'/k$ , equation in Step 4 simply is not the statement $1-R=T$ — bare $|t|^2$ balances the books only in the accidental case $k'=k$ . *Verified: dj/dx = 0 was checked symbolically, and the step-potential state (below) reproduces identical currents at sampled points on both sides to 12 decimal places, with $R+T-1<10^{-12}$ .*

The step, worked as the showcase

The potential step is where the velocity factor stops being pedantry and becomes visible. The solve itself — matching $\psi$ and $\psi'$ at the jump — belongs to Potential Step and Reflection; we quote its amplitudes for a step up of height $V_0$ with $E>V_0$ :

r=\frac{k-k'}{k+k'},\qquad t=\frac{2k}{k+k'} \qquad (4)

where: $k$ and $k'$ — the wavenumbers before and after the step. Both are real here, so $r$ and $t$ are real too. Notice $t>1$ whenever $k'<k$ : the transmitted wave is taller than the incident one. Squaring it gives a number bigger than 1 — obviously not a probability. Now apply the definitions:

R=\left(\frac{k-k'}{k+k'}\right)^{2}, \qquad T=\frac{k'}{k}\left(\frac{2k}{k+k'}\right)^{2}=\frac{4kk'}{(k+k')^{2}} \qquad (5)

where: the velocity factor has tamed $|t|^2$ into a genuine fraction. Check the books: $(k-k')^2+4kk'=(k+k')^2$ , so $R+T=1$ identically — the general theorem, landing in one concrete lap. Below the step top ( $E<V_0$ ) the transmitted wave becomes evanescent, carries zero steady current, and the step reflects everything: $|r|=1$ , $R=1$ , $T=0$ , with the subtleties of the penetrating tail again in Potential Step and Reflection.

The chapter's ledger

Every scattering result of this chapter, in current-honest form — each derived in its own article, none re-derived here:

obstacle	result	where derived
step up, $E>V_0$	$R=\left(\frac{k-k'}{k+k'}\right)^2$ , $T=\frac{4kk'}{(k+k')^2}$	Potential Step and Reflection
step, $E<V_0$	$R=1$ , $T=0$ (evanescent tail, no steady flux)	Potential Step and Reflection
rectangular barrier, $E<V_0$	$T=\left[1+\frac{V_0^2\sinh^2(\kappa a)}{4E(V_0-E)}\right]^{-1}$	Rectangular Barrier and Quantum Tunnelling
rectangular barrier, $E>V_0$	$T=\left[1+\frac{V_0^2\sin^2(k_2a)}{4E(E-V_0)}\right]^{-1}$ ; $T=1$ at $k_2a=n\pi$	Rectangular Barrier and Quantum Tunnelling
delta spike $\pm\alpha\,\delta(x)$ (well or barrier)	$T=\left[1+\frac{m\alpha^2}{2\hbar^2E}\right]^{-1}$ — same for attraction and repulsion	Delta-Function Potential Well and Bound State · Delta-Function Barrier and Scattering

Two structural remarks the table invites. First, every entry respects $R+T=1$ — the theorem above guarantees it wholesale. Second, the delta-spike row is the $V_0\to\infty$ , $V_0a=\alpha$ limit of the barrier row (verified symbolically for this edition), a reminder that the chapter's obstacles are one family. The resonant spikes of the barrier row get their own story in Resonant Transmission and Tunnelling Resonances.

Limiting cases

case	result	meaning
$E\gg V_0$ (any obstacle here)	$T\to1$ , $R\to0$	fast particles ignore the potential; for the step, $R\approx(V_0/4E)^2$
step threshold, $E\to V_0^+$	$T\to0$ like $4k'/k\propto\sqrt{E-V_0}$	the transmitted channel opens with zero speed — almost everything reflects despite $E>V_0$
equal wavenumbers, $k'=k$	$T=\|t\|^2$	velocity factor silently equals 1 — the barrier/delta case
swap sides, $k\leftrightarrow k'$	same $R$ , same $T$	reciprocity: uphill and downhill scattering are equally reflective
complex (absorbing) potential	$R+T<1$	Step 2 of the conservation proof fails by design — flux is eaten

Worked examples

Example 1 — electron onto a step (confidence builder). An electron with $E=10\,\mathrm{eV}$ meets a step up of $V_0=6\,\mathrm{eV}$ . Find $r$ , $t$ , $R$ , $T$ .

