Physics A2 - Lecture 9: Barrier Penetration and Tunneling - Huynh Quang Linh

Nội dung text: Physics A2 - Lecture 9: Barrier Penetration and Tunneling - Huynh Quang Linh

1.  “All of modern physics is governed by that magnificent and thoroughly confusing discipline called quantum mechanics It has survived all tests and there is no reason to believe that there is any flaw in it .We all know how to use it and how to apply it to problems; and so we have learned to live with the fact that nobody can understand it.”  Murray Gell-Mann
2. Lecture 9: Barrier Penetration and Tunneling nucleus x U(x) U(x) U0 E A B C B A 0 L x 0 x
3. Content  How quantum particles tunnel  Nuclear Decay  Solar Fusion  NH3 Maser
4. “Leaky” Particles: Revisited  Due to “barrier penetration”, the electron density of a metal actually extends outside the surface of the metal! x Vo Work function F EF Occupied levels x = 0 Assume that the work function (i.e., the energy difference between the most energetic conduction electrons and the potential barrier at the surface) of a certain metal is F = 5 eV. Estimate the distance x outside the surface of the metal at which the electron probability density drops to 1/1000 of that just inside the metal. 2  (x) 1 1 1 e 2Kx x ln 0.3nm  (0) 2 1000 2K 1000 22mm5eV using K ee V E 2 F 2 11.5 nm 1 20 h 21.505 eV nm 2
5. Application: Tunneling Microscopy  Due to the quantum effect of “barrier x Metal penetration,” the electron density of a tip material extends beyond its surface: One can exploit this material STM tip to measure the electron density on a ~ 1 nm material’s surface: material STM tip Real STM tip Na atoms DNA Double on metal: Helix: STM images www-aix.gsi.de/~bio
6. Tunneling Through a Barrier (1) U(x)  What is the “Transmission Coefficient T”, the probability U=Uo I III an incident particle tunnels II through the barrier?  Consider a barrier (II) in the middle of a very wide infinite U=0 square well. 0 L x  To get an “exact” result describing how quantum particles penetrate this barrier, we write the proper wavefunction in each of the three regions shown in Figure: E > U: oscillatory solution Region I: I (x) A1 sin kx A2 coskx Kx Kx E U: oscillatory solution Region III: III (x) C1 sin kx C2 coskx  Next we would need to apply the “continuity conditions” for both  and d/dx at the boundaries x = 0 and x = L to determine the A, B, and C coefficients.
7. Tunneling Through a Barrier (2) U(x)  In general the tunneling coefficient T U0 can be quite complicated (due to the E contribution of amplitudes “reflected” off the far side of the barrier). 0 L x  However, in many situations, the barrier width L is much larger than the ‘decay length’ 1/K of the penetrating wave; in this case (KL >> 1) the tunneling coefficient simplifies to: EE 2m 2KL where T Ge G 16 1 K 2 U0 E UU00  This is nearly the same result as in the “leaky particle” example! Except for G: • G slightly modifies the 4 3 transmission probability G 2 • G arises from the fact 1 that the amplitude at 0 x = 0 is not a maximum 0 0.25 0.5 0.75 1 E/U0
8. Tunneling Through a Barrier (3) U(x) 2KL T Ge U0 EE 2m E where G 16 1 K U0 E UU 2 00 0 L x  By far the dominant effect is the decaying exponential*: T e 2KL  T depends on the energy below the barrier (U0-E) and on the barrier width L. 0.6 T E=2/3 U0 0.4  The plot illustrates how the E=1/3 U transmission coefficient T changes 0.2 0 as a function of barrier width L, 0 for two different values of the 0.5 0.75 1 1.25 1.5 particle energy. L *In fact, some references (wrongly) completely omit G (including Phys. 214 before 2006!). We will state when you can ignore G.
