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Note that ε = 2m ∗ E/ ¯
h 2 and L 2 is the second-order differential operator
L 2 =
d 2
dx 2 + η(x)
d
dx
+ γ (x).
(6)
Let us remark that η(x) is given by
η(x) = 2
k +
e
c ¯
h
A(x)
,
(7)
thus it is directly related with B(x) as follows:
B(x) =
c ¯
h
2e
dη(x)
dx
.
(8)
For the time being, the form of the function γ (x) is not relevant here. We will see
later on that it can be expressed in terms of η(x).
The coupled system of Eqs. (5) and the form of the operator L 2 in Eq. (6) suggest
to use the second-order SUSY QM to address the problem.
3 Second-Order SUSY QM
Let us consider two Schrödinger Hamiltonians
H j = −
d 2
dx 2 + V j (x), j = 0, 2.
(9)
They are called second-order SUSY partners if there exists a second-order operator
L 2 intertwining them
H 2 L 2 = L 2 H 0 ,
(10)
with L 2 being given by Eq. (6). Equation (10) produces a set of relations among
η(x), γ (x), V 0 (x), V 2 (x), which after some work leads to
V 2 (x) = V 0 (x) + 2η
(x),
(11)
γ (x) =
η 2 (x)
2
−
η (x)
2
− V 0 (x) +
1 + 2
2
,
(12)
V 0 (x) =
η (x)
2η(x)
−
(η (x)) 2
4η 2 (x)
− η
(x) +
η 2 (x)
4
+
1 + 2
2
+
1 − 2
2η(x)
2
,
(13)