230
3 Quantum Mechanics – II
=
1
/ 2 m ω
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
2mω 2
(2)
Put X = x +
qE
mω 2 ; then we can write
d
dx
=
d
dX
and
d
2
dx 2 =
d
2
dX 2
Equation (1) becomes
−
2
2m
d
2
dX 2 +
1
2
mω
2 X
2
ψ n (X ) =
E n +
q
2 E
2
2mω 2
ψ n (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
E n +
q
2 E
2
2mω 2 =
n +
1
2
ω
or
E n =
n +
1
2
ω − q
2 E
2
/2mω
2
(4)
3.99 The matrix of H
is
(H
) =
H
11 H
12
H
21 H
22
, with H
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 − E
H
12
H
21
E 0 + H
22 − E
= 0
E 1 =
1
/ 2
2E 0 + H
11 + H
22
+
1
/ 2
H
11 − H
22
2 + 4
H
12
2
1/2
E 2 =
1
/ 2
2E 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 Hamiltonian H =
H 0 + H
. The perturbation theory requires finding the eigen values of H
and
adding them to E 0 , which gives an exact result.
3.100 The nuclear charge seen by the electron is Z and not 2. This is because of
screening the effective charge is reduced.
The smallest value of E(Z ) = −
e
2
2a 0
27
4
Z − 2Z
2
must be determined
which is done by setting