Introduction to nanotechnology

kα is the velocity of an electron in the state ψkα. It is now easy to derive an expression for the current IL→R running from the left reservoir L into the right reservoir R through the nanostructure. We consider the energy 74 CHAPTER 6. TRANSPORT IN NANOSTRUCTURES ε = εkα and multiply three factors: (1) the Fermi-Dirac probability n L F(ε) that the right- moving state ψkα is occupied from the left reservoir; (2) the current Ikα carried by that state; and (3) the transmission probability Tα(ε) of an electron at energy ε making it through to the reservoir to the right. Then we sum over all possible electron channels α and wavenumbers k: IL→R = ∑ α ∑ k [ nLF(ε) ]× [− 2 e L vkα ]× [Tα(ε)]. (6.51) First we convert the k-sum into a k-integral by use of Eq. (3.10), which in 1D reads∑ k → (L/2π) ∫ dk. Then we convert the k-integral to an ε-integral: ∫ dk → ∫ dε ∂k∂ε . But from Eq. (2.8b) we know that ∂k∂ε = 1
v . So collecting all this yields ∑ k → (L/2π) ∫ dε 1
v and Eq. (6.51) becomes IL→R = − 2e h ∑ α ∫ ∞ −∞ dε nLF(ε) Tα(ε). (6.52) Note how all references to geometry, i.e. L, and velocity have vanished from the expression. The current IR→L flowing from the right to the left is obtained in a similar way, IR→L = − 2e h ∑ α ∫ ∞ −∞ dε nRF (ε) Tα(ε). (6.53) The total current I flowing in the system is therefore I = IL→R − IR→L = − 2e h ∑ α ∫ ∞ −∞ dε Tα(ε) [ nLF(ε) − nRF (ε) ] . (6.54) 6.6.3 The conductance formula for nanostructures We now have a relatively simple expression for the current, so to obtain the conductance we need to find the voltage dependence. In our set-up in Fig. 6.1b the voltage V is applied to the left reservoir. If we restrict ourselves to consider only small voltages (the linear response limit) it follows from Eq. (6.2) that nLF(ε) − nRF (ε) = nF(ε, µ0 − eV )− nF(ε, µ0) ≈ nF ∂µ ∣∣∣∣ µ0 (−eV ) = ( − nF ∂ε ) (−eV ). (6.55) With this Eq. (6.54) becomes I = 2e2 h ∑ α ∫ ∞ −∞ dε Tα(ε) ( − nF ∂ε ) V. (6.56) Finally, we obtain the temperature dependent conductance G(T ), G(T ) = I V = 2e2 h ∑ α ∫ ∞ −∞ dε Tα(ε) ( − nF ∂ε ) . (6.57) 6.7. QUANTIZED CONDUCTANCE 75 In the limit of very low temperature (temperatures around 1 K are routinely obtained in the lab cooling with liquid helium) Eqs. (3.28a) and (3.29) state that − nF∂ε = δ(εF − ε). Inserting this in Eq. (6.56) yields the zero temperature conductance G(T = 0) = 2e2 h ∑ α Tα(εF). (6.58) Eq. (6.57) is the conductance formula for nanostructures. We derived it by consid- ering scattering states, i.e. the scattering matrix formalism. We note that the integral is dimensionless, so that means that the prefactor G0 = e 2/h, depending entirely on universal constants, is some kind of conductance quantum, a natural unit for measuring conductance, G0 = e2 h = 3.87404614 × 10−5 S, R0 = 1 G0 = h e2 = 2.58128056 × 104 kΩ. (6.59) The zero temperature conductance Eq. (6.58) is simply the conductance quantum 2e 2 h times the sum of the transmission coefficients for each channel α evaluated at the Fermi energy set by the metal reservoirs. All information about the nanostructure lies in the transmission coefficient Tα(ε). In the following section we shall see a spectacular consequence of the scattering wave nature of the conductance of nanostructures. 6.7 Quantized conductance The conductance quantum is directly observable in the beautiful experiment shown in Fig. 6.9 based on the quantum point contact depicted in Fig. 6.1a. In the following we give a simple explanation of the observed quantization of the conductance. The quantum point contact is fabricated on a GaAs-GaAlAs heterostructure. As explained in Exercise 3.4 all conduction electrons of this structure are bound to move at the interface between GaAs and GaAlAs, all having the same wavefunction ζ0(z) in the z direction (perpendicular to the interface). By etching techniques a narrow wire is created in the x direction. As can be seen in Fig. 6.1a the width w of the wire in the y direction varies with position, w(x), being of the order 100 nm at the narrowest point. The actual width can be controlled during the experiment by changing the gate voltage Vg on the side electrodes as sketched in Fig. 6.9a. Let us model the potential Vx(y) of the wire at each position x as a simple potential box stretching w/2 on both sides of the x axis: Vx(y) = { 0, for − 12 w(x) < y < 12 w(x), ∞, for |y| > 12 w(x). (6.60) We further imagine that the change of width as we move along the x axis is so slow that a simple product wavefunction fitting to the width at any given position is a good solution 76 CHAPTER 6. TRANSPORT IN NANOSTRUCTURES



















































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































g 0V < g 0V <












































































































































































































































































































































































































































































































































































































































