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One-Dimensional Scattering Revisited

One-dimensional scattering is often introduced as a collection of boundary-matching exercises: a step, a barrier, a well, or a delta interaction. The same problems become more systematic when recast in the language of asymptotic channels, flux-normalized amplitudes, an SS-matrix, analytic continuation, and poles.

The essential simplification is also the essential difference from three dimensions. At a fixed energy, there is no continuum of outgoing angles. There are only two asymptotic propagation directions, left and right. Reflection and transmission are therefore channel probabilities rather than area-valued cross sections.

This page supplies the graduate-level bridge. The canonical calculations remain at Reflection and Transmission Coefficients, Transfer Matrix Method, Scattering from a Delta Potential, and Resonant Transmission. Here the goal is to show how their amplitudes fit into one scattering structure.

Assume a real, time-independent potential with asymptotic limits

V(x)⟶VL(x→−∞),V(x)⟶VR(x→+∞).\begin{aligned} V(x)&\longrightarrow V_L && (x\to-\infty), \\ V(x)&\longrightarrow V_R && (x\to+\infty). \end{aligned}

At an energy above both asymptotic thresholds, define

kL=2m(E−VL)ℏ,kR=2m(E−VR)ℏ,\begin{aligned} k_L &= \frac{\sqrt{2m(E-V_L)}}{\hbar}, \\ k_R &= \frac{\sqrt{2m(E-V_R)}}{\hbar}, \end{aligned}

and

vL=ℏkLm,vR=ℏkRm.v_L=\frac{\hbar k_L}{m}, \qquad v_R=\frac{\hbar k_R}{m}.

The general asymptotic wave has four traveling-wave coefficients:

ψ(x)∼AL+eikLx+AL−e−ikLx,x→−∞,ψ(x)∼AR+eikRx+AR−e−ikRx,x→+∞.\begin{aligned} \psi(x) &\sim A_L^+e^{ik_Lx} + A_L^-e^{-ik_Lx}, &&x\to-\infty, \\ \psi(x) &\sim A_R^+e^{ik_Rx} + A_R^-e^{-ik_Rx}, &&x\to+\infty. \end{aligned}

The superscript ++ means right-moving and −- means left-moving. It does not mean incoming and outgoing everywhere:

  • AL+A_L^+ and AR−A_R^- are incoming amplitudes.
  • AL−A_L^- and AR+A_R^+ are outgoing amplitudes.

If E<VRE<V_R, then kRk_R is imaginary and the right side is a closed channel. An evanescent tail may remain near the interaction region, but it carries no outgoing flux to +∞+\infty.

For

ψ(x)=Ae±ikx,\psi(x)=Ae^{\pm ikx},

the probability current is

j=±v∣A∣2.j = \pm v|A|^2.

Raw coefficients therefore carry unequal flux when vL≠vRv_L\ne v_R. Define incoming and outgoing flux amplitudes by

aL=vL AL+,aR=vR AR−,bL=vL AL−,bR=vR AR+.\begin{aligned} a_L&=\sqrt{v_L}\,A_L^+, & a_R&=\sqrt{v_R}\,A_R^-, \\ b_L&=\sqrt{v_L}\,A_L^-, & b_R&=\sqrt{v_R}\,A_R^+. \end{aligned}

Then ∣aL∣2|a_L|^2 and ∣aR∣2|a_R|^2 are incoming fluxes, while ∣bL∣2|b_L|^2 and ∣bR∣2|b_R|^2 are outgoing fluxes. The scattering matrix is the channel map

(bLbR)=S(E)(aLaR).\begin{pmatrix} b_L\\ b_R \end{pmatrix} = S(E) \begin{pmatrix} a_L\\ a_R \end{pmatrix}.

Incoming and outgoing left and right channels surrounding a one-dimensional interaction region

The right-moving and left-moving waves exchange their incoming or outgoing role across the interaction region. Flux normalization turns the four asymptotic coefficients into a two-channel map b=Sa\mathbf b=S\mathbf a.

