Quantum Mechanics in One Dimension. Solutions of Selected Problems

Transcription

1 Chapter 6 Quantum Mechanics in One Dimension. Solutions of Selected Problems 6.1 Problem 6.13 (In the text book) A proton is confined to moving in a one-dimensional box of width.2 nm. (a) Find the lowest possible energy of the proton. (b) What is the lowest possible energy of an electron confined to the same box? (c) How do you account for the large difference in your results for (a) and (b)? Solution (a) The energy of a particle of mass min a on dimensional box of length L is: E n n2 π 2 2 2mL 2 n2 π 2 (h/2π) 2 2mL 2 n2 h 2 8mL 2 The lowest energy (n 1) of a proton with mass m p kg in a box with L m w get:

2 2 CHAPTER 6. QUANTUM MECHANICS IN ONE DIMENSION. SOLUTIONS OF SELECTED PROBLEMS E 1 h 2 8m p L 2 ( J s) kg (2 1 1 m) J J J/eV ev (b) The lowest energy of an electron in a similar box is: E 1 h 2 8m e L 2 ( ) (2 1 1 ) J 9.4 ev (c) The electron energy is much larger than the proton energy because the proton mass m p 2m e.

3 6.2. PROBLEM 6.2 (IN THE TEXT BOOK) Problem 6.2 (In the text book) Consider a particle with energy E bound to a finite square well of height U and width 2L situated on L x +L. Because the potential energy is symmetric about the midpoint of the well, the stationary state waves will be either symmetric or antisymmetric about this point. (a) Show that for E < U, the conditions for smooth joining of the interior and exterior waves lead to the following equation for the allowed energies of the symmetric waves: k tan kl α (symmetric case) where α (2m/ 2 )(U E) and k 2mE/ 2 is the wavenumber of oscillation in the interior. (b) Show that the energy condition found in (a) can be rewritten as k sec kl 2mU Apply the result in this form to an electron trapped at a defect site in a crystal, modeling the defect as a square well of height 5 ev and width.2 nm. Solve the equation numerically to find the ground-state energy for the electron, accurate to ±.1eV. Solution The Schrödinger equation for such a problem can written as: ( ) d 2 ψ 2m dx {U(x) E} ψ(x) 2 2 Inside the well U(x) and outside the well U(x) U independent of x, so Schrödinger equation becomes: ( ) 2m d 2 ψ {U E} ψ(x) for x > +L and x < L (exterior) dx 2 2 ( 2m 2 ) Eψ(x) for L < x < +L (interior) Using k 2 2mE/ 2 and α 2 (2m/ 2 )(U E), Schrödinger, then becomes:

4 4 CHAPTER 6. QUANTUM MECHANICS IN ONE DIMENSION. SOLUTIONS OF SELECTED PROBLEMS { d 2 ψ α 2 dx ψ(x) for x > +L and x < L (exterior) 2 k 2 ψ(x) for L < x < +L (interior) Solution of the interior equation are cos kx and sin kx. The cos kx solution is symmetric around x, while the sin kx is antisymmetric around x. Since we are considering the symmetric case, then the sin kx solution must be discarded. So we then have: ψ(x) A sin kx L < x < +L The solutions of the exterior equation are of the form e ±αx. Since the wavefunction must be zero ant, we then must discard the e +αx and the exterior solution becomes: ψ(x) Ce α x x > L or x < L where A and C are constants. (a) At x L, the the interior and exterior wavefunctions and their derivatives must match, i.e. Dividing Equation (6.2) by Equation (6.1) we get: A cos kl Ce αl (6.1) Ak sin kl Cαe αl (6.2) k tan kl α (6.3) (b) Note that: from which we get: k 2 + α 2 2mE 2 + 2m 2mU (U E) 2 2 α 2 2mU 2 k 2 Using Equation (6.3) we get:

