《无机化学 Chemical Theory》课程教学资源（讲稿）Part I Basic concepts of Quantum mechanics Chapter 2 Some Simple Cases

Chapter 2.Some simple cases1. Translational motion (free particles)2. Particle in a box3.Theharmonicoscillator

Chapter 2. Some simple cases 1. Translational motion (free particles) 2. Particle in a box 3. The harmonic oscillator

2.1FreeparticlesClassically, a particle moving in one dimension without external forces has constantmomentum.QuantummechanicallyweexpecttobeabletofindstatesofdefinitemomentumIf the motion is along the x axis we need to look for solutions of the eigenvalue equationp.y (x)= py/ (x)That is,dit-y=pydxThesolutiontothis equationis,=exp(ipx / h)Remember that p is an eigenvalue - a constant with (in this case) dimensions of momentum

2.1 Free particles Classically, a particle moving in one dimension without external forces has constant momentum. Quantum mechanically we expect to be able to find states of definite momentum. If the motion is along the x axis we need to look for solutions of the eigenvalue equation That is, The solution to this equation is Remember that p is an eigenvalue — a constant with (in this case) dimensions of momentum. p x p x ˆ x       d d i p x      p =exp / ipx 

TheHamiltonian for a free particle contains only the kinetic energy term:H==-_ d?2m dx?2mThe wavefunction , is an eigenfunction of this operator too:? d?h?ipx/hjpx/hHy, 2m dx?2m21Its energy is p? /2m, just as we would expect for a particle with momentum p.Notice that a particle with momentum -p i.e. with wavefunctionΦ-p = exp(-ipx/h) — has the same energy p2 /2m

The Hamiltonian for a free particle contains only the kinetic energy term: The wavefunction ψp is an eigenfunction of this operator too: Its energy is 𝑝 2 /2𝑚, just as we would expect for a particle with momentum p. Notice that a particle with momentum −p — i.e. with wavefunction 𝜓−𝑝 = exp(−𝑖𝑝𝑥/ℏ) — has the same energy 𝑝 2 /2𝑚. 2 2 2 2 ˆ d ˆ 2 2 d p H T m m x     2 2 2 2 2 / / 2 d e e 2 d 2 2 ipx ipx p p ip p H m x m m             

Thetime-independent Schrodingerequationist? d'yHy=Ey2m dx?andthegeneralsolutionofthisisy = aeikr + be-ikrwhere k = V2mE/h.If E = p?/2m, then k = p/h and we arrive at the resulty = ay,+by-pThis superposition or linear combination of two wavefunctions, both with energy p?/2m.is also an eigenfunction of H with energy p?/2m, for any values of the constants a and bHowever it is not an eigenfunction of the operator px, unless a = O or b = O, so it doesn't have adefinitemomentum. We can write the same wavefunction in the formy = Asin(px/h)+ Bcos(px/h)

The time-independent Schrödinger equation is and the general solution of this is where 𝑘 = 2𝑚𝐸/ℏ. If 𝐸 = 𝑝 2 /2𝑚, then 𝑘 = 𝑝/ℏ and we arrive at the result This superposition or linear combination of two wavefunctions, both with energy 𝒑 𝟐 /𝟐𝒎, is also an eigenfunction of H with energy 𝒑 𝟐 /𝟐𝒎, for any values of the constants a and b. However it is not an eigenfunction of the operator px , unless a = 0 or b = 0, so it doesn’t have a definite momentum. We can write the same wavefunction in the form 2 2 2 d 2 d H E m x       e e ikx ikx  a b   p p    a b       A px B px sin / cos /    

2.2 Particle in a boxConsider a particle in a box': suppose that the potential is zero for O < x < a and infiniteoutsidethisrange.TheSchrodingerequationish?dHy=(↑+V)y=+VW=Ey2m dx2Outside the box, where Vis infinite, the only solution is y = O.Inside the box, possiblesolutions are exp(ipx/h) and exp(-ipx/h) both with energy E = p?/2m.However the wavefunction has to be continuous, so it must be zero at both ends of the boxWe can achieve this by using the wavefunction Asin(px/h) + Bcos(px/h)If the wavefunction is to be zero when x = O, then B = 0. If it is to be zero when x = a, thenpa=n元for integer nh

