Читать книгу Solid State Physics - Philip Hofmann - Страница 39

Covalent bonding is based on the sharing of electrons between different atoms. The simplest case is that of the hydrogen molecule, which we will discuss quantitatively below. In solids, covalent bonding is often found for elements with a roughly half‐filled outer shell. A prominent example is carbon, which forms solids such as diamond, graphene, and graphite as well as complex molecules such as buckminsterfullerene or carbon nanotubes. The covalent bonds in diamond are constructed from linear combinations of the 2s orbital and three 2p orbitals. This results in four so‐called orbitals that stick out in a tetrahedral configuration from the carbon atoms. In graphene and graphite, the 2s orbital is combined with only two 2p orbitals, giving three orbitals, all in one plane and separated by angles of , and one p orbital oriented perpendicular to this plane. This linear combination of orbitals already reveals an important characteristic of covalent bonding: It is highly directional. In addition to this, it is also very stable and the cohesive energies for covalently bonded solids are typically several electron volts per atom.

We study covalent bonding by the simplest example, the diatomic hydrogen molecule , with its two electrons. We denote the positions of the nuclei of the two hydrogen atoms as and . In the following, we keep the nuclei fixed at these positions and denote the distance in between them by . The Hamiltonian for the hydrogen molecule can then be written as

(2.3)

where and are the coordinates of the electrons belonging to nucleus A and B, respectively. The first two terms describe the kinetic energies of the two electrons. The operators and act only on the coordinates and , respectively. The electrostatic term contains the repulsion between the two nuclei and the repulsion between the two electrons as well as the attraction between each electron and each nucleus.

Calculating the energy eigenvalues and wave functions for this Hamiltonian is a formidable problem, mostly because of the interactions between the two electrons (the second term in the curly brackets). We shall return to this problem later. For now, we exploit the fact that the essence of covalent bonding can already be understood by considering just one electron, i.e. by simplifying Eq. (2.3) so that it describes the molecular ion:

(2.4)

This Hamiltonian is that of an electron moving in the Coulomb potential of two protons separated by a distance . The electrostatic repulsion of the nuclei (the term) does not depend on the position of the electron and just leads to an energy offset that could be treated separately. We choose to leave it in the Hamiltonian, though, as we want to inspect the dependence of the resulting energy levels on later.

We can calculate an approximate ground‐state solution to the Schrödinger equation by writing as a linear combination of the atomic 1s wave functions of the two atoms, and . This approach is commonly known as linear combination of atomic orbitals (LCAO). Our ansatz is thus

(2.5)

where and are constants. Multiplying this equation from the left with either or and integrating gives two algebraic equations

(2.6)

where we have introduced the so‐called overlap integral , as well as the abbreviations and correspondingly for and . As the two nuclei at and are completely equivalent, we can simplify this by noticing that and . With this, Eq. (2.6) becomes

(2.7)

This equation system has a nontrivial solution when the determinant of the coefficient matrix vanishes. This gives two possible solutions

(2.8)

With these solutions, we can calculate the relation between the coefficients and from Eq. (2.7), and we find that for the “” solution and for the “” solution (see Problem 3 and note that ).

The energy levels and wave functions from this calculation are illustrated in Figure 2.2a. For a large distance between the nuclei, the overlap integral and the matrix element vanish and the only solution is . This makes sense as the second proton is far away from , so is approximately the ground state energy of atomic hydrogen. A corresponding solution does, of course, exist for the electron located on the other proton. For a short distance , on the other hand, this energy level splits up into the bonding and antibonding molecular states . The single electron will occupy the lower level and this will result in the energy gain of covalent bonding. The wave function corresponding to this is the one for which , which increases the electron density between the nuclei. The wave function corresponding to the antibonding upper energy level is constructed by choosing , and it shows the characteristics of an antibonding state with a node between the two nuclei.

Figure 2.2 (a) Formation of bonding and antibonding energy levels in the ion. The electronic energy level of an isolated H atom splits into the levels according to Eq. (2.8). The radial wave function of an isolated H atom is shown at the left, and the bonding and antibonding wave functions along the molecular axis of the ion are shown at the right (for proper normalization see Problem 3). (b) Bonding and antibonding energy levels as a function of internuclear distance . Zero energy corresponds to the ground state energy of a free hydrogen atom.