9. Example: Barrier Tunneling in an STM  Let’s consider a simple problem: U(x) An electron with a total energy of E=6 eV U0 approaches a potential barrier with a E height of Uo = 12 eV. If the width of the metal STM tip barrier is L=0.18 nm, what is the 0 L x probability that the electron will tunnel air through the barrier? gap
10. Example: Barrier Tunneling in an STM  Let’s consider a simple problem: U(x) An electron with a total energy of E=6 eV U0 approaches a potential barrier with a E height of Uo = 12 eV. If the width of the metal STM tip barrier is L=0.18 nm, what is the 0 L x probability that the electron will tunnel air through the barrier? gap 2KL EE 11 T Ge G 16 1 16 1 4 UU00 22 22mm 6eV KUEUE ee 2 2 12.6 nm 1 200h 21.505 eV-nm 2 T 4e 2(12.6)(0.18) 4(0.011) 4.3% Question: What will T be if we double the width of the gap?
11. Lecture 9, exercise 1 Consider a particle tunneling through a barrier. U(x) 1. Which of the following will increase the U0 likelihood of tunneling? E a. decrease the height of the barrier b. decrease the width of the barrier 0 L x c. decrease the mass of the particle 2. What is the energy of the particles that have successfully “escaped”? a. initial energy
12. Lecture 9, exercise 1 Consider a particle tunneling through a barrier. U(x) 1. Which of the following will increase the U0 likelihood of tunneling? E a. decrease the height of the barrier b. decrease the width of the barrier 0 L x c. decrease the mass of the particle Reducing m or U0 will reduce K T e 2KL 2. What is the energy of the particles that have successfully “escaped”? a. initial energy Although the amplitude of the wave is smaller after the barrier, no energy is lost in the tunneling process.
13. Example: Al wire contacts  “Everyday” problem: U(x) You’re putting the electrical wiring in your new U0 house, and you’re considering using Aluminum E outlet wiring, which is cheap and a good conductor. Al wire However, you also know that aluminum tends to contact form an oxide surface layer (Al2O3) which can be 0 L x as much as several nanometers thick. Al2O3 This layer could cause a problem in making electrical contacts with outlets, for example, since it presents a barrier of roughly 10 eV to the flow of electrons in and out of the Al. Your requirement is that your transmission coefficient across any contact must be T > 10-10, or else the resistance will be too high for the high currents you’re using, causing a fire risk. Should you use aluminum wiring or not? (You can neglect G here.)
14. Example: Al wire contacts  “Everyday” problem: U(x) You’re putting the electrical wiring in your new U0 house, and you’re considering using Aluminum E outlet wiring, which is cheap and a good conductor. Al wire However, you also know that aluminum tends to contact form an oxide surface layer (Al2O3) which can be 0 L x as much as several nanometers thick. Al2O3 This layer could cause a problem in making electrical contacts with outlets, for example, since it presents a barrier of roughly 10 eV to the flow of electrons in and out of the Al. Your requirement is that your transmission coefficient across any contact must be T > 10-10, or else the resistance will be too high for the high currents you’re using, causing a fire risk. Should you use aluminum wiring or not? (You can neglect G here.) Compute L: 1 10 Oxide is thicker 2KL 10 L ln 10 0.72nm T e 10 2K than this, so go with Cu wiring! (Al wiring in 22mmee 10eV 1 KUEUE 2 00 2 2 2 2 16nm houses is illegal h 1.505eV-nm for this reason)
15. Nature 434, 361 - 364 (17 March 2005) Current measurement by real-time counting of single electrons JONAS BYLANDER, TIM DUTY & PER DELSING ~40 fA ~80 fA ~120 fA Electrons that successfully tunnel through the 50 junctions are detected using a fast single electron transistor (SET).
16. Tunneling and Radioactivity In large atoms (e.g., Uranium), the nucleus nucleus can be unstable to the emission of an alpha particle (2p + 2n), which is U(x) a 4He nucleus. This form of radioactivity is a tunneling process, involving transmission of a 4He nucleus from a low-energy valley through a A B C B A barrier to a lower energy outside. 0 L  The “trapped” particle tunnels through region B out to region A.  The particle will not return.  The magnitude of the part left in C decays exponentially in time. [If you look at any one radioactive nucleus, you see it decay at some particular random time. An ensemble of many nuclei is needed to see the exponential decay.]
