1000 Solved Problems in Modern Physics

230 3 Quantum Mechanics – II

=^1 / 2 mω^2 [(x+qE/mω^2 )^2 −q^2 E^2 /m^2 ω^4 ]

=^1 / 2 mω^2

(

x+

qE

mω^2

) 2

−

q^2 E^2

2 mω^2

(2)

PutX=x+mqEω 2 ; then we can write

d

dx

=

d

dX

and

d^2

dx^2

=

d^2

dX^2

Equation (1) becomes

(

−

^2

2 m

d^2

dX^2

+

1

2

mω^2 X^2

)

ψn(X)=

(

En+

q^2 E^2

2 mω^2

)

ψn(X)(3)

Left hand side of (3) is the familiar Hamiltonian for the Simple harmonic

oscillator. The modified eigen values are then given by the right hand side

En+

q^2 E^2

2 mω^2

=

(

n+

1

2

)

ω

or

En=

(

n+

1

2

)

ω−q^2 E^2 / 2 mω^2 (4)

3.99 The matrix ofH′is

(H′)=

(

H′ 11 H′ 12

H′ 21 H′ 22

)

,withH 12 ′ =H 21 ′∗

The matrix of H is

<H>=

[

E 0 +H′ 11 H′ 12

H′ 21 E 0 +H′ 22

]

∣

∣

∣

∣

E 0 +H′ 11 −EH′ 12

H′ 21 E 0 +H 22 ′ −E

∣

∣

∣

∣=^0

E 1 =^1 / 2

(

2 E 0 +H 11 ′ +H 22 ′

)

+^1 / 2

[(

H 11 ′ −H 22 ′

) 2

+ 4

∣

∣H 12 ′

∣

∣^2

] 1 / 2

E 2 =^1 / 2

(

2 E 0 +H 11 ′ +H 22 ′

)

−^1 / 2

[(

H 11 ′ −H 22 ′

) 2

+ 4

∣

∣H 12 ′

∣

∣^2

] 1 / 2

These are the energy levels of a two state system with HamiltonianH=

H 0 +H′. The perturbation theory requires finding the eigen values ofH′and

adding them to E 0 , which gives an exact result.

3.100 The nuclear charge seen by the electron isZand not 2. This is because of
screening the effective charge is reduced.
The smallest value ofE(Z) =−e
2
2 a 0

( 27

4 Z−^2 Z

2 )must be determined

which is done by setting