Figure 2.2b shows how the energy levels vary as a function of the distance between the nuclei. The antibonding energy increases monotonically as the nuclei approach each other, but the bonding energy has a minimum of eV at approximately twice the Bohr radius . With the single electron placed into the bonding state , the total energy gain resulting from formation of the covalent bond is thus 1.77 eV. It is tempting to extend this model to the two electrons in the molecule by placing the second electron in the same energy level. This electron would then also lower its energy by 1.77 eV, amounting to a total energy gain of 3.54 eV for the formation of the hydrogen molecule. Handling the additional electron in this way is too simplistic, as we shall see below, but the energy gain thus computed is at least similar to the experimental value of eV. It is also clear that no additional electrons can be placed in the energy level, as this would violate the Pauli principle. A third electron would need to occupy the level, destabilizing the molecule as compared to with two electrons. In covalently bonded solids, each atom forms bonds to several neighbors and the total energy gain per atom can be higher than for . In silicon, for example, the energy gain per atom (or cohesive energy) is 4.6 eV.

Apart from a high cohesive energy, a characteristic feature of covalent bonding is the directional nature of the bonds. The preference for certain bond directions governs the crystal structures of covalently bonded solids, and these are more complex than the close‐packed structures typically encountered in metals. The treatment of the molecular ion has demonstrated the idea of constructing bonding and antibonding molecular orbitals from a linear combination of atomic orbitals, but since we combined two isotropic 1s orbitals in this case, the directional character of covalent bonding did not emerge.

Figure 2.3 demonstrates the consequences of using directional orbitals, such as p orbitals, for the formation of wave functions in molecules or solids. Figure 2.3a represents the case that we have just treated. Two s orbitals are combined to produce a bonding and an antibonding orbital by being added either in phase or out of phase (as indicated by the shading and the sign on the wave function). Due to the symmetry of the s orbital, the bond direction is not important. Combining an s orbital with a p orbital along the interatomic axis works in the same way (see Figure 2.3b). The p orbital is strongly anisotropic, but still the direction between the atoms does not matter in this case, because we define the intermolecular axis as the axis and then use the orbital aligned in this direction to form bonding and antibonding states. However, in this arrangement, no bonding interactions are possible between the s orbital of the left atom and the and orbitals of the right atom. This is illustrated in Figure 2.3c. No matter what linear combination coefficients we use, the overlap of the wave functions contains equal and mutually canceling bonding and antibonding contributions. This dilemma is the source of the directional preference in covalent bonding. In order to achieve the highest energy gain, it is often favorable to use linear combinations of the orbitals in one atom before combining them with other atoms. Examples for this are the and hybrid orbitals found in carbon‐based solids such as diamond or graphite, which directly explain the preferred bonding directions shown in Figure 1.7. The orbitals in graphene are also shown in Figure 6.15a. All these orbitals are highly directional.

Figure 2.3 Linear combination of orbitals on neighboring atoms. (a) Two s orbitals as in the molecular ion. (b) An s orbital and a orbital (the choice of the direction is arbitrary). (c) An s orbital with a or orbital. Due to symmetry, the total overlap is zero and no bonding or antibonding orbitals can be formed.

In more complex solids, several bonding types can coexist. For instance, the diamond structure of Figure 1.7 is also found for materials that mix group III and group V elements, such as GaAs, or group II and group VI elements, such as InP. In these, the different atoms occupy alternating sites but still adopt the bonding arrangement. This is accompanied by an electron transfer from the higher‐group atom to the lower‐group atom, resulting in a bonding situation with both covalent and ionic contributions.

We now return to the Schrödinger equation for the full Hamiltonian from Eq. (2.3). This has to be solved by a wave function containing the coordinates and of two electrons (we still keep the positions of the nuclei fixed). This makes the problem much harder and we can only begin to imagine the difficulty of finding a wave function describing all the electrons in a solid! Fortunately, even in a solid, most cases can be described quite well by considering one electron moving in the potential of the ions and some averaged potential arising from the other electrons. In this book, there are only two cases where this simple description will fail: magnetism and superconductivity. The following discussion will mainly be useful for the treatment of magnetism in Chapter 8. Understanding these details is not crucial at this point, and you could decide to jump to Section 2.4 and return here later.

Solving the Schrödinger equation for the Hamiltonian from Eq. (2.3) would be greatly simplified if we could somehow “switch off” the electrostatic interaction between the two electrons, because then the Hamiltonian could be written as the sum of two parts, one for each electron. Indeed, the Hamiltonian in Eq. (2.3) could essentially be the sum of two Hamiltonians similar to the one in Eq. (2.4), one for each electron (but with the term appearing only once). If a Hamiltonian containing two electronic coordinates could be separated into a sum of two Hamiltonians that contain only one electronic coordinate each, the corresponding Schrödinger equation could be solved by a product of the two wave functions that are solutions to the two individual Hamiltonians. We could therefore start an attempt to solve Eq. (2.3) based on what we have already learned for the ion. However, we will start from an even simpler point of view: Without any interaction between the electrons, and for a large distance , the electron near nucleus A will not feel the potential of nucleus B and vice versa. In this case, Eq. (2.3) would simply turn into the sum of the Hamiltonians for two hydrogen atoms and we could approximate the two‐electron wave function by , with and being the wave functions for atomic hydrogen.