17. a-Radiation: Example 1 Consider a very simple model of a-radiation: Assume the alpha particle (m = 6.64 x 10-27 kg) is trapped in the nucleus of a Polonium-212 atom (Z = 84), which presents a square barrier of width L = 9.4 x 10-15 m (9.4 fermi) and height Uo = 26 MeV. What is the tunneling probability for an alpha particle with energy ~ 9 MeV each time the particle hits the barrier? What is the approximate lifetime of the 212Po?
18. a-Radiation: Example 1 Consider a very simple model of a-radiation: Assume the alpha particle (m = 6.64 x 10-27 kg) is trapped in the nucleus of a Polonium-212 atom (Z = 84), which presents a square barrier of width L = 9.4 x 10-15 m (9.4 fermi) and height Uo = 26 MeV. What is the tunneling probability for an alpha particle with energy ~ 9 MeV each time the particle hits the barrier? What is the approximate lifetime of the 212Po? [For this order of magnitude calculation you may neglect G.] 2KL 2m T e KUE 2 0 27 2 6.64 10 kg 6 19 15 1 2 17 10eV 1.6 10 J / eV 1.8 10 m 1.05  10 34 Js T exp 2 9.4 10 15 m 1.8 10 15 m 1  2 10 15 The particle makes about 1021 attempts at the barrier per second (~nuclear diameter/velocity), so the probability of escape is about 2 x 106 per second. Decay time for an a-particle is about (2 x 106 s-1)-1 = 0.5 x 10-6 seconds = 0.5 μs.
19. Lecture 9, exercise 2 We just looked at Polonium, with an effective barrier width of ~10 fermi, and saw a tunneling probability of ~10-15. Now consider Uranium, which has a similar barrier height, but an effective width of about ~20 fermi. Estimate the tunneling probability in Uranium: a. 10-30 b. 10-14 c. 10-7
20. Lecture 9, exercise 2 We just looked at Polonium, with an effective barrier width of ~10 fermi, and saw a tunneling probability of ~10-15. Now consider Uranium, which has a similar barrier height, but an effective width of about ~20 fermi. Estimate the tunneling probability in Uranium: a. 10-30 -14 b. 10 Think of it this way – there is a 10-15 chance to c. 10-7 get through the first half of the barrier, and a 10-15 chance to then get through the second half. Alternatively, when we double L in T e 2KL this is equivalent to squaring the transmission T. Now let’s do it for real
21. a-Radiation: Example 2 Consider a very simple model of a-radiation: Assume the alpha particle (m = 6.64 x 10-27 kg) is trapped in the nucleus of a Uranium-235 atom (Z = 92), which presents a square barrier of width L ~ 21 x 10-15 m (21 fermi) and height Uo = 28 MeV. What is the tunneling probability for an alpha particle with energy ~4.5 MeV each time the particle hits the barrier? What is the approximate lifetime of the 235U? [For this order of magnitude calculation you may neglect G.]
22. a-Radiation: Example 2 Consider a very simple model of a-radiation: Assume the alpha particle (m = 6.64 x 10-27 kg) is trapped in the nucleus of a Uranium-235 atom (Z = 92), which presents a square barrier of width L ~ 21 x 10-15 m (21 fermi) and height Uo = 28 MeV. What is the tunneling probability for an alpha particle with energy ~4.5 MeV each time the particle hits the barrier? What is the approximate lifetime of the 235U? [For this order of magnitude calculation you may neglect G.] 2m 2KL KUE T e 2 0 27 2 6.64 10 kg 6 19 15 1 2 23.5 10eV 1.6 10 J / eV 2.1 10 m 1.05  10 34 Js T exp 2 21 10 15 m 2.1 10 15 m 1  5 10 39 The particle makes about 1021 attempts at the barrier per second (~nuclear diameter/velocity), so the probability of escape is about 5 x 10-18 per second. Decay time for an a-particle is about (5 x 10-18 s-1)-1 = 2 x 1017 seconds ~ 1010 yrs!