V 0I V =g 0 V =g 0


























































































































































































































































































































































































































































































































































































































V 0I
  Figure 6.9: (a) The principle in measuring the quantized conductance of a point contact. The channel width is controlled during the experiment by tuning the voltage Vg on the gate electrodes. (b) Measurements of the conductance G versus gate voltage Vg at T = 1.3, 11, 20, and 31 K. The curves are displaced vertically for clarity. to the Schro¨dinger equation. This approximation which effectively neglects the derivatives w′(x) and w′′(x) is known af the adiabatic approximation. The explicit wavefunctions are ψεn0(x, y, z) = 1√ L eikεn(x)x √ 2 w(x) sin ( nπ [ y w(x) + 1 2 ]) ζ0(z). (6.61) The wavefunction is fully characterized by the energy ε, the number n of standing half- waves in the wire and the number 0 reminding us that all electrons have the same wave ζ0(z) in the z direction. The index α used prior is here a double index α = n0. The total energy ε of the state ψεn0 can be written as ε = εk + εn(x) + εz =
2kεn(x) 2 2m + π2
2 2m n2 w(x)2 . (6.62) Here we have chosen to put the zero point of the energy scale at the value εz of the quantum energy of ζ0(z), i.e. εz = 0. We can interpret Eq. (6.62) as the energy eqaution for a particle moving along the x-axis with a position dependent kinetic energy εk = 1 2m
2kεn(x)2 and a position dependent potential energy εn(x) given by εn(x) = π2
2 2m n2 w(x)2 . (6.63) As the particle moves towards the narrowest point of the point contact it must convert more and more of its kinetic energy in the forward x direction into potential energy (really kinetic energy in the transverse y direction). To find out whether the particle actually has enough kinetic energy to pass through it is of course now highly relevant to know what 6.7. QUANTIZED CONDUCTANCE 77 is the maximal potential energy εmaxn along the wire. The smallest width of the wire is denoted wmin, then according to Eq. (6.63) we have εmaxn = π2
2 2m n2 w2min . (6.64) The x-dependent wavenumbe kεn(x) is given by kεn(x) = √ 2m
2 ( ε− εn(x) ) . (6.65) We see that in our simple model we obtain full transmission if kεn(x) remains real through- out the passage, i.e. if ε > εmaxn . If however we arrive at ε < εmaxn then kεn(x) becomes imaginary corresponding to tunneling through the effectigve potential barrier εn(x). As a rough model for the transmission coefficient Tn of channel n, i.e. all electrons having a transverse wave consisting of n half-waves, we simply take Tn(ε) = { 1, for ε > εmaxn 0, for ε < εmaxn . (6.66) With the transmission coefficient Eq. (6.66) inserted into Eq. (6.58) we obtain an explanation of the zero temperature conductance of the quantum point contact. The resulting formula is namely G(T = 0, Vg) = 2e2 h ∑ n Tn,Vg(εF), (6.67) where we explicitly write that the conductance must depende on the gate voltage Vg that according to Fig. 6.9a controls the actual minimal width of the wire, i.e. wmin = wmin(Vg). If we initially have a large negative Vg the channel will be pinched off entirely. No electrons can pass and G = 0. As we gradually increase Vg the wire gets wider and wider. At some point electrons in channel n = 1 will be able to pass, since they have the lowest threshold potential εmaxn=1, and the conductance jumps from 0 up to 2e 2/h. As Vg is increased further we reach a point where also the electrons in the n = 2 channel can pass, and the conductance increases from 2e2/h to 4e2/h. This process goes on as we increase Vg more and more, and we thus trace out a conductance curve like the lowest one (T = 1.3 K) in Fig. 6.9. This result is a dramatic deviation from classical physics. The conductance does not increase linearly with decreasing width, but instead it increases in discrete steps of height 2e2/h, the conductance quantum. Another deviation from Ohm’s law is achieved if one places two quantum point contacts in series close to one another. If we assume that point contact 1 is more narrow than point contact 2, then electron waves can only pass through the second if they are able to pass through the first. So the total conductance G of the system is entirely determined by one of the point contacts, G = min{G1, G2} = G1. In terms of the electrical resistance R we have R = max{R1, R2}. But this is radically 78 CHAPTER 6. TRANSPORT IN NANOSTRUCTURES different form the usual classical formula R = R1 + R2. We only recover the classical result if the distance between the point contacts is so large that the quantum waves are destroyed in the region between them, and the whole electron wave picture thus breaks down. Chapter 7 Scanning Tunneling Microscopy (STM) Although treated first in these lecture notes, the AFM was actually preceded by the scanning tunneling microscope (STM). Invented in 1981 the STM was the first of many scanning probe microscopes (SPM). In 1986 the inventors Gert Binnig and Heinrich Rohrer from the IBM Research Laboratory in Ru¨schlikon (Switzerland), were awarded the Nobel Prize in Physics. Nowadays, SPMs can be found in many academic and industrial physics, chemistry and biology laboratories, and as we have already seen, they are used both as standard analysis tools and as high-level research instruments. 