5 6.2. PROBLEM 6.2 (IN THE TEXT BOOK) 5 k tan kl kl tan kl 2mU k 2 2 2mUL 2 k 2 L 2 2 (kl) 2 tan 2 kl 2mUL2 2 k 2 L 2 (kl) 2 (tan 2 kl + 1) 2mUL2 2 It can be shown that (tan 2 kl + 1) sec 2 kl, the last equation than becomes: (kl) 2 sec 2 kl 2mUL2 kl sec kl k sec kl 2 2mU L (6.4) 2mU (6.5) For an electron in a a well of height U 5 ev and width 2L.2 nm, using Equation (6.4) we get: kl sec kl kl cos kl kl cos kl 2me UL 2me c 2 UL c (.1) (6.6) Equation (6.6) can be solved numerically, using sophisticated methods. However, it is not very difficult to solve it by try and error. Trying kl 1 rad makes the right had side of Equation (6.6) larger than , trying kl 1.5 rad gives even larger value. So, obviously kl < 1 rad trying values less than one quickly gives a solution with kl.799, which gives k 7.99 nm 1, the energy is then given by: E 2 k 2 2 c 2 k 2 ( ) ev 2m e 2m e c

6 6 CHAPTER 6. QUANTUM MECHANICS IN ONE DIMENSION. SOLUTIONS OF SELECTED PROBLEMS 6.3 Problem 6.29 (In the text book) An electron is described by the wavefunction ψ(x) where x is in nanometers and C is a constant (a) Find the value of C that normalizes ψ. { for x < Ce x (1 e x ) for x > (b) Where is the electron most likely to be found; that is, for what value of x is the probability for finding the electron largest? (c) Calculate < x > for this electron and compare your result with its most likely position. Comment on any differences you find. Solution (a) Normalization requires: 1 ψ 2 dx C 2 e 2x (1 e x ) 2 dx C 2 e 2x (1 2e x + e 2x ) dx C 2 (e 2x 2e 3x + e 4x ) dx { 1 C } 4 C2 12 C 12

7 6.3. PROBLEM 6.29 (IN THE TEXT BOOK) 7 (b) The most likely location for the electron is where the probability ψ 2 of finding the electron is largest, and that is where the wavefunction ψ is largest,i.e. dψ dx d dx C(e x e 2x ) C 2 { e x + 2e 2x} C 2 e x { 2e x 1 } The right hand side of the last equation vanishes when x and when 2e x 1, or x ln nm. The most likely location of the the electron is x mp.693 nm. (c) The average location of the electron is given by: < x > x ψ 2 dx C 2 xe 2x (1 2e x + e 2x ) dx o C 2 x(e 2x 2e 3x + e 4x ) dx o { 1 C } 16 ( ) 13 C nm < x > is larger than x mp due to the fact that values of x larger that x mp have higher probability than those lower than x mp.

8 8 CHAPTER 6. QUANTUM MECHANICS IN ONE DIMENSION. SOLUTIONS OF SELECTED PROBLEMS 6.4 Problem 6.32 (In the text book) Calculate < x >, < x 2 >, and x for a quantum oscillator in its ground state. Hint: Use the integral formula x 2 e ax2 dx 1 π 4a a a > Solution The quantum oscillator wavefunction is given by ψ(x) Ce ax2, where C 4 a/π and a mω/2. The expectation value of x is then: < x > x ψ 2 dx C 2 xe 2ax2 dx The integrand in the above equation is an odd function of x so integration from to + gives zero, so < x >. Now, < x 2 > is: Now x is given by: < x 2 > x 2 ψ 2 dx 2C 2 x 2 e 2ax2 dx 2C 2 1 π 8a 2a a 1 π 2 π 8a 2a 1 4 2a 2 2mω x < x 2 > (< x >) 2 < x 2 > 2 2mω