23. FYI: The sun! The solar nuclear fusion process starts when two protons fuse together (they eventually become a helium nucleus, which fuses with another one, releasing two energetic protons). The temperature of the sun is some 1015 K. This corresponds to an average kinetic energy: -16 3/2 kBT = 3 x 10 J (kB = Boltzman’s constant) How much energy would they need classically to “touch” (approach to a distance the size of the nucleus, ~1 fm)? They are both charged, and therefore repel each other: 2 9 -19 2 -15 -13 U(r) = (1/4 e0)xe /r = (9x10 )x(1.6x10 C) /10 m = 2 x 10 J Thus, classically, the protons in the sun do not have enough energy to overcome their coulomb repulsion. How do they fuse then? By tunneling through the coulomb barrier!
24. Another “Tunneling” Example: The NH3 Molecule  The following example will bring together several things you’ve learned up to this point. Consider the ammonia (NH3) molecule: H Plane of N hydrogen atoms. The N atom in the ammonia molecule (NH3) can have two equal configurations: (Look along the H-plane.) NH3 Model U(x) x 0 The nitrogen atom can tunnel between these two equivalent positions.
25. Application: “Tunneling” & the NH3 Maser Question 1: What are the two lowest eigenstates of this ‘double-well’ potential? First consider two separate U(x) wells: U(x) 0 x H.O. potential  (x,t)  (x) x Ground state x The double-well states are superpositions x of these two single-well wavefunctions.
26. Application: “Tunneling” & the NH3 Maser Question 1: What are the two lowest eigenstates of this ‘double-well’ potential? First consider two separate U(x) wells: U(x) E2 E1 0 x H.O.  (x) potential n=2  (x) x Difference Ground state Sum n=1 x The double-well states are superpositions of these two single-well wavefunctions.
27. Application: “Tunneling” & the NH3 Maser Question 2: Given the energy difference between the ground and first excited states, E2 - E1 = 1.8x10-4 eV, estimate how long it U(x) takes for the N atom to “tunnel” E from one side of the NH molecule 2 3 E1 to the other? 0 x Two-well simulation Qmdbw
28. Application: “Tunneling” & the NH3 Maser Question 2: Given the energy difference between the ground and first excited states, E2 - E1 = 1.8x10-4 eV, estimate how long it U(x) takes for the N atom to “tunnel” E from one side of the NH molecule 2 3 E1 to the other? 0 x As discussed in Lect. 9, this takes a half an oscillation period, T = h/(E2-E1): T h 4.136  10 15 eV sec t 1.1 10 11 sec o 4 2 2 E21 E 2 1.8 10 eV Two-well simulation Qmdbw
29. Application: “Tunneling” & the NH3 Maser Question 3: (Stimulated) Emission of radiation between these two lowest energy configurations of ammonia, E1 =0.004637 eV and E2 =0.004655 eV (DE = 1.8x10-4 eV) was used to create U(x) the Ammonia Maser, by C. Townes in E2 1954 (for which he won the Nobel E1 prize in 1964). What wavelength of 0 x radiation does the maser emit? Microwave Laser! (Maser)
30. Application: “Tunneling” & the NH3 Maser Question 3: (Stimulated) Emission of radiation between these two lowest energy configurations of ammonia, E1 =0.004637 eV and E2 =0.004655 eV (DE = 1.8x10-4 eV) was used to create U(x) the Ammonia Maser, by C. Townes in E2 1954 (for which he won the Nobel E1 prize in 1964). What wavelength of 0 x radiation does the maser emit? Solution: By energy conservation, E2-E1 = Ephoton, where Ephoton = hc/l E2-E1 = hc/l Microwave Laser! -4 l = hc/(E2-E1) = 1240 eV•nm/1.8x10 eV = 6888 mm (Maser) = 6.88 mm
31. Application: “Tunneling” & the NH3 Maser Astronomical N Masers: Plane of hydrogen atoms. U(x) E2 E1 0 x
32. Lecture 9, exercise 3 You are trying to make a laser that emits violet light (400 nm), based on the transition an U0 electron makes between the ground and first- E1 excited state of a double quantum well as E shown. Your first sample emitted at 390 nm. 0 What could you modify to shift the wavelength to 400 nm? a. decrease the height of the barrier b. increase the height of the barrier c. decrease the width of the barrier
33. Lecture 9, exercise 3 You are trying to make a laser that emits violet light (400 nm), based on the transition an U0 electron makes between the ground and first- E1 excited state of a double quantum well as E shown. Your first sample emitted at 390 nm. 0 What could you modify to shift the wavelength to 400 nm? a. decrease the height of the barrier b. increase the height of the barrier c. decrease the width of the barrier The wavelength of the emitted photon was too low the frequency of the photon was too high the frequency of the electron oscillating between the left and right well was too high the probability to “tunnel” was too high! You can reduce this by increasing the barrier height.
