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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, for 0 < x < L, • but rises to infinity on the outside of the box, that is V , x ≤ 0 and x ≥ L Page 1 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 2 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 to be determined by the properties of the wavefunction as well as boundary and normalization conditions. Page 3 Particle in a one dimensional Box (infinite square well potential) 1. Sin(x) and Cos(x) are finite and single-valued functions 2. Continuity condition is: 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). kL = nπ 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 k 2 2mE 2 Page 4 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 7 n 2 h 2 n 2 2 2 En 2 8mL 2mL2 Thus the energy of the particle bounded in a box is quantized. Another conclusion for the motion of particle in a box can also be drawn that the 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. If the particle is bound in 1D box of width L. The particle can not have zero kinetic energy, because 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 ia an absurd result because there should be node at the boundary for the bounded particle. Page 8 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 4 E4 E4 = L L nx 2 2 n (x) Asin L To find the value of A, we use the normalization 3 L condition 2 2 | n (x) | dx 1 E3 L A 2 sin 2 0 nx dx 1 L A2 L / 2 1 L 2 E2 2 E2 =L L 2 E1 =2L A 2/L 2 nx sin L L E1 2L n=1,2,3,... n Initial 2wavefunctions and Energy 2 n The normalized eignfunctions of the particle n (x) E3 =2L 3 k=2π/λ=nπ/L n 2h 2 En for the first four states in 8mL2 a one-dimensional particle in a box Page 11 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 12 No. of antinodes in eigenfunction = n, No. of nodes =No. of antinodes+1 Page 13 Electron in the 10nm Wide Well with Infinite Barriers n 2h 2 2 En n E1 2 8mL h2 2 2 where E1 2 8mL 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 14 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 15 Particle in the Infinite Potential Well For the n th state 2 nx n sin L L n 2 2 2 En 2mL2 En n 2 E1 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 16 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 17 Page 18 Exercise: Determine < x > for a particle in a box of length ‘L’. 2 nx n (x) sin L L L x n (x) * x n (x)dx 0 2 nx x sin 2 dx L0 L L 2 2L L L4 2 Exercise: Determine < p > for a particle in a box of length ‘L’. 2 n nx nx p * (x) (x)dx sin cos dx i x i L L 0 L L 0 L L L 2 1 k sin(kx) cos(kx)dx 0 (sin ce sin ax cos ax dx sin 2 ax) Li 0 2a 2 nx 2 2 p sin 0 (sin 0 sin n 0, n 1, 2,3,....) iL L 0 L Page 19