For incidence from the left, choose unit raw incoming amplitude:

ψL(x)∼eikLx+rLe−ikLx,x→−∞,ψL(x)∼tLeikRx,x→+∞.\begin{aligned} \psi_L(x) &\sim e^{ik_Lx} + r_L e^{-ik_Lx}, &&x\to-\infty, \\ \psi_L(x) &\sim t_L e^{ik_Rx}, &&x\to+\infty. \end{aligned}

For incidence from the right,

ψR(x)∼tRe−ikLx,x→−∞,ψR(x)∼e−ikRx+rReikRx,x→+∞.\begin{aligned} \psi_R(x) &\sim t_R e^{-ik_Lx}, &&x\to-\infty, \\ \psi_R(x) &\sim e^{-ik_Rx} + r_R e^{ik_Rx}, &&x\to+\infty. \end{aligned}

Introduce flux-normalized transmission amplitudes

τL=vRvL tL,τR=vLvR tR.\tau_L = \sqrt{\frac{v_R}{v_L}}\,t_L, \qquad \tau_R = \sqrt{\frac{v_L}{v_R}}\,t_R.

In the left/right incoming basis,

S(E)=(rLτRτLrR).S(E) = \begin{pmatrix} r_L & \tau_R\\ \tau_L & r_R \end{pmatrix}.

The observable probabilities are

RL=∣rL∣2,TL=∣τL∣2=vRvL∣tL∣2,RR=∣rR∣2,TR=∣τR∣2=vLvR∣tR∣2.\begin{aligned} R_L&=|r_L|^2, & T_L&=|\tau_L|^2 = \frac{v_R}{v_L}|t_L|^2, \\ R_R&=|r_R|^2, & T_R&=|\tau_R|^2 = \frac{v_L}{v_R}|t_R|^2. \end{aligned}

When VL=VRV_L=V_R, the velocities agree and τL,R=tL,R\tau_{L,R}=t_{L,R}. The familiar rule T=∣t∣2T=|t|^2 is therefore a special equal-velocity case, not the fundamental definition.

For a real conservative potential with both asymptotic channels included,

S†S=I.S^\dagger S=I.

The two column norms give

∣rL∣2+∣τL∣2=1,∣rR∣2+∣τR∣2=1.|r_L|^2+|\tau_L|^2=1, \qquad |r_R|^2+|\tau_R|^2=1.

The columns must also be orthogonal:

rL∗τR+τL∗rR=0.r_L^*\tau_R + \tau_L^*r_R =0.

Thus unitarity constrains phases as well as probabilities. Checking only R+T=1R+T=1 does not test the full matrix.

For spinless time-reversal-invariant scattering, channel phases can be chosen so that

S=ST.S=S^{\mathsf T}.

Hence

τL=τR.\tau_L=\tau_R.

When the asymptotic velocities are equal, this becomes tL=tRt_L=t_R. An asymmetric real potential can still have different reflection amplitudes rLr_L and rRr_R, although unitarity and reciprocity constrain their magnitudes and relative phases. If the potential also has parity symmetry, then

rL=rR.r_L=r_R.

A complex optical potential can absorb or amplify flux, so the reduced two-channel matrix is then nonunitary. Additional physical channels can produce the same apparent deficit if they have simply been omitted.

The scattering matrix groups amplitudes by incoming versus outgoing status. The transfer matrix instead relates coefficients on opposite sides:

(AR+AR−)=M(E)(AL+AL−).\begin{pmatrix} A_R^+\\ A_R^- \end{pmatrix} = M(E) \begin{pmatrix} A_L^+\\ A_L^- \end{pmatrix}.

Write

M=(M11M12M21M22).M = \begin{pmatrix} M_{11}&M_{12}\\ M_{21}&M_{22} \end{pmatrix}.

This convention is valuable because interfaces and propagation regions compose in spatial order. If a structure consists of successive elements,

Mtotal=MNMN−1⋯M2M1.M_{\mathrm{total}} = M_NM_{N-1}\cdots M_2M_1.

