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Quantum Mechanics Lecture-8 Concept of Modern Physics by “A. Beiser” Page 1 Exercise: A Particle limited to the x-axis has the wave function =ax between x=0 and x=1; =0 elsewhere. (a)Find the probability that the particle can be found x=0.45 and x=0.55. (b)Find the expectation value < x > of the particle’s position. (a) 0.0251 a2 (b) a2/4 Page 2 Particle in a one dimensional Box (infinite square well potential) • A particle is confined to a one-dimensional region of space between two impenetrable walls separated by distance L – This is a one- dimensional “box” • The particle is bouncing elastically back and forth between the walls – As long as the particle is inside the box, the potential energy does not depend on its location. We can choose this energy value to be zero • V= 0, 0 < x < L, V , x ≤ 0 and x ≥ L Page 5 Particle in a one dimensional Box (infinite square well potential) • Since the walls are impenetrable, there is zero probability of finding the particle outside the box. Zero probability means that ψ(x) = 0, for x < 0 and x > L • The wave function must also be 0 at the walls (x = 0 and x = L), since the wavefunction must be continuous – Mathematically, ψ(0) = 0 and ψ(L) = 0 2 2 x V x E x 2m x 2 • In the region 0 < x < L, where V = 0, the Schrödinger equation can be expressed in the form 2 x E x 2m x 2 2 Page 6 Particle in a one dimensional Box (infinite square well potential) • We can re-write it as 2 x 2mE 2 x 2 x 2 x 2 k x 2 x k 2 2mE 2 The most general solution to this differential equation is ψ(x) = A sin kx + B cos kx – A and B are constants determined by the properties of the wavefunction as well as boundary and normalization conditions Page 7 Particle in a one dimensional Box (infinite square well potential) Since the particle cannot have infinite energy, it cannot exist outside the box. Therefore, the wave function must be zero outside the box. must be also zero at the walls, that is, at x = 0 and x = L, for otherwise there would be discontinuities at the walls. 1. Sin(x) and Cos(x) are finite and single-valued functions 2. Continuity: ψ(0) = ψ(L) = 0 • ψ(0) = A sin(k0) + B cos(k0) = 0 B = 0 Not admissible ψ(x) = A sin(kx) • ψ(L) = A sin(kL) = 0 sin(kL) = 0 kL = nπ, n =0, ±1, ±2… ( n = 0 is not admissible because it yields zero everywhere which means that the particle is no where). kn 2 L n 2 kn En 2m 2 ( n) 2 2 2 2 2 h 2 L n n En 2 2 2m 2mL 8mL Page 8 Particle in a one dimensional Box (infinite square well potential) The particle can not have arbitrary energy , but can have certain discrete energy corresponding to n=1,2,3,… n2h2 En 8mL2 Each permitted energy is called ‘eigen value’ of the particle and constitute the energy level of the system, and the integer ‘n’ that specifies an energy level En is called its principal quantum number. The wave function corresponding to each eigen value is called eigen functions. Lowest level n = 1, energy not zero why ? Page 11 Conclusions n 2 h 2 n 2 2 2 En 2 8mL 2mL2 Energy of the particle bounded in a box is quantized. Particle can not have zero energy but has minimum energy and called as zero point energy. The state corresponding to this energy is called ground state. The particle can not have zero kinetic energy, because from Heisenberg uncertainty principle, the uncertainty in position of the trapped particle in a box is Δx = L, hence Δp as well as velocity of the particle and their kinetic energy can not be zero. According to classical mechanics, when a particle is placed in a box, it can have zero energy or continuous kinetic energy. Thus the quantization of energy is a specific result derived from quantum mechanics. If one assumes v = 0 for a particle in a box, the de-Broglie wave associated with it will be λ =(h/mv) = , which is an absurd result because there should be node at the boundary for the bounded particle. Page 12 An e- is in a box 0.10 nm across, which is the order of magnitude of atomic dimensions. Find its permitted energies. n 2h 2 En 8mL2 n 2 (6.63 1034 J.s) 2 En 6.0 1018 n 2 J 31 10 2 8(9.1 10 kg)(1.0 10 m) 38n 2eV The minimum energy of the e- can have 38eV, corresponding to n=1.the sequence of energy levels continues with E2=152 eV, E3=342 eV, and so on. If such box existed, the quantization of a trapped e-’s energy would be a prominent feature of the system. ( and indeed energy quatization is prominent in the case of an atomic election.) A 10 g marble is in a box 10 cm across. Find its permitted