However, this simple approach is physically incorrect because such a wave function does not obey the Pauli principle. Since the electrons are fermions, the total wave function must be antisymmetric with respect to particle exchange, and the simple product wave function does not meet this requirement. The total wave function consists of a spatial part and a spin part, and thus there are two possibilities for forming an antisymmetric wave function – we can either combine a symmetric spatial part with an antisymmetric spin part or vice versa. The spatial part of the wave function can be constructed in two ways,

(2.9)

(2.10)

The plus sign in Eq. (2.9) returns a symmetric spatial wave function, which we can combine with an antisymmetric spin wave function with the total spin equal to zero (the so‐called singlet state); the minus in Eq. (2.10) results in an antisymmetric spatial wave function to be combined with a symmetric spin wave function with the total spin equal to 1 (the so‐called triplet state).

The antisymmetric wave function in Eq. (2.10) vanishes for – that is, the two electrons cannot be at the same location simultaneously. This leads to a depletion of the electron density between the nuclei and hence to an antibonding state. For the symmetric case, on the other hand, the electrons have opposite spins and can be at the same place, which leads to a charge accumulation between the nuclei and hence to a bonding state (see Figure 2.4).

Figure 2.4 The energy changes and for the formation of a hydrogen molecule. The dashed lines represent the approximation for long distances. The two insets show grayscale images of the corresponding electron probability density.

An approximate way to calculate the eigenvalues of Eq. (2.3) was suggested by W. Heitler and F. London in 1927. Their approach was to use the known single‐particle 1s wave functions for atomic hydrogen for and to form a two‐electron wave function , in the form of either Eq. (2.9) or Eq. (2.10). These wave functions might not be entirely correct because the atomic wave functions will certainly be modified by the presence of the other atom. However, even if they are only approximately correct, we can obtain the molecular energy levels as

(2.11)

According to the variational principle in quantum mechanics, the resulting energy will always be higher than the correct ground‐state energy, but it will approach it for a good choice of the trial wave functions.

The detailed calculation is quite lengthy and shall not be given here.¹ The resulting ground‐state energies for the singlet and triplet states can be written as

(2.12)

(2.13)

is the ground‐state energy for one hydrogen atom; it is multiplied by two because we start with two atoms. The energies and are also shown in Figure 2.4. is always larger than zero and does not lead to any chemical bonding. , on the other hand, shows a minimum with negative energy at approximately . This is the bonding state.

For long distances between the nuclei, Eqs. (2.12) and (2.13) can be rewritten in the form

(2.14)

where the and signs apply to the singlet and triplet state, respectively. In this representation, the energy change upon bonding contains two contributions, of which one depends on the relative spin orientations of the electrons () and the other does not (). The energy difference between the two states is then given by , where is called the exchange energy. In the case of the hydrogen molecule, the exchange energy is always negative. Equation (2.14) is a remarkable result, because it means that the energy of the system depends on the relative orientation of the spins, even though these spins are not explicitly mentioned in the Schrödinger equation.

We will encounter similar concepts in the chapter about magnetism, where the underlying principle for magnetic ordering is very similar to what we see here: Through the exchange energy, the total energy of a system of electrons depends on their relative spin orientations, and therefore a particular ordered spin configuration is energetically favored. For two electrons, the “magnetic” character is purely given by the sign of . For negative , the coupling with two opposite spins is favorable (the “antiferromagnetic” case), whereas a positive would lead to a situation where two parallel spins give the lowest energy (the “ferromagnetic” case).

Finally, it is interesting to compare the single‐electron solution for the ion to the two‐electron Heitler‐London solution for . The curves for the bonding energy levels in Figures 2.2 and 2.4 are strikingly similar, especially if we expect a second electron in the ion to show approximately the same energy gain as the first one (and hence a minimum of the total energy that is twice as deep). The similarity suggests that the interaction between the two electrons is not too important. Qualitatively, we can understand this in the following way: For , there is no second electron and for the singlet state of , the symmetry of the wave function ensures that the two electrons are never at the same place, reducing their interaction. A noticeable difference between and is the position of the energy minimum, i.e. the equilibrium distance between the nuclei, which is significantly smaller in the case of . This does not come as a surprise. After all, two electrons contribute to the attraction. Despite the similarity between the curves, one needs to keep in mind that they represent two entirely different things: Figure 2.2 shows the energy levels of one single electron and we are not allowed to multiply the curves by a factor of two to obtain the energy for two electrons (even though we just contemplated this). The curves in Figure 2.4 represent the energy levels for two electrons. We cannot split this into contributions for the separate energies of two single electrons.