9 6.5. PROBLEM 6.37 (IN THE TEXT BOOK) Problem 6.37 (In the text book) Nonstationary states. Consider a particle in an infinite square well described initially by a wave that is a superposition of the ground and first excited states of the well: Ψ(x, ) C [ψ 1 (x) + ψ 2 (x)] (a) Show that the value C 1/ 2 normalizes this wave, assuming ψ 1 and ψ 2 are themselves normalized. (b) Find ψ(x, t) at any later time t. (c) Show that the superposition is not a stationary state, but that the average energy in this state is the arithmetic mean (E 1 + E 2 )/2 of the ground- and first excited-state energies E 1 and E 2. Solution (a) The normalized wavefunctions and energies of a particle in a an infinite square well, with width of L, are given by: 2 ( nπx ) ψ n (x) L sin L E n n2 π 2 2 2mL 2 for o < x < L and n 1, 2 3, The normalization of the superpositioned wave function requires: 1 Ψ 2 dx C 2 {ψ1 + ψ2} {ψ 1 + ψ 2 } dx { C 2 ψ 1 2 dx + ψ 2 2 dx + ψ1ψ 2 dx + } ψ2ψ 1 dx The first two integrals on the right hand side are unity each since each wave function is normalized, while the last two integral are actually the same since both ψ 1 and ψ 2 are real. Using the actual form of the each wavefunction we get:

10 1 CHAPTER 6. QUANTUM MECHANICS IN ONE DIMENSION. SOLUTIONS OF SELECTED PROBLEMS ψ 1 ψ 2 dx 2 L L ( πx ) ( ) 2πx sin sin L L Using sin θ 1 sin θ {sin(θ 1 θ 2 ) cos(θ 1 + θ 2 )}, the last equation becomes: ψ 1 ψ 2 dx 1 L L { 1 L L π 1 { L L π 1 L { ( πx cos [ sin [ sin ) ( )} 3πx cos dx L L ( πx )] L L [ ( )] } L 3πx sin L o 3π L ( )] πl L [ ( )]} 3πL sin L 3π L } { L π sin π L sin 3π 3π The normalization condition becomes: 1 C 2 { ψ 1 2 dx + ψ 2 2 dx + ψ1ψ 2 dx + } ψ2ψ 1 dx C 2 ( ) 2C 2 C 1 2 (b) The time dependent wavefunction is: Ψ(x, t) C { ψ 1 (x)e iω 1t + ψ 2 (x)e iω 2t } since ω E/, Ψ(x, t) C { ψ 1 (x)e ie 1t/ + ψ 2 (x)e ie 2t/ } (c) The time dependent wavefunction Ψ(x, t) is a stationary state only if it is an eigenfunction of the enrgy operator, [E] i, i.e. if: t [E]Ψ(x, t) EΨ(x, t) (6.7)

11 6.5. PROBLEM 6.37 (IN THE TEXT BOOK) 11 Let us see if the above condition in Equation (6.7) can be satisfied: { ( ) ( ) } ie1 [E]ψ C i ψ 1 e ie1t/ ie2 + i ψ 2 e ie 2t/ C { E 1 ψ 1 e ie1t/ + E 2 ψ 2 e } ie 2t/ Since E 1 E 2, then obviously, the stationary state condition in Equation (6.7) is not satisfied, and Ψ(x, t) is not a stationary state. the average energy < E > of the state is: < E > Ψ [E]Ψ dx C 2 { ψ 1(x)e +ie 1t/ + ψ 2(x)e +ie 2t/ } [E] { ψ 1 (x)e ie 1t/ + ψ 2 (x)e ie 2t/ } dx C { 2 ψ1(x)e +ie1t/ + ψ2(x)e } { +ie 2t/ E 1 ψ 1 (x)e ie1t/ + E 2 ψ 2 (x)e } ie 2t/ dx { } C 2 E 1 ψ 1 2 dx + E 2 ψ 2 2 dx The cross product terms vanish as in part (a). Since ψ 1 and ψ 2 are both normalized and C 2 1 2, then: < E > 1 2 (E 1 + E 2 )