This method was first introduced by Schr¨ odinger 7–9 and Dirac 10 and was further developed by Infeld and Hull 11 and Green. 12 The spirit of the factorization method is to write the second-order differential operator H as the product of two first-order differential operators a and a † , plus a real constant E . The form of these operators depends on the form of the potential V ( x ) and the factorization energy. The procedure for finding the ladder operators and the eigenfunctions is as follows. 13 We find operators a$_{1}$, a$_{2}$, a$_{3}$, . . . and real constants E$_{1}$, E$_{2}$, E$_{3}$, . . . from the recursive relations 13 More generally Also assume that there exists a null eigenfunction (root function) | ξ$_{n}$ 〉 with zero eigenvalue for each a$_{n}$ , namely, Hence, E$_{n}$ is the n th eigenvalue of the Hamiltonian with the corresponding eigenfunction 13 (up to a normalization coefficient) Figure 2 shows a schematic diagram describing the relation between root functions, the stationary states, and the ladder operators. As can be seen from Eq. (6), a † $_{1}$a$_{1}$ is equal to H except for a constant. Thus, a$_{1}$ must have a linear momentum term to be consistent with the kinetic energy part of the Hamiltonian. According to Eq. (7), each of the annihilation operators a$_{n}$ should also have a a linear momentum term. Thus, a$_{n}$ can be written as where P is the momentum operator and f$_{n}$ ( x ) is a real function of x . Although these operators are not hermitian ( a † n = 1 √$_{2}$$_{m}$ ( P − if$_{n}$ ( x )) = a$_{n}$ ), their product is hermitian a † $_{n}$a$_{n}$ = 1 2 m P 2 + 1 2 m f 2 n + ¯ h 2 m df$_{n}$ dx . (11) 4 a$_{n}$ = 1 √$_{2}$$_{m}$ [ P + if$_{n}$ ( x )] , (10) | E$_{n}$ 〉 = a † $_{1}$a † $_{2}$. . . a † n − 1 | ξ$_{n}$ 〉 . (9) a$_{n}$ | ξ$_{n}$ 〉 = 0 . (8) a † n $_{+1}$a$_{n}$$_{+1}$ + E$_{n}$$_{+1}$ = a$_{n}$a † n + E$_{n}$, ( n = 1 , 2 , . . . ) . (7) a † $_{3}$a$_{3}$ + E$_{3}$ = a$_{2}$a † 2 + E$_{2}$, . . . . (6c) a † $_{2}$a$_{2}$ + E$_{2}$ = a$_{1}$a † 1 + E$_{1}$, (6b) a † $_{1}$a$_{1}$ + E$_{1}$ = H, (6a)