For left incidence,

(tL0)=M(1rL),\begin{pmatrix} t_L\\ 0 \end{pmatrix} = M \begin{pmatrix} 1\\ r_L \end{pmatrix},

which gives

rL=−M21M22,tL=det⁡MM22.r_L = -\frac{M_{21}}{M_{22}}, \qquad t_L = \frac{\det M}{M_{22}}.

For right incidence,

(rR1)=M(0tR),\begin{pmatrix} r_R\\ 1 \end{pmatrix} = M \begin{pmatrix} 0\\ t_R \end{pmatrix},

so

rR=M12M22,tR=1M22.r_R = \frac{M_{12}}{M_{22}}, \qquad t_R = \frac{1}{M_{22}}.

For equal asymptotic velocities, the raw-amplitude scattering matrix is therefore

S=1M22(−M211det⁡MM12).S = \frac{1}{M_{22}} \begin{pmatrix} -M_{21}&1\\ \det M&M_{12} \end{pmatrix}.

For the ordinary Schrödinger equation with equal asymptotic wave numbers,

det⁡M=1,\det M=1,

so tL=tRt_L=t_R. If the asymptotic wave numbers differ, the raw transfer matrix need not have unit determinant; flux normalization restores the physically unitary channel description.

Transfer-matrix conventions differ. Some references propagate right to left or reverse the coefficient ordering, moving the relevant denominator from M22M_{22} to M11M_{11}. The invariant check is to impose the stated boundary conditions and rederive rr and tt rather than memorizing an index.

Transfer matrices compose elegantly but can be numerically ill-conditioned. An evanescent region introduces factors

eκaande−κa.e^{\kappa a} \quad\text{and}\quad e^{-\kappa a}.

For a thick barrier or long multilayer stack, the growing factor can overflow while the physical transmission is exponentially small. Stable scattering-matrix recursion, logarithmic derivatives, or rescaled propagation avoids multiplying enormous and tiny numbers in the same matrix.

The same denominator that controls transmission also remembers the bound spectrum. Continue the amplitudes from real positive kk into the complex momentum plane. For equal asymptotic thresholds, a bound state has

k=iκ,κ>0.k=i\kappa, \qquad \kappa>0.

At x→−∞x\to-\infty, normalizability keeps only eκxe^{\kappa x}; at x→+∞x\to+\infty, it keeps only e−κxe^{-\kappa x}. In the traveling-wave notation, this means

AL+=0,AR−=0,A_L^+=0, \qquad A_R^-=0,

while AL−A_L^- and AR+A_R^+ may be nonzero. These are outgoing-only boundary conditions after analytic continuation.

The transfer relation then becomes

(AR+0)=M(k)(0AL−).\begin{pmatrix} A_R^+\\ 0 \end{pmatrix} = M(k) \begin{pmatrix} 0\\ A_L^- \end{pmatrix}.

A nontrivial solution requires

M22(k)=0.M_{22}(k)=0.

Since tR=1/M22t_R=1/M_{22}, the same condition is a pole of the transmission amplitude. The direct bound-state matching equation and the scattering-pole equation are not separate facts; they are the same homogeneous boundary-value condition viewed on different parts of the complex energy surface.

For a short-range potential with threshold at E=0E=0, the momentum-plane locations have different physical meanings.

Pole locationInterpretationSpatial behavior
k=iκk=i\kappa, κ>0\kappa>0bound statedecays at both infinities
k=−iκk=-i\kappa, κ>0\kappa>0virtual or antibound statenon-normalizable continuation
Re⁡k>0\operatorname{Re}k>0, Im⁡k<0\operatorname{Im}k<0resonanceoutgoing Gamow condition

For a resonance,

Epole=ER−i2Γ,E_{\mathrm{pole}} = E_R-\frac{i}{2}\Gamma,

with Γ>0\Gamma>0. Its analytically continued wave grows spatially, so it is not a Hilbert-space eigenstate; the negative imaginary energy instead encodes temporal decay.

