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 =Bsinx (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