University Physics V · Finite Potential Wells and Quantum Tunneling · 6.9
Scanning Tunnelling Microscopy & the LDOS
An STM is one tunnel barrier with a knob on it. Everything hard about the instrument follows from a single exponent, and everything subtle about its pictures follows from what sits in front of that exponent. Keep the two apart: the exponent buys the sensitivity, the prefactor decides what a bright spot means.
Build the model
Connect the measurement to the mechanism.
An STM is a tunnel junction whose barrier width you set to a picometre, and two facts organise the whole subject. First the exponent. A sample state at the Fermi level leaks into vacuum as exp(−κz) with κ = √(2mΦ)/ħ, so the current carries the square, I ∝ exp(−2κz); for a metal work function near 4 eV that is κ = 1.02 Å⁻¹ and a factor of 7.8 — nine tenths of a decade — per angstrom of gap. Every claim about vertical sensitivity is that number and nothing else.
Second the prefactor. Bardeen's transfer Hamiltonian refuses to solve the joint problem: it diagonalises tip and sample apart and treats the overlap of their vacuum tails as a first-order perturbation, giving I ∝ ∫ dε ρₜ ρₛ |M|² with M a surface integral of a current-like operator taken anywhere in the gap. Tersoff and Hamann then model the apex as one s-wave orbital, which collapses |M|² onto |ψᵥ(r₀)|² and leaves I ∝ V ρₜ(EF) ρₛ(r₀, EF). A constant-current image is therefore a contour of local density of states at the Fermi level, evaluated several angstroms out in the vacuum — not a map of nuclei, not even of total charge.
The cost is a stack of assumptions: weak coupling, elastic tunnelling, a structureless tip, small bias, and a barrier the tip's own presence does not distort. Each fails somewhere, and those failures are the interesting part.
- Simple definition
- In an STM a sharp metal tip is held a few angstroms above a conducting surface, and the tunnel current, which falls off as exp(−2κz), is used either as the signal or as the feedback variable that fixes the tip's height.
- Example
- With Φ = 4.0 eV, κ = √(2mΦ)/ħ = 1.02 Å⁻¹ and 2κ = 2.05 Å⁻¹, so retracting the tip by 1.00 Å cuts the current by exp(2.05) = 7.8, and a 1 pm ripple in height is a 2.0 % ripple in current.
Φ = 4.0 eV gives κ = 1.02 Å⁻¹, so the current falls 7.8× for every angstrom the gap opens.
Φ is the apparent barrier height in eV, roughly the mean of the two work functions; κ in Å⁻¹
The slope of ln I against z is the only routine barrier-height probe the instrument has.
ħ²/8m = 0.9525 eV Ų, so a log-slope of −2.05 Å⁻¹ returns Φₐₚₚ = 4.00 eV
First order in the tip–sample coupling: the two electrodes are diagonalised apart, then joined.
ρ in states per eV per atom, M in eV; valid at T → 0, weak coupling and elastic transfer
A probability-current overlap, so it inherits the exp(−κz) tail of each electrode.
S is any surface lying wholly inside the vacuum gap; the value does not depend on which one
Every tip property leaves the lateral dependence, so constant current is a constant-LDOS contour.
R is the apex radius and r₀ its centre of curvature, a distance R + z from the surface plane
Fine detail decays faster than the mean current, so the image is the LDOS through a Gaussian low-pass.
G = 2π/a is the surface reciprocal-lattice vector; z is measured from the surface plane
One exponent sets every number in the instrument
Ask what a sample state at EF does in the gap. There the Hamiltonian is p²/2m + V₀ with V₀ − EF = Φ, so the stationary equation reads ψ″ = κ²ψ with κ = √(2mΦ)/ħ, and the admissible solution decays: ψ ∝ e(−κz). The current is a probability current built from two such tails, one from each electrode, so it carries the square: I ∝ e(−2κz). Numerically κ = 0.5123 √(Φ/eV) Å⁻¹, so a 4.0 eV barrier gives κ = 1.02 Å⁻¹ and 2κ = 2.05 Å⁻¹ — the current changes by e2.05 = 7.8, or 0.89 of a decade, per angstrom. That single number is the entire vertical performance of the microscope. A loop holding the current to 1 % holds the height to δz = 0.01/2κ = 5 × 10⁻³ Å, half a picometre, across a junction whose resistance at 1 nA and 1 V is 1 GΩ. Nothing about lateral resolution follows from it; that is a separate and much weaker argument.
