(Solution Manual) Quantum Mechanics I The Fundamentals

Rajasekar,Velusamy, 2

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Solution Manual for Quantum Mechanics I The Fundamentals, The downloadable file consists of 396 pages, and the total size of the file is 3.6 MB. Purchase now only for highly discounted price.

Author(s): Rajasekar, Velusamy

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1 Why was Quantum Mechanics Developed? 1.1 Calculate the energy of a photon of 253 .6 nm light. State the energy in eV (1eV = 1 .6×10−19J). We obtain E=hc/λ =6.626×10−34Js×3×108ms−1 253.6×10−9m = 7 .83832 ×10−19J =7.83832 ×10−19 1.6×10−19eV = 4 .89895 eV . 1.2 Calculate the wavelength of a quantum of electromagnetic radiation with energy 4000 eV. We have λ=hc/E. We have hc= 6 .626×10−34Js×3×108ms−1. Since Eis in eV we express hcin units of eV. The conversion factor is 1J = 1 /(1.6×10−19) eV. So, hc=6.626×10−34eV×3×108m 1.6×10−19= 1242 .375×10−9eVm = 1242 .375 eVnm . Then the wavelength of a quantum of electromagnetic radiation with energy 4000eV is λ=1242.375 eVnm 4000 eV= 3.106×10−10m = 3 .106˚A. 1Solutions Manual for Quantum Mechanics I The Fundamentals, 2e by S. Rajasekar, R. Velusamy (All Chapters)2 Why was Quantum Mechanics Developed? 1.3 The Planck’s radiation formula for an ideal black body is E(ν)dν=8πν2dν c3hν ehν/kBT−1, whereE(ν)dνis the energy density in the frequency interval be- tweenνandν+ dν. Determine λmax. Substitution of ν=c/λ, dν=−(c/λ2)dλandE(ν)dν=−p(λ)dλ in the given formula. The result is p(λ) =8πhc λ51 ehc/(λkBT)−1. The wavelength λat whichp(λ) becomes maximum can be found from the condition d p(λ)/dλ= 0. We get λ=hc kBTehc/(λkBT) 5/parenleftbig ehc/(λkBT)−1/parenrightbig =hc kBT1 5/parenleftbigg1 +hc/(λkBT) hc/(λkBT)/parenrightbigg That is, λmax = 0.25hc kBT. 1.4 Consider the exercise 1.3. Examine the spectral shift according to Wien’s displacement law λmaxT= constant if the value of his doubled. We got the result λmax = 0.25hc kBT. Whenhbecomes 2hthenλmaxwould become doubled. Then the solar surface temperature TSalso change. We have 4 πR2σT4 S=G, σ= 2π5k4 B/(15c2h3) andM/R≈2k2 Bq4/(Gh2). SinceR∝h2, σ∝h−3andG∝h−3. We have TS∝h−1. Therefore, the peak wavelength of the sun will follow that λmax,sun= 0.25hc/(kBT)∝ h2. As a result the main radiation from the sun would be shifted to the infrared region. As pointed out in [P.K. Yang, Eur. J. Phys. 37:055406, 2016] the creatures on earth might develop eyes capa ble of detecting infrared radiation.3 1.5 A light of wavelength 1000 nm produces photoelectrons from a metal. The threshold potential V0is 0.5 V. Calculate the work func- tionφ0. The work function φ0is given by φ0=hc λ−|e|V0. hcis calculated as hc= 6.626×10−34Js×3×108ms−1 =6.626×10−34×3×108 1.6×10−19eVm = 1242.375 eVnm. An electron accelerated by 1 V will have 1 eV of kinetic energy. Therefore, an electron accelerated by 0 .5 V will have the kinetic energy of 0.5 eV. Then φ0=1242.375 eVnm 1000 nm−0.5 eV = 0.742375 eV. 1.6 Consider the previous exercise. Calculate the threshold wavelen gth λc. We obtain λc=hc E=1242.375 eVnm 0.742375 eV= 1674 nm . 1.7 Calculate the frequency of light required to produce electrons o f ki- netic energy 5 eV from illumination of a material whose work func- tion is 0.5 eV. According to conservation of energy Energy of a photon = K .E.of an emitted electron + Work function . We obtain the energy of the incident light as E=1 2mv2 max+φ0= 5eV + 0.5eV = 5.5 eV. Then ν=E h=5.5 eV 6.626×10−34Js. Since 1 eV = 1 .6×10−19J we get ν=5.5×1.6×10−19J 6.626×10−34Js= 1.3281×1015Hz.4 Why was Quantum Mechanics Developed? 1.8 Light of wavelength 500 nm is incident on a metal with work func- tion (φ0) of 0.75 eV. What is the maximum kinetic energy ( Kmax) of the ejected photoelectrons? Kmaxishν−φ0. We obtain Kmax =hν−φ0=hc λ−φ0 =6.626×10−34Js×3×108ms−1 500×10−9m−0.75×1.6×10−19J = 2.7756×10−19J. 1.9 Suppose you are standing in front of a 40 W incandescent light bulb 5 m away. If the diameter of your pupils is about 2 mm, about how many photons, n, enter your eye every second? Intensity at the eye is I=40πd2 4πr2=10d2 r2W, wheredis the diameter of the pupil and ris the distance between the eye and light source. We get I=10×(0.002)2 52W = 1.6×10−6Js−1. The number of photons that enters the eye every second is obtain ed as n=I hν=1.6×10−6Js−1 6.626×10−34Js×ν=0.2414729×1028 νs−2. Ifνis stated in Hz then n=2.414729×1027 νs−1. 