Physics A2 - Lecture 7: Schrödinger’s Equation and the Particle in a Box - Huynh Quang Linh

Nội dung text: Physics A2 - Lecture 7: Schrödinger’s Equation and the Particle in a Box - Huynh Quang Linh

1. Lecture 7: Schrửdinger’s Equation and the Particle in a Box y(x) U= U= n=1 n=3 0 L x n=2
2. Content  Particle in a “Box” matter waves in an infinite square well  Wavefunction normalization  General properties of bound-state wavefunctions
3. Last lecture: The time-independent SEQ (in 1D) 2 d 2 y (x) U(x)y (x) Ey (x) KE term 2m dx2 Total E term PE term  Notice that if U(x) = constant, this equation has the simple form: d2y C y(x) dx 2 2m where C ( U E ) is a constant that might be positive or negative. 2 For positive C, what is the form of the solution? a) sin kx b) cos kx c) eax d) e-ax For negative C, what is the form of the solution? a) sin kx b) cos kx c) eax d) e-ax
4. Last lecture: The time-independent SEQ (in 1D) 2 d 2 y (x) U(x)y (x) Ey (x) KE term 2m dx2 Total E term PE term  Notice that if U(x) = constant, this equation has the simple form: d2y C y(x) dx 2 2m where C ( U E ) is a constant that might be positive or negative. 2 For positive C, what is the form of the solution? a) sin kx b) cos kx c) eax d) e-ax For negative C, what is the form of the solution? a) sin kx b) cos kx c) eax d) e-ax
5. Constraints on the form of y(x) 2  y(x) corresponds to a physically meaningful quantity – the probability of finding the particle near x. Therefore, in a region of finite potential: y(x) must be finite, continuous and single-valued. (because probability must be well defined everywhere) dy/dx must be finite, continuous and single valued. (because dy/dx is related to the classical momentum) There is usually no significance to the sign of y(x). (it goes away when we take the absolute square) {In fact, we will see that y(x) can even be complex!}
6. Exercise 1 1. Which of the following hypothetical wavefunctions for a particle in some realistic potential U(x) is acceptable? (a) y(x) (b) y(x) (c) y(x) x x x 2. Which of the following wavefunctions corresponds to a particle more likely to be found on the left side? (a) (b) (c) y(x) y(x) y(x) 0 x 0 x 0 x
7. Solution 1. Which of the following hypothetical wavefunctions for a particle in some realistic potential U(x) is acceptable? (a) y(x) (b) y(x) (c) y(x) x x x (a) Not acceptable (b) Acceptable (c) Not acceptable y(x) is not Both y(x) and dy/dx dy/dx is not continuous at x=0. are continuous continuous at x=0 everywhere dy not defined. dx
8. Solution 2. Which of the following wavefunctions corresponds to a particle more likely to be found on the left side? (a) (b) (c) y(x) y(x) y(x) 0 x 0 x 0 x None of them! (a) is clearly completely symmetric. y2 (b) might seem to be “higher” on the left than on the right, but it is only the absolute square the 0 x determines the probability.