Bound States and Scattering Poles owns the general analytic classification and sheet structure.

For V(x)=λδ(x)V(x)=\lambda\delta(x), define

γ=mλℏ2.\gamma=\frac{m\lambda}{\hbar^2}.

The exact transmission amplitude is

t(k)=kk+iγ.t(k)=\frac{k}{k+i\gamma}.

If λ<0\lambda<0, then γ<0\gamma<0 and the pole lies at

k=i∣γ∣,k=i|\gamma|,

the positive imaginary axis. It reproduces the unique attractive-delta bound state. The complete matching derivation belongs at Scattering from a Delta Potential; Delta Potential Scattering audits the lead-channel swap, reduced parity eigenvalues, and pole residue in one fixed convention.

Consider two barriers separated by a classically allowed region of length LL and wave number kk. Let t1t_1 transmit through the first barrier from the left, t2t_2 transmit through the second, r1′r_1' reflect from the first barrier as seen from inside the middle region, and r2r_2 reflect from the second barrier as seen from inside.

The transmitted amplitude is a coherent sum over all round trips:

ttot=t1t2eikL∑n=0∞(r1′r2e2ikL)n=t1t2eikL1−r1′r2e2ikL.\begin{aligned} t_{\mathrm{tot}} ={}& t_1t_2e^{ikL} \sum_{n=0}^{\infty} \left( r_1'r_2e^{2ikL} \right)^n \\ ={}& \frac{ t_1t_2e^{ikL} }{ 1-r_1'r_2e^{2ikL} }. \end{aligned}

The n=0n=0 term is the direct path through both barriers; terms with n≥1n\ge1 include one or more round trips. What matters is that the amplitudes form a geometric series whose denominator is produced by repeated coherent reflection.

A resonance occurs when the round-trip phase nearly satisfies

2kL+arg⁡(r1′r2)≈2πn,2kL + \arg(r_1'r_2) \approx 2\pi n,

and ∣r1′r2∣|r_1'r_2| is close to one. The wave then builds up between the barriers like a leaky standing wave. High transmission does not mean that either barrier has ceased to tunnel; it means the full structure has arranged destructive interference in reflection and constructive interference in transmission.

Near an isolated resonance, a common probability form is

T(E)≈ΓLΓR(E−ER)2+Γ2/4,Γ=ΓL+ΓR.\begin{aligned} T(E) &\approx \frac{ \Gamma_L\Gamma_R }{ (E-E_R)^2+\Gamma^2/4 }, \\ \Gamma &= \Gamma_L+\Gamma_R. \end{aligned}

At E=ERE=E_R,

T(ER)≈4ΓLΓR(ΓL+ΓR)2.T(E_R) \approx \frac{ 4\Gamma_L\Gamma_R }{ (\Gamma_L+\Gamma_R)^2 }.

A symmetric lossless structure can reach unit transmission. Unequal leakage rates lower the peak. Resonant Transmission owns the double-barrier calculation, while Breit–Wigner Form owns the general isolated-pole parameterization and its limits.

If

V(x)=V(−x)V(x)=V(-x)

and the asymptotic velocities agree, then

S=(rttr).S = \begin{pmatrix} r&t\\ t&r \end{pmatrix}.

Its eigenvectors are the even and odd incoming combinations,

12(11),12(1−1),\frac{1}{\sqrt2} \begin{pmatrix} 1\\ 1 \end{pmatrix}, \qquad \frac{1}{\sqrt2} \begin{pmatrix} 1\\ -1 \end{pmatrix},

with eigenvalues

Se=r+t,So=r−t.S_{\mathrm e}=r+t, \qquad S_{\mathrm o}=r-t.

Unitarity gives ∣Se∣=∣So∣=1|S_{\mathrm e}|=|S_{\mathrm o}|=1. These eigenvalues can be described by parity phase shifts. There is a convention subtlety: in the left/right channel basis, the free-particle matrix is

Sfree=(0110),S_{\mathrm{free}} = \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix},

whose even and odd eigenvalues are +1+1 and −1-1. Some phase-shift conventions absorb the odd-channel minus sign so both free phase shifts vanish. Apparent sign disagreements should be traced to this free-channel convention before comparing formulas.