7.1 The basic principle of the STM The basic principle of the STM is sketched in Fig. 7.1. In contrast to the atomic force microscope the STM can only scan conducting surfaces. The reason is that the instrument relies on the small tunneling current, typically in the sub-nA regime, that runs between the substrate and the metal (tungsten) tip. Through a computer controlled feed-back loop and a piezo-element this current controls the height d in the z direction over the substrate. Scans across the surface in the xy plane are controlled by a piezo-electric element. The feed-back voltage controlling the height in the z direction together with the xy scan voltage is used as output. With the STM one can achieve a resolution of around 0.1 nm. This is better than the AFM, and the reason for this is the exponential dependence of the distance d in the transmission coefficient, T (ε) ≈ exp[−4κd], that were derived in Eq. (6.40). For a typical parameter values the tunneling current reduces by a factor 10 for every 0.1 nm increase in d. This means that over a typical atomic diameter of e.g. 0.3 nm, the tunneling current changes by a factor 1000. This is what makes the STM so sensitive. The tunneling current depends so strongly on the distance that it is dominated by the contribution flowing between the last atom of the tip and the nearest atom in the specimen. The scan range for a STM is typically up to 1 µm with a scan speed of the order mm/min. The scan speed depends on the mode used for the STM. 79 80 CHAPTER 7. SCANNING TUNNELING MICROSCOPY (STM) Figure 7.1: The principle of the STM. The instrument relies on the tunneling current between the conducting substrate and the metallic scanning tip. Through a computer controlled feed-back loop and a piezo-element this current controls the height d over the substrate. Scanning across the surface is controlled by a piezo-robot. The feed-back voltage is used as output. To the right is shown a 3 nm by 3 nm surface scan with atomic resolution of a graphite surface. The constant height mode is the fastest scan mode. Here the tip is kept at a fixed vertical position during the scan. The changes in the tunneling current thus reflects the electronic topology of the surface. Using this mode there is a risk that the tip will bump into unexpected high regions on the surface and be destroyed. The constant current mode is perhaps the most widely used STM mode. In this mode the feed-back mechanism ensures that the tunneling current is kept constant by displacing the tip vertically as the scan proceeds. This is slower than the constant height mode, but without its risk. 7.2 The piezo-electric element and spectroscopy The principle of the piezo-electric element is sketched in Fig. 7.2. The principal effect is that for crystal lacking inverison symmetry an applied voltage in one direction makes the element longer or shorter in another direction. A simple tripod allowing for control of the tip in all three spatial directions can be made by mounting three piezo-rods perpendicular to each other Fig. 7.2(b). However, in most modern scanning probe microscopes, one uses a tube geometry, like the one shown on Fig. 7.2(c). Each of the four indicated sections can be made longer or shorter individually. If all four sections are made longer or shorter by the same amount, the tip moves in the z direction. If the X+ side is made longer, and at the same time the X− side is made shorter by the same amount, the tube tilts a little bit to one side, as indicated. For small deformations, this makes the tip move primarily in the x direction. The same can be done in the y direction. 7.3. THE LOCAL ELECTRONIC DENSITY OF STATES 81 Figure 7.2: The principle of the piezo-element. (a) Elongation of a simple piezo-electric by applying a voltage. (b) The tripod consisting of three mutually perpendicular piezo-rods. (c) The tube geometry that is widely used in many STM setups. 7.3 The local electronic density of states It is important to realize that the STM does not measure the topology but rather in some sense the electronic topology. From the discussion of quantum conductance in Sec. 6.6 we know that the current running between two reservoirs is given by counting how many electrons that are transferred per unit time. However, in our previous analysis we tacitly assumed that there always were states available in both reservoirs. This is true if we are dealing with bulk metals. This assumption is not generally true, and in fact we need to introduce the concept of the local electronic density of states d(r, ε). The density of state function contains the information about how many quantum states of energy ε there are available in the neighborhood of the point r on the surface. Therefore the expression for the tunneling current I(r), when the tip is placed at the position r is modified to I(r) = CT (ε) d(r, ε) ≈ A exp[−4κd] d(r, ε). (7.1) Consequently great care must be taken when interpreting STM scan pictures. Imag- ine using the constant height mode to scan a metallic surface covered with some small non-conducting particles. Such particles have a very small electronic density of states. Therefore according to Eq. (7.1) the tunneling current would be very low