la medición de la sobrecualificación en el observatorio bancaja-Ivie

In this section we make slightly more quantitative some of the idea of the previous section. Let us write a Hamiltonian for two Hydrogen atoms. Since the nuclei are heavy compared to the electrons, we will fix the nuclear positions and solve the Schroedinger equation for the electrons as a function of the distance between the nuclei. This fixing of the position of nuclei is known as a

48 CHAPTER 5. CHEMICAL BONDING

0 0

10 10

Figure 5.4: Particle in a box picture of covalent bonding. Two separated hydrogen atoms are like two different boxes each with one electron in the lowest eigenstate. When the two boxