130_notes.dvi

(Frankie) #1

The angular part of the integral can be done. All the terms of the wavefunction contain aY 00
andrdoes not depend on angles, so the angular integral just gives 1.


〈ψ|r|ψ〉=

1

2

∫∞

0

(R 10 −R 20 e−i(E^2 −E^1 )t/ ̄h)∗r(R 10 −R 20 e−i(E^2 −E^1 )t/ ̄h)r^2 dr

The cross terms are not zero because of ther.


〈ψ|r|ψ〉=

1

2

∫∞

0

(

R^210 +R^220 −R 10 R 20

(

ei(E^2 −E^1 )t/ ̄h+e−i(E^2 −E^1 )t/h ̄

))

r^3 dr

〈ψ|r|ψ〉=

1

2

∫∞

0

(

R^210 +R 202 − 2 R 10 R 20 cos

(

E 2 −E 1

̄h

t

))

r^3 dr

Now we will need to put in the actual radial wavefunctions.


R 10 = 2

(

1

a 0

) (^32)
e−r/a^0


R 20 =

1


2

(

1

a 0

)^32 (

1 −

r
2 A 0

)

e−r/^2 a^0

〈ψ|r|ψ〉 =

1

2 a^30

∫∞

0

(

4 e−^2 r/a^0 +

1

2

(

1 −

r
a 0

+

r^2
4 a^20

)

e−r/a^0

− 2


2

(

1 −

r
2 a 0

)

e−^3 r/^2 a^0 cos

(

E 2 −E 1

̄h

t

))

r^3 dr

=

1

2 a^30

∫∞

0

(

4 r^3 e

−a 2 r

(^0) +


1

2

r^3 e

−ar

(^0) −


1

2 a 0

r^4 e

−ar

(^0) +


1

8 a^20

r^5 e

−ar
0

+

(

− 2


2 r^3 e

− 2 a 3 r

(^0) +



2

a 0

r^4 e

− 2 a 3 r
0

)

cos

(

E 2 −E 1

̄h

t

))

dr

=

1

2 a^30

[

24

(a
0
2

) 4

+ 3a^40 −

1

2 a 0

24 a^50 +

1

8 a^20

120 a^50

+

(

− 2


26

(

2 a 0
3

) 4

+


2

a 0

24

(

2 a 0
3

) 5 )

cos

(

E 2 −E 1

̄h

t

)]

=

a 0
2

[

3

2

+ 3−12 + 15 +

(

− 12


2

16

81

+


2

a 0

24

32

243

)

cos

(

E 2 −E 1

̄h

t

)]

=

a 0
2

[

3

2

+ 3−12 + 15 +

(



2

64

27

+

256


2

81

)

cos

(

E 2 −E 1

̄h
t

)]

= a 0

[

15

4

+

32


2

81

cos

(

E 2 −E 1

̄h

t

)]
Free download pdf