Introduction to Quantum Mechanics Coursework - 1100 Words

Introduction to Quantum Mechanics (Coursework Sample) Content: INTRODUCTION TO QUANTUM MECHANICS(Student Name)(Course No.)(Lecturer)(University)(City State)(Date) 1 Solution * The function f(x) = e-i(3x+2y) will be an Eigen function if we apply the operator d2/dx2 to the function and the function is regenerated.d2/dx2 f(x) = d/dx [d/dxe-i(3x+2y)]= d/dx-3ie-i(3x+2y)= -9e-i(3x+2y)= -9f(x)Thus, acting on the function f(x) = e-i(3x+2y) with the operator d2/dx2 regenerates the function, and the Eigen value in this case is -9.We can say that e-i(3x+2y) is an Eigen function of the operator d2/dx2 with -9 as the Eigen value. * f(x) = x2+y2(1/x)(x2+y2) d/dx[f(x,y)] = (1/x)(x2+y2) d/dxx2+y2= (1/x)(x2+y2) d/dxx2+y2 = (1/x)(x2+y2)[1/2(2x)( x2+y2)-1/2]= (1/x)(x2+y2)[x(x2+y2)-1/2] = (1/x)(x2+y2)[x/ (x2+y2)1/2]= (x2+y2)1/2Thus, the function f(x) = x2+y2 is the Eigen function of the operator (1/x)(x2+y2) d/dx and the Eigen value is 1. * f(ÃŽÂ ¸)=SinÃŽÂ ¸CosÃŽÂ ¸Applying the operator Sin ÃŽÂ ¸ddÃŽÂ ¸(SinÃŽÂ ¸ddÃŽÂ ¸) + 6 Sin 2ÃŽÂ ¸ we have;= Sin ÃŽÂ ¸ddÃŽÂ ¸SinÃŽÂ ¸CosÃŽÂ ¸(SinÃŽÂ ¸ddÃŽÂ ¸SinÃŽÂ ¸CosÃŽÂ ¸) + 6 Sin2ÃŽÂ ¸= Sin ÃŽÂ ¸[CosÃŽÂ ¸ddÃŽÂ ¸SinÃŽÂ ¸+ SinÃŽÂ ¸ddÃŽÂ ¸ CosÃŽÂ ¸] {Sin ÃŽÂ ¸[CosÃŽÂ ¸ddÃŽÂ ¸SinÃŽÂ ¸+ SinÃŽÂ ¸ddÃŽÂ ¸ CosÃŽÂ ¸]} + 6 Sin2ÃŽÂ ¸= Sin ÃŽÂ ¸[Cos2 ÃŽÂ ¸ - Sin2ÃŽÂ ¸] {Sin ÃŽÂ ¸[Cos2 ÃŽÂ ¸ - Sin2ÃŽÂ ¸]} + 6 Sin2ÃŽÂ ¸From this equation it is clear that the function f(ÃŽÂ ¸)=SinÃŽÂ ¸CosÃŽÂ ¸ cannot be generated. Thus, the function is not an Eigen function of the operatorSin ÃŽÂ ¸ddÃŽÂ ¸(SinÃŽÂ ¸ddÃŽÂ ¸) + 6 Sin2ÃŽÂ ¸ 2 Solutionf(x) = e-12ÃŽx^2 and f(x) = (2ÃŽx2 à ¢Ã¢â€š ¬ 1) e-12ÃŽx^2 , -à ¢Ã… ¾Ãƒ ¢xà ¢Ã… ¾To find if the two wave functions are orthogonal. We find the integral of their product over the indicated range. If it is zero then they are orthogonal CITATION Eri12 \l 1033 (Smith Fellows, 2012).Therefore;-à ¢Ã… ¾Ãƒ ¢Ã… ¾(2ÃŽx^2 à ¢Ã¢â€š ¬ 1) e(-12ÃŽx^2)^2dx= -à ¢Ã… ¾Ãƒ ¢Ã… ¾(2ÃŽx^2 à ¢Ã¢â€š ¬ 1) e1/4ÃŽ^2x^4dx=-à ¢Ã… ¾Ãƒ ¢Ã… ¾[2ÃŽx2 e1/4ÃŽ^2x^4 - e1/4ÃŽ^2x^4]dx=-à ¢Ã… ¾Ãƒ ¢Ã… ¾2ÃŽx2 e1/4ÃŽ^2x^4dx - -à ¢Ã… ¾Ãƒ ¢Ã… ¾e1/4ÃŽ^2x^4dxIntegrating the first part of the equation by parts;Let u = 2ÃŽx2 and dv = e1/4ÃŽ^2x^4dxdu = 4 ÃŽx dx v = 15 e1/4ÃŽ^2x^4= 25ÃŽx2 e1/4ÃŽ^2x^4 - -à ¢Ã… ¾Ãƒ ¢Ã… ¾4/5ÃŽx e1/4ÃŽ^2x^4dxIntegrating the second part of the equation by parts;Let u = 4/5ÃŽx and dv = e1/4ÃŽ^2x^4dxdu = 4/5ÃŽdx v = 1/5 e1/4ÃŽ^2x^4= 4/25ÃŽx e1/4ÃŽ^2x^4 - -à ¢Ã… ¾Ãƒ ¢Ã… ¾4/25ÃŽ e1/4ÃŽ^2x^4dxIntegrating the second part of the equation;4/25ÃŽx e1/4ÃŽ^2x^4 - 4/125ÃŽ e1/4ÃŽ^2x^4= [25ÃŽx2 e1/4ÃŽ^2x^4 - 4/25ÃŽx e1/4ÃŽ^2x^4 + 4/125ÃŽ e1/4ÃŽ^2x^4 - 1/5 e1/4ÃŽ^2x^4]= à ¢Ã… ¾-à ¢Ã… ¾+à ¢Ã… ¾ - à ¢Ã… ¾ = 0Thus, the two wave functions are orthogonal. 3 SolutionFrom 0aà ¢Ã‚ £ÃƒÅ½Ã‚ ¨(x)à ¢Ã‚ £2dx = 1 , we will have;=0aA2 (Xa)41-xa2dx=1=A40a(xa)4[1-xa]4dx=1=0a(xa-x2)4dx=(ax)4= a59-2a39+6a79-a29+a99=(aA)4A4=a-5A = 1a54 4 SolutionFor the particle moving along a circle covers some radius r from center of circle, expression for angular momentum is L= rÃÆ' p CITATION Eri12 \l 1033 (Smith Fellows, 2012)Where p is linear momentum with operator à ¡Ã‚ ¹(x, y) = ÄiddxdyHence, operator for angular momentum is à ¡Ã‚ ¸Ã‚ ¼= à ¡Ã‚ ¹ ÃÆ' à ¡Ã‚ ¹, = à ¡Ã‚ ¹ÃƒÆ'= ÄiddxdyThis is for the particle in circular motion along xy-plane.Applying wavefunction; à ¡Ã‚ ¸Ã‚ ¼= à ¡Ã‚ ¹ÃƒÆ' Äiddxdy(exp-2ià )Hence the position à  is in the xy-planeThat is; à ¡Ã‚ ¸Ã‚ ¼= à ¡Ã‚ ¹ ÃÆ' Äiddà (exp-2ià ) = - à ¡Ã‚ ¹2iÄexpà ¢Ã‚ Ã‚ ¡(-2ià )i, = -2ià ¡Ã‚ ¹ÃƒÅ½Ã‚ ¨(à ).The angular momentum is therefore L= -2ià ¡Ã‚ ¹ÃƒÅ½Ã‚ ¨(à ) 5 SolutionFor normalized wavefunction; ʃ [ÃŽÂ ¨(à )]2dà =1 * ʃ [à ¢Ã… ¡(12à ¢Ã‚ )expià ]2dà = ʃ 12à ¢Ã‚ ex pà ¢Ã‚ Ã‚ ¡(2ià )dà Assuming spherical nature of the light atom, we use spherical polar coordinates;x = rsinÓcosÃ’Â ¨,y = rsinÓsinÃ’Â ¨z = rcosÓdà  = r2sinÓdrdÓdÃ’Â ¨Where r= 0 toà ¢Ã… ¾Ãƒâ€œ= 0 to à ¢Ã‚ Ãƒâ€™Ã‚ ¨= 0 to à ¢Ã‚ Hence the probability can be obtained from 12à ¢Ã‚ [r33ÃÆ'(-cosÓ)ÃÆ'exp2iÃ’Â ¨2i=0 after applying the respective limits. * Applying the limits we have;à ¢Ã‚ 23à ¢Ã‚ 212à ¢Ã‚ exp2ià dà =12à ¢Ã‚ [exp2ià 2i] With limits of 3à ¢Ã‚ 2 to à ¢Ã‚ 2This comes to;= 14à ¢Ã‚ i[e6à ¢Ã‚ i2-e4à ¢Ã‚ i2]= eà ¢Ã‚ i4à ¢Ã‚ i 6 SolutionFor harmonic oscillator;Zero Point Energy (ZPE) = Potential Energy V(x)But V(x) = 0.5kf x2The Potential Energy given above is from the relation F=ke and F= maSince ZPE = V(x)ZPE = 0.5kf x2But each mass has extension of 2.5ÃÆ'10725= 0.03448mHence for combination of two masses, total extension = 0.03445/2=0.01724mHere the gravitational intensity is assumed to be 10N/K gHence;ZPE = 0.5ÃÆ'725ÃÆ'0.0003= 0.10875JComparing the ZPE with Thermal Energy kT at 298KWhere k is the Boltzmann constant given as 1.3806503x10-23J/KTherefore; Thermal Energy = kT=1.3806503x10-23 J/K x 298K= 4.114x10-21 JOn comparing, we find that the ZPE is much higher than the Thermal Energy.This shows that the population of particles in the energy level with similar amount of energy as zero point energy is very minimal.Converting the zero point energy to translational energy, we obtain the speed of the two masses as follows;1.18ÃÆ'10-40ÃÆ'n2max = 0.10875Where; nmax=à ¢Ã… ¡ (0.108751.18ÃÆ'10-40)= 3.0358x1019 7 Solution * Rotational Constant is given as 11.007cm-1Mass of 19F = 18.9984032amuMass of 2D = 2.0141018amuThe Bond length in 2D19F is calculated as follows;Rotational Constant is given by B = à ¢Ã‚ 2/2ÂR2Where à ¢Ã‚  is the Reduced Planckà ¢Ã¢â€š ¬s Constant given as 1.0546 x 10-27cm2g/sAnd R is the Bond lengthÂ, which is the reduced mass, is given by;Æš = m1m2/ m1+m2where m1 = 18.9984032amum2 = 2.0141018amu = 18.9984032 x 2.014101818.9984032+ 2.0141018= 38.2647221.012505 = 1.821045From the equation of Rotational Constant, making R the subject of the equation we have R = à ¢Ã… ¡(à ¢Ã‚ 2/2ÂB) = à ¢Ã‚ Ãƒ ¢Ã… ¡2ÂB= 1.0546 x 10^-27à ¢Ã… ¡[21.82104511.007]= 1.0546 x 10^-27à ¢Ã… ¡[3.642090.09085]= 6.0663 x 10-28cm * At ground state n = 0Therefore; B0 = B à ¢Ã¢â€š ¬ (0+1/2) ÃŽBut; B = 6.551cm-1andÃŽ = 0.183cm-1Hence; B0 = 6.551 -  ½(0.183)= 6.551 à ¢Ã¢â€š ¬ 0.0915 = 6.4595cm-1Bond length; R = à ¢Ã‚ Ãƒ ¢Ã… ¡2ÂB0Reduced mass is first calculated from m1 = 1.00794amu and m2 = 126.90447amu = 1.00794 x 126.904471.00794+ 126.90447= 127.9120915127.91241 = 0.9999 = 1R = 1.0546 x 10^-27à ¢Ã… ¡[26.45951] = 1.8953 x 10-27cmAnd for n = 3B3 = B à ¢Ã¢â€š ¬ (3+1/2) ÃŽ= 6.551 à ¢Ã¢â€š ¬ 3/2(0.183)= 6.551 à ¢Ã¢â€š ¬ 0.2745= 6.2765cm-1Bond Length; R = 1.0546 x 10^-27à ¢Ã… ¡[26.27651] = 1.8952 x 10-27cm 8 Solution * Assume the potential for bond breaking is harmonicThen V(x) = 0.5kfx2Where; kf = force c...