Bardeen: diagonalise the electrodes apart, then couple them
You cannot solve the tip-plus-sample Hamiltonian, and Bardeen's move is not to try. Take Hₜ and Hₛ separately, each with a complete set of eigenstates, and treat the overlap of their tails as a perturbation transferring an electron from |μ⟩ in the tip to |ν⟩ in the sample. Fermi's golden rule gives a rate ∝ |Mμν|² δ(Eμ − Eν), and summing over the window the bias opens gives I = (4πe/ħ) ∫₀(eV) ρₜ(EF − eV + ε) ρₛ(EF + ε)|M|² dε at low temperature. The matrix element is not a potential integral but a surface integral of a current-like operator, M = (ħ²/2m)∫S(ψμ*∇ψν − ψν∇ψμ*)⋅dS, over any surface lying wholly inside the gap; that the answer does not depend on which surface is the statement that no current is lost in the vacuum. Three assumptions are now fixed and stay fixed: the coupling is weak enough for first order, the transfer is elastic, and each electrode is the one you solved, unperturbed by the other.
Tersoff and Hamann make the tip vanish into a prefactor
Bardeen leaves M depending on the tip you happen to have, which is the one thing an experiment does not know. Tersoff and Hamann close that gap by modelling the apex as a single s-wave orbital of radius R centred at r₀, so ψμ in the vacuum is a spherical decaying wave. Evaluating the surface integral with that choice collapses |M|² onto |ψᵥ(r₀)|², and the current becomes I ∝ V ρₜ(EF) R² e(2κR) Σᵥ |ψᵥ(r₀)|² δ(Eᵥ − EF). Every tip property is now a multiplicative constant, and the only laterally varying factor is the sample's local density of states at the Fermi level evaluated at the tip's centre of curvature, ρₛ(r₀, EF). Holding I fixed while scanning therefore traces a surface of constant ρₛ(r₀, EF), at a point R + z out in the vacuum where that quantity is exponentially small and keeps no memory of where the nuclei sit.
Reading heights: an LDOS ratio is an apparent height
Constant current means ρₛ(r₀, EF) e(−2κz) is held fixed, so between two sites the tip moves by Δz = ln(ρ₁/ρ₂)/2κ and by nothing else. At Φ = 4.0 eV that turns a factor of 2 in LDOS into 34 pm of apparent height, a factor of 3 into 54 pm, a factor of 10 into 112 pm. None of it is geometry, and the consequences are routine and counter-intuitive. CO on Cu(111) carries its carbon roughly 1.9 Å above the metal plane and still images as a depression at small bias, because it pushes the metal's Fermi-level LDOS down; oxygen on Ni and on Al does the same. Subsurface impurities several layers down can appear, because they modulate ρₛ(EF) in the vacuum above them. Change the bias and the picture changes with it, since the integration window changes — which is the basis of tunnelling spectroscopy, where dI/dV at fixed z tracks ρₛ(EF + eV) up to a transmission factor usually divided out as (dI/dV)/(I/V).
Lateral resolution is a Gaussian filter — graphite breaks it
Decompose the surface LDOS into lateral Fourier components. A component with in-plane wavevector G must satisfy the same vacuum equation, so it decays not with κ but with κG = √(κ² + G²); relative to the mean it is suppressed by e(−(κ_G − κ)z), which for small G is e(−G²z/2κ). That is a Gaussian low-pass, and the image is the surface LDOS convolved with a real-space Gaussian of width σ = √((R + z)/κ). With R = 5 Å, z = 5 Å and κ = 1.02 Å⁻¹ that is σ = 3.1 Å, wider than the lattice constant of any close-packed metal — Tersoff–Hamann says you should not resolve atoms on Al(111) at all. On graphite a = 2.46 Å gives G = 2.55 Å⁻¹ and κG − κ = 1.69 Å⁻¹, so a 5 Å gap suppresses the corrugation by 2.2 × 10⁻⁴ and the model predicts about 0.012 pm. Experiments see 0.5 to 3 Å. The exponential is not what fails; the s-wave tip is.
The five assumptions and where each one dies
Weak coupling: below about 100 kΩ of junction resistance the electrodes hybridise, first order loses its small parameter, and the measured d ln I/dz collapses from 2κ towards zero — the apparent barrier height falls from 4 eV to a fraction of an eV as the gap closes past about 3 Å. Elastic transfer: inelastic channels open at thresholds set by vibrational modes, and the kinks in d²I/dV² are the signal in inelastic tunnelling spectroscopy. A structureless tip: real apices carry their own resonances, and a tip state at EF prints itself onto every image. Small bias: at 1 V and above the barrier is no longer rectangular and ρₜ can no longer be pulled outside the integral. No forces: chemical and van der Waals forces between apex and surface reach the nanonewton scale and elastically relax both electrodes, which is what AFM measures and STM ignores. The instrument still works; what changes is what the picture is a picture of.
Change one variable at a time
Make the relationship visible.
Hold Φ at 4 eV and a at 9 Å, then drag the gap from 3 Å to 9 Å: κ never changes, yet the trace shrinks from 32 pm to 8 pm peak to peak. Now pull a down to 3 Å and it dies to nothing — the corrugated component decays with √(κ² + G²), not with κ.
DECAY CONSTANT κ1.025 Å⁻¹
CURRENT DROP PER Å7.76 ×
SURVIVING MODULATION u0.2536
CORRUGATION PEAK-PEAK25.31 pm
Live interpretationDECAY CONSTANT κ: 1.025 Å⁻¹. CURRENT DROP PER Å: 7.76 ×. SURVIVING MODULATION u: 0.2536. CORRUGATION PEAK-PEAK: 25.31 pm
Catch the common trap
Explain before calculating.
