Scanning Tunneling Microscopy/Spectroscopy

Scanning Tunneling Microscopy/Spectroscopy 0

Scanning Tunneling Microscope 1

Scanning Tunneling Microscope 2

Scanning Tunneling Microscope 3

Typical STM talk or paper... The differential conductance di/dv is proportional to local density of states (LDOS) of the sample...... up to a matrix element. however... Later when we look into CDWs, high-t c superconductivity, topological insulators, heavy-fermion states, interference of quantum mechanical electronic states will play a key role in understanding those measurements. And the statement above will be inadequate. 4

Today The basic concept of tunneling. Tunneling Bardeen Formula Bardeen s approximation of the matrix element (HW). Tersoff-Hamman Projecting out the tip. Quick overview of the basic instrumentation. Typical modes of operation with examples: Constant current imaging. Differential conductance: local density of states. Conductance mapping. 5

Tunneling Basic principles 6

Calculating the transmission probability through a potential barrier in 1D is a well known text book problem... Transmission Coefficient T 16k. / κ / / k. / κ / / / e3/4 56 V = φ Tunneling decays exponentially with barrier length E φ ik / = κ / = 2mφ ħ / φ is the work function. V = 0 x = 0 x = a 7

How sharp must the tip be to see atoms? Tip Consider a very blunt tip, with a lucky atom hanging from the end. For most transition metals φ 5eV so κ / 1A 3.. Suppose a W tip, then we approximate the extra separation to tunnel into the body of the tip versus the hanging atom to one lattice constant Δz 3.2 A. Then the suppression in tunneling is e 3/4 5DE 0.0017. Not the entire story, see HW for more... 8

Bardeen s tunneling. 9

Bardeen s tunneling formula Tunneling current as a function of bias differential between the tip and the sample (ev). I(V) = 2πe ħ K f(e M) 1 f E N + ev M MN / δ(e M E N ) M,N Bardeen s Matrix element contains all the relevant tunneling information. M MN = ħ/ 2m S ds V ψ M ψ N ψ N ψ M Homework! John Bardeen S is a closed surface separating the tip and the sample. ψ N,M are tip or sample states (more precisely defined later). 10

Set-up of the problem Assumptions: 1. Perturbation theory is applicable. 2. Tip and sample states are nearly orthogonal. 3. Ignore electron-electron interactions. 4. Tip and sample are always in electrochemical equilibrium. Using A3 we can assume a single-particle Hamiltonian for the STM system: Hψ = ħ/ 2m / + V ψ When isolated from each other, we can define tip (φ) and sample (χ) states. H N φ = ħ/ 2m / + V N φ H M χ = ħ/ 2m / + V M χ We are using a simpler notation for now but ψ N,M = φ, χ. 11

Strategy for deriving Bardeen s tunneling formula 1. Work in time-dependent perturbation theory (1 st order) to obtain probability of tunneling through vacuum, when systems are near each other. I(V) = 2πe ħ K f(e M) 1 f E N + ev M MN / δ(e M E N ) M,N 2. Project out unknown potentials (V M and V N ) to obtain the simple expression for M MN. M MN = ħ/ 2m S ds V ψ M ψ N ψ N ψ M 12

Suppose that at t = 0, we have ψ t = 0 = χ. Then in t > 0 : ψ t = χ e 3^N_` ħ + K a b (t)φ b And φ b are the bound states of the tip: b reminder: ψ t solves the STM SE. H N φ b = E b φ b Assuming near orthogonality between the tip and sample eigenstates : Tells you the probability that a sample state φ b χ 0 φ b ψ t = a b (t) becomes a tip state. To learn about a b (t), we plug it into the SE of the STM system: Hψ r, t = H χ e 3^N_` ħ + K a b t Hφ b b = e 3^N_` ħ H M + (H H M ) χ + K a b t (H N + (H H N ))φ b b = e 3^N_` ħ E M χ + e 3^N_` ħ H H M χ + K a b t (E b φ b + H H N φ b ) And the time derivative is: b iħ t ψ r, t = E M χ e 3^N_` ħ + iħ K d dt a b t φ b b 13

Equate the two sides of the SE to obtain: iħ K d dt a b t φ b b = e 3^N_` ħ H H M χ + K a b t (E b φ b + H H N φ b ) b As usual, take the inner product with φ f to solve for a b t iħ d dt a f t = e 3^N_` ħ φ f H H M χ + E f a f t + K a b t φ f H H N φ b Keeping only first order terms in the perturbation... b iħ d dt a f t = e 3^N_` ħ φ f H H M χ + E f a f t In the privacy of your home you can verify the solution, for a f t = 0 = 0: a f t = e3^n_` ħ e 3^N_ g ħ E M E f φ f H H M χ 14

