Chapter 11.Find the de Broglie wavelength for each of the following cases:(a)a 70kg man traveling at 60 km/h;Solution:λ===0.568m;(b)a 1kg stone traveling at 10 m/s;Solution:λ==m=6.63m;(c)a g particle of dust moving at 1 m/s;Solution:λ==m=6.63m;(d)an electron with 3 eV energy;Solution:===m=0.709m(e)a helium with kinetic energy of E=KT(K is the Boltzmann constant) at T=1.0K.Solution:===m=m=1.265m;2.A pare of positron and electron can be produced by two photons under certain conditions .If the two photons have the same energy ,please find out the maximum wavelength of the photons in order to produce a pare of positron and electron?Solution:When both positron and electron are stationary ,the wavelength of photons is maximumSo 2h2hhλ==2.43nm=2.43nm3.A particle with mass m moves in the field V(x).Please verify theprobability conservation law of +=0. Here and are probability density and current density ,respectively.Solution:If Ψ(,t) is the wave function of the particleSo the probability of the particle==+ (1)from the Schrödinger equation we get+U(r) (2)andU(r) (3)Substituting equation(2) and (3) into equation (1) we get=()=().where =() is the probability flux.4.what kinetic energy (in electron volts) should neutrons and electrons have if they are to be diffracted from crystals?(Takingλ=10),Note: Appreciable diffraction willoccur if the de Brogile wavelength of the particle is if the same order of magnitude as the intet-atomic distance.Solution:A：neutronλ=;that is E =8.2B ：electron λ=;that is E =Chapter 21. An electron is confined in the ground state in a one-dimensional box of width 10-10 m. Its energy is 38 e V . Calculate: a) The energy of the electron in its first excited state.b) The average force on the walls of the box when the electron isin the ground state. Solution:a ） the energy eigenvalues ofelectron is22222n n E ma π=The energy of the electron in the ground state is2212382E eVma π==Then the nergy of the electron in the excited state is222241522E eVma π==b) Because the electron is confined in the ground state in aone-dimensional box of width 1010m - the ground state of electronis 1x aπψ= 1010a m -=so when0x =or 1010x -=, 10ψ= , 210ψ=there is not an electron at 0x =or 1010x -=, so no force on thewalls of the box, the average force on walls is zero . 2. Suppose the wave function is in the form of/21)(ipx p e x πψ=for a one dimensional particle.a) Please verify the equation )()2/ˆ()(ˆ2x m p x H p pψψ=holds for the wave function)(x p ψ and Hamilton operator222222/ˆˆdx d m m pH -==.b) Please find out the specific form of ),(t x p ψ if )()0,(x x p p ψψ=. Solution: a)thedimensional particle′s energy equation shouldbe :()()()p p r r E r ψψ∧H =()()()()222222ipx ipx p p p p px x x e e x m m ψψψ∧∧H ====2()()2p p p E r r mψψ=()()()p p x x E x ψψ∧∴H = So it has proved the question .b) the specific form of (),p x t ψ is ()2222,i p ipx ip tipxt m m p x t e ψ--=∙=then we can verify ()(),0ipxp p x x ψψ=istight.3. A particle is confined in a 1-D box with walls of infinite height at x =-a and x = +a . Suppose the particle is in the first excited state and its wave function is ψ(x) = Asin(πx/a) for |x | ≤a , ψ(x) = 0 otherwise.(a) Find the value of A such that the wave function is correctly normalized.(b) If a measurement of the position of the particle is made, atwhat value of x is it most likely to be found?(c) What is the probability of finding the particle between x = 0 and x = a /2 ?(d) What would be the average value of x if many measurementswere made on the particles all in this same state? What is the probability density of finding the particle at this particular value of x ? (note: the average value of x at the state of ψ(x ) is defined as dx x xx x ba ⎰=)(ˆ)(*ψψ). Solution:a)()21aax dx ψ-=⎰ then ,22sin 1aax A dx a π-⎛⎫= ⎪⎝⎭⎰so A =b) ()x x a πψ⎛⎫= ⎪⎝⎭ then , ()'0x x a πψ⎛⎫== ⎪⎝⎭so, we can kown2xk aπππ=+ k z ∈,a x a -≤≤ 2ax ∴=± c) ()22220011sin 4a a x x dx dx a a πψ⎛⎫== ⎪⎝⎭⎰⎰ d)()()21sin 0aaaax x x x x dx x dx aa πψψ-*--⎛⎫=== ⎪⎝⎭⎰⎰x x a πψ--⎛⎫⎛⎫⎪∴== ⎪ ⎪⎝⎭⎝⎭∴the probability density offinding the particle at this particular value ofx-is20x ψ-⎛⎫= ⎪⎝⎭. 