content_image/369710.jpg content_image/369711.jpg content_image/369712.jpg
where we have taken into account the fact that the mass may depend on the position coordinate and prime denotes the derivative with respect to x . Eliminating χ ( x ) using Eq. (5a), we obtain for φ ( x ) the second order equation: 

<img src='content_image/369710.jpg'>

Solving this second order differential equation one can obtain the χ ( x ) component via Eq. (5a) and then reconstruct the upper u$_{1}$ ( x ) and lower u$_{2}$ ( x ) components of the complete spinor solution ψ ( x ). Eq. (7) reduces to the one studied in [24] if m ′ = 0, i.e. if the mass reduces to a constant. Thus we see that keeping a position dependence in the mass introduces two new terms: one which multiplies φ $^{′}$( x ) while the other enters the φ ( x ) term. These new terms must be appropriately constrained in order to be able to solve the equation in terms of the Hypergeometric function. 

Let us make a final remark before to discuss the details of the computations. The attentive reader might wonder what would happen if one were to derive a second order equation for the χ ( x ) eliminating instead the φ ( x ) component and then computing it via Eq. 5b. The second order equation for the χ ( x ) component turns out to be: 

<img src='content_image/369711.jpg'>

It is easily checked that Eq. 8 can be obtained from Eq. 7 using the map E → − E and V ( x ) → − V ( x ). This can be interpreted as the negative energy solution corresponding to the charge conjugate particle (antiparticle). Since V ( x ) is the temporal component of a four-potential the change V ( x ) → − V ( x ) amounts to reversing the charge of the particle. Indeed with E → − E and V ( x ) → − V ( x ) we have χ → φ , and by using Eq. 5a, φ → − χ and by using the inverse of Eqs. (4a,4b) we have in turn u$_{1}$ → iu$_{2}$ and u$_{2}$ → iu$_{1}$ which amounts, up to inessential phase factors, to the charge conjugation symmetry of the Dirac equation [24]. 

## III. EFFECTIVE-MASS DIRAC SCATTERING PROBLEM 

The form of the Woods-Saxon potential (illustrated in Fig. 1) is given by (see also ref. [24]): 

<img src='content_image/369712.jpg'>

where W is a positive parameter in the scattering problem (potential barrier) and negative in the bound state problem (potential well); a and L are two real and positive parameters. 