1.10 Calculate the minimum voltage required for an electron to give all of its energy in a collision with a target and produce an electromag- netic radiation of wavelength 0 .05 nm. When an electron is accelerated through a potential difference V, it acquires an energy |e|V. If this energy goes into producing a photon of energyEwith associated wavelength λ, we getE=hc/λ . So E=6.626×10−34Js×3×108ms−1 0.05×10−9m =6.626×10−34×3×108 1.6×10−19×0.05×10−9eV = 24847.5 eV.5 Now, V=E |e|=24847.5×1.6×10−19J 1.6×10−19C = 24847.5 JC−1 = 24847.5 V. 1.11 Show that the spacing between two minima in the Franck–Hertz curve increases linearly with the order n. LetEais the lowest excitation energy of atoms and δiis the ad- ditional energies gained by the electrons in the ith collision. In n collisions the energy gained by the electrons is En=n(En+δn). The mean free path λof the electrons is generally much shorter than the distance Lbetween the grids in the F-H tube, λ≪L. So δn≪Eaandδn=nλEa/L. Then the spacing between two minima in the F-H curve is ∆En=En−En−1=/bracketleftbigg 1 +λ L(2n−1)/bracketrightbigg Ea. ThusδEnincreases linearly with n. 1.12 Find the wavelength of photons emitted by mercury atoms in Franck–Hertz experiment if an electron of energy 6 eV passed through mercury vapour with an energy 1 eV. We obtain λ=hc ∆E =6.626×10−34×3×108eV nm 1×1.60218×10−19eV = 248.137 nm = 2481.37˚A. 1.13 Assume that the resolving power of a microscope is equal to the wavelength of the light used. If electrons are used as the light sour ce calculate the kinetic energy of electrons needed to have the resolv ing power as 10−11m. Fromλ=h/p=h/mv we get (using 1J = 1 kgm2/s2) v=h mλ =6.626×10−34Js 9.11×10−31kg×10−11m = 7.273326×107m/s.6 Why was Quantum Mechanics Developed? Then the kinetic energy E=mv2/2 required is E=1 2×9.11×10−31kg×/parenleftbig 7.273326×107m/s/parenrightbig2 ×1 eV 1.6×10−19J = 15060.33 eV. 1.14 Suppose a laser provides 100 TW of power in 1ns pulses at a wave- length of 0 .26µm. How many photons are contained in a single pulse? From dW/dt=Pwe obtain dW=Pdt= 100×1012W×1×10−9s = 105J. That is,nhν= 105J ornhc/λ = 105J. So n=λ ch105photons =0.26×10−6m×105J 3×108ms−1×6.626×10−34Jsphotons = 130797 ×1018photons. 1.15 Express the change in wavelength in terms of Compton waveleng th of electron when light is scattered through an angle π/4 by a proton. For a proton ∆λ=λcme mp/parenleftbigg 1−1√ 2/parenrightbigg = 0.29289me mpλc. 1.16 A beam of X-rays of wavelength λ= 0.738˚A is Compton scattered by electrons through an angle π/2. Calculate the wavelength and the energy of the scattered radiation. Change in wavelength for the given problem is ∆λ=λ′−λ=λc(1−cosφ) =λc =h mec. Substituting the values of h,meandcwe get ∆λ=6.626×10−34Js 9.11×10−31kg×3×108ms−1 = 2.42444×10−12m = 0.02424 ˚A.7 Thenλ′=λ+λc= 0.738˚A + 0.02424 ˚A = 0.76224 ˚A. Next, E=hc λ =6.626×10−34Js×3×108m/s 0.76224×10−10m = 2.60784×10−15J. 1.17 Show that the de Broglie wavelength λdassociated with photons is equal to the electromagnetic wavelength λ. The momentum pof a photon of energy E(=hν) is p=E c=hν c. Sincecisνλwe get λd=h p=h hν/c=c ν=νλ ν=λ. 1.18 A 5.78 MeVα-particle (assume that its velocity is ≪c) is emitted in the decay of radium. If the diameter of the radium nucleus is 2×10−14m, how many α-particle de Broglie wavelengths fit inside the nucleus? We obtain E=p2 2m= 5.78×106eV = 9.248×10−13J. Thenp=√ 2mEgives p=/bracketleftbig 2×6.62×10−27×9.248×10−13/bracketrightbig1/2kgm/s = 11.06542×10−20kgm/s. Next, the de Broglie wavelength of α-particle is λ=h p=6.626×10−34 11.06542×10−20m = 0.5988×10−14m. Then n=2×10−14 λ=2×10−14 0.5988×10−14= 3.34.8 Why was Quantum Mechanics Developed? 