d. The fraction of systems observed in Yn is Ja/2. The possible energies measured are En and Em. The probabilities of measuring each of these energies is Ja/2 and (b/2 e. Once the system is observed in pn, it stays in pn f P(En)=<Yn> =lc, Si x(L-x)dx x(Lx)Sn(L丿 0 0 These integrals can be evaluated to give NIX nπx|n2ax2x2 SI 0 L6 Sin(nπ)-Sin(0) L2 LCos(nπ) 72( Sin(nT)-OSIn(O) 21
21 d. The fraction of systems observed in Yn is |a|2. The possible energies measured are En and Em. The probabilities of measuring each of these energies is |a|2 and |b|2. e. Once the system is observed in Yn, it stays in Yn. f. P(En) = ï ï ï ï <Yn|Y> 2 = |cn| 2 cn = õ ô ó 0 L 2 L Sinè ç æ ø ÷ npxö L 30 L5 x(L-x)dx = 60 L6õ ô ó 0 L x(L-x)Sinè ç æ ø ÷ npxö L dx = 60 L6 ë ê ê é û ú ú ù Lõ ô ó 0 L xSinè ç æ ø ÷ npxö L dx - õ ô ó 0 L x2Sinè ç æ ø ÷ npxö L dx These integrals can be evaluated to give: cn = 60 L6 ë ê é û ú ù Lè ç æ ø ÷ ö L2 n2p2Sinè ç æ ø ÷ npxö L - Lx npCosè ç æ ø ÷ npxö L ï ï ïL 0 - 60 L6 ë ê ê é û ú ú ù è ç æ ø ÷ ö 2xL2 n2p2Sinè ç æ ø ÷ npxö L - è ç æ ø ÷ ö n2p2x2 L2 - 2 L3 n3p3Cosè ç æ ø ÷ npxö L ï ï ïL 0 cn = 60 L6 { L3 n2p2( ) Sin(np) - Sin(0) - L2 np( ) LCos(np) - 0Cos0 ) - ( 2L2 n2p2( ) LSin(np) - 0Sin(0)
n2兀2-2 2)n33 Cos(nπ) 2 Cn=L-360-nt Cos(nx)+(n272. 2) 23 Cos(nr) 2 3I nr/1) Cn=n573(+1)+1) 4(60) I nl=n66 )(()n If n is even then Cn=0 If n is odd then (4)(6 042-2 The probability of making a measurement of the energy and obtaining one of the eigenvalues, given by n22n2 En=2mL2 IS P(En)=0 if n is even )=n6,6 if n is odd
22 - ( ) n2p2 - 2 L3 n3p3 Cos(np) + è ç æ ø ÷ ö n2p2(0) L2 - 2 L3 n3p3 Cos(0))} cn = L-3 60 {- L3 np Cos(np) + ( ) n2p2 - 2 L3 n3p3 Cos(np) + 2L3 n3p3 } cn = 60è ç æ ø ÷ ö - 1 np (-1)n + ( ) n2p2 - 2 1 n3p3(-1)n + 2 n3p3 cn = 60è ç æ ø ÷ ö è ç æ ø ÷ -1 ö np + 1 np - 2 n3p3 (-1)n + 2 n3p3 cn = 2 60 n3p3 )( ) -(-1)n + 1 |cn| 2 = 4(60) n6p6 )( ) -(-1)n + 1 2 If n is even then cn = 0 If n is odd then cn = (4)(60)(4) n6p6 = 960 n6p6 The probability of making a measurement of the energy and obtaining one of the eigenvalues, given by: En = n2p2-h2 2mL2 is: P(En) = 0 if n is even P(En) = 960 n6p6 if n is odd
<-凹y X(L-x)dx x(L-xdx2/(xL-x2)dx 0 mL5)X(L-x)(-2)dx 30h xL-x2dx 302 mL5人12-3 30n25h mL2 =∑ceh<出H>eh
