648
F Review of Solid State Concepts
C 1 E = E 0 C 1 − AC 2 ,
(F.1.12)
C 2 E = −AC 1 + E 0 C 2
(F.1.13)
which can have a nontrivial solution for C 1 and C 2 only if the determinant of their
coefficients vanishes, i.e., provided that
E 0 − E −A
−A E 0 − E
= 0
(F.1.14)
or
(E − E 0 )
2
− A
2
= 0
⇒
E = E 0 ± A .
(F.1.15)
From this result we see that the original energy E 0 of the uncoupled states has been
replaced in the presence of the coupling by a symmetric splitting of the energy into
the pair of energies E 0 ±A. Whenever coupling occurs, the energy splits. For our H
+
2
example, if the H atom electron is a 1s electron, the two energy levels generated for
H
+
2 will be what are referred to as the 1sσ g bonding and 1sσ u antibonding molecular
orbital energy levels that you may have discussed in a general chemistry course.
Now we ask how A and E 0 behave for our model system. We know that the larger
the coupling (i.e., magnitude of A), the larger the energy splitting will be, so that
A must somehow be related to the probability that the electron can tunnel through
the barrier between the two protons. Now, as we know that tunneling probabilities
typically vary exponentially with the distance into the barrier, we may expect that A
varies as on the left-hand side of Fig. F.1. Similarly, as E 0 consists of the potential
plus kinetic energies of the electron, together with the potential energies of the two
protons (for any given fixed distance between them), we may argue that when the
two protons are far apart, their potential energies are essentially zero (or very very
small), while the electron energy, because it is bound to one of the protons, is quite
negative (recall that a 1s-electron in an H atom is bound by 13.6 eV). This means
that E 0 is negative for large interproton separations, but rises rapidly when the
interproton separation is less than the average distance of the (fluctuating) electron
from the protons: E 0 then behaves as on the left-hand side of Fig. F.1. If we form
the two linear combinations E = E 0 ± A and display these values as functions of
the interproton separation, we obtain the results displayed on the right-hand side of
Fig. F.1.
F.2 Generalization of the Feynman Model to Solids
We shall now utilize the Feynman model to see how band structures occur in
solids. We begin by asking what happens when N atoms are brought close together.
Generalizing upon our discussion of the H
+
2 molecular ion, it is not unreasonable
to expect that there will be an N-fold split in the energy levels: we should then
F Review of Solid State Concepts
C 1 E = E 0 C 1 − AC 2 ,
(F.1.12)
C 2 E = −AC 1 + E 0 C 2
(F.1.13)
which can have a nontrivial solution for C 1 and C 2 only if the determinant of their
coefficients vanishes, i.e., provided that
E 0 − E −A
−A E 0 − E
= 0
(F.1.14)
or
(E − E 0 )
2
− A
2
= 0
⇒
E = E 0 ± A .
(F.1.15)
From this result we see that the original energy E 0 of the uncoupled states has been
replaced in the presence of the coupling by a symmetric splitting of the energy into
the pair of energies E 0 ±A. Whenever coupling occurs, the energy splits. For our H
+
2
example, if the H atom electron is a 1s electron, the two energy levels generated for
H
+
2 will be what are referred to as the 1sσ g bonding and 1sσ u antibonding molecular
orbital energy levels that you may have discussed in a general chemistry course.
Now we ask how A and E 0 behave for our model system. We know that the larger
the coupling (i.e., magnitude of A), the larger the energy splitting will be, so that
A must somehow be related to the probability that the electron can tunnel through
the barrier between the two protons. Now, as we know that tunneling probabilities
typically vary exponentially with the distance into the barrier, we may expect that A
varies as on the left-hand side of Fig. F.1. Similarly, as E 0 consists of the potential
plus kinetic energies of the electron, together with the potential energies of the two
protons (for any given fixed distance between them), we may argue that when the
two protons are far apart, their potential energies are essentially zero (or very very
small), while the electron energy, because it is bound to one of the protons, is quite
negative (recall that a 1s-electron in an H atom is bound by 13.6 eV). This means
that E 0 is negative for large interproton separations, but rises rapidly when the
interproton separation is less than the average distance of the (fluctuating) electron
from the protons: E 0 then behaves as on the left-hand side of Fig. F.1. If we form
the two linear combinations E = E 0 ± A and display these values as functions of
the interproton separation, we obtain the results displayed on the right-hand side of
Fig. F.1.
F.2 Generalization of the Feynman Model to Solids
We shall now utilize the Feynman model to see how band structures occur in
solids. We begin by asking what happens when N atoms are brought close together.
Generalizing upon our discussion of the H
+
2 molecular ion, it is not unreasonable
to expect that there will be an N-fold split in the energy levels: we should then