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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 [email protected] 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 [email protected] 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