when the tip is just above one such particle even though the distance is small, while the current would be much higher for positions of the tip above the metallic substrate, even though the distance is larger. With a naive interpretation of the resulting STM picture one would get the impression that the sample has a lot of holes, completely contrary to the real situation. With a proper understanding, the appearance of the electronic density of states in Eq. (7.1) can be advantageous. The STM can namely be used as a spectrometer for measuring electronic energies. Consider a STM tip placed at a fixed position r0. By changing the applied voltage V between the STM tip and the sample the energy ε of the electrons leaving the STM tip is changed by the amount ε = ε0 − eV , where ε0 is the energy at zero applied voltage. In an energy range where the energy dependence of 82 CHAPTER 7. SCANNING TUNNELING MICROSCOPY (STM) Figure 7.3: An example of a STM from Aarhus University, Denmark. On the figure is shown: (1) Sample, (2) Sample holder, (3) Clamps, (4) Tip, (5) Tip holder, (6) Piezo- electric scanner tube, (7) Approach motor rod, (8) Motor mount, (9) Approach mount, (10) Quartz balls, and (11) Zener Diode. For more explanations see the text in Sec. 7.4 the transmission coefficient is negligible T (ε) = T , measuring the tunneling current as a function of aplied voltage V gives directly an energy scan of the density of states, I(r0, V ) = CT d(r0, ε − eV ), (7.2) and thus a direct measurements of at which energies there are electronic states available in the sample. This technique has been used in numerous investigations on materials on the nanometer scale. 7.4 An example of a STM We end this chapter on the STM by showing an concrete example of a STM. It is taken from the Department of Physics and Astronomy, Aarhus University, Denmark (www.ifa.au.dk/camp/home.htm). A sketch of the instrument is shown in Fig. 7.3 The setup is described as follows. The sample (1) is placed in a tantalum holder (2), which may be removed from the STM, and which is normally held down on the STM top by springs (3). The top plate is thermally and electrically insulated from the STM body by three quartz balls (10). The top plate is mounted on a 0.6 kg aluminum block which may be cooled to −160◦C or heated to +130◦C. The tip (4) is held by a macor holder (5), which is affixed to the top of the scanner tube (6). The scanner tube is 4 mm long with an outer/inner diameter of 3.2/2.1 mm and is mounted to the rod (7), which together with 7.4. AN EXAMPLE OF A STM 83 the piezo tube (9) forms a small inchworm motor used for coarse approach. The electrode of the tube is divided into three rings. In the tube two bearings are placed under the upper and the lower electrode with an extremely good fit to the rod (7). Applying a positive voltage to an electrode will clamp that electrode to the rod whereas a negative voltage will free that electrode from the rod. A voltage applied to the center electrode will cause it to elongate or contract. With the right sequence of voltages applied to the three electrodes the rod will move up or down since the tube is fixed to the STM body by the macor ring (8). The motor may work in steps of down to 2 A˚, but at full speed it moves around 2 mm/min. The scan range is up to +-1 um when using antisymmetrical scan voltages of +-200 V. The Zener diode BZY93C75 (11) is used to counter-heat the STM body during cooling. 84 CHAPTER 7. SCANNING TUNNELING MICROSCOPY (STM) Appendix A Exercises Exercises for Chap. 1 Exercise 1.1 Moore’s law for lengths of gate electrodes. An exponential law cannot be maintained indefinitely. According to Moore’s law in Fig. 1.3 when will the length of a gate electrode equal the diameter of a hydrogen atom? Discuss the result. Exercise 1.2 The minimum photolithographic line width. The usual diffraction limit for resolu- tion of light with wavelength λ is roughly given by wdiff ≈ λ. Discuss this in comparison with the minimum photolithographic width wmin given by Eq. (1.1). Exercise 1.3 Exposure time of electron beam lithography. Use Eq. (1.3) to estimate the exposure time for an electron beam lithography process on a 100 mm diameter wafer with a pattern density of 10%, when the clearing dose is D = 250 µC cm−2 and the beam current is I = 20 pA. Exercise 1.4 The de Broglie wavelength of an electron. Derive the expression Eq. (1.2) for the de Broglie wavelength λ of an electron accelerated by a voltage drop U . Use the information given just above the equation and verify the result λ = 0.012 nm for U = 10 kV. Exercises for Chap. 2 Exercise 2.1 Frequency, energy, and temperature. Calculate the energy in J and eV and the 85 86 EXERCISES FOR CHAP. 2 corresponding temperatures in kelvin of the characteristic photons coming from (a) a FM- radio transmitter, (b) a mobile phone, (c) a candle, and (d) an X-ray machine at the dentist. Use that hf = E = kBT , where h is Planck’s constant and kB is Boltzmann’s constant. Exercise 2.2 de Broglie wave lengths. From p = h/λ find the de Broglie wave lengths for the following particles: (a) a nitrogen molecule in air at room temperature, (b) an electron in copper given a velocity of 1.57 × 106 m/s, (c) an electron in gallium-arsenide, given a velocity of 1.12 × 105 m/s and an effective mass of m∗ = 0.067me, and (d) a rubidium atom in a cold atomic trap with a temperature of 40 nK. Exercise 2.3 A simple estimate of the radius of the hydrogen atom. Due to the wave nature of particles it costs energy to localize them. This can be used to estimate the size of a hydrogen atom. (a) Convince yourself that the classical energy given in Eq. (2.14) is correct. (b) Use the Heisenberg uncertainty principle in the extreme quantum limit Eq. (2.11) p =