9. Application of SEQ: “Particle in a Box”  Recall, from last lecture, the time-independent SEQ in one dimension: 2 d 2 y (x) U(x)y (x) Ey (x) KE term 2m dx2 Total E term PE term  As a specific important example, consider a quantum particle confined to a small region, 0 < x < L, by infinite potential walls. We call this a “one-dimensional (1D) box”. U(x) This is a basic problem in “Nano-science”. It’s a simplified (1D) model for an electron confined in a quantum structure (e.g., “quantum dot”), which scientists/engineers make, e.g., at the UIUC Microelectronics Laboratory! (www.micro.uiuc.edu) ‘Quantum 0 L dots’ U = 0 for 0 < x < L U = everywhere else (www.kfa-juelich.de/isi/) (newt.phys.unsw.edu.au)
10. Waves: Boundary conditions  Boundary condition: Constraints on a wave where the potential changes  E = 0 at surface of a metal film Displacement = 0 for wave on string E = 0  If both ends are constrained (e.g., for a cavity of length L), then only certain wavelengths l are possible: n l 1 2L 2 L nl = 2L 3 2L/3 n = 1, 2, 3 ‘mode index’ 4 L/2 L Rope Demo
11. Particle in a Box (1)  Solving the SEQ for the particle-in-a-box: (a basic boundary-value problem)  There are 2 distinct regions, (I) outside, and (II) inside the well Region I: When U = , what is y(x)? U(x) d 2 y (x) 2m (E U)y (x) 0 I II I dx2 2 0 L U = 0 for 0 < x < L U = everywhere else
12. Particle in a Box (1)  Solving the SEQ for the particle-in-a-box: (a basic boundary-value problem)  There are 2 distinct regions, (I) outside, and (II) inside the well Region I: When U = , what is y(x)? U(x) d 2 y (x) 2m (E U)y (x) 0 dx2 2 I II I yI yI For U = , the SEQ can only be satisfied if: 0 L yI(x) = 0 U = 0 for 0 < x < L U = everywhere else (otherwise, the energy would have to be infinite, to cancel U.) Note: The infinite well is an idealization. On the atomic scale there are no infinitely high and sharp barriers. y(x) and dy(x)/dx go to zero continuously near a boundary.
13. Particle in a Box (2) U(x) Region II: When U = 0, what is y(x)? d 2y (x) 2m (E U )y (x) 0 II dx2 2 d 2y (x) 2mE y (x) 0 L 2 2 dx  U = 0 for 0 < x < L U = everywhere else
14. Particle in a Box (2) U(x) Region II: When U = 0, what is y(x)? II d 2y (x) 2m (E U )y (x) 0 dx2 2 y d 2y (x) 2mE y (x) 0 L dx2 2 The general solution to this is: 2 where, k y (x) B1 sin kx B2 coskx l Oscillatory solutions! Plugging y(x) into 2k2 h2 Consistent with DeBroglie’s the SEQ, we find hypothesis: E 2 the requirement: 2m 2ml p = h/l = ħk and E = p2/2m B1 and B2 are coefficients to be determined by the boundary conditions.
15. Particle in a Box (3) U(x)  Matching wavefunctions at the boundaries: I The wavefunction takes on the following form: I II y y II Region I: y I ( x ) 0 I yI 0 L Region II: y II( x ) B1 sin kx B2 cos kx Key point: The total wavefunction y(x) must be continuous at all boundaries !
16. Particle in a Box (3) U(x)  Matching y at the first boundary (x = 0) I The wavefunction takes on the following form: I II y y II y Region I: y I ( x ) 0 I I 0 L Region II: y II( x ) B1 sin kx B2 cos kx Key point: The total wavefunction y(x) must be continuous at all boundaries ! At x = 0: y I (x 0) y II (x 0) 0 B1 sin(0) B2 cos(0) 0 B2 because cos(0) = 1 and sin(0) = 0 So this “boundary condition” requires that there be no cos(kx) term!
17. Particle in a Box (4) U(x)  Matching y at the second boundary (x = L) I II I At x = L: y I (x L) y II (x L) y y II y 0 B1 sin (kL) I I 0 This constraint forces k to have special values ! L n 2 knn , 1, 2, Using k we find : nl 2L L l This is precisely the condition we found for confined waves, e.g., EM waves in a laser cavity: n l (= c/f) 4 L/2 For matter waves, the 3 2L/3 wavelength is related to the particle energy: 2 2 2 L En = h /2ml 1 2L Therefore
18. In a confining potential, Particle Energy is Quantized. • The discrete En are known as “energy eigenvalues”: electron n l En nln 2L 4 L/2 h2 1.505 eV nm 2 En 2 2 3 2L/3 2mln ln h2 2 L E E n2 with E n 1 1 8mL 2 1 2L E1 U= U= y(x) En U= U= n=1 n=3 n=3 n=2 n=1 0 L x n=2 0 L x Allowed wavefunctions have an integral # of half-wavelengths that precisely “fit” in the well.