4. The lowest energy eigenfunction of the time-independent Schrödinger equation for a simple harmonic oscillator is2/2)(yAe y -=ψwhere x m y 2/10)/( ω=, x is the displacement of the oscillator from equilibrium, m is the mass, ω0 is the angular frequency of the oscillator, and A is a constant.(a) Write the wave function in terms of x and verify, by explicitsubstitution in the time-independent Schrödinger equation that the associated energy is2/ω =E .(b) Find a value of the constant A which normalizes the wavefunction. Write out the probability density as a function of x . (c) Find out the average value of x 2 in this state, and thencalculate the average potential energy.(d) Consider a carbon-hydrogen bond in a molecule. Thestretching frequency of the bond is ν = 1.0×1014 Hz. The mass is roughly considered as that of the hydrogen atom since the carbon atom remains nearly fixed. For the lowest vibrational energy eigenstate, find (i) the associated zero-point energy, and (ii) the average value of x 2 of the hydrogen atom from equilibrium.(⎰+∞∞--=a dx ea x π22/,2/3/222⎰+∞∞--=a dx ex a x π)Solution: a) ()22y y Aeψ-= and12m y x ω⎛⎫= ⎪⎝⎭()22m x x Aeωψ-∴=Accoding to ()()x x ψψ∧H =E Then , ()()()22222122d x m x x x m dx ψωψψ-+=E()()()()222222122m m x x m x x x m ωωψψωψψ⎛⎫∴--++=E ⎪⎝⎭12ω∴E =b) ()21x dx ψ+∞-∞=⎰221m x A edx ω+∞--∞∴=⎰then,1A = so,14m A ωπ⎛⎫= ⎪⎝⎭we get the ()x ψ is ()2142m x m x eωωψπ-⎛⎫= ⎪⎝⎭c)()()1222222m x m x x x x dx x edx m ωωψψπω+∞+∞---*-∞-∞⎛⎫===⎪⎝⎭⎰⎰sotheaveragepotentialenergy :()()221124V x m x x dx ψωψω+∞-*-∂∞==⎰d) (i) we can take the hydrogen atom as the linear harmonic oscillator .the energy eigenvalues of the harmonic oscillator is12n n ω⎛⎫E =+ ⎪⎝⎭0,1,2,3n =If 0n =,then the associated zero-point energyIs210115.271022v Jω-E ===⨯(ii) according to (C),2223.151022x m m mvω--===⨯5. A particle moves in an infinite potential well⎩⎨⎧><∞≤≤=a x x ax x V ,000)(. Suppose it is at the ground state (n=1) with energy 22212/ma E π=. At the time t =0, the width of the well expands to 2a suddenly, sofast that the wave function could not change, i.e.,ax a x x πψψsin 2)()0,(1== to the expanded well⎩⎨⎧><∞≤≤=ax x ax x V 2,0200)(. If the wave function )0,(x ψ is still the energy eigenstate, what is the probability to obtain the value of E 1 when the energy is measured?(Hint: the energy eigenvalue and eigenstate are 22228/ma n n πε= and ax n ax n 2sin 1)(πϕ=, respectively. 12E =ε. Expand the wavefunction )0,(x ψ with )(x n ϕ: )()0,(x C x n nn ϕψ∑=, and then find2nC )Solution:Before the well expands to 2asuddenly, the ground state withenergy22122maπE =and the wave function is()()1,0xx x aπψψ==When the well expands to2a ,the energy eigenvalueand eigenstate are22228n n ma πε=and ()2n n xx aπϕ=Because the well expands so fast that the wave functioncould not change, then , 1n ε=E , we get the value of n ,2n =Expand the wave function (),0x ψ with ()n x ϕ :()(),0n n nx c x ψϕ=∑ and 2n =.So ()()22202a ax c x x dx dx a πϕψ*===⎰The probability to obtain the value of 1E is2212c =. Chapter 31. Show that if two operators A and B commute and ψ is an eigenvector of A, then B ψ is also an eigenvector of A with the same eigenvalue.Solution:Because Aˆand B ˆ commute,therefore []0ˆˆˆˆˆ,ˆ=-=A B B A B A(1) Bacause ψ is an eigenvector of Aˆ,we have λψψ=A ˆ (2) ThenψλλψψB B A Bˆˆˆˆ== (3) (1)(3)⇒ ψλψψB A B B Aˆˆˆˆˆ== (4)Then ψB is also an eigenvector of Aˆ with the same eigenvalue λ. 2. Show that 22))((B A B A B A +=-+ only of A and B commute.