/a to derive the quantum energy given in Eq. (2.14). (c) Verify that the Bohr radius a0 indeed is the radius that minimizes the quantum energy. (d) Verify the expression for the ground state energy E0 = E(a0) given in Eq. (2.16). Although the estimates for a0 and E0 are based on an approximate quantum theory, Iit turns out that by chance the estimates are exact. Exercise 2.4 A simple estimate of the maximal height of mountains. The quantum pressure of particle waves also determines the maximal height of mountains on Earth. A mountain cannot exceed the height H for which the potential energy MgH of the last added molecule, say SiO2, equals the energy Ebend that it takes to bend the electrons orbitals forming the mineral structure at the bottom of the mountain. Now, to remove an electron completely from a hydrogen atom costs E0 = 13.6 eV, but electrons in minerals are bound more loosely, typically with an energy 0.1E0, and furthermore we only need to bend the electron orbitals not destroy them alltogether, so we take Ebend ≈ 0.1 × 0.1 ×E0. Given this, find an estimate for the maximal height H of mountains. Exercise 2.5 Size-quantization. The minimal quantized kinetic energy for a particle in a cubic box is given in Eq. (2.12). (a) Find the similar expression for a box with unequal side lengths Lx, Ly, and Lz. (b) In general the standing wave in the box can contain any integer number nx, ny, and nz of half wavelengths in each of the three directions. Find the general expression for the size-quantized kinetic energy E(nx, ny, nz). EXERCISES FOR CHAP. 3. 87 Exercise 2.6 The energy spectrum of a particle in a box. (a) Write down the triplets (nx, ny, nz) corresponding to the lowest 10 energy levels. (b) If a number g of triplets lead to the same energy, that energy level is said to be g-fold degenerate or to have a degeneracy of g. What are the degenracies of the 10 lowest energy levels? (c) Find the difference in energy between the two lowest states for an electron in each of the three cubes with side lengths L = 0.1 nm, L = 10 nm, L = 1 cm. (d) What are the corresponding temperatures? Exercise 2.7 A semiconductor nanowire. In modern computer chips the smallest wires are only 100 nm wide. Consider a silicon nanowire in the x-direction with a square cross section Ly = 100 nm and Lz = 100 nm. The effective mass of an electron in silicon is m∗ = 0.2me. (a) What is the energy difference between the two lowest quantum states for an electron in the nanowire with px = 0 kg m/s? (b) Same question for a 10× 10 nm2 quantum wire. Exercise 2.8 The harmonic oscillator. The classical 1D harmonic oscillator with the equation of mo- tion mx¨ = −Kx has the energy E = p22m + 12Kx2. (a) Express the oscillation frequency ω in terms of m and K. (b) Now use quantum theory and write down the time-independent Schro¨dinger equation for the oscillator. (c) Show that ψ0(x) = exp[−x2/(22)] is a solution to this equation if the characteristic length  is chosen properly. (d) Determine the solu- tions ψ1(x) = (x + B) exp[−x2/(22)] and ψ2(x) = (x2 + Bx + C) exp[−x2/(22)] keeping  the same. (e) Express the corresponding eigenenergies E0, E1, and E2 in terms of ω. Exercise 2.9 1D wavefunctions. Draw the three lowest eigenstates for the 1D box and the 1D har- monic oscillator. Discuss similarities and differences between the two sets of wavefunctions. Exercises for Chap. 3 Exercise 3.1 The momentum of a standing wave. Calculate and discuss the expectation value 〈pˆ〉k of the momentum operator pˆ for the standing wave ψ˜k = √ 2 L sin ( π xL ) . Exercise 3.2 Kinetic energy and momentum of a travelling wave. Prove that the travelling wave Eq. (3.6) indeed is an eigenstate of both the kinetic energy operator and the momentum operator with the appropriate eigenvalues as stated in Eqs. (3.7a) and (3.7b). 88 EXERCISES FOR CHAP. 3 Exercise 3.3 The microscopic parameters of metallic iron. Iron (Fe) in its metallic state has valence II, and X-ray measurements have revealed that it forms a body-centered-cubic (BCC) crystal. Each cubic unit cell in the BCC crystal has side length a = 0.287 nm, while one atom sits in each corner and one in the center. Find the density n of the resulting gas of valence electrons, and determine kF, εF, vF, and λF. Exercise 3.4 A 2D electron gas. In GaAs-heterojunctions the conduction electrons are caught at the interface between GaAs at one side and GaAlAs at the other. Many nano-devices are based on this structure. The interface lies in the xy plane. The wavefunction of the electrons has the special form ψkxkyn(r) = (1/A)ei(kxx+kyy)ζn(z), where A is the area and ζn(z) describes the confinement to the interface (see figure). In the ground state all electrons have the same wavefunction ζ0(z) in the z-direction so motion in that direction is locked and can be left out of the problem. The electrons are only free to move in the xy-plane described by (1/A)ei(kxx+kyy), and one therefore calls this system for the 2D electron gas (2DEG). Eqs. (3.12) and (3.16) relates kF and Ekin/N to the density n in 3D. Derive the analogous expressions for the 2DEG using ψk(x, y) = (1/A)ei(kxx+kyy).