1.19 A diatomic gas molecule consisting of two atoms of mass msepa- rated by a fixed distance dis rotating about a central axis perpen- dicular to the line joining the two atoms. Assuming that its angular momentum is quantized as in the Bohr atom, determine the pos- sible angular velocities ( ω) and the possible quantized rotational energies (E). We haveL=IωandL=n/planckover2pi1. These give ω=n/planckover2pi1/I. Then E=L2 2I=n2/planckover2pi12 2I, n = 1,2,···. 1.20 In a particle diffraction experiment a beam of 54 eV electrons are perpendicularly diffracted to a single crystal having spacing of the planes isa= 0.91˚A. The angle of incidence and scattering relative to the family of Bragg planes is θ= 65◦. Determine the wavelength of the diffraction electron from nλ= 2asinθwithn= 1 andλ= h/(mv). What do you observe? We havenλ= 2asinθ. Withn= 1,a= 0.91˚A andθ= 65◦we obtain the wavelength of the diffraction electron as λ= 2×0.91˚A×sin 65◦= 1.64948 ˚A. (1.1) Next, consider λ=h/(mv). To compute mvconsiderT=mv2/2 whereTis kinetic energy. Then mv =√ 2mT =/parenleftbig 2×9.1×10−31kg×54 eV×1.6×10−19J/eV/parenrightbig1/2 = 3.96545×10−24kg.m/sec. (1.2) Then λ=h mv=6.626×10−34 3.96545×10−24= 1.67093 ˚A. (1.3) λgiven by Eqs. (1.1) is in agreement with the de Broglie wave- length. That is, Davisson and Germer experiment indeed provides a verification of de Broglie hypothesis.2 Schr¨odinger Equation and Wave Function 2.1 Write the Schr¨ odinger equation for a particle in a potential sin x. The Schr¨ odinger equation is i/planckover2pi1∂ψ ∂t=−/planckover2pi12 2m∂2ψ ∂x2+ sinxψ. 2.2 For a particle carrying a charge qin a uniform time-dependent elec- tric field E(t) the force experienced by it is F=−∇(−qE(t)·X). Write the Schr¨ odinger equation of the system. The Schr¨ odinger equation is i/planckover2pi1∂ψ ∂t=−/planckover2pi12 2m∇2ψ−(qE(t)·X)ψ. 2.3 Starting from the equation for ψ, obtain the equation for ψ∗. The Schr¨ odinger equation for ψis i/planckover2pi1∂ψ ∂t=−/planckover2pi12 2m∇2ψ+V(X,t)ψ. Taking complex conjugate of the above equation we get −i/planckover2pi1∂ψ∗ ∂t=−/planckover2pi12 2m∇2ψ∗+V∗(X,t)ψ∗. This is the Schr¨ odinger equation for ψ∗. 2.4 What is the Schr¨ odinger equation of a system of two particles of massesm1andm2carrying charges q1andq2respectively with H=1 2m1p2 1+1 2m2p2 2+q1q2 r12andr12=|r1−r2|? The Schr¨ odinger equation i /planckover2pi1∂ψ ∂t=Hψbecomes i/planckover2pi1∂ψ ∂t=−/planckover2pi12 2m1∇2 1ψ−/planckover2pi12 2m2∇2 2ψ+q1q2 r12ψ, 910 Schr¨odinger Equation and Wave Function where ∇2 i=∂2 ∂x2 i+∂2 ∂y2 i+∂2 ∂z2 i, i= 1,2. 2.5 Write the Schr¨ odinger equation for a charged particle moving in a n electromagnetic field with H=1 2m/parenleftig p−q cA(X,t)/parenrightig2 +qφ(X,t). Rewrite the Schr¨ odinger equation for A= (Bz,0,0) andφ=−Ez. The Schr¨ odinger equation is i/planckover2pi1∂ψ ∂t=1 2m/parenleftig p−q cA/parenrightig2 ψ+qφψ =1 2m/parenleftbigg −/planckover2pi12∇2ψ+q2 c2A2ψ−p/parenleftigq cAψ/parenrightig −q cA·pψ/parenrightbigg +qφψ =1 2m/bracketleftbigg −/planckover2pi12∇2ψ+ 2i/planckover2pi1q cA·∇ψ+ i/planckover2pi1q cψ∇·A+q2 c2A2ψ/bracketrightbigg +qφψ. ForA= (Bz,0,0) andφ=−Ezwe have i/planckover2pi1∂ψ ∂t=1 2m/bracketleftbigg −/planckover2pi12∇2ψ+ 2i/planckover2pi1q cBzψz+q2 c2B2z2ψ/bracketrightbigg −qEzψ. 2.6 IfE1andE2are the eigenvalues and φ1andφ2are the correspond- ing eigenfunctions, respectively, of a Hamiltonian then find whether the energy corresponding to the superposition state φ1+φ2is equal toE1+E2or not. We haveHφ1=E1φ1andHφ2=E2φ2. Then H(φ1+φ2) =Hφ1+Hφ2 =E1φ1+E2φ2 ∝ne}ationslash= (E1+E2) (φ1+φ2).11 2.7 Express/bracketleftbigg −/planckover2pi12 2m∇2+V(x,y,z )/bracketrightbigg ψ(x,y,z ) =Eψ(x,y,z ) in the spherical polar coordinates defined by x=rsinθcosφ,y= rsinθsinφ,z=rcosθand in the parabolic coordinates defined byξ=r(1−cosθ),η=r(1 + cosθ),φ=φcosθ. In spherical polar coordinates the Schr¨ odinger equation is −/planckover2pi12 2m/bracketleftbigg1 r2∂ ∂r/parenleftbigg r2∂ ∂r/parenrightbigg +1 r2sinθ∂ ∂θ/parenleftbigg sinθ∂ ∂θ/parenrightbigg +1 r2sin2θ∂2 ∂φ2+V(r,θ,φ )/bracketrightbigg ψ(r,θ,φ ) =Eψ(r,θ,φ ). In the parabolic coordinates the Schr¨ odinger equation is −/planckover2pi12 2m/bracketleftigg 4 ξ+η/braceleftbigg∂ ∂ξ/parenleftbigg ξ∂ ∂ξ/parenrightbigg +∂ ∂η/parenleftbigg η∂ ∂η/parenrightbigg +1 ξη∂2 ∂φ2/bracerightbigg +V(ξ,η,φ )/bracketrightigg ψ=Eψ. 2.8 Find the condition under which both ψ(X,t) andψ∗(X,−t) will be the solutions of the same time-dependent Schr¨ odinger equation. The Schr¨ odinger equations for ψ(X,t) andψ∗(X,−t) are given by i/planckover2pi1∂ψ ∂t=−/planckover2pi12 2m∇2ψ+V(X,t)ψ, i/planckover2pi1∂ψ∗ ∂t=−/planckover2pi12 2m∇2ψ∗+V∗(X,−t)ψ respectively. The above two equations are identical if V(X,t) = V∗(X,−t), that is if Vis real and an even function of t. 2.9 What is the major difference between a real and complex ψ? The probability current density Jis given by J=/planckover2pi1 2mi[ψ∗∇ψ−ψ∇ψ∗]. Ifψis real then J= 0. Ifψis complex then Jneed not be zero. 2.10 What is the difference between the wave function ψ1= ei(kx−ωt) andψ2= ei(k·X−ωt)? ψ1is a plane (probability)wave travelling along a one-dimensional line.ψ2is a three-dimensional (probability)wave.12 Schr¨odinger Equation and Wave Function 2.11 Write the Hamiltonian of a photon. For a photon H=E=mc2=mcc=pc. 2.12 What is the physical meaning of ∝an}b∇acketle{tx∝an}b∇acket∇i}ht= 0? ψis symmetric or antisymmetric with respect to x= 0. 2.13 Write the operator forms of kinetic energy and angular moment um L=r×p. The operator form of kinetic energy mv2/2 is (K.E.)op=1 2mv2=p2 2m=−/planckover2pi12 2m∇2. ForL=r×pwe have with r=ix+jy+kz L=r×p=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglei j k x y z pxpypz/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle =i(ypz−zpy) +j(zpx−xpz) +k(xpy−ypx). The components of Lare Lx= i/planckover2pi1/parenleftbigg −y∂ ∂z+z∂ ∂y/parenrightbigg = i/planckover2pi1/parenleftbigg z∂ ∂y−y∂ ∂z/parenrightbigg , Ly= i/planckover2pi1/parenleftbigg −z∂ ∂x+x∂ ∂z/parenrightbigg = i/planckover2pi1/parenleftbigg x∂ ∂z−z∂ ∂x/parenrightbigg , Lz= i/planckover2pi1/parenleftbigg −x∂ ∂y+y∂ ∂x/parenrightbigg = i/planckover2pi1/parenleftbigg y∂ ∂x−x∂ ∂y/parenrightbigg . 2.14 Consider the one-dimensional Schr¨ odinger equation ψxx+ (λ− g(x))ψ= 0,λ= 2mE//planckover2pi12andg= 2mV//planckover2pi12. Obtain the Schr¨ odinger equation under the change of variable u(x) =−d dxlnψ(x). The general transformation is u=−ψx/ψorψx=−uψ. Then ψxx=−uxψ−uψx. Now, the Schr¨ odinger equation is rewritten as (withλ= 2mE//planckover2pi12andg= 2mV//planckover2pi12) −uxψ−uψx+ (λ−g)ψ= 0 or −uxψ+u2ψ+ (λ−g)ψ= 0. That is,ux−u2−λ+g= 0.13 2.15 Find the conditions to be satisfied by the functions fandgsuch that under the transformation ψ=f(x)F(g(x)) the Schr¨ odinger equationψxx+ (E−V)ψ= 0 can be written as Fgg+Q(g)Fg+R(g)F(g) = 0. (2.1) DetermineE−V. In terms of fandFthe Schr¨ odinger equation is written as fF′′+ 2f′F′+f′′F+ (E−V)fF= 0, (2.2) where prime denotes differentiation with respect to x. SinceFis F(g(x)) we have F′=Fgg′, F′′=Fgg(g′)2+Fgg′′. (2.3) Then Eq. (2.2) becomes Fgg+/parenleftbigg2f′ g′f+g′′ g′2/parenrightbigg Fg+/bracketleftbiggf′′ g′2f′2+(E−V) g′2/bracketrightbigg F= 0. (2.4) Comparison of Eqs. (2.1) and (2.4) gives Q=2f′ g′f+g′′ g′2, R =f′′ g′2f′2+E−V g′2. (2.5) The conditions on fandgare given by Eqs. (2.5). From Eqs. (2.5) we obtain f′ f=g′Q 2−g′′ 2g′(2.6a) E−V=g′2R−f′′ f. (2.6b) From (2.6a) we find f′′/fand substituting it in (2.6b) we obtain d dx/parenleftbiggf′ f/parenrightbigg =f′′ f−f′2 f2. (2.7) That is, f′′ f=d dx/parenleftbiggf′ f/parenrightbigg +f′2 f2 =d dx/parenleftbiggg′Q 2−g′′ 2g′/parenrightbigg +/parenleftbiggg′Q 2−g′′ 2g′/parenrightbigg2 =1 2Qg−/parenleftbiggg′′ 2g′/parenrightbigg +g′2Q2 4+/parenleftbiggg′′ 2g′/parenrightbigg2 . (2.8)14 Schr¨odinger Equation and Wave Function Substituting the above in Eq. (2.6b) we obtain E−V=g′2/bracketleftbigg R−1 2Qg−Q2 4/bracketrightbigg +/parenleftbiggg′′ 2g′/parenrightbigg′ −/parenleftbiggg′′ 2g′/parenrightbigg2 . (2.9) 2.16 What are the effects of addition of a constant