Step 1. Wavenumbers: $k=\sqrt{2m_eE}/\hbar=1.62\times10^{10}\,\mathrm{m}^{-1}$ ; beyond the step the kinetic energy is $4\,\mathrm{eV}$ , so $k'=1.02\times10^{10}\,\mathrm{m}^{-1}$ and $k'/k=\sqrt{4/10}=0.632$ .

Step 2. Amplitudes from equation (4): $r=(1-0.632)/(1+0.632)=0.225$ and $t=2/(1+0.632)=1.225$ . The transmitted wave stands 22.5% taller than the incident one.

Step 3. Coefficients: $R=r^2=0.051$ and $|t|^2=1.501$ — bigger than one, as promised. The velocity factor rescues it: $T=0.632\times1.501=\boxed{0.949}$ .

Step 4. Books: $R+T=0.051+0.949=1.000$ exactly. About 5% of the electrons reflect off a step they classically would not even notice.

Example 2 — reflection going downhill (exam level). An electron with $E=4\,\mathrm{eV}$ reaches the edge of a region where the potential drops by $6\,\mathrm{eV}$ (a cliff, not a wall). Classically it accelerates over the edge, transmission certain. Quantum mechanically?

Step 1. Wavenumbers: incident side $k=\sqrt{2m_e\cdot4\,\mathrm{eV}}/\hbar=1.02\times10^{10}\,\mathrm{m}^{-1}$ ; beyond the cliff the kinetic energy is $4+6=10\,\mathrm{eV}$ , so $k'=1.62\times10^{10}\,\mathrm{m}^{-1}$ . Note these are Example 1's numbers with the roles swapped.

Step 2. Amplitude: $r=(k-k')/(k+k')=-0.225$ . The sign flip (relative to Example 1) records the half-turn phase shift of reflection off a faster region — but $R$ cares only about the magnitude.

Step 3. $R=0.051$ : 5.1% of the electrons bounce off a downhill edge. And $T=(k'/k)|t|^2=1.588\times(0.775)^2\cdots$ — no need to grind: $T=1-R=0.949$ .

Step 4. The two examples together demonstrate reciprocity: $R$ depends on the pair $\{k,k'\}$ , not on which side is uphill. Impedance mismatch, not slope, causes reflection. This is why anti-reflection engineering (in optics and in quantum devices alike) works by grading the transition instead of sharpening it.

The bookkeeper of scattering

Origins · 1 minEugene Paul Wigner and how this was discovered

Eugene Paul Wigner

1902–1995 · Princeton

Wigner built much of the machinery that turns "what fraction gets through?" into deep theory. With Gregory Breit he wrote the resonance formula (1936) that governs every peak in a transmission curve; with Leonard Eisenbud he founded R-matrix theory (1947), the systematic accounting of scattering amplitudes that nuclear and atomic physics still run on; and in 1955 he proved that the energy-derivative of a scattering phase defines a time delay — how long the particle lingers in the obstacle — turning the stationary coefficients of this article back into statements about time. His 1963 Nobel Prize cited symmetry principles, but scattering theory is where his bookkeeping instincts shine plainest.

History

1921

— Ramsauer finds argon nearly transparent to slow electrons — a transmission resonance measured before the theory existed

1926

— Born's paper on collision processes introduces the probability interpretation of the wavefunction — scattering is where quantum probability was born

1936

— Breit and Wigner publish the resonance formula describing peaks in transmission and cross-section curves

1947

— Wigner and Eisenbud's R-matrix theory systematizes scattering amplitudes for nuclear reactions

1955

— Wigner's time-delay theorem connects the energy-dependence of scattering phases to the duration of the collision

1957

— Landauer proposes that electrical conductance is transmission: G proportional to T for electrons crossing a disordered region

1988

— van Wees and Wharam observe conductance quantization in point contacts — T jumping in integer steps, watched on an ammeter

Going further

References. Griffiths, Introduction to Quantum Mechanics, 3rd ed., §2.5–2.6 · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., §2.4 and Ch. 6 · Shankar, Principles of Quantum Mechanics, 2nd ed., §5.4 · Cohen-Tannoudji, Quantum Mechanics, Vol. 1, Complement H_I and Ch. I · Townsend, A Modern Approach to Quantum Mechanics, 2nd ed., Ch. 6