    
    

 
  
        

  

  
  
    !  "  #$ (a) A sketch of a GaAs heterostructure with a 2DEG at the GaAs-Ga1−xAlxAs interface in the xy plane. (b) The potential profile in the z direction near the interface (GaAs for z > 0). The wavefunctions ζn(z) and their energies εn for n = 0, 1 and 2 are also shown. Exercise 3.5 Thermal smearing of the Fermi sphere. Consider a Fermi-sphere that contains N electrons and that have the Fermi energy εF. As the temperature is raised from zero to T some of the ∆N electrons that started out with energies ε in the interval εF−kBT < ε < εF can, roughly speaking, receive thermal energy kicks by energies up to kBT . Thereby they are thermally excited. Excitations of any of the other electrons in the Fermi sphere are blocked by the Pauli exclusion principle. (a) Express the ratio ∆NN of the number ∆N of thermally excitable electrons over the total number N of electrons in terms of εF and kBT . (b) What is this ratio for copper at room temperature? (c) Do the electrons contribute significantly to the specific heat of copper at room temperature? EXERCISES FOR CHAP. 4. 89 Exercise 3.6 The stability of metals. Prove from Eq. (3.23) that the stability point for metals indeed is given by Eq. (3.24a). What is in SI-units the density corresponding to r∗s? Exercise 3.7 Youngs modulus for metals in the jellium model. Hooke’s law F = K ∆ for an ordinary spring states that the reaction force F is proportional to the stretch ∆. The proportionality constant is called the spring constant K. This law is generalized for any elastic solid to a relation between the reaction force per cross section area A (called stress) and the relative stretch ∆/0 (called strain): F A = Y ∆ 0 , where 0 is the length before external forces are applied. The “spring constant” Y is called Young’s modulus of the material. Consider in the jellium model the volume 1/n that contains one electron. The length can be taken to be 0 = r∗s a0, where a0 = 0.053 nm is the Bohr radius, and the cross section area to be A = 4π3 20. (a) Expand the energy E(rs)/N of the electron Eq. (3.23) around r∗s to second order in ∆rs = rs − r∗s . (b) The resulting expression has the form of the potential energy of a harmonic oscillator. Find an expression for the spring constant. Note that rs and ∆rs are dimensionless, so it is preferable to introduce  = rs a0 and ∆ = ∆rs a0. (c) Find an expression for Young’s modulus of the model by considering F/A. (d) Calculate Young’s modulus in SI units and discuss the result in comparison with the table below. metal Al Cu Fe Ni Pb Y /[1010 Pa] 7.0 11.0 21.0 21.0 1.6 Exercises for Chap. 4 Exercise 4.1 Spherical coordinates (a) Write in the 3D coordinate system to the right the proper definitions of cartesian coordinates (x, y, z) and spherical coordinates (r, θ, φ). (b) In cartesian coordinates the volume element is dx dy dz. What is it in spherical coordinates? Exercise 4.2 The expectation value of 1/r in the hydrogen atom 90 EXERCISES FOR CHAP. 4 Calculate the expectation value 〈1 r 〉100 = ∫ dr ψ∗100(r) 1 r ψ100(r) in the ground state ψ100(r) of the hydrogen atom. Exercise 4.3 Eigenstates of the angular momentum operator. Prove by direct calculation that Y 00 (θ, φ), Y 0 1 (θ, φ), Y 1 2 (θ, φ) are eigenstates to both Lˆ 2 and Lz. What are the corresponding eigenvalues? Exercise 4.4 The angular part of the Laplacian. Prove Eq. (4.35) using Eq. (4.34). Exercise 4.5 Radial wavefunctions Plot the six radial wavefunctions in Eq. (4.25) and discuss the resulting graphs. Exercise 4.6 Spherical harmonics Select five of the spherical harmonics in Eq. (4.14) and plot their absolut values |Y ml (θ)| in a polar plots. Discuss the resulting graphs Exercise 4.7 Counting of orbitals Go through the argument in Secs. 4.5 and 4.6 leading from a single carbon atom with the 2s22p2 configuration to the σ and π orbitals in graphene. Make a list over the orbitals present at each stage and their occupancy. Exercise 4.8 The chiral vector in carbon nanotube Consider the chiral vector c = na1 + ma2 in Fig. 4.6. What are the values of n and m in this case? Describe the resulting carbon nanotube. EXERCISES FOR CHAP. 5. 91 Exercises for Chap. 5 Exercise 5.1 An actual AFM-setup. Use the Internet or the library to find an example of an actual AFM-setup. What kind of tip and cantilever was used? Which mode was used during the AFM-scan? What was the sample? Exercise 5.2 The effective mass for an oscillating cantilever. In the circular shape approximation the displacement in the z direction of an oscillating cantilever is written as z(x, t) = z0(x/L)2 cos(ωt). The cantilever has a homogeneous mass density ρ = m/L, where m is the total mass of the cantilever and L its length. (a) What is the velocity of the end point x = L? (b) If the entire mass was situated at the end point what would then be the kinetic energy? (c) What is the velocity of an arbitrary point between 0 and L? (d) Calculate the kinetic energy of the cantilever and show why it is reasonable to say that the effective mass is meff = 15 m. Exercise 5.3 A copper cantilever. Young’s modulus and the mass density of copper is given by YCu = 1.1 × 1011 Pa and ρCu = 8.96 × 103 kg m−3. Consider a rectangular cantilever with the dimensions L = 100 µm, w = 10 µm, and h = 1 µm. Assume the a force Ftip is deflecting the tip ∆z ≈ 1 A˚. (a) Calculate the spring constant K, the tip-force Ftip, and the resonance frequency ω0 for the system. (b) Compare with the results Eqs. (5.7a) – (5.7c) for silicon and discuss the differences. Exercise 5.4 Atomic Polarization. As stated in Eq. (2.14) and in Fig. 5.4b an approximate expression for the energy E(a) of the hydrogen atom as a function of the average distance a from the electron to the nucleus is given by E(a) =
