
Quantum core · institution-dependent · Interactive lesson
Scanning tunnelling microscopy and 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.
UPV-06.09 · Lesson 9 of 9 in this unit. Position is not a completion score.Quantum core · institution-dependent. Your university determines the required depth of these quantum topics, including the use of operators, differential equations and computation. Institution-dependent does not automatically mean optional.
Build the model
A tunnel junction you can drag across a surface.
The key idea
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.
The complete physical model and assumptions
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.
1One 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.
2Bardeen: 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.
3Tersoff 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 only a smoothed, indirect memory of where the nuclei sit.
4Reading heights: an LDOS ratio is an apparent height
Constant current means ρₛ(r₀, EF) is held fixed, and that is ρ e(−2κz) with ρ the Fermi-level LDOS referred to the surface, 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).
5Lateral resolution is a Gaussian filter — graphite breaks it
Decompose each Fermi-level sample state into lateral Fourier components. A component with in-plane wavevector G obeys the same vacuum equation, so its amplitude decays not with κ but with κG = √(κ² + G²). For a state near k∥ = 0 the LDOS corrugation is the cross term of that component with the uniform one, falling as exp[−(κ + κG)z] against e(−2κz) for the mean, so relative to the mean it is suppressed by exp[−(κG − κ)z], which for small G is e(−G²z/2κ). That is a Gaussian low-pass, and with the LDOS sampled at the tip centre, R + z out, 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 2.6–2.9 Å atomic spacing of close-packed Cu, Al or Au — Tersoff–Hamann says you should not resolve atoms on Al(111) at all. They are resolved, and there the s-wave tip is what fails: a pz or dz² apex reweights the high-G components (Chen's derivative rule). Graphite fails differently. Its hexagonal lattice, a = 2.46 Å, has G = 4π/(√3 a) = 2.95 Å⁻¹; at Φ = 4.5 eV, κG − κ = 2.06 Å⁻¹, so even a point tip 5 Å out sees the corrugation suppressed by 3.4 × 10⁻⁵, and a 60 % surface modulation becomes 0.002 pm. Experiments see 0.5 to 3 Å. The exponential is not what fails; the k∥ = 0 assumption is. Graphite's Fermi-level states sit at the zone corner, where the three leading Fourier components, one at each equivalent corner, have the same |k∥| and so the same decay constant, and the corrugation they make does not decay at all.
6The 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, each adding a step to dI/dV, and the peaks those steps make 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.
Key equations
Read the units, assumptions and limits before substituting.
Vacuum decay constant
κ = √(2mΦ)/ħ = 0.5123 √(Φ/eV) Å⁻¹
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⁻¹Gap law and apparent barrier height
I(z) = I₀ e(−2κz)Φₐₚₚ = (ħ²/8m)(d ln I/dz)²
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 eVBardeen tunnel current
I = (4πe/ħ) ∫₀(eV) ρₜ(EF − eV + ε) ρₛ(EF + ε) |M|² dε
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 transferBardeen matrix element
Mμν = (ħ²/2m) ∫S (ψμ* ∇ψν − ψν ∇ψμ*) ⋅ dS
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 oneTersoff–Hamann s-wave limit
I ∝ V ρₜ(EF) R² e(2κR) ρₛ(r₀, EF)ρₛ(r₀, E) = Σᵥ |ψᵥ(r₀)|² δ(Eᵥ − E)
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 planeCorrugation decay and lateral filtering
Δz(x) = ln[1 + u cos(Gx)]/2κu ∝ exp[−(κG − κ)z], κG = √(κ² + G²)
Fine detail decays faster than the mean current, so the image is the LDOS through a Gaussian low-pass.
G = 2π/a for a one-dimensional period a, 4π/(√3 a) for a hexagonal lattice; z is the height of the tip centre r₀, the gap itself for a point tip; assumes Fermi-level states near k∥ = 0Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. 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.
