
5.3 Non-stationary perturbation theory 147
If the perturbation
ˆ
V (t) is small and weakly changes during the time interval
T
mn
= 2π/ω
mn
, then the magnitude of the integral in Eq. (5.51) and, correspond-
ingly, the probability of transition are small. In the limit of infinitesimally slow
changes of the perturbation, the probability of any transition between the energy
levels tends to zero. In another limiting case, that of very fast (sudden) switching
on of the perturbation, the derivatives dV
mn
/dt tend to infinity at the instant of
switching on of the perturbation. Taking this into account, we can take the slowly
changing factor e
iω
mn
t
out of the integral in Eq. (5.67). After this the integral can
be easily evaluated and as a result we obtain the following expression:
P
mn
=
1
h
-
2
ω
2
mn
|
V
mn
|
2
. (5.68)
The transition probability of a system at a sudden switching on of the pertur-
bation can also be found in cases when the perturbation is not small. Let the
system be in one of the quantum states ψ
(0)
n
of the unperturbed Hamiltonian
ˆ
H
0
.
If the change of the Hamiltonian happens during a time that is small in compari-
son with the time of transition, 2π/ω
mn
, then the wavefunction of the system
does not have enough time to change and will stay the same as it was before
the switching on of the perturbation. Thus, it will not be an eigenfunction of
the new Hamiltonian,
ˆ
H =
ˆ
H
0
+
ˆ
V , and, therefore, the initial state ψ
(0)
n
will not
be stationary. The probability of the transition of a system to a new stationary
state is defined according to Eq. (5.44) by the corresponding expansion coef-
ficient of the function ψ
n
, which is decomposed over the set of unperturbed
eigenfunctions ψ
(0)
n
:
P
mn
=
ψ
∗
m
ψ
(0)
n
dx
2
. (5.69)
Example 5.2. Under the influence of a suddenly switched-on homogeneous
electric field E the charged oscillator makes a transition from the ground state
to the excited state. Find the probability of such a transition if the mass of the
oscillator is equal to m and its charge is equal to e.
Reasoning. The potential energy of the charged oscillator in a homogeneous
electric field can be written as follows:
U (x ) =
mω
2
2
x
2
− e
|
E
|
x =
mω
2
2
(x − x
0
)
2
−
e
2
|
E
|
2
2mω
2
, (5.70)
where the coordinate x
0
is equal to x
0
= e
|
E
|
/(mω
2
). Equation (5.70) has the
form of the potential energy of a linear oscillator with shifted equilibrium position
x
0
. If the wavefunction of the unperturbed oscillator has the form ψ
(0)
n
(x), then
for the perturbed oscillator this function must have the form ψ
n
(x − x
0
). The
wavefunctions of the unperturbed harmonic oscillator (4.89)havetheform
ψ
(0)
n
(ξ) = A
n
H
n
(ξ)e
−ξ
2
/2
,