19. In a confining potential, Particle Energy is Quantized. • The discrete En are known as “energy eigenvalues”: electron n ln En nln 2L 4 L/2 16E h2 1.505 eV nm 2 1 E n 2ml2 l2 3 2L/3 9E1 n n 2 2 h 2 L 4E1 E E n with E n 1 1 8mL 2 1 2L E1 U= U= y(x) En U= U= n=1 n=3 9E1 n=3 4E1 n=2 E 0 L x 1 n=1 n=2 0 L x Allowed wavefunctions have an integral # of half-wavelengths that precisely “fit” in the well.
20. Particle-in-Box: Example  Calculate ground state energy and the photon energy for a transition. An electron is trapped in a “quantum wire” that is L = 4 nm long. Assuming that the potential seen by the electron is approximately that of an infinite square well, estimate the ground (lowest) state energy of the electron. What photon energy is required to excite the trapped electron to the next available energy level (i.e., n = 2)? U= U= En n=3 n=2 n=1 0 L x
21. Particle-in-Box: Example  Calculate ground state energy and energy for a transition. An electron is trapped in a “quantum wire” that is L = 4 nm long. Assuming that the potential seen by the electron is approximately that of an infinite square well, estimate the ground (lowest) state energy of the electron. Allowed energies for electron in a 1D box: 2 2 h 1.505 eV nm l 2L/n En 2 2 n 2mln ln 2 2 E = ground state 2 h 1.505 eV nm 1 E E n with E energy (n = 1) n 1 1 8mL2 4L2 1.505 eV nm2 E 0.0235 eV 1 4( 4nm )2 What photon energy is required to excite the trapped electron to the next available energy level (i.e., n = 2)? U= U= En n=3 n=2 n=1 0 L x
22. Particle-in-Box: Example  Calculate ground state energy and energy for a transition. An electron is trapped in a “quantum wire” that is L = 4 nm long. Assuming that the potential seen by the electron is approximately that of an infinite square well, estimate the ground (lowest) state energy of the electron. Allowed energies for electron in a 1D box: 2 2 h 1.505 eV nm l 2L/n En 2 2 n 2mln ln h2 1.505 eV nm2 E = ground state E E n2 with E 1 n 1 1 8mL2 4L2 energy (n = 1) 1.505 eV nm2 E 0.0235 eV 1 4( 4nm )2 What photon energy is required to excite the trapped electron to the next available energy level (i.e., n = 2)? U= U= En 2 E 22 12 E n=3 En n E1 ( ) 1 n=2 So, energy difference 3E1 n=1 between n = 2 and n = 1 = 0.071 eV 0 L x levels:
23. Lecture 7, exercise 2 1. An electron is in a quantum “dot”. If U= U= we decrease the size of the dot, the En ground state energy of the electron will n=3 n=2 a) decrease n=1 b) increase 0 L x c) stay the same 2. If we decrease the size of the dot, the difference between two energy levels (e.g., between n = 2 and 3) will a) decrease b) increase c) stay the same
24. Lecture 7, exercise 2 1. An electron is in a quantum “dot”. If U= U= we decrease the size of the dot, the En ground state energy of the electron will n=3 n=2 a) decrease n=1 h2 b) increase 0 L x E1 c) stay the same 8mL2 2. If we decrease the size of the dot, the difference between two energy levels (e.g., between n = 2 and 3) will a) decrease ΔE ~ E1 b) increase E3 - E2 = (9 – 4)E1 = 5E1 c) stay the same Since E1 increases, so does ΔE
25. M. Nayfeh (UIUC) : New photonic and electronic material Discrete uniform Si nanoparticles • Transition from bulk to molecule-like in Si • A family of magic sizes of hydrogenated Si nanoparticles • No magic sizes > 20 atoms for non- hydrogenated clusters • Small clusters glow: color depends on size (“quantum confinement”) • Used to create Si nanoparticle 1 nm 1.67 nm 2.15 nm 2.9 nm Blue Green Yellow Red microscopic laser:
26. Probabilities What we measure in an experiment is the probability density, |y(x)|2. n Wavefunction = 2 2 2 n Probability per y n (x) N sin x Probability amplitude y n (x) N sin x unit length L L (in 1-dimension) y y2 U= U= n=1 0 L x 0 L x y y2 n=2 0 L x 0 L x y y2 n=3 0 L x 0 L x
