Physics A2 - Lecture 7 - Huynh Quang Linh

Nội dung text: Physics A2 - Lecture 7 - Huynh Quang Linh

1. Lecture 7: Schrödinger’s Equation and the Particle in a Box y(x) U= U= n=1 n=3 0 L x n=2
2. 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) dx2 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
3. Constraints on the form of y(x) y(x)2 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!}
4. 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
5. 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)
6. 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
7. 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
8. 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 sinkx + B2 coskx Key point: The total wavefunction y(x) must be continuous at all boundaries !
9. 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
10. 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nm2 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 8mL2 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.
11. 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
12. 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
13. 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:
14. 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 sin2 x dx = N 2 L 2 y2 0 N2 Integral under the curve = 1 N = n=3 0 L x
15. 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.)
16. 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%
17. 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
18. 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
19. 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)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).
20. 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
21. 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
22. 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
23. 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
24. 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?