For a smooth barrier with turning points x1x_1 and x2x_2, define the forbidden-region action

K(E)=∫x1x22m[V(x)−E] dx.K(E) = \int_{x_1}^{x_2} \sqrt{2m\left[V(x)-E\right]} \,dx.

The leading WKB transmission probability is

TWKB≈e−2K(E)/ℏ.T_{\mathrm{WKB}} \approx e^{-2K(E)/\hbar}.

This formula captures the dominant suppression when K/ℏ≫1K/\hbar\gg1. It does not by itself provide the complete complex amplitude needed for coherent composition.

MethodMain objectStrengthMain limitation
direct matchingwavefunction coefficientsexact for solvable profilesbecomes cumbersome for many regions
transfer matrixspatial coefficient map MMexact composition for layered modelsevanescent factors can be ill-conditioned
scattering matrixincoming-to-outgoing map SSunitary and numerically stable in open channelsdoes not compose by naive multiplication in space
leading WKBaction K(E)K(E)exposes exponential scale for smooth barriersloses important prefactors and phases

Why probabilities cannot be multiplied coherently

Section titled “Why probabilities cannot be multiplied coherently”

For two separated barriers, multiplying the one-barrier probabilities,

Tnaive=T1T2,T_{\mathrm{naive}}=T_1T_2,

discards the phases of all paths that bounce between them. It therefore cannot produce resonant transmission. A semiclassical treatment can recover resonances only if it keeps complex reflection amplitudes, propagation phases, and the repeated-round-trip denominator.

The WKB barrier action still has an important role near a narrow resonance. It estimates the exponentially small leakage through each barrier and therefore the partial widths ΓL\Gamma_L and ΓR\Gamma_R. The phase accumulated in the intermediate allowed region determines the approximate resonance energy.

Leading WKB also becomes unreliable near a barrier top, at coalescing turning points, for abrupt discontinuities without separate matching, and whenever K/ℏK/\hbar is not large. Barrier Penetration and Tunneling owns the action formula and connection-rule caveats.

The S(E)S(E) matrix is defined at a sharp energy, but a physical incoming particle is a packet. For a left-incident momentum envelope g(k)g(k), the late-time reflected and transmitted probabilities are, for equal asymptotic thresholds,

PR=∫0∞dk ∣g(k)∣2∣rL(k)∣2,P_R = \int_0^\infty dk\, |g(k)|^2|r_L(k)|^2,

and

PT=∫0∞dk ∣g(k)∣2∣tL(k)∣2.P_T = \int_0^\infty dk\, |g(k)|^2|t_L(k)|^2.

If the packet is narrow around k0k_0, then PR≈R(k0)P_R\approx R(k_0) and PT≈T(k0)P_T\approx T(k_0). A broad packet samples energy-dependent magnitudes and phases, so it can distort, split, or acquire a delay.

The phase of a scattering amplitude matters even when a stationary transmission probability does not show it. Near a resonance, rapid phase variation is associated with temporary probability storage in the interaction region. Wave Packets and Scattering owns the time-dependent construction.