2 2m 1 a2 − e 2 4π 0 1 a . Show in detail how to get from this expression to Eq. (5.11) for the atomic dipole ed: ed = (a0e) 2 2E0 E . Exercise 5.5 The relative dielectric constant of a gas. In classical electromagnetism the relative dielectric constant r is defined by the relation D = r 0E between the displacement field 92 EXERCISES FOR CHAP. 5 D and the electric field E . This can also be expressed in terms of the polarization P, which is nothing but the number N of dipole moments ed per volume V: D = r 0E = 0E + P = 0E + N V ed. (a) Combine this with Exercise 5.4 to show that r = 1 + 8πN a30 V . (b) In a gas at room temperature and atmospheric pressure Avogadro’s number of molecules takes up a volume of 22.4 L. Calculate r for hydrogen H2 and compare with the experi- mental value r = 1.00026. Exercise 5.6 The index of refraction of a liquid. The speed of light in a liquid is given by c = c0/nliq, where nliq is the index of refraction of the liquid. We know from electromagnetic theory that the speed of light is given by c = 1/√µrµ0 r 0, where µ0 is the magnetic susceptibility of vacuum and µr the relative susceptibility (which is 1 for non-magnetic media). (a) Check this formula for vacuum where µr = r = 1. (b) Show that for a non-magnetic medium we have nliq = √ r. (c) Give an estimate of the density N/V of molecules in a liquid. (d) Calculate nliq assuming that each molecule has the dipole moment ed of Exercise 5.4, and compare the result to the experimental value for water. Exercise 5.7 The electric dipole. Go carefully through the arguments leading to Eqs. (5.12) and (5.13). Exercise 5.8 The harmonic oscillator solved with complex numbers. Consider a harmonic oscillator given by the equation of motion mx¨(t) = −Kx(t), where m is the mass and K the spring constant. (a) Assume x(t) = A cos(ωt) and determine the resonance frequency ω = ω0 in terms of K and m. (b) Solve the same problem now assuming x(t) = A exp(iω0t). (c) What is the relation between the real-number solution and the complex-number solution? (d) Is it possible to determine the amplitude A from the equation of motion? Exercise 5.9 The damped harmonic oscillator. To the equation of motion of the harmonic oscillator in Exercise 5.8 we now add a friction term −mγx˙(t) that damps the oscillator. (a) Assume the complex solution x(t) = A exp(iωt) and show that ω must fulfill the equation −ω2 = −ω20 − iγω. EXERCISES FOR CHAP. 5 93 (b) If γ  ω0 we obtain ω ≈ ω0, and therefore ω2 ≈ ω20 + iγω0. Use this expression in x(t) = A exp(iωt), and sketch x(t). Exercise 5.10 The externally driven, damped harmonic oscillator. To the damped harmonic oscillator of Exercise 5.9 we now add an external driving force F (t) = F0 exp(iωt). The equation of motion the becomes mx¨(t) = −K x(t)−mγ x˙(t) + F0 exp(iωt). (a) Assume the complex solution x(t) = A exp(iωt), which contains the same frequency ω as the driving force, and determine the amplitude A. (b) Sketch |A(ω)|2, i.e. the absolute square of the amplitude as a function of the driving frequency ω. Exercise 5.11 Maximal amplitude and half with of the oscillator resonance line. Consider the externally driven, damped harmonic oscillator of Exercise 5.9. For small damping, γ  ω0, the amplitude curve |A(ω)|2 changes rapidly in a narrow range around ω0. For frequencies ω in that range we can therefore make the following approximations: ω20 − ω2 = (ω0 + ω)(ω0 − ω) ≈ 2ω0(ω0 − ω) and γω ≈ γω0. (a) Use these approximations to show that |A(ω)|2 = F 2 0 4m2ω20 1 (ω0 − ω)2 + 14γ2 . (b) For which ω is |A(ω)| maximal? (c) Determine the maximal value of |A(ω)|. (d) For which values of ω is |A(ω)| exactly half of its maximal value? (e) The Q-factor of a resonance peak is defined as Q = ω0/∆ω, where ω0 is the resonance frequency and ∆ω is the full width at half maximum. Use (d) to write down an expression for the Q-factor of the resonance Exercise 5.12 A cantilever based mass sensor. In a collaboration between MIC and institutes in Barcelona a cantilever based mass sensor with ultra high sensitivity has been developed (see the course web page for the original papers). The cantilever forms one plate in an on-chip capacitor, while an high frequency AC-voltage can be applied to the other near-by capacitor plate. The AC-voltage forces the cantilever to oscillate. By tuning the 94 EXERCISES FOR CHAP. 6 AC-frequency f = ω/2π and observing the amplitude of the oscillation (either directly or through the capacitance of the system) the resonance frequency f0 = ω0/2π of the cantilever can be measured. In an active chemical environment the mass of the cantilever chages as molecules are deposited on its surface. In the following we use the theory in Sec. 5.2 to analyze how sensitive the system is to changes of the mass of the cantilever. (a) State the relationship between the resonance frequency f0, the mass m of the cantilever, and the spring constant K (the latter is assume to remain constant as molecules are adhering to the