27. Probability Density 2 2 2 n Probability per y n( x ) N sin x unit length L (in 1-dimension) For a classical particle bouncing back and forth in a well: The probability of finding the particle is equally likely throughout the well. For a quantum particle in a stationary state, the probability distribution is NOT uniform there are “nodes” where the probability is ZERO! Reflects INTERFERENCE effects caused by the wave-like character of quantum particles! The “normalization factor” N is determined by setting the total probability of finding the particle in the well equal to one: L 2 n L Total probability y dx N 2 sin 2 x dx N 2 L 2 y2 0 N2 Integral under the curve = 1 N = n=3 0 L x
28. Probability Density – Example problem Consider an electron trapped in a 1D well with L = 5 nm. Let’s say that the electron is in the following state: y2 N2 2 2 2 n y n( x ) N sin x L 0 5 nm x a) What is the energy of the electron in this state (in eV)? b) What is the value of the normalization factor squared N2? c) Estimate the probability of finding the electron within ±0.1 nm of the center of the well? (No integral required. Do it graphically.) Discussed in section.
29. Probability Density – Example problem Consider an electron trapped in a 1D well with L = 5 nm. Let’s say that the electron is in the following state: y2 N2 2 2 2 n y n( x ) N sin x L 0 5 nm x a) What is the energy of the electron in this state (in eV)? 1.505 eV nm2 n 3, E 32 0.135 eV 3 2 4( 5nm) Answer = 0.135 eV b) What is the value of the normalization factor squared N2? c) Estimate the probability of finding the electron within ±0.1 nm of the center of the well? (No integral required. Do it graphically.)
30. Probability Density – Example problem Consider an electron trapped in a 1D well with L = 5 nm. Let’s say that the electron is in the following state: y2 N2 2 2 2 n y n( x ) N sin x L 0 5 nm x a) What is the energy of the electron in this state (in eV)? 1.505 eV nm2 n 3, E 32 0.135 eV 3 2 4( 5nm) Answer = 0.135 eV b) What is the value of the normalization factor squared N2? (Area under y 2 ) N 2L / 2 1 Answer = 0.4 nm-1 c) Estimate the probability of finding the electron within ±0.1 nm of the center of the well? (No integral required. Do it graphically.)
31. Probability Density – Example problem Consider an electron trapped in a 1D well with L = 5 nm. Let’s say that the electron is in the following state: y2 N2 2 2 2 n y n( x ) N sin x L 0 5 nm x a) What is the energy of the electron in this state (in eV)? 1.505 eV nm2 n 3, E 32 0.135 eV 3 2 4( 5nm) Answer = 0.135 eV b) What is the value of the normalization factor squared N2? (Area under y 2 ) N 2L / 2 1 Answer = 0.4 nm-1 c) Estimate the probability of finding the electron within ±0.1 nm of the center of the well? (No integral required. Do it graphically.) Probability y 2( x ) N 2( 0.2nm ) 0.08 Answer = 0.08 = 8%
32. Properties of Bound States Several trends exhibited by the particle-in-box states are generic to bound eigenstate wavefunctions in any 1D potential (even complicated ones). y(x) (1) The overall curvature of the wavefunction increases with increasing kinetic energy. U= U= 2 d 2 y (x) p2 n=1 n=3 like 2 2m dx 2m 0 L x n=2
33. Properties of Bound States Several trends exhibited by the particle-in-box states are generic to bound eigenstate wavefunctions in any 1D potential (even complicated ones). (1) The overall curvature of the wavefunction y(x) increases with increasing kinetic energy. 2 d 2 y (x) p2 U= U= like n=1 n=3 2 2m dx 2m (2). The lowest energy bound state always 0 has finite energy a “zero-point” energy L x n=2 Even the lowest energy bound state requires some wavefunction curvature (kinetic energy) to satisfy boundary conditions
34. Properties of Bound States Several trends exhibited by the particle-in-box states are generic to bound eigenstate wavefunctions in any 1D potential (even complicated ones). (1) The overall curvature of the wavefunction y(x) increases with increasing kinetic energy. U= 2 d 2 y (x) p2 U= like n=1 n=3 2 2m dx 2m (2). The lowest energy bound state always 0 x has finite energy a “zero-point” energy L n=2 Even the lowest energy bound state requires some wavefunction curvature (kinetic energy) to satisfy boundary conditions (3). The n-th wavefunction (eigenstate) has (n-1) zero-crossings. Larger n means larger E, which means more wiggles in the wavefunction.