energies. n 2 (6.63 1034 J.s) 2 En 8.28 1065 n 2 J 2 1 2 8(1.0 10 kg)(1.0 10 m) The minimum energy of the marble can have 8.2810-65 J, corresponding to n=1. A marble with this K.E. has a speed of only 1.310-31 m/s and therefore can not be experimentally distinguished from a stationary marble. Hence in the domain of everyday experience, quantum effects are unnoticeable, which accounts for the success of Newtonian Mechanics in this domain. (mL2>>h2) Page 13 Particle in a one dimensional Box (infinite square well potential) 2 x 2mE 2 x 2 x 2 x 2 k x 2 x (x) Asin(kx) Bcos(kx) (0) Bcos(0) 0 (L) Asin(kL) 0 k n L n kn n L 2mE n n 2 L 2 kn2 n 2h 2 En 8mL2 n (x) Asin( x) L Page 14 Particle in a one dimensional Box (infinite square well potential) Let us now consider the eigenfunctions of the particle. Substituting B=0 and k=n/L in the equation of general solution. Solutions of the Schrödinger equation are E4 E4 = L nx 2 n (x) Asin L To find the value of A, we use the normalization condition 2 | n E3 E3 =2L 3 E2 E2 =L E1 E1 =2L (x) | dx 1 L A 2 sin 2 0 nx dx 1 L A2 L / 2 1 A 2/L The normalized eignfunctions of the particle n (x) 2 nx sin L L 2L n=1,2,3,... n Initial 2wavefunctions and Energy 2 n n 2h 2 En for the first four states in 8mL2 a one-dimensional particle in a box Page 15 Particle in a one dimensional Box (infinite square well potential) Wavefunctions 2 nx n (x) sin L L n=1,2,3,... Probability: *n n n=1,2,3,... Note particle most likely to be found in the middle for n=1 Energy: (a) The first two wavefunctions, (b) the corresponding probability distributions, and (c) a representation of the probability distribution in terms of the darkness of shading. n 2h 2 E= 8mL2 n=1,2,3,... Page 16 No. of antinodes in eigenfunction = n, No. of nodes =No. of antinodes+1 Page 17 Electron in the 10nm Wide Well with Infinite Barriers 2 2 2 En n E1 , where E1 2mL2 • Calculate E1=? 3.142 (1.05 1034 )2 E1 2 9.11031 (10 109 )2 E1 6 1022 J 0.00375 eV 3.75 meV • Assume that a photon is absorbed, and the electron is transferred from the ground state (n = 1) to the second excited state (n = 3) What was the wavelengths of the photon? Page 18 Electron in the 10nm Wide Well with Infinite Barriers • E1 = 3.75 meV Eground E1 0.00375 eV Third excited state is E3 E3 3 E1 9 0.00375 eV 0.0338 eV 2 (h ) E3 E1 0.0338 0.00375 0.030 eV 34 c 6.63 10 3 10 h 0.030eV, 0.030 1.6 1019 1240 λ 41333 nm 41 μm 0.03 8 Page 19 Particle in the Infinite Potential Well For the n th state 2 x n sin n L L 2 2 2 En n 2 2mL En n 2E1 Probability to Find particle in the Right Half of the Well L L 2 2 | (x) | dx [ sin kx] dx L/2 L/2 L 2 L 2 2 L1 1 sin 2 (kx)dx [ ] L L/2 L 22 2 Page 20 Some trajectories of a particle in a box according to Newton's laws of classical mechanics (A), and according to the Schrödinger equation of quantum mechanics (BD). In (B-D), the horizontal axis is position, and the vertical axis is the real part (blue) or imaginary part (red) of the wave function. The states (B,C,D) are energy eigenstates. Page 21 Page 22 Problems: Normalization. 1. Determine normalization constant i ) ( x) N sin(nx / L), 0 x L Ans : N 2 L ii) ( x) N exp kx , 0 x Ans : N 2k x2 iii) ( x) N exp 2 ix , x Ans : N 2a iv) ( x) N cos x , x Ans : N 2 2 2 1 a 8 3 2 1 8 i ) Ans : N , ii ) Ans : N 2 k iii ) Ans : N , iv ) An : N L 3 a 23 Page 23 Hint for solution of (iii) problem * ( x ) ( x )dx 1 N2 x2 exp 2 dx 1 a 2aN 2 Say x t , dx adt if x , t a exp t 2 dt 1 2aN 1 2 1 N a 2 Page 24 Problems: Particle in a box. 1. An particle is confined to an onedimensional infinity potential well of width 0.2 nm. It is found that when the energy of the particle is 230 eV, its eigenfunction has 5 antinodes. Find the mass of the particle and show that it can never have energy equal to 1 keV. Answer m =9.1 x10E-31 Kg n = 10.4 Page 25 Problems: Particle in a box. 1. Find the probability that a particle trapped in a box L wide can be found between 0.45L and 0.55L for the ground and first excited state. 2. Write down Schrödinger equation for a one dimensional box, obtain the expression for eigen function and eigen values. If length of the box ix 25Å, calculate the probability of finding the particle with in an interval of 5Å at the centre of the box when it is in the state of least energy. 3. A particle is in a cubic box with infinitely hard walls whose edges are L long. The wave function of the particle is given by by 8 n x n yy n zz ( x, y, z ) A sin x sin sin Ans : A 3 L L L L Find the normalization constant A. 26 Page 26