cantilever). (b) Find, say by differentiation, the relation between a small change dm of mass and the accompanying change df0 in the resonance frequency. (c) In one of the papers an cantilever of length L = 48 µm, height h = 1 µm, and width w = 780 nm is studied. Assume that the cantilever is made of pure silicon and that the resonance frequency can be measured with an accuracy of 1 Hz. Find the smallest detectable change of mass, and comment on the result. (d) In one of the original papers the resonance peak of the cantilever is measured directly using an AFM. It is displayed in panel (e) of the figure. Estimate the observed Q-factor of the system (see Exercise 5.11). Exercises for Chap. 6 Exercise 6.1 Electron reservoirs Set µ = 1 and plot the Fermi-Dirac function nF(ε, µ = 1, V = 0, T ) Eq. (6.2) as a function of ε for kBT = 0.01, 0.1, 1, and 10. Do the same for the derivative −∂nF/∂ε. These functions are important when dealing with the electron reservoirs connected to nanostructures. Exercise 6.2 Right- and left-moving waves Prove that ψ±k(x, t) in Eqs. (6.5a) and (6.5b) are solution to the time-dependent Schro¨dinger equation Eq. (2.17). Explain why ψ+k(x, t) is a right-moving wave and ψ−k(x, t) a left- moving wave. Exercise 6.3 The current density J What are the units of the current density J(r) =
m Im [ ψ∗(∇ψ) ] ? Exercise 6.4 The electrical current I Prove Eq. (6.10) from Eqs. (6.7), (6.8), and (6.9). EXERCISES FOR CHAP. 6 95 Exercise 6.5 The eigenstates for the potential step Show that the wavefunction ψε(x) given by Eq. (6.18) with the coefficients B and C specified in Eq. (6.20) is a solution to the time-independent Schro¨dinger equation Eq. (2.32) with the potential V (x) given by Eq. (6.17). Exercise 6.6 The transfer matrix for a potential jump Consider the potential barrier problem defined by Eqs. (6.28) and (6.29). Prove the correctness of the transfer matrix equation Eq. (6.31). Exercise 6.7 The transmission coefficient in the extreme tunneling regime Prove expression Eq. (6.40) for the transmission coefficient T (ε) by taking the appropriate limit of the general expression Eq. (6.39). Exercise 6.8 The probability of electron tunneling between metals Two copper cubes of side length 1 cm are placed 1 nm from each other. Use the simple model of metals in Chap. 3 and the transmission coefficient Eq. (6.40) in the extreme tunneling regime to estimate how often an electron tunnels from one cube to the other. Exercise 6.9 Above barrier reflection In Fig. 6.6 it is clearly seen that T < 1 for a slightly above-barrier energy ε = V +. Find an analytical expression for the reflection coefficient R(ε = V +) by expanding the sin-function in Eq. (6.42) to first order. Compare the result with the graph. Exercise 6.10 Resonant transmission Show from Eq. (6.42) that unity transmission is obtained when the resonance condition 2k2a = nπ is fullfilled. Calculate the position of the first two resonance peaks, n = 1, 2 and compare with graph (Ea = 0.1 V ). Exercise 6.11 The unit of the conductance quantum Show that the unit of the conductance quantum G0 = e2/h indeed is siemens (S). 96 EXERCISES FOR CHAP. 7 Exercise 6.12 The conductance formula Go carefully through the proof of the conductance formula Eq. (6.57). Exercises for Chap. 7 Exercise 7.1 The wavelength of an STM electron Let the energy of an electron be given by the applied voltage of U = 1 V that is typically used in a STM. What is the de Broglie wavelength of this electron? Could we observe atoms using the acceleration voltage in an electron microscope? Exercise 7.2 Electrons inside tunneling barriers The work function Φ is the distance from the unperturbed Fermi energy to the top of the barrier that keeps electrons inside a metal. The probability P (x) of an electron to have penetrated the distance x into the barrier is given by the square of its wavefunction ψ(x), P (x) = |ψ(x)|2 = |ψ(0)|2 e−2κx, where κ = 1
√ 2m(V − ε). How far does the electron penetrate into the barrier for platinum, Φ(Pt) = 5.7 eV, and for tungsten, Φ(Pt) = 4.8 eV? Exercise 7.3 Tunneling frequncy and time How many electrons are present in the STM tunnel gap at the same time? Do they influence each other? Time τ is not an observable in quantum mechanics, so for the barrier V (x) we make a semiclassical WKB analysis, τ = ∫ x2 x1 √ m 2[V (x) − ε] dx, where x1 and x2 are the classical turning points where ε = V (xi). Take a rectangular barrier of height V0. Assume a barrier width of 0.4 nm and an effective height of 4 eV. Compare the result to typical STM currents of 1 nA. Exercise 7.4 Speed of the STM tip A typical STM scan covers a square of 10 nm by 10 nm divided into 256 by 256 pixels. The scanning time is of the order tscan = 10 s. What is the average horizontal speed of the tip? And how long time would it take to move the tip 1 m at that speed? EXERCISES FOR CHAP. 7 97 Exercise 7.5 Local vacuum in the STM tunnel gap Assume a distance d = 0.5 nm between the tip and the sample. Give an estimate of the number of air molecules present in the gap volume at room temperature and a 1 atm pressure. Exercise 7.6 The exponential decrease in tunneling current Use the typical values of the STM parameters from the previous exercises to verify the statement in Sec. 7.1 that by increasing the distance from the substrate by 0.1 nm the tunneling current decreases by a factor of 10.