10. Appendix

Solution to Problem 9.14 669
Again by symmetrizing u, v, x and y as in the case of k 0 one can simplify
the 4 × 4 determinant into two 2 × 2 determinants from which the eigenvalues
and eigenvectors can be calculated. The parity of the modes can be deduced
from the eigenvectors afterwards. This is left as an exercise for the readers.
Solution to Problem 9.14
The double barrier structure relevant to Fig. 9.34 is shown schematically below:
V d1
d2
d2
The height of the barrier V 1.2 eV. The widths of the barrier (d2) and of
the well (d1) are equal to 2.6 and 5 nm, respectively. The zero bias transmission
coefficient T(E) can be calculated with the transmission matrix method
described in section 9.5.1.
To apply the transfer matrix technique we will divide the potential into 5
regions to be labeled as 1, ..., 5 from left to right. To define these 5 regions we
will label the horizontal axis as the x-axis and define the five regions by:
x [∞, d2 (d1/2)], [d2 (d1/2), (d1/2)], [(d1/2), (d1/2)],
[(d1/2), d2 (d1/2)], [d2 (d1/2), ∞].
We will choose the origin for the potential such that the potential Vi is equal
to 0 inside the regions i 1, 3 and 5 and equal to V in the regions 2 and 4.
The incident wave is assumed to arrive in region 1 from the left while the
transmitted wave emerges into region 5. Let Ai and Bi be the amplitudes of
the incident and reflected wave in region i. In region i 1, 3 and 5 we can
define the generalized wave vector k1 by:
2 k 2 1 /2m1 E (9.14a)
where E is the energy of the incident electron and is assumed to be less than
V in this problem. In regions i 2 and 4 we will define k2 by:
2 k 2 2 /2m2 E V (9.14b)
m1 and m2 are, respectively, the electron masses in the well and in the barrier.
The wave vector k2 in Eq. (9.14b) is imaginary since E is smaller than the
barrier V. The wave amplitudes An1 and Bn1 in the final region is related