23 g. <Y|H|Y> = õ ô ô ó 0 L è ç æ ø ÷ 30ö L5 1 2 x(L-x)è ç æ ø ÷ ö -h-2 2m d2 dx2 è ç æ ø ÷ 30ö L5 1 2 x(L-x)dx = è ç æ ø ÷ 30ö L5 è ç æ ø ÷ ö -h-2 2m õ ô ó 0 L x(L-x)è ç æ ø ÷ ö d2 dx2 ( xL-x ) 2 dx = è ç æ ø ÷ ö -15h-2 mL5 õó 0 L x(L-x)(-2)dx = è ç æ ø ÷ ö 30h-2 mL5 õó 0 L xL-x2dx = è ç æ ø ÷ ö 30h-2 mL5 è ç æ ø ÷ ö L x2 2 - x3 3 ï ï ïL 0 = è ç æ ø ÷ ö 30h-2 mL5 è ç æ ø ÷ L3 ö 2 - L3 3 = è ç æ ø ÷ ö 30h-2 mL2 è ç æ ø ÷ 1 ö 2 - 1 3 = 30h-2 6mL2 = 5h-2 mL2 10. <Y|H|Y> = å ij Ci *e iEi t -h <Yi |H|Yj> e -iEj t -h Cj
Since <YiHYD>=Ejoij I(EP-Ei)t <HⅣP=∑C℃CEet 平H>=∑CGF( not time dependent For other properties <A>=∑ceh<A>ehc but, <pilA i> does not necessarily =aj dij because the Y, are not eigenfunctions of A unless [A, H=0 A>=∑Cceh<A> Therefore, in general, other properties are time dependent a. The lowest energy level for a particle in a 3-dimensional box is when n1 =1, n2=1 and n3=1. The total energy(with L1=L2=L3) will be h2n12+n2+n3)8mL2 Etotal-8mL 4
24 Since <Yi |H|Yj> = Ejdij <Y|H|Y> = å j Cj *CjEje i(Ej -Ej )t -h <Y|H|Y> = å j Cj *CjEj (not time dependent) For other properties: <Y|A|Y> = å ij Ci *e iEi t -h <Yi |A|Yj> e -iEj t -h Cj but, <Yi |A|Yj> does not necessarily = ajdij because the Yj are not eigenfunctions of A unless [A,H] = 0. <Y|A|Y> = å ij Ci *Cje i(Ei -Ej )t -h <Yi |A|Yj> Therefore, in general, other properties are time dependent. 11. a. The lowest energy level for a particle in a 3-dimensional box is when n1 = 1, n2 = 1, and n3 = 1. The total energy (with L1 = L2 = L3) will be: Etotal = h2 8mL2( ) n1 2 + n2 2 + n3 2 = 3h2 8mL2
Note that n=0 is not possible. The next lowest energy level is when one of the three quantum numbers equals 2 and the other two equal 1 n1=1,n2=1,ny=2 n1=1,n2=2,n3=1 n2=1,n3 Each of these three states have the same energy Etotal=8mL2(n12+n2+n32-8mL2 Note that these three states are only degenerate if L1=L2=L3 distortion 2=L3 ≠ For L1=L2=L3 V=LIL2L3=L3 Etotal L1)=281 +8 2 2 For L3+ LI=L2. V=L L2L3=L12L3, L3=V/L12 Etotal ( L1)=281+E2
25 Note that n = 0 is not possible. The next lowest energy level is when one of the three quantum numbers equals 2 and the other two equal 1: n1 = 1, n2 = 1, n3 = 2 n1 = 1, n2 = 2, n3 = 1 n1 = 2, n2 = 1, n3 = 1. Each of these three states have the same energy: Etotal = h2 8mL2( ) n1 2 + n2 2 + n3 2 = 6h2 8mL2 Note that these three states are only degenerate if L1 = L2 = L3. b. ¾ ¾ ¾¾ ¾¾ ¾¾¾¾® distortion ¾¾ ¾¾ ¾ ¾ ¾¾¯ ¾¾¯ L1 = L2 = L3 L3 ¹ L1 = L2 For L1 = L2 = L3, V = L1L2L3 = L1 3, Etotal(L1) = 2e1 + e2 = 2h2 8mè ç æ ø ÷ ö 12 L1 2 + 12 L2 2 + 12 L3 2 + 1h2 8mè ç æ ø ÷ ö 12 L1 2 + 12 L2 2 + 22 L3 2 = 2h2 8mè ç æ ø ÷ ö 3 L1 2 + 1h2 8mè ç æ ø ÷ ö 6 L1 2 = h2 8mè ç æ ø ÷ ö 12 L1 2 For L3 ¹ L1 = L2, V = L1L2L3 = L1 2L3, L3 = V/L1 2 Etotal(L1) = 2e1 + e2