35. Properties of Bound States Several trends exhibited by the particle-in-box states are generic to bound state wavefunctions in any 1D potential (even complicated ones). y (1) The overall curvature of the wavefunction (x) increases with increasing kinetic energy. U= 2 d 2 y (x) p2 U= like n=1 n=3 2 2m dx 2m (2). The lowest energy bound state always 0 L x has finite energy a “zero-point” energy n=2 Even the lowest energy bound state requires some wavefunction curvature (kinetic energy) to satisfy boundary conditions (3). The n-th wavefunction (eigenstate) has (n-1) zero-crossings. Larger n means larger E, which means more wiggles in the wavefunction. (4). If the potential U(x) has a center of symmetry (such as the center of the well above), the eigenstates will be (alternating) even or odd functions about that center of symmetry
36. Bound State Properties: Example  Let’s reinforce your intuition about the properties of bound state wavefunctions with this example: Through “nano-engineering,” you are able to create a “step” in the potential “seen” by an electron trapped in a 1D structure, as shown below. You’d like to estimate the wavefunction for an electron in the 5th energy level of this potential, without solving the SEQ. Qualitatively sketch the 5th wavefunction: Things to consider: U= U= (1) 5th wavefunction has _ zero-crossings. E5 (2) Wavefunction must go to zero at x = _ and x = _. Uo (3) Kinetic energy is ___ on right side of 0 L x well, so the curvature of y is ___ there y (wavelength is longer). (4) Kinetic energy is ___ on right side, so amplitude of y is ___ there. (Classically, x the particle spends more time there because it is moving more slowly).
37. Bound State Properties: Example  Let’s reinforce your intuition about the properties of bound state wavefunctions with this example: Through “nano-engineering,” you are able to create a “step” in the potential “seen” by an electron trapped in a 1D structure, as shown below. You’d like to estimate the wavefunction for an electron in the 5th energy level of this potential, without solving the SEQ. Qualitatively sketch the 5th wavefunction: Things to consider: U= U= (1) 5th wavefunction has _4 zero-crossingscrossings. E5 (2) Wavefunction must go to zero at x = _0 and x = _.L. Uo (3) Kinetic energy is ___lower on right sideside of 0 L x well, so the curvature of y is ___smallertherethere y (wavelength is longer). (4) Kinetic energy is ___lower on right side,side, so amplitude of y is ___larger there.there. x (Classically, the particle spends more time there because it is moving more slowly).
38. Bound State Properties: Example  Let’s reinforce your intuition about the properties of bound state wavefunctions with this example: Through “nano-engineering,” you are able to create a “step” in the potential “seen” by an electron trapped in a 1D structure, as shown below. You’d like to estimate the wavefunction for an electron in the 5th energy level of this potential, without solving the SEQ. Qualitatively sketch the 5th wavefunction: Things to consider: U= U= (1) 5th wavefunction has _4 zero-crossingscrossings. E5 (2) Wavefunction must go to zero at x = _0 and x = _. Uo and x = L. (3) Kinetic energy is ___lower on right sideside of 0 L x well, so the curvature of y is ___smallertherethere y (wavelength is longer). (4) Kinetic energy is ___lower on right side,side, so amplitude of y is ___larger there.there. x (Classically, the particle spends more time there because it is moving more slowly).
39. Lecture 7, exercise 3 The wavefunction below describes a quantum particle in a range x: y(x) In what energy level is the particle? n = (a) 7 (b) 8 x (c) 9 What is the approximate shape of the potential U(x) in which this x particle is confined? U(x) U(x) U(x) (a) (b) (c) E E E x x x
40. Lecture 7, exercise 3 The wavefunction below describes a quantum particle in a range x: y(x) In what energy level is the particle? n = (a) 7 (b) 8 x (c) 9 What is the approximate shape of the potential U(x) in which this x particle is confined? U(x) U(x) U(x) (a) (b) (c) E E E x x x
41. Lecture 7, exercise 3 The wavefunction below describes a quantum particle in a range x: y(x) In what energy level is the particle? Smaller l, Larger l, n = larger KE smaller KE (a) 7 Symmetric (b) 8 distribution x (c) 9 What is the approximate shape of the potential U(x) in which this x particle is confined? U(x) U(x) U(x) (a) (b) (c) E E E x x x
42. Lecture 7, exercise 3 The wavefunction below describes a quantum particle in a range x: y(x) In what energy level is the particle? Smaller l, Larger l, n = larger KE smaller KE (a) 7 Symmetric (b) 8 distribution x (c) 9 What is the approximate shape of the potential U(x) in which this x particle is confined? U(x) U(x) U(x) (a) (b) (c) E E E not symmetric KE smaller in middle x x x
43. Supplementary Problem 1  Suppose a ball of mass 1 g trapped in our (1-d) box - 1 cm wide - has an energy of 10-10 J. What is its “quantum number”, n? (Hint:  = 1 x 10-34 J•s) Answer: 1025
44. Supplementary Problem 1  Suppose a ball of mass 1 g trapped in our (1-d) box - 1 cm wide - has an energy of 10-10 J. What is its “quantum number”, n? (Hint: h = 6.626 x 10-34 J•s, p = h/l =  k ) 2 22 2 pn h h n En 2  2m 2mln 2m 2L 8mE 8 10 3 kg 10 10 J n2  L 2  10 4 m 2 h2 43.9 10 68 J 2 s 2 n ~ 1025 ! 2 22k 2 n n -34 or use: En  ħ = 1.0546 x 10 J•s  2m 2m L 2 2 n n n Note : k therefore y (x) = Nsin(kx) =Nsin x l 2L L L
45. Supplementary Problem 2 Consider a particle in the n = 2 state of a box. y(x) a) Where is it most likely to be found? b) Where is it least likely to be found? U= U= c) What is the ratio of probabilities for the particle to be near x = L/3 and x = L/4? n=2 0 L/4 L/3 L x
46. Supplementary Problem 2 Consider a particle in the n = 2 state of a box. y(x) a) Where is it most likely to be found? b) Where is it least likely to be found? U= U= c) What is the ratio of probabilities for the particle to be near x = L/3 and x = L/4? n=2 0 L/4 L/3 L x Solution: a) x = L/4 and x = 3L/4. Maximum probability is at max y. b) x = 0, x = L/2, and x = L. Minimum probability is at the nodes. c) y(x) = Nsin(2 x/L) The sine wave must have nodes at x = 0, x = L, and, because n = 2, at x = L/2 as well. Prob(L/3) / Prob(L/4) Prob(x) = |y(L/3)|2 / |y(L/4)|2 = sin2(2 /3) / sin2( /2) = 0.8662 = 0.75 0 L x
47. Homework # 4  Why is the Heisenberg uncertainty principle not more readily apparent in our daily observation ?  Does a photon have a de Broglie wavelength ? Explain.  Why the wave nature of matter not more apparent in our dialy observation? 