Taking the modulus squared, and summing over all tip states we obtain the transition probability: K a f t / f = K 4 sin / E f E M t 2ħ f φ f H H M χ / E f E M / The RHS can be written as: K 4 f t 2ħ sin/ t E f E M 2ħ φ f H H M χ / (t/2ħ) E f E M / The summation over states can also be a summation over energy [we will do this step later]. As long as φ f H H M χ / is approximately constant in the E M /ħ < E f < E M + /ħ interval we can take the limit of N _ g3_` 1. A typical STM N N current is 10pA. If we take the conservative estimate that only one electron is responsible for the current, the energy interval for which the matrix element must be constant is approximately 40neV. A high STM current is 10nA, requiring constancy over 40 μev, still very small. ħ 15

A well-known identity is: sin / αt lim N s α / = π δ(α) t So, in the limit when N _ g3_` /ħ K a f t / = K 2πt ħ f f 1 φ f H H M χ / δ(e f E M ) To calculate the tunneling current, we want to obtain the transition rate: d dt K a f t / f = K 2π ħ φ f H H M χ / δ(e f E M ) f A few steps still needed... 1. Sum over the sample states. 2. Introduce the Fermi-Dirac distribution to account for the probability that an electron at E M is occupied and can tunnel to an electron state of the tip that is unoccupied at E f. 3. Multiply the expression by the electron charge to obtain a physical current. 4. Switch indices to s and t. 5. Introduce a bias voltage between tip and sample: E M E M + ev. Note that at ev = 0 the sample and the tip share the same ground. Their chemical potentials are equal. 16

If you do all those little bookkeeping steps: I M N (V) = 2πe ħ K f(e M) 1 f E N M,N M MN / δ(e N E M ev) The derivations has been tip-sample symmetric so far, so: I N M (V) = 2πe ħ K f(e N) 1 f E M M,N M MN / δ(e N E M ev) The difference between the two currents is (after a couple of variable changes): I(V) = 2πe ħ K f E M [1 f E N + ev ] M MN (E M, E N + ev) / δ(e N E M ) M,N 17

I(V) = 2πe ħ K f E M [1 f E N + ev ] M MN (E M, E N + ev) / δ(e N E M ) M,N Measured Response = Sample X Probe 18

So what about the density of states... I(V) = 2πe ħ K f E M [1 f E N + ev ] M MN / δ(e N E M ) M,N Turn the summation to an integral over energy: ρ M,N E M,N de M,N Where ρ M,N E M,N I(V) = 2πe ħ Typically the following is done... M,N s 3s is the density of states per energy. After a little manipulation... s S deρ M E ρ N E + ev 3s And M / is a strange looking object: f E M [1 f E N + ev ] M / M / / K φ N H H M χ M W(s, t) M,N Where W = 1 whenever E N = E M + ev, and W = 0 otherwise. 19

Typically the following is done... continued I(V) = 2πe ħ s S deρ M E ρ N E + ev 3s f E [1 f E + ev ] M / The Fermi-Dirac functions limit the range of the integral, and in the low-temperature limit: I(V) = 2πe } S deρ ħ M E ρ N E + ev M / f E 3~ f E [1 f E + ev ev E 1 f E + ev ev E E 20

Typically the following is done... continued I(V) = 2πe ħ s S deρ M E ρ N E + ev 3s f E [1 f E + ev ] M / The Fermi-Dirac functions limit the range of the integral, and in the low-temperature limit: I(V) = 2πe } S deρ ħ M E ρ N E + ev M / 3~ ρ M E V = 0 V = V } V = V } dv E ρ N E ev } 21

Typically the following is done... continued I(V) = 2πe ħ } S deρ M E ρ N E + ev 3~ Formally, one can take the derivative w.r.t. V and assume M, ρ N constant over E: di 2πe/ (V) = dv ħ ρ M ev ρ N 0 M / M / ρ M E V = 0 V = V } V = V } dv E ρ N E ev } 22

Typical STM talk or paper... The differential conductance di/dv is proportional to local density of states of the sample...... up to a matrix element. di dv (V) = 2πe/ ħ ρ M ev ρ N 0 M / di dv V ~ ρ M ev 23

Let s contemplate what just happened We started with this: I(V) = 2πe ħ K f E M [1 f E N + ev ] M MN (E M, E N + ev) / δ(e N E M ) M,N Any information about the sample was inside this: M MN = φ N H H M χ M Instead of doing something to project χ M out of M MN (to obtain information about the sample), we changed the summation variable from the indices s, t to energy E and completely ignored the matrix element. Eventually we got this: di dv V ~ ρ M ev 24

Where is the spatial resolution??? di dv V ~ ρ M ev It is implicit that there is an energy-independent, but spatially dependent matrix element multiplying the sample s density of state. So that at each position of the tip on top of the sample we have: di dv V, r ~ M r ρ M ev di dv V, r ~ ρ M ev, r 25

Bardeen s approximation for the matrix element. Goal = Simplify this formula. M MN = φ N H H M χ M T 1 M MN = S dr φ N r H H M χ M (r) T 2 Bardeen s approximation is to separate the integral over all space in 2 volumes. T 1 : The space where H H N = 0 T 2 : The space where H H M = 0... (details in the homework). M MN = ħ/ 2m S ds V ψ M ψ N ψ N ψ M S is an arbitrary surface separating T 1 and T 2. 26