Provenance. All formulas cross-checked against Griffiths 3rd ed. §2.5–2.6, Cohen-Tannoudji Vol. 1 Complement H_I, and Sakurai 2nd ed. §2.4. A verifier should: (1) recompute the step result $(k-k')^2+4kk'=(k+k')^2$ to confirm $R+T=1$ ; (2) verify the barrier simplification $(k^2+\kappa^2)^2/(4k^2\kappa^2)=V_0^2/[4E(V_0-E)]$ using $k^2=2mE/\hbar^2,\ \kappa^2=2m(V_0-E)/\hbar^2$ ; (3) check the thick-barrier limit prefactor $16E(V_0-E)/V_0^2$ from $\sinh^2\kappa a\approx \tfrac14 e^{2\kappa a}$ . All three checks, the current computations, the conservation proof, the delta-limit of the barrier row, and every worked number on this page were re-verified symbolically (sympy) and numerically for this edition. (Base-content note: this article's AllQFT source arrived with systematically doubled LaTeX escapes — 358 occurrences — which rendered its mathematics unreadable; the present edition is a verified full rewrite.)

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

Leads to: Resonant Transmission and Tunnelling Resonances · Wave Packets and Their Construction · Group Velocity versus Phase Velocity · the Scattering Theory chapter (S-matrix, partial waves)

Why do we define probabilities from currents when the Born rule already gives probabilities from amplitudes?

The Born rule gives the probability of finding the particle at a place; scattering asks for the probability of the particle arriving — a rate, integrated over the experiment. Rates are currents. The two notions coincide for wave packets: build a normalizable packet, let it collide, and the Born-rule weight of the far-side packet at late times equals the plane-wave $T$ of its central energy. The plane-wave current computation is the shortcut to that packet answer, valid whenever the packet is narrow in energy — the honest construction lives in Wave Packets and Their Construction.

How can a single stationary state describe a particle that is "incident and then reflected"? Nothing moves in a stationary state.

Right — a stationary state describes a steady experiment, not a movie of one particle. Think of a beam that has been running long enough that the incoming flow, the reflected flow, and the transmitted flow all coexist in steady state, like a fountain in equilibrium. The time-resolved single-particle story (packet arrives, splits, two packets separate) requires superposing many energies. Both descriptions yield the same $R$ and $T$ ; the stationary one is simply the easier calculation.

Is the reflected wave's phase ever physically meaningful, since R only uses its magnitude?

Very much so. The phase of $r$ shifts the positions of the interference fringes the incident and reflected waves form together — measurable in neutron and atom interferometry. Its energy-derivative is the Wigner time delay, telling you how long the collision detains the particle. And below a step top, where $|r|=1$ and $R=1$ carries no information at all, the phase of $r$ is the only output of the scattering problem — it encodes the penetration depth of the evanescent tail. Coefficients are the headline; phases are the fine print that later chapters (and the scattering-theory track) cash in.

If the transmitted amplitude can exceed 1, can the wavefunction's norm be violated somewhere?

No. The scattering state is not normalized to begin with (plane waves never are), so amplitudes are all relative — what is conserved is not any single amplitude but the flux ledger. $|t|>1$ says only that the probability density beyond a slow step is higher than the incident density, the pile-up of the gorge picture. Density can pile up freely; it is the current that must balance, and it does: $(\hbar k/m)(1-|r|^2)=(\hbar k'/m)|t|^2$ holds exactly in every solved case.

Do R and T depend on where I put the coordinate origin or how I normalize the incident wave?

Neither. Shifting the origin multiplies $r$ and $t$ by pure phases (their magnitudes survive), and rescaling the incident amplitude rescales all three currents together, cancelling in the ratios. That invariance is the sign that $R$ and $T$ are physical observables while $r$ , $t$ , and $\psi$ itself are representation choices. The same lesson recurs in 3D as the difference between scattering amplitudes and cross-sections.

What exactly does a real-valued wavefunction carrying zero current mean for bound states?

Bound states (Bound States versus Scattering States) can always be chosen real, so they carry zero net current everywhere: nothing flows, which is what "stationary and confined" should mean. Scattering states cannot be chosen real — the boundary condition "incident from the left only" is intrinsically complex, breaking left–right symmetry. The imaginary part of $\psi$ is, in this precise sense, where the motion lives; that is why the current formula reads the imaginary part of $\psi^*\psi'$ .

Does R + T = 1 still hold when there are bound states in the potential, like a well that could capture the particle?

Yes — at any fixed scattering energy $E>0$ the theorem holds unconditionally for real potentials, wells included; the delta well row of the ledger balances exactly like the barrier row. True capture (the particle ending up bound) cannot happen in elastic 1D scattering because it would need the particle to shed energy, and there is nothing to shed it to. Add a mechanism — photon emission, a second particle, an absorbing optical potential — and flux really can leave the elastic channel; the accountant's entry for that is precisely $R+T<1$ .