题目中应为“Show that 22ˆˆ)ˆˆ)(ˆˆ(B A B A B A-=-+ only of A ˆand B ˆ commute,” 而不是“Show that 22ˆˆ)ˆˆ)(ˆˆ(B A B A B A+=-+ only of A ˆand B ˆ commute,” Proof : )ˆˆ()ˆˆ)(ˆˆ(22B A B A B A---+ )ˆˆ(ˆˆˆˆˆˆ2222B A B A B B A A---+-=]ˆ,ˆ[ˆˆˆˆB A A B B A-=+-= Only of Aˆand B ˆ commute,we have 0)ˆˆ()ˆˆ)(ˆˆ(22=---+B A B A B A . So 22ˆˆ)ˆˆ)(ˆˆ(B A B A B A-=-+ only of A ˆand B ˆ commute, 3. Show that if A and B are both Hermitian matrices, AB+BA and i(AB-BA) are also Hermitian. Note that Hermitian matrices are defined such that *ji ij A A = where * denotes complex conjugationProof :Because Aˆand B ˆ are both Hermitian matrices,we have +=A Aˆˆand +=B B ˆˆ. Then A B B A B A A B B A A B A B B Aˆˆˆˆˆˆˆˆˆˆˆˆ)ˆˆˆˆ(+=+=+=++++++ )ˆˆˆˆ()ˆˆˆˆ()ˆˆˆˆ()]ˆˆˆˆ([A B B A i B A A B i B A A B i A B B Ai -=--=--=-+++++ So A B B Aˆˆˆˆ+ and )ˆˆˆˆ(A B B A i - are also Hermitian matrices. 4. Show that the differential operatordxdi p-=ˆ Is linear and hermitian operator in the space of all differentiable wavefunction )(x φ,say, which vanish at both ends of an interval ),(b a .Solution:(ⅰ))]()([)(ˆ22112211x c x c dxdi c c pψψψψ+-=+ )()(2211x dxdi c x dx d i c ψψ --=)(ˆ)(ˆ2211x p c x pc ψψ+= (1) Where )(1x ψ and )(2x ψ are two arbitrary wave functions, 1c and 2c are two numbers thatare independent of argument x .So the differential operator dxdi p-=ˆ is linear. (ⅱ)dx x dxdx i dx x p x baba)()()(ˆ)(**ψϕψϕ⎰⎰-=])()()()([**dx x dxd x x x i babaϕψψϕ⎰--= dx x x dxd i ba)()(*ψϕ⎰= dx x x dxdi ba )())((*ψϕ⎰-=dx x x pba)())(ˆ(*ψϕ⎰= So the differential operator dxdi p-=ˆ is also a hermitian operator. 5 . Assume Fˆ and G ˆ are two hermitian operators, Their eigenequations of them are respectively, i i i F φλφ=ˆ, i i i G ϕμϕ=ˆ. Then the representation Fis refer to the space spanned by states },,1),({n i x i =φ and the representation G ˆ to the space spanned by },,1),({m i x i =ϕ. Suppose the spectrums of Fˆ and G ˆare all discrete (or discontinuous). Please(1) write out the representation of an arbitrary state ψ in representation F ˆ. (2) Write out the representation of an arbitrary quantum observable Aˆ in representation Gˆ (3) Write out the transformation from representation Fˆ to representation G ˆ. Solution :(1)An arbitray state ψ can be expand in the subspace spanned by states},...,2,1),({n i x i =φ)()(1x a x i ni i φψ∑-= with τψφd x x a i i )()(*⎰=.So the representation of an arbitray state ψ in representation Fˆ is=ψ⎪⎪⎪⎪⎪⎭⎫ ⎝⎛n a a a ...21 with τψφd x x a i i )()(*⎰=(2) The representation of an arbitray quantum Aˆ in representation G ˆ is ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡=mm m m m m a a a a a a a a a A ...... (2)12222111211,with τϕϕd x a x a j i ij )(ˆ)(*⎰= (3) the transformation from representation Fˆ to representation G ˆ is ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡=mn m m n n s s s s s s s s s S (2)12222111211,with τϕφd x x s j iij )()(**⎰=6 . A free particle of mass m moves in one dimensional space. At time 0=t the normalized wave function of the particle is)4/exp()2(),0,(224122x x x x x σπσσψ-=-,Where>=<22x x σ.Let the momentum spread is defined as22><-><=p p p σ, please provexp σσ2 =.Proof:Because )4exp()2(),0,(224122xx x xx σπσσψ-=-We havedx dxdi p ψψ)( -=⎰* dx x x i xx x xxx )4exp()2)(42())(4exp()2(22412222412σπσσσπσ----=--⎰0)42e x p (21)2(222212=-⋅=⎰-x d x x i xxx σσπσdxxd d pψψ2222)( -=⎰*dxxxxxx xxxx )4exp()2]()2(42[))(4exp()2(22412222222412σπσσσσπσ-+---=--⎰ dx x xxxxx )2exp(])2(21[)2(222222122σσσπσ-+--=⎰-2224)4222(21x x x x σσπσπσπ=+--= Soxx p ppσσσ2042222=-=-=7.Particles in the state]4e x p [)21()(220212ξπξψx x p i x -= Where ξ is a constant, show the average momentum value of the particle and calculation of uncertainty relations ?)()(22=∆∙∆p x这样由归一化条件12)2()2exp()2()(212222122=⋅=-=--⎰⎰ξππξξπξψdx xx才成立.否则由归一化条件条件就着能得出一个特定的πξ21=，而不是一任意实数，当然这也不能说错，只是失去了题目的一般性质，且给解题带来不便。