34. Lecture 9, exercise 3 You are trying to make a laser that emits violet light (400 nm), based on the transition U0 an electron makes between the ground and E1 first-excited state of a double quantum well as E shown. Your first sample emitted at 390 nm. 0 What could you modify to shift the wavelength to 400 nm? a. decrease the height of the barrier b. increase the height of the barrier c. decrease the width of the barrier The wavelength of the emitted photon was too low the frequency of the photon was too high the frequency of the electron oscillating between the left and right well was too high the probability to “tunnel” was too high! You can reduce this by increasing the barrier height. The wavelength of the emitted photon was too low the frequency of the photon was too high the energy splitting between the ground and first-excited state was too large. Raising the barrier makes these two states look more “degenerate” (= “same energy”). Why?
35. Lecture 9, exercise 3,cont. As we raise the height of the central barrier, the coupling between the two wells decreases. In the limit of an infinite barrier, it looks like two independent wells same wavefunction curvature for both the symmetric (ground state) and anti- symmetric (1st excited state) wavefunctions same kinetic energy, i.e., degenerate solutions. 0 1 1 0
36. FYI: If QM is lossless, how do we get dissipation, heating, “collapse” of the wavefunction, etc.???  We discussed an example of tunneling (or lack thereof) leading to heating of an aluminum contact. But no energy is lost in the tunneling process (or in any quantum process!), so how is it that we do see heating, and “friction”?  Answer: In our simple model we were neglecting other parts of the system. For example, we ignored all impurities in the material that the electron could scatter off. We also neglected “phonons”, the quantum mechanical excitations of lattice vibrations (i.e., quantized sound).  While none of these interactions actually loses energy energy is ALWAYS conserved they can take energy away from the particle we are considering (this is dissipation), distributing it to the unobserved degrees of freedom.  In fact, even if the particle becomes coupled to these other degrees of freedom without losing energy, they can still cause “decoherence”, and a “collapse” of the wavefunction.
37. Example Problem Suppose an electron of KE = 0.1 eV approaches a barrier. a) What is the wavelength of this electron? b) For what barrier height will it have a 50% chance of penetrating 1 nm into the forbidden region? What about 1 mm? Solution: a) In the allowed region: Use the De Broglie relation, p = h/l, and the kinetic E = h2/2ml2 = 1.505eVnm2/l2 energy E = p2/2m. = 0.1 eV (for an electron) l = 3.9 nm b) In the forbidden region: U is the unknown barrier height and E = 0.1 eV. The 2 in e-2KL results from probability being |Y|2. K = (2m(U-E)) / e-2KL = 1/2 Note that all that really matters is U-E. You can U = E + ( ln2/2L)2 / 2m also use h2/2m = 1.505eVnm2 for an electron. = 0.1 eV + 7.4 10-22 J L = 1 nm. = 0.105 eV L = 1 mm. Penetration by a significant distance isn’t or = 0.1 eV + 7.4 10-28 J = 0.100000005 eV possible unless the energy is nearly allowed.