to a potential on the time-independent Schr¨ odinger equation and the energy levels? The time-independent Schr¨ odinger equation with the addition of a constantαto a potential is given by /parenleftbigg −/planckover2pi12 2m∇2+V+α/parenrightbigg ψn=Enψn. This can be rewritten as /parenleftbigg −/planckover2pi12 2m∇2+V/parenrightbigg ψn= (En−α)ψn=E′ nψn, whereE′ n=En−α. Therefore ψnremains the same but the energy eigenvalues of the new system are En−α. 2.17 Which of the following wave functions are admissible in quantum mechanics? State the reasons. (a) e−x2. (b) sechx. (c) e−x. (d) tanhx. (e) sinx, 0<x< 2π. (f) sinx,−∞<x<∞. (g)√ e−x2. (h) tanx. (i) secx. (j)xe−x2. The conditions to be satisfied by a wave function are: (i) ψshould be normalizable. (ii) It should be single-valued. (iii) It must be finite at every point. (iv) It and its first partial derivatives must be continuous. The functions (a), (b), (e), and (j) satisfy all the above conditio ns and hence they are admissible wave functions (verify). The functions (c), (d), (f), (h), (i) are nonnormalizable. The fun c- tion (g) is multivalued. Therefore, they are not admissible wave functions. 2.18 Normalize the wave function ψ=Neikx−x2/(2σ2). From the normalization condition we obtain 1 =/integraldisplay∞ −∞ψ∗ψdx=N2σ/integraldisplay∞ −∞e−y2dy=N2σ√π. ThusN= 1/(σ√π)1/2.15 2.19 Find the value of Nfor which the wave function ψ(x) =Nfor |x|<aand 0 for |x|>ais normalized. The normalization condition gives 1 =/integraldisplaya −aψ∗ψdx=N2/integraldisplaya −adx=N2x|a −a = 2aN2. Thus,N= 1/√ 2a. 2.20 Normalize the wave function ψ= e−|x|sinαx. It is given that/integraldisplay∞ 0e−xsinαxdx=α2/(1 +α2). The normalization condition gives 1 =N2/integraldisplay∞ −∞e−2|x|sin2αxdx =N2/integraldisplay0 −∞e2xsin2αxdx+N2/integraldisplay∞ 0e−2xsin2αxdx = 2N2/integraldisplay∞ 0e−2xsin2αxdx =−N2 2e−2x|∞ 0−N2 2/integraldisplay∞ 0e−xcosαxdx =N2 2−N2 2/bracketleftbigg −e−xcosαx|∞ 0−α/integraldisplay∞ 0e−xsinαxdx/bracketrightbigg =N2 2−N2 2/bracketleftbigg 1−α3 1 +α2/bracketrightbigg =N2α3 2(1 +α2). That is,N=/radicalbig 2(1 +α2)/α3.16 Schr¨odinger Equation and Wave Function 2.21 A particle of mass mmoves in a one-dimensional box of length L with origin as the centre. If the wave function of this particle is ψ(x) =Nx(1−x2) for|x|<L/ 2 and 0 otherwise find the factor N. Ncan be determined from normalization condition. We obtain 1 =N2/integraldisplayL/2 −L/2x2/parenleftbig 1−x2/parenrightbig2dx =N2/integraldisplayL/2 −L/2/parenleftbig x2−2x4+x6/parenrightbig dx =N2/bracketleftbiggL3 12−L5 160+L7 448/bracketrightbigg Thus, N=/bracketleftbiggL3 12−L5 160+L7 448/bracketrightbigg−1/2 . IfLis very small then N≈/radicalbig 12/L3. 2.22 A quantum mechanical particle moving in one-dimension has the wave function ψ(x) =cxe−|x|/b,−∞< x <∞wherecandbare constants ( b > 0). Find the probability that the position of the particle lies in the region −∞<x≤b. First, we normalize the wave function. We obtain 1 =/integraldisplay∞ −∞ψ∗ψdx =c2/integraldisplay∞ −∞x2e−2|x|/bdx =c2/integraldisplay0 −∞x2e2x/bdx+c2/integraldisplay∞ 0x2e−2x/bdx = 2c2/integraldisplay∞ 0x2e−2x/bdx. Substituting 2 x/b=yin the integral, we get 1 =b3c2 4/integraldisplay∞ 0y2e−ydy. Integrating by parts we get 1 =b3c2 2/integraldisplay∞ 0ye−ydy=b3c2 2/integraldisplay∞ 0e−ydy=b3c2 2.17 That isc2= 2/b3orc=/radicalbig 2/b3. The probability P(−∞< x≤b) is P=/integraldisplayb −∞ψ∗ψdx =2 b3/bracketleftigg/integraldisplay0 −∞x2e2|x|/bdx+/integraldisplayb 0x2e−2x/bdx/bracketrightigg =2 b3b3 8/bracketleftigg −/integraldisplay0 ∞y2e−ydy+/integraldisplayb 0y2e−ydy/bracketrightigg =1 4/bracketleftigg/integraldisplay∞ 0y2e−ydy+/integraldisplayb 0y2e−ydy/bracketrightigg Integrating by parts we get P= 1−e−b 4/parenleftbig b2+ 2b+ 2/parenrightbig . 2.23 What is the probability current density corresponding to ψ(x) = Ae−αxwhereAandαare constants? The probability current density is given by J=/planckover2pi1 2mi[ψ∗∇ψ−ψ∇ψ∗]. Since the given ψis real,ψ∗=ψ, J=/planckover2pi1 2mi[ψ∇ψ−ψ∇ψ] = 0. 2.24 A free particle in one-dimension is in a state described by ψ= Aei(pxx−Et)//planckover2pi1+Be−i(pxx+Et)//planckover2pi1wherepxandEare constants. Find the probability current density. Jis given by J=/planckover2pi1 mIm(ψ∗∇ψ) =/planckover2pi1 mIm/braceleftig/bracketleftig A∗e−i(pxx−Et)//planckover2pi1+B∗ei(pxx+Et)//planckover2pi1/bracketrightig ×ipx /planckover2pi1/bracketleftig Aei(pxx−Et)//planckover2pi1+Be−i(pxx+Et)//planckover2pi1/bracketrightig/bracerightbigg =px m/bracketleftbig |A|2−|B|2/bracketrightbig .18 Schr¨odinger Equation and Wave Function 2.25 Ifψ(x,0) =Ne−ikx−(x2/2a2)for a particle then calculate the prob- ability density and current density. First, we normalize the given wave function. We obtain 1 =/integraldisplay∞ −∞ψ∗ψdx= 2N2a/integraldisplay∞ 0e−y2dy=N2a√π. That isN= (1/(a2π))1/4. The probability density is obtained as P(x) =1 a√πe−x2/a2. Then the current density is determined as J(x) =/planckover2pi1 2mi/parenleftbigg ψ∗dψ dx−ψdψ∗ dx/parenrightbigg =/planckover2pi1k m1 a√πe−x2/a2 =/planckover2pi1k mP(x). 2.26 For the Gaussian wave function ψ(x) =/parenleftbigg1 σ√π/parenrightbigg1/2 e−x2/2σ2calcu- late the probability current density. The probability current density is given by J=/planckover2pi1 2mi/parenleftbigg ψ∗dψ dx−ψdψ∗ dx/parenrightbigg . For any real ψ ψ∗dψ dx−ψdψ∗ dx= 0. ThenJ= 0. Therefore, for the real Gaussian function J= 0. 2.27 Verify whether the wave function ψ=Neikx−x2/(2a2)satisfies the continuity equation or not. The continuity equation is given by∂ρ ∂t+dJ dx= 0. For the given functionρ=ψ∗ψ=N2e−x2/a2and∂ρ ∂t= 0. Further J=/planckover2pi1 mIm/parenleftbigg ψ∗dψ dx/parenrightbigg =/planckover2pi1 mIm/parenleftig N2e−x2/a2/parenleftbig ik−x/a2/parenrightbig/parenrightig =N2/planckover2pi1k me−x2/a2.19 Then dJ dx=−2N2/planckover2pi1k ma2xe−x2/a2. Now the continuity equation becomes xe−x2/a2= 0 which is true only ifx= 0. Hence, the given ψdoes not satisfy the continuity equation. 2.28 The wave function of a linear harmonic oscillator with potential V=mω2x2/2 isψ=xe−mωx2/(2/planckover2pi1). Find its energy. The Schr¨ odinger equation for a potential Vis ψxx+2m /planckover2pi12(E−V)ψ= 0. For the given potential and ψthe above equation becomes −3mω /planckover2pi1x+m2ω2 /planckover2pi12x3+2mE /planckover2pi12x−m2ω2 /planckover2pi12x3= 0. From the above equation we get E= 3/planckover2pi1ω/2. 2.29 Ifψ=/radicalbig 1/Lcos(πx/(2L)) for the system confined to the potential V(x) = 0 for |x|<Land∞for|x|>Lthen calculate E. The Schr¨ odinger equation for the given system is ψxx+2m /planckover2pi12Eψ= 0. Then Eψ =−/planckover2pi12 2mψxx=/planckover2pi12 2m√ L/parenleftigπ 2L/parenrightig2 cosπx 2L=/planckover2pi12π2 8mL2ψ. Thus.E=/planckover2pi12π2/(8mL2). 2.30 The wave function of a particle confined to a box of length Lis/radicalbig 2/Lsin(πx/L ) in the region 0 <x<L and 0 otherwise. Calculate the probability of finding the particle in the region 0 <x≤L/2. We obtain P(0<x≤L/2) =/integraldisplayL/2 0ψ∗ψdx =2 L/integraldisplayL/2 0sin2πx Ldx =1 L/integraldisplayL/2 0/parenleftbigg 1−cos/parenleftbigg2πx L/parenrightbigg/parenrightbigg dx=1 2.20 Schr¨odinger Equation and Wave Function 2.31 Write the law of conservation of energy H=T+Vin terms of expectation values. We write ∝an}b∇acketle{tH∝an}b∇acket∇i}ht=∝an}b∇acketle{tT∝an}b∇acket∇i}ht+∝an}b∇acketle{tV∝an}b∇acket∇i}ht. 2.32 Are the wave functions ψ1=/parenleftbigg1 πa3 0/parenrightbigg1/2 e−r/a0andψ2=/parenleftbigg1 32πa3 0/parenrightbigg1/2/parenleftbigg 2−r a0/parenrightbigg e−r/(2a0) of the electron in hydrogen atom orthogonal? The orthogonality integral is calculated as /integraldisplay∞ 0ψ∗ 1ψ2dτ∝/integraldisplay∞ 0/parenleftbigg 2−r a0/parenrightbigg r2e−3r/2a0dr ∝6/integraldisplay∞ 0y2e−ydy−2/integraldisplay∞ 0y3e−ydy ∝6/integraldisplay∞ 0y2e−ydy−6/integraldisplay∞ 0y2e−ydy = 0. Since/integraldisplay∞ 0ψ∗ 1ψ2dτ= 0, the given two wave functions are orthogonal. 2.33 Find the potential corresponding to the following wave function s: (a)ψ=1 π1/4e−x2/2,E=/planckover2pi1ω/2. (b)ψ=Asin(πx/(2L)) for|x|< L and 0 for |x|> L,E= /planckover2pi12π2/(8mL2). (a) The eigenvalue equation Hψ=Eψis −/planckover2pi12 2md2ψ dx2+Vψ=Eψ orVψ=/planckover2pi12 2mψxx+Eψ. Substituting the given ψandEin the above equation we get Vψ=/planckover2pi12 2md dx(−x)ψ+Eψ=/planckover2pi12 2m/parenleftbig −ψ+x2ψ/parenrightbig +Eψ. Then V=/planckover2pi12 2m/parenleftbig −1 +x2/parenrightbig +E=/planckover2pi12 2m/parenleftbig x2−1/parenrightbig +/planckover2pi1ω 2.21 (b) For the given wave function we obtain Vψ =/planckover2pi12 2md dxπ 2LAcosπx 2L+Eψ =−/planckover2pi12π2 8mL2ψ+Eψ =−Eψ+Eψ = 0. Therefore,V(x) = 0 for |x|<Land∞for|x|>L. 2.34 A particle of mass mis confined in the potential V= 0 for 0<x≤L andV=∞otherwise. It has the normalized stationary state φn(x) and eigenvalues En=n2/planckover2pi12π2/(2mL2). Its wave function at time t= 0 is given by ψ(x,0) = (φ1(x)+φ2(x))/√ 2 . What is the smallest positive time τfor whichψ(x,t) will be orthogonal to ψ(x,0)? The wave function ψ(x,t) is ψ(x,t) =1√ 2/parenleftig φ1e−iE1t//planckover2pi1+φ2e−iE2t//planckover2pi1/parenrightig . Att=τthe condition for ψ(x,0) andψ(x,t) to be orthogonal is /integraldisplayL 0ψ∗(x,τ)ψ(x,0) dx= 0. We get /integraldisplayL 0/parenleftig φ∗ 1eiE1τ//planckover2pi1+φ∗ 2eiE2τ//planckover2pi1/parenrightig (φ1+φ2) dx= 0. Since/integraltextL 0φmφndx=δmnwe get eiE1τ//planckover2pi1+ eiE2τ//planckover2pi1= 0. That is cosE1τ /planckover2pi1+ i sinE1τ /planckover2pi1+ cosE2τ /planckover2pi1+ i sinE2τ /planckover2pi1= 0. Equating the real and imaginary parts to zero separately we get cosE1τ /planckover2pi1=−cosE2τ /planckover2pi1,sinE1τ /planckover2pi1=−sinE2τ /planckover2pi1. We rewrite the above conditions as cosE1τ /planckover2pi1= cosπ+E2τ /planckover2pi1,sinE1τ /planckover2pi1= sinπ+E2τ /planckover2pi1. Thus,τ=/planckover2pi1π/(E1−E2). SinceE2>E1we writeτ=/planckover2pi1π/(E2−E1).22 Schr¨odinger Equation and Wave Function 2.35 Show that ∝an}b∇acketle{txpx+pxx∝an}b∇acket∇i}htis real. Consider ∝an}b∇acketle{tpxx∝an}b∇acket∇i}ht. We obtain ∝an}b∇acketle{tpxx∝an}b∇acket∇i}ht=/integraldisplay∞ −∞ψ∗pxxψdx =/integraldisplay∞ −∞(p† xψ)∗xψdx =/parenleftbigg/integraldisplay∞ −∞pxψx∗ψ∗dx/parenrightbigg∗ =/parenleftbigg/integraldisplay∞ −∞ψ∗xpxψdx/parenrightbigg∗ =∝an}b∇acketle{txpx∝an}b∇acket∇i}ht∗. Now,∝an}b∇acketle{txpx+pxx∝an}b∇acket∇i}ht=∝an}b∇acketle{txpx∝an}b∇acket∇i}ht+∝an}b∇acketle{tpxx∝an}b∇acket∇i}ht=∝an}b∇acketle{txpx∝an}b∇acket∇i}ht+∝an}b∇acketle{txpx∝an}b∇acket∇i}ht∗= 2Re∝an}b∇acketle{txpx∝an}b∇acket∇i}ht. Thus∝an}b∇acketle{txpx+pxx∝an}b∇acket∇i}htis real. 2.36 Show that ∝an}b∇acketle{tAn∝an}b∇acket∇i}ht=∝an}b∇acketle{tA∝an}b∇acket∇i}htnin its eigenstates. LetAψ=αψ. Then we obtain ∝an}b∇acketle{tAn∝an}b∇acket∇i}ht=/integraldisplay∞ −∞ψ∗Anψdτ =/integraldisplay∞ −∞ψ∗An−1Aψdτ =α/integraldisplay∞ −∞ψ∗An−1ψdτ =αn/integraldisplay∞ −∞ψ∗ψdτ =αn. We find ∝an}b∇acketle{tA∝an}b∇acket∇i}ht=/integraldisplay∞ −∞ψ∗Aψdτ=α/integraldisplay∞ −∞ψ∗ψdτ=α. Hence, ∝an}b∇acketle{tAn∝an}b∇acket∇i}ht= αn=∝an}b∇acketle{tA∝an}b∇acket∇i}htn. 2.37 Calculate ∝an}b∇acketle{tp2 x∝an}b∇acket∇i}htforψ=√ ke−k|x|. We obtain ∝an}b∇acketle{tp2 x∝an}b∇acket∇i}ht=/integraldisplay∞ −∞ψ∗p2 xψdx=−/planckover2pi12/integraldisplay∞ −∞e−k|x|d2 dx2e−k|x|dx. The above equation can be rewritten as ∝an}b∇acketle{tp2 x∝an}b∇acket∇i}ht=−/planckover2pi12k/bracketleftbigg/integraldisplay0 −∞ekxd2 dx2ekxdx+/integraldisplay∞ 0e−kxd2 dx2e−kxdx/bracketrightbigg .23 Then we obtain ∝an}b∇acketle{tp2 x∝an}b∇acket∇i}ht=−/planckover2pi12k3/bracketleftbigg/integraldisplay0 −∞e2kxdx+/integraldisplay∞ 0e−2kxdx/bracketrightbigg =−/planckover2pi12k2. 2.38 A particle of mass min the potential V(x) = 0 for 0 ≤x≤Land ∞otherwise is in a state with ψ(x) =Nx(L−x) whereNis the normalization constant. Determine ∝an}b∇acketle{tE∝an}b∇acket∇i}htin this state. The normalization condition gives 1 =N2/integraldisplayL 0x2(L−x)2dx=N2L5 30. That is,N=/radicalbig 30/L5. Next, ∝an}b∇acketle{tE∝an}b∇acket∇i}ht=/integraldisplayL 0ψ∗Hψdx =/integraldisplayL 0ψ/parenleftbigg −/planckover2pi12 2md2 dx2/parenrightbigg ψdx =/planckover2pi12 mN2/integraldisplayL 0/parenleftbig xL−x2/parenrightbig dx =5/planckover2pi12 mL2. 2.39 Show thatd dt∝an}b∇acketle{tx2∝an}b∇acket∇i}ht=2 m∝an}b∇acketle{txpx∝an}b∇acket∇i}ht−i/planckover2pi1 m. ∝an}b∇acketle{tx2∝an}b∇acket∇i}htis given by ∝an}b∇acketle{tx2∝an}b∇acket∇i}ht=/integraldisplay∞ −∞ψ∗x2ψdx. Then d dt∝an}b∇acketle{tx2∝an}b∇acket∇i}ht=d dt/integraldisplay∞ −∞ψ∗x2ψdx =/integraldisplay∞ −∞∂ψ∗ ∂tx2ψdx+/integraldisplay∞ −∞ψ∗x2∂ψ ∂tdx. Using the Schr¨ odinger equation for∂ψ ∂tand∂ψ∗ ∂twe get d dt∝an}b∇acketle{tx2∝an}b∇acket∇i}ht=−1 i/planckover2pi1/integraldisplay∞ −∞/parenleftbiggp2 xψ∗ 2m+Vψ∗/parenrightbigg x2ψdx +1 i/planckover2pi1/integraldisplay∞ −∞ψ∗x2/parenleftbiggp2 x 2mψ+Vψ/parenrightbigg dx =i 2m/planckover2pi1/integraldisplay∞ −∞ψ∗p2 x/parenleftbig x2ψ/parenrightbig dx−i 2m/planckover2pi1/integraldisplay∞ −∞ψ∗x2p2 xψdx.24 Schr¨odinger Equation and Wave Function Expanding the first integral we obtain d dt∝an}b∇acketle{tx2∝an}b∇acket∇i}ht=i 2m/planckover2pi1/integraldisplay∞ −∞ψ∗/parenleftbig 2pψpxx2+ψp2x2/parenrightbig dx =i 2m/planckover2pi1/integraldisplay∞ −∞ψ∗/parenleftbig −4i/planckover2pi1xpxψ+ 2ψ(−i/planckover2pi1)2/parenrightbig dx =2 m∝an}b∇acketle{txpx∝an}b∇acket∇i}ht−i/planckover2pi1 m. 2.40 Show thatd2 dt2∝an}b∇acketle{tx2∝an}b∇acket∇i}ht=2 m2∝an}b∇acketle{tp2 x∝an}b∇acket∇i}ht+2 m∝an}b∇acketle{txF∝an}b∇acket∇i}ht. We haved dt∝an}b∇acketle{tx2∝an}b∇acket∇i}ht=2 m∝an}b∇acketle{txpx∝an}b∇acket∇i}ht−i/planckover2pi1 m. Then d2 dt2∝an}b∇acketle{tx2∝an}b∇acket∇i}ht=2 md dt∝an}b∇acketle{txpx∝an}b∇acket∇i}ht =2 md dt/integraldisplay∞ −∞ψ∗(xpx)ψdx =2 m/integraldisplay∞ −∞∂ψ∗ ∂txpxψdx+2 m/integraldisplay∞ −∞ψ∗xpx∂ψ ∂tdx. Using the Schr¨ odinger equation d2 dt2∝an}b∇acketle{tx2∝an}b∇acket∇i}ht=i m2/planckover2pi1/integraldisplay∞ −∞p2 xψ∗xpxψdx+2i m/planckover2pi1/integraldisplay∞ −∞Vψ∗xpxψdx −i m2/planckover2pi1/integraldisplay∞ −∞ψ∗xp3 xψdx−2i m/planckover2pi1/integraldisplay∞ −∞ψ∗xpx(Vψ) dx =i m2/planckover2pi1/integraldisplay∞ −∞ψ2/bracketleftbig px(pxxpxψ+xp2 xψ)−xp3 xψ/bracketrightbig dx −2i m/planckover2pi1/integraldisplay∞ −∞ψ∗xψ/parenleftbigg −i/planckover2pi1dV dx/parenrightbigg dx. Substituting F=−dV/dxwe get d2 dt2∝an}b∇acketle{tx2∝an}b∇acket∇i}ht=i m2/planckover2pi1/integraldisplay∞ −∞ψ∗/parenleftbig 2pxxp2 xψ/parenrightbig dx+2 m∝an}b∇acketle{txF∝an}b∇acket∇i}ht =2i m2/planckover2pi1/integraldisplay∞ −∞ψ∗(−i/planckover2pi1)p2 xψdx+2 m∝an}b∇acketle{txF∝an}b∇acket∇i}ht =2 m2∝an}b∇acketle{tp2 x∝an}b∇acket∇i}ht+2 m∝an}b∇acketle{txF∝an}b∇acket∇i}ht.25 2.41 Show thatd dt∝an}b∇acketle{tp2 x∝an}b∇acket∇i}ht=∝an}b∇acketle{t2Fpx+pxF∝an}b∇acket∇i}ht. By definition ∝an}b∇acketle{tp2 x∝an}b∇acket∇i}ht=/integraldisplay∞ −∞ψ∗p2 xψdτ. Then d dt∝an}b∇acketle{tp2 x∝an}b∇acket∇i}ht=/integraldisplay∞ −∞∂ψ∗ ∂tp2 xψdτ+/integraldisplay∞ −∞ψ∗p2 x∂ψ ∂tdτ . Using the Schr¨ odinger equation we obtain d dt∝an}b∇acketle{tp2 x∝an}b∇acket∇i}ht=−1 i/planckover2pi1/integraldisplay∞ −∞/parenleftbiggp2 xψ∗ 2m+Vψ∗/parenrightbigg p2 xψdτ +1 i/planckover2pi1/integraldisplay∞ −∞ψ∗p2 x/parenleftbiggp2 xψ 2m+Vψ/parenrightbigg dτ . That is, d dt∝an}b∇acketle{tp2 x∝an}b∇acket∇i}ht=i /planckover2pi1/integraldisplay∞ −∞Vψ∗p2 xψdτ−i /planckover2pi1/integraldisplay∞ −∞ψ∗p2 x(Vψ) dτ =−i /planckover2pi1/bracketleftbigg 2/integraldisplay∞ −∞ψ∗pxVpxψdτ+/integraldisplay∞ −∞ψ∗/parenleftbig p2 xV/parenrightbig ψdτ/bracketrightbigg =−2i /planckover2pi1(−i/planckover2pi1)/integraldisplay∞ −∞ψ∗∇Vpxψdτ −i /planckover2pi1(−i/planckover2pi1)/integraldisplay∞ −∞ψ∗(px∇V)ψdτ = 2/integraldisplay∞ −∞ψ∗Fpxψdτ+/integraldisplay∞ −∞ψ∗pxFψdτ =∝an}b∇acketle{t2Fpx+pxF∝an}b∇acket∇i}ht. 2.42 Show thatd dt∝an}b∇acketle{tpxx∝an}b∇acket∇i}ht=1 m∝an}b∇acketle{tp2 x∝an}b∇acket∇i}ht+∝an}b∇acketle{txF∝an}b∇acket∇i}ht. The expectation value of pxxis ∝an}b∇acketle{tpxx∝an}b∇acket∇i}ht=/integraldisplay∞ −∞ψ∗pxxψdx =−i/planckover2pi1/integraldisplay∞ −∞ψ∗ψdx+/integraldisplay∞ −∞ψ∗xpxψdx =−i/planckover2pi1+/integraldisplay∞ −∞ψ∗xpxψdx. Now d dt∝an}b∇acketle{tpxx∝an}b∇acket∇i}ht=d dt/integraldisplay∞ −∞ψ∗xpxψdx =/integraldisplay∞ −∞∂ψ∗ ∂txpxψdx+/integraldisplay∞ −∞ψ∗xpx∂ψ ∂tdx.