2.2 Particle in a box Consider a ‘particle in a box’: suppose that the potential is zero for 0 < x < a and infinite outside this range. The Schrödinger equation is Outside the box, where V is infinite, the only solution is ψ = 0. Inside the box, possible solutions are exp(𝑖𝑝𝑥/ℏ) and exp(−𝑖𝑝𝑥/ℏ) both with energy 𝐸 = 𝑝 2 /2𝑚. However the wavefunction has to be continuous, so it must be zero at both ends of the box. We can achieve this by using the wavefunction 𝐴𝑠𝑖𝑛(𝑝𝑥/ℏ) + 𝐵𝑐𝑜𝑠(𝑝𝑥/ℏ). If the wavefunction is to be zero when x = 0, then B = 0. If it is to be zero when x = a, then   2 2 2 d ˆ ˆ 2 d H T V V E m x                 pa  n for integer n

V(x)=00V(x)=00IIh?d?H=-+VIII18元m dx2V(x)= 0x=lx-0h?I, II:ay+Vy=Ey8元2m 0°xa'y8元°mVy=0: (V=00)..V- E-Va?xh?a'yh?=0W8元2mVa?x

2 2 2 2 ˆ ˆ 8 m x h d H V  d    I, III: 2 2 2 2 8 h V E m x          2 2 2 2 0 8 h x mV         2 2 2 2 8 0 ( ) m V V V E V x h            

II: V=0V(x)=00V(x)=00IIIIIV(x)= 0x=x=0Hy= Eyh?d?d'y8元^mH= 8元'max +E=0h?d?xd'y+βy= 0d2y = Acos Bx+ Bsin Bxβ2_ 8元°mEh?

II: V=0 H E ˆ    2 2 2 2 ˆ ˆ 8 m x h d H V  d    2 2 2 2 8 0 d m E d x h      2 2 2 2 2 2 0 8 = d d x mE h             A x B x cos sin

Boundary condition and continuous condition: y(O)-0, y(a)-0Hence, y(O)=AcosO +BsinoA=0, B±0y=Bsinβx (a)=Bsinβx=Bsin βa=O,Thus, βa-n元,β=n元/a8元mEh?n'h?E:(n=1,2,3...)8ma2Normalizationn元sin? nRy=Bsin- xdx = 1aan元sin0a

Boundary condition and continuous condition: (0)=0, (a)=0 Hence, (0) =Acos0 + Bsin0 A=0, B≠0 =Bsinx  (a) =Bsin  x =Bsin a=0, Thus, a=n,  =n/a 2 2 2 2 2 2 2 2 2 8 ( 1, 2,3.) 8 mE n h a n h E n ma        sin n B x a    2 2 0 sin 1 a n B xdx a    Normalization 2 B a  2 sin n x a a   

2.Theproperties ofthe solutionsh?2元xEn=1Wsin8ma?aa22元x4h?n=2E2siny228maaa29h?3元×n=3E,sin38ma?aa1.The particle can exist in many states2.quantization energy3. The minimum energy (h?2/8ma?)

2. The properties of the solutions 1. The particle can exist in many states 2. quantization energy 3. The minimum energy (h 2 /8ma 2 ) 2 1 2 8 h E ma  1 2 sin x a a    2 2 2 4 8 h E ma  2 2 2 sin x a a    2 3 2 9 8 h E ma  3 2 3 sin x a a    n=1 n=2 n=3

Boundary conditions and quantizationSo the allowed wavefunctions for the particle in a box aren元2sin8ma2for integer n > O, and the corresponding energies areE,=P_nh?2m8ma山We see that the imposition of boundary conditions leadsto quantization: only certain values of the energy are31possible.0Wavefunctionsforaparticleinabox

Boundary conditions and quantization So the allowed wavefunctions for the particle in a box are for integer n > 0, and the corresponding energies are We see that the imposition of boundary conditions leads to quantization: only certain values of the energy are possible. 2 2 2 2 2 8 n p n h E m ma   Wavefunctions for a particle in a box 2 sin n x a a   