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D. O. Campa et al.
4 Solvable Cases
4.1 Case with j=0
Let us consider now a constant magnetic field
B = (0, 0, B 0 ). In the Landau gauge
the corresponding vector potential becomes
A = (0, xB 0 , 0). Equation (7) implies
that η(x) = 2k + ωx, where ω = 2eB 0 /c ¯
h. By using Eqs. (11)–(13) the SUSY
partner potentials become
V 0 (x) =
ω 2
4
x +
2k
ω
2
−
ω
2
,
(19)
V 2 (x) =
ω 2
4
x +
2k
ω
2
+
3
2
ω.
(20)
It is seen that these potentials are two shifted harmonic oscillators differing from
each other by a constant. In the language of SUSY QM it is said that they are shape
invariant potentials, in the sense that one of them can be obtained from the other
by changing just some parameters and, perhaps, displacing the energy by a global
quantity, see Fig. 2.
The energy levels of V 0 (x) and V 2 (x) are
E
(0)
0 = 0, E
(0)
1 = ω, E
(0)
n = E
(2)
n = nω, n = 2, 3, 4, . . . ,
(21)
i.e., Sp(H 0 ) = {E
(0)
0 , E
(0)
1 , Sp(H 2 )}. The factorization energies were taken in this
case as 1 = E
(0)
1 and 2 = E
(0)
0 . The eigenfunctions of H 0 and H 2 can be expressed
in terms of Hermite polynomials
-10
-5
5
x
5
10
15
V(x)
V 0 (x)
V 2 (x)
eB(x)/ch
Fig. 2 Second-order SUSY partner potentials V 0 (x), V 2 (x) as functions of x and the constant
magnetic field (scaled) inducing them