Live results
- DECAY CONSTANT κ
- 1.025 Å⁻¹
- CURRENT DROP PER Å
- 7.76 ×
- SURVIVING MODULATION u
- 0.2536
- CORRUGATION PEAK-PEAK
- 25.31 pm
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Barrier height Φ: 4.00 eV; Tip–sample gap z: 4.00 Å; Surface period a: 9.0 Å
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The bright spots are not atoms, and they are not the charge density either. A constant-current image is a contour of the sample's local density of states at the Fermi level, sampled several angstroms out in the vacuum and smoothed by the tip's own decay length, so the only thing a height difference measures is a ratio of Fermi-level LDOS: Δz = ln(ρ₁/ρ₂)/2κ. At Φ = 4.0 eV a factor of three in LDOS makes 54 pm of apparent relief with no geometry behind it, and the sign can run the wrong way — CO on Cu(111) stands 1.9 Å proud of the metal and images as a hole, because it depletes ρₛ(EF) above itself. Change the bias and the same surface can invert. Quote apparent heights as apparent, name the bias, and never read a topograph as a map of nuclei.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
An 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.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- 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.
Answer. κ = 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.
MediumExample 2
A 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.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- 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 = −2.137 Å⁻¹. The middle point checks it: ln 0.343 = −1.070 at half the interval.
- Hence κ = 2.137/2 = 1.069 Å⁻¹.
- Apparent barrier height: Φₐₚₚ = ħ²κ²/2m with ħ²/2m = 3.810 eV Ų, so Φₐₚₚ = 3.810 × 1.069² = 3.810 × 1.143 = 4.35 eV — consistent with a clean gold junction.
- Apparent height: constant current fixes ρ e(−2κz), so Δz = ln(2.5)/2κ = 0.9163/2.137 = 0.429 Å = 43 pm.
Answer. Φₐₚₚ = 4.35 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.
HardExample 3
Graphite's surface lattice is hexagonal with a = 2.46 Å, and the apparent barrier height is Φ = 4.5 eV. Take the LDOS corrugation in the surface plane as 60 % of the mean. Working in the Tersoff–Hamann limit with a point s-wave tip, and assuming as for a free-electron metal that the Fermi-level states sit near k∥ = 0, find (a) κ and the decay constant κG of the first-star lateral Fourier component, (b) the fraction of the corrugation surviving 5.0 Å from the surface, (c) the peak-to-peak apparent corrugation, and (d) compare it with the 0.5–3 Å routinely measured and name the assumption that fails.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- κ = 0.5123 √4.5 = 0.5123 × 2.1213 = 1.087 Å⁻¹. The shortest reciprocal-lattice vectors of a hexagonal lattice have G = 4π/(√3 a) = 4π/(√3 × 2.46) = 2.949 Å⁻¹; the one-dimensional 2π/a = 2.554 Å⁻¹ would understate it.
- A lateral component G of a k∥ = 0 state still obeys ψ″ = (κ² + G²)ψ in the vacuum, so it decays with κG = √(κ² + G²) = √(1.181 + 8.698) = √9.879 = 3.143 Å⁻¹.
- In |ψ|² the corrugation is the cross term of that component with the uniform one, falling as exp[−(κ + κG)z] against e(−2κz) for the mean, so it is suppressed by exp(−(κG − κ)z) = exp(−2.056 × 5.0) = exp(−10.28) = 3.43 × 10⁻⁵, and u = 0.60 × 3.43 × 10⁻⁵ = 2.06 × 10⁻⁵.
- Apparent corrugation: Δzₚₚ = ln[(1 + u)/(1 − u)]/2κ ≈ u/κ = 2.06 × 10⁻⁵/1.087 = 1.9 × 10⁻⁵ Å = 0.0019 pm.
- Measured corrugations of 0.5–3 Å are four to five orders of magnitude larger. The exponential is not in doubt; the k∥ = 0 assumption is. Graphite's Fermi-level states sit at the zone corner K, and the three equivalent corners have the same |k∥|, hence one decay constant, so the corrugation of |ψ|² does not decay relative to the mean at all (Tersoff's resolution of the puzzle). Tip–surface forces of order 1 nN at these gaps then amplify it by deforming the soft layer.
Answer. κ = 1.087 Å⁻¹ and κG = 3.143 Å⁻¹; only 3.4 × 10⁻⁵ of the modulation survives 5.0 Å, giving 0.0019 pm peak to peak against 0.5–3 Å measured. The failing assumption is k∥ = 0: graphite's Fermi-level states sit at the zone corner, where the corrugation does not decay.
Quick check
Select an answer and explain it before checking.
Review & continue
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.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.