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Region II

The solution in region II must solve the differential equation (2) and match the appropriate boundary conditions. The boundary conditions at x=-a, as discussed in the previous section, are that the slope and value of the wavefunction at x=-a must match those of the wavefunction in region I,

 

In region I the TISE reads

 

where we have introduced a new constant

 

with a minus sign in (10) to remind us of the sign of the left hand side of (2).

Just as in region I, this is a second order linear differential equation with constant coefficients, and so again, this equation must also have solutions of the form . In this case . The general solution to (10) is thus , where and are the two independent constant parameters. Again, it is good form to write the solution as centered at the boundary where and . In this case, one final redefinition of constants is convenient. Because the TISE is a real differential equation, we expect it to possesses two linearly independent real solutions. We bring this out explicitly by writing new constants and giving us the form

 

Again, one may arrive at 12 directly. In a classically allowed region (where we define k's instead of 's), the solutions are always sinusoidal and direct substitution verifies that (12) solves (10).

With the most general solution to the differential equation region II, it remains only to adjust the coefficients so that the boundary conditions are matched. With the function written in this form, the algebra of this is simple,

and

Comparing this with with the value and slope carried across, ``injected'' from region I (9) we have , . With the constants determined, we have the final solution in region II,



next up previous
Next: Boundary Between Region Up: Finite Square Well Previous: Boundary Between Region



Prof. Tomas Alberto Arias
Thu Oct 12 21:20:39 EDT 1995