  1. State the asymptotic potentials and determine which left and right channels are open.
  2. Fix whether amplitudes are raw wave coefficients or flux normalized.
  3. Define the ordering and propagation direction of the transfer matrix before multiplying layers.
  4. Extract both left- and right-incidence amplitudes as a convention check.
  5. Verify S†S=IS^\dagger S=I for a real conservative model.
  6. Test reciprocity and, when applicable, parity symmetry.
  7. Locate bound or resonance states from outgoing-only boundary conditions or zeros of the transfer denominator.
  8. Use WKB only in a controlled smooth-barrier regime, retaining phases for multibarrier interference.
  9. Fold stationary amplitudes with the incident packet or experimental energy distribution.
  • Calling RR or TT a three-dimensional cross section; they are dimensionless channel probabilities.
  • Treating the ++ and −- traveling-wave labels as incoming and outgoing on both sides.
  • Writing T=∣t∣2T=|t|^2 when the asymptotic velocities differ.
  • Comparing transfer-matrix entries across sources without matching coefficient ordering and propagation direction.
  • Multiplying transfer matrices through thick evanescent regions without checking numerical conditioning.
  • Checking only R+T=1R+T=1 and ignoring the phase orthogonality required by full SS-matrix unitarity.
  • Assuming asymmetric real potentials have identical reflection amplitudes from both sides.
  • Identifying every transmission maximum as a resonance without checking phase motion, internal buildup, or pole structure.
  • Treating a resonance pole state as a normalizable bound state.
  • Multiplying single-barrier transmission probabilities and thereby erasing resonant interference.
  • Using the leading WKB exponent near a barrier top or as a substitute for an exact phase-sensitive amplitude.

Starting from raw left- and right-incidence amplitudes rL,tL,rR,tRr_L,t_L,r_R,t_R with asymptotic velocities vLv_L and vRv_R, construct the flux-normalized scattering matrix and identify the four channel probabilities.

Solution

The incoming flux vector is

a=(vLAL+vRAR−),\mathbf a = \begin{pmatrix} \sqrt{v_L}A_L^+\\ \sqrt{v_R}A_R^- \end{pmatrix},

and the outgoing flux vector is

b=(vLAL−vRAR+).\mathbf b = \begin{pmatrix} \sqrt{v_L}A_L^-\\ \sqrt{v_R}A_R^+ \end{pmatrix}.

Therefore

S=(rLvL/vR tRvR/vL tLrR).S = \begin{pmatrix} r_L & \sqrt{v_L/v_R}\,t_R \\ \sqrt{v_R/v_L}\,t_L & r_R \end{pmatrix}.

The probabilities are

RL=∣rL∣2,TL=vRvL∣tL∣2,RR=∣rR∣2,TR=vLvR∣tR∣2.\begin{aligned} R_L&=|r_L|^2, & T_L&=\frac{v_R}{v_L}|t_L|^2, \\ R_R&=|r_R|^2, & T_R&=\frac{v_L}{v_R}|t_R|^2. \end{aligned}

For a unitary two-channel problem, RL+TL=RR+TR=1R_L+T_L=R_R+T_R=1.

2. Convert a transfer matrix to scattering amplitudes

Section titled “2. Convert a transfer matrix to scattering amplitudes”

Let

(AR+AR−)=(M11M12M21M22)(AL+AL−).\begin{pmatrix} A_R^+\\ A_R^- \end{pmatrix} = \begin{pmatrix} M_{11}&M_{12}\\ M_{21}&M_{22} \end{pmatrix} \begin{pmatrix} A_L^+\\ A_L^- \end{pmatrix}.

Derive rL,tL,rR,tRr_L,t_L,r_R,t_R.

Solution

For left incidence,

(tL0)=M(1rL).\begin{pmatrix} t_L\\ 0 \end{pmatrix} = M \begin{pmatrix} 1\\ r_L \end{pmatrix}.

The lower row gives

rL=−M21M22,r_L=-\frac{M_{21}}{M_{22}},

and substitution into the upper row gives

tL=det⁡MM22.t_L=\frac{\det M}{M_{22}}.

For right incidence,

(rR1)=M(0tR).\begin{pmatrix} r_R\\ 1 \end{pmatrix} = M \begin{pmatrix} 0\\ t_R \end{pmatrix}.

Hence

tR=1M22,rR=M12M22.t_R=\frac{1}{M_{22}}, \qquad r_R=\frac{M_{12}}{M_{22}}.

If det⁡M=1\det M=1, the two transmission amplitudes agree.