The Tersoff-Hamann Approximation PRB 31, 805 (1985) 27

T-H basic assumptions M MN = ħ/ 2m S ds V ψ M ψ N ψ N ψ M How to get the density of states out of the expression above? Propose a specific model for the tip, tailored to factorize the tip states out of the problem. In T-H the tip is modeled as a spherical object with characteristic radius R at a distance d from the sample surface. The sample states are modeled as Bloch states with a characteristic decay into the vacuum. 28

Tersoff-Hamman continued... Sample states: Periodic Bloch wave functions. Exponential decay into the vacuum. Tip states: 29

Tersoff-Hamman continued... Plugging in the (non-general) tip states and the sample states into the Bardeen matrix element allowed T-H to obtain the following formula. di dv V, r 0.1R/ e /4 ρ M E = ev, r Measured Response = Probe x Sample ρ M E, r = K χ M r / δ(e M E) M LOCAL DENSITY OF STATES 30

STM instrumentation How all this is actually done. 31

The piezoelectric effect E.g. Quartz (SiO 2 ) Si O 32

The piezoelectric effect E.g. Quartz (SiO 2 ) Dipole moment induced by pressure. Si O 33

The piezoelectric effect E.g. Quartz (SiO 2 ) Dipole moment induced by pressure. Therefore, the pressure induces surface charge and voltage. 34

The piezoelectric effect E.g. Quartz (SiO 2 ) Dipole moment induced by pressure. Therefore, the pressure induces surface charge and voltage. The converse is true. If one applies a voltage to the surface, the material will stretch/compress. This is the piezoelectric effect. 35

The tip tube scanner Four disconnected outer electrodes. One whole inner electrode. Image from Introduction to Scanning Tunneling Microscopy by C. Julian Chen, 2 nd edition. 36

The tip tube scanner Lateral motion is achieved by applying a voltage across a horizontal direction. Image from Introduction to Scanning Tunneling Microscopy by C. Julian Chen, 2 nd edition. 37

The tip tube scanner Vertical motion is achieved by applying a voltage between inner and outer electrodes. Image from Introduction to Scanning Tunneling Microscopy by C. Julian Chen, 2 nd edition. 38

Scanning control Most imaging measurements are performed in a constant current condition. Tip Bias V Vacuum Surface As the tip moves laterally, the height of the tip is adjusted to maintain a constant current. The height profile is recorded. The control is achieved using a PID negative feedback control of the current, which adjusts the voltages applied to the piezo 39

Obtaining clean surfaces for STM measurements. STM operation requires Ultra High Vacuum (~ 10-10 Torr) Surfaces need to be prepared in-situ. For example by ion sputtering. 40

Obtaining clean surfaces for STM measurements. STM operation requires Ultra High Vacuum (~ 10-10 Torr) Surfaces need to be prepared in-situ. For example by cleaving the sample. 41

Example: constant current imaging in CeCoIn 5. 100Å The record of the height profile in a constant current image is often called a topograph, topography, or topographic image. Images from Aynajian, da Silva Neto, et al. Nature (2012), JPSJ (2014). 42

Example: constant current imaging in CeCoIn 5. Surface A Surface B pm pm pm x 1 x 1 6 6 5 0.5 4 x 1 6 Surface C 4 4 4 2 2 2 0 0 0 0 0 0-2 -2-2 -4-4 -4-5 -6 45 Å x 45 Å -0.5-6 -6-4 45 Å x 45 Å 75 Å x 75 Å Some materials cleave on multiple chemical terminations. What is happening on surface C? Images from Aynajian, da Silva Neto, et al. Nature (2012), JPSJ (2014). 43

Example: constant current imaging in CeCoIn 5. A C A C A B B 50 Å Often, because it is difficult to visualize surfaces at multiple heights, the derivative of the topography is presented. The height profile of the different layers from the topography is different than the information from X-ray diffraction. WHY? Images from Aynajian, da Silva Neto, et al. Nature (2012), JPSJ (2014). 44

Measuring di/dv E.g. Gap in BCS Superconductor I(V) V Two options: 1. At a constant height (feedback off) Measure I(V) and numerically take the derivative. 2. Use a Lock-in Amplifier (more common, better data). di dv 45 V

di/dv Mapping 1. Determine a grid (e.g. 256x256 pixels). 2. Go to the first point. 3. Turn off the feedback control and measure di/dv. 4. Turn on the feedback and go to the second point. 5. Go to step 3. V = 80 mev V = -20 mev V = -120meV c b a This is an example of a quasiparticle interference measurement (QPI) in Cu 111. This kind of measurements have a direct connection to ARPES measurements. Yazdani, da Silva Neto & Aynajian, Ann. Rev. Cond Mat. Phys. (2016) 46

Take-away message. The differential conductance di/dv is proportional to local density of states of the sample...... up to a matrix element. but remember... TUNNELING IS INTERFERENCE χ M6 ~ φ / / / N^ χm6 ~ φn^ 47