An STM scanning Cu(111) at constant current and small bias crosses a CO molecule, and the feedback lowers the tip by 30 pm. Take Φ = 4.0 eV, so κ = 1.02 Å⁻¹. What has actually been measured?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn STM tip of work function 4.5 eV scans a gold surface of work function 5.1 eV. Take the apparent barrier height as the mean of the two. Find κ, the factor by which the tunnel current changes when the gap is opened by 1.00 Å, and the height change that corresponds to a 1.0 % change in current.
- Apparent barrier height: Φ = (4.5 + 5.1)/2 = 4.8 eV.
- Decay constant: κ = √(2mΦ)/ħ = 0.5123 √(Φ/eV) Å⁻¹ = 0.5123 × 2.1909 = 1.122 Å⁻¹, so 2κ = 2.245 Å⁻¹.
- Opening the gap by Δz = 1.00 Å multiplies the current by exp(−2κΔz) = exp(−2.245) = 0.106: it falls to 10.6 % of its value, a factor of 9.4.
- For small changes dI/I = −2κ dz, so a 1.0 % change in current is dz = 0.010/2.245 = 4.5 × 10⁻³ Å = 0.45 pm.
Answerκ = 1.12 Å⁻¹; a 1.00 Å retraction cuts the current to 10.6 %, a factor of 9.4; and 1.0 % of current is 0.45 pm of height. Sub-picometre vertical resolution comes free from the exponent.
MediumA tip is stepped over Au(111) at fixed bias and the current recorded: 1.00 nA at z = 5.00 Å, 0.343 nA at 5.50 Å, 0.118 nA at 6.00 Å. Find the apparent barrier height. Then a molecule adsorbs that raises the local Fermi-level LDOS by a factor 2.5 with no change in geometry; find how tall it looks in the constant-current image.
- The gap law I = I₀ e(−2κz) makes ln I linear in z with slope −2κ. End points: ln(0.118/1.00) = −2.137 over Δz = 1.00 Å, so d ln I/dz = −2.14 Å⁻¹. The middle point checks it: ln 0.343 = −1.070 at half the interval.
- Hence κ = 1.070 Å⁻¹.
- Apparent barrier height: Φₐₚₚ = ħ²κ²/2m with ħ²/2m = 3.810 eV Ų, so Φₐₚₚ = 3.810 × 1.070² = 3.810 × 1.1449 = 4.36 eV — consistent with a clean gold junction.
- Apparent height: constant current fixes ρ e(−2κz), so Δz = ln(2.5)/2κ = 0.9163/2.140 = 0.428 Å = 43 pm.
AnswerΦₐₚₚ = 4.36 eV, and the molecule images 43 pm high although nothing moved geometrically — the whole apparent relief is the factor 2.5 in Fermi-level LDOS.
HardGraphite has a surface period a = 2.46 Å and an apparent barrier height Φ = 4.5 eV. Take the LDOS corrugation in the surface plane as 60 % of the mean. Working in the Tersoff–Hamann limit, find (a) κ and the decay constant κG of the first lateral Fourier component, (b) the fraction of the corrugation surviving a 5.0 Å gap, (c) the peak-to-peak apparent corrugation, and (d) compare it with the 0.5–3 Å routinely measured.
- κ = 0.5123 √4.5 = 0.5123 × 2.1213 = 1.087 Å⁻¹, and the first reciprocal-lattice vector is G = 2π/a = 2π/2.46 = 2.554 Å⁻¹.
- A lateral component G still obeys ψ″ = (κ² + G²)ψ in the vacuum, so it decays with κG = √(1.087² + 2.554²) = √(1.181 + 6.524) = √7.705 = 2.776 Å⁻¹.
- Relative to the mean current the corrugated part is suppressed by exp(−(κG − κ)z) = exp(−1.689 × 5.0) = exp(−8.445) = 2.15 × 10⁻⁴, so u = 0.60 × 2.15 × 10⁻⁴ = 1.29 × 10⁻⁴.
- Apparent corrugation: Δzₚₚ = ln[(1 + u)/(1 − u)]/2κ ≈ u/κ = 1.29 × 10⁻⁴/1.087 = 1.19 × 10⁻⁴ Å = 0.012 pm.
- Measured corrugations reach 0.5–3 Å, four to five orders of magnitude larger. The barrier is not in doubt; the s-wave tip is. A pz or dz² apex reweights the high-G components (Chen's derivative rule), and at these gaps forces of order 1 nN elastically deform both electrodes.
Answerκ = 1.087 Å⁻¹ and κG = 2.776 Å⁻¹; only 2.2 × 10⁻⁴ of the modulation survives 5.0 Å, giving 0.012 pm peak to peak against 0.5–3 Å measured. The s-wave tip is the assumption that fails.