3. Recover the attractive-delta bound state from a pole

Section titled “3. Recover the attractive-delta bound state from a pole”

For

t(k)=kk+iγ,γ=mλℏ2,t(k)=\frac{k}{k+i\gamma}, \qquad \gamma=\frac{m\lambda}{\hbar^2},

locate the pole for λ<0\lambda<0 and find its energy.

Solution

For an attractive interaction, γ=−∣γ∣\gamma=-|\gamma|. The denominator vanishes at

k+iγ=k−i∣γ∣=0,k+i\gamma = k-i|\gamma| =0,

so

k=i∣γ∣.k=i|\gamma|.

This lies on the positive imaginary axis and gives

E=ℏ2k22m=−ℏ2γ22m=−mλ22ℏ2.E = \frac{\hbar^2k^2}{2m} = -\frac{\hbar^2\gamma^2}{2m} = -\frac{m\lambda^2}{2\hbar^2}.

The pole energy is exactly the bound-state eigenvalue obtained by solving the normalizable boundary-value problem directly.

For equal asymptotic velocities, take

S=(rLttrR).S = \begin{pmatrix} r_L&t\\ t&r_R \end{pmatrix}.

Derive the conditions imposed by unitarity and explain why R+T=1R+T=1 is not the whole statement.

Solution

The diagonal elements of S†S=IS^\dagger S=I give

∣rL∣2+∣t∣2=1,∣rR∣2+∣t∣2=1.|r_L|^2+|t|^2=1, \qquad |r_R|^2+|t|^2=1.

The off-diagonal element gives

rL∗t+t∗rR=0.r_L^*t+t^*r_R=0.

The first two relations imply ∣rL∣=∣rR∣|r_L|=|r_R|, but the third fixes a relative phase relation. Two sets of amplitudes can obey R+T=1R+T=1 while failing to form a unitary scattering matrix if their phases violate the off-diagonal condition.

A wave crosses a first barrier with amplitude t1t_1, propagates through a middle region with phase eikLe^{ikL}, and crosses a second barrier with amplitude t2t_2. Each complete round trip multiplies the amplitude by

z=r1′r2e2ikL.z=r_1'r_2e^{2ikL}.

Sum the transmitted paths and state the resonance condition.

Solution

The direct path contributes

t1t2eikL.t_1t_2e^{ikL}.

Paths with nn round trips contribute an additional factor znz^n, so

ttot=t1t2eikL∑n=0∞zn=t1t2eikL1−r1′r2e2ikL,t_{\mathrm{tot}} = t_1t_2e^{ikL} \sum_{n=0}^{\infty}z^n = \frac{ t_1t_2e^{ikL} }{ 1-r_1'r_2e^{2ikL} },

provided ∣z∣<1|z|<1 on the physical real-energy axis. Transmission is resonantly enhanced when the denominator is small:

2kL+arg⁡(r1′r2)≈2πn,2kL+\arg(r_1'r_2) \approx 2\pi n,

with ∣r1′r2∣|r_1'r_2| close to one.

6. Why the leading WKB probability misses a resonance

Section titled “6. Why the leading WKB probability misses a resonance”

Explain why using

Tj≈e−2Kj/ℏT_j\approx e^{-2K_j/\hbar}

for each of two barriers and then multiplying T1T2T_1T_2 cannot predict unit resonant transmission.

Solution

Each TjT_j contains only a probability and therefore discards the complex reflection and transmission phases. A double barrier admits infinitely many paths that differ by round trips in the middle region. Those amplitudes must be added before squaring:

ttot∝e−(K1+K2)/ℏeiΦ1−r1′r2e2iΦ.t_{\mathrm{tot}} \propto \frac{ e^{-(K_1+K_2)/\hbar} e^{i\Phi} }{ 1-r_1'r_2e^{2i\Phi} }.

Near a resonance, the denominator can compensate for the small numerator and produce large or unit transmission in a symmetric lossless structure. Multiplying T1T2T_1T_2 keeps only the direct path and cannot represent that interference. WKB can still estimate the leakage widths if its complex amplitudes and connection phases are retained.

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