Corrections Quantum Mechanics: Concepts and Applications Nouredine Zettili (Modi…ed last on September 30, 2010)
Note: This list of corrections pertains to: N. Zettili, Quantum Mechanics: Concepts and Applications, (Chichester: John-Wiley, 2009); 2nd edition, ISBN: 978-0-470-02678-6 (Hardcover), 978-0-470-02679-3 (Paperback). Chapter 2 Page 151, replace Eq. (2.471) by ^2 H
=
2
(j
=
2
(j
1 ih 2 1 ih 1
j+j
2 ih 1
j+j
2 ih 2
j) (j j) ;
1 ih 2
j+j
2 ih 1
j)
Page 8, Page 151, replace Eq. (2.472) by (
2
^ 2 )2 H
=
(j
= j
1 ih 1
1 ih
1
j+j
j+j
2 ih 2
2 ih
2
j) (j
j=
1 ih 1 2 ^2
H :
j+j
2 ih 2
j)
Chapter 3 Replace Problem 3.5 and its solution on Page 196 by Problem 3.5 Assuming that the system of Problem 3.4 is initially in the state j 3 i, what values for the energy and the observable A will be obtained if we measure: (i)H …rst then A, (ii) A …rst then H? ^ and A^ are compatible. (b) Compare the results obtained in (i) and (ii) and infer whether H ^ ^ Calculate [H; A]j 3 i. ^ whose action on j n i is de…ned by Bj ^ ni = (c) Consider now an other operator B ^ nb0 j n+1 i. Repeat Questions (a) and (b) for the operator B. Solution
^ (a) (i) The measurement of H …rst then A is represented by A^Hj 2 ^ ^ Hj n i = n E0 j n i and Aj n i = (n + 1)a0 j n i, we have ^ A^Hj
3i
^ = 9E0 Aj
1
3i
= 36E0 a0 j
3 i:
3 i.
Using the relations (1)
(ii) Measuring A …rst and then H, we will obtain ^ Aj ^ H
^ = 4a0 Hj
3i
3i
= 36E0 a0 j
3 i:
(2)
^ and H ^ A^ yield the same result. (b) Equations (1) and (2) show that the actions of A^H ^ ^ This means that H and A commute; hence they are compatible. We can thus write ^ A]j ^ [H;
3i
= (36
36)E0 a0 j
3i
= 0:
(3)
^ Hj ^ (c) (i) The measurement of H …rst then B is represented by B 2 ^ ^ Hj n i = n E0 j n i and Bj n i = nb0 j n+1 i, we have ^ Hj ^ B
3i
^ = 9E0 Bj
3i
= 27E0 b0 j
3 i.
Using the relations
4 i:
(4)
4 i:
(5)
(ii) Measuring B …rst and then H, we will obtain ^ Bj ^ H
3i
^ = 3b0 Hj
4i
= 48E0 b0 j
^H ^ and H ^B ^ yield di¤erent results. This Equations (4) and (5) show that the actions of B ^ and B ^ do not commute; hence they are not compatible. We can thus write means that H ^ B]j ^ [H;
3i
= (48
27)E0 b0 j
4i
= 17E0 b0 j
4 i:
(6)
Replace the …rst two lines of Problem 3.11 on Page 204 by: Consider a system whose initial state j (0)i and Hamiltonian are given by 0 1 0 1 3 0 0 3 1@ A 0 ; H = @ 0 0 5 A; j (0)i = 5 0 5 0 4 where
has the dimensions of an energy.
Replace the third line from the bottom of Page 204 by: A measurement of the energy yields the values E1 =
5 , E2 = 3 , E3 = 5 ; the
Replace Eq. (3.222) on Page 205 by: p 2 2 j (t)i = e 5
3 iE1 t=h j 1i + e 5
p 2 2 iE2 t=h e j 2i + 5
0 1 iE3 t=h j 3i = @ 5
1 3e 3i t=h 4i sin (5 t=h) A : 4 cos (5 t=h)
Replace Eq. (3.223) on Page 205 by: 9 ^ h jHj 25 2 8 9 8 27 ( 5) + (3) + (5) = : 25 25 25 25
^ (0)i = E(0) = h (0)jHj =
8 h 25
^
1 jHj 1 i
2
+
2i
+
8 h 25
^
3 jHj 3 i
Replace Eq. (3.224) on Page 205 by: ^ (0)i = E(0) = h (0)jHj
25
3
0
4
0
3 @ 0 0
0 0 5
10 1 0 3 27 5 A@ 0 A = : 25 0 4
Replace Eq. (3.225) on Page 205 by: E(0) =
2 X
P (En )En =
n=1
9 8 27 8 ( 5) + (3) + (5) = : 25 25 25 25
Replace Eq. (3.226) on Page 205 by: 8 iE1 t=h iE1 t=h ^ 1 i + 9 eiE2 t=h e iE2 t=h h 2 jHj ^ 2i e e h 1 jHj 25 25 8 ^ 3 i = 8 ( 5) + 9 (3) + 8 (5) = 27 = E(0): + eiE3 t=h e iE3 t=h h 3 jHj 25 25 25 25 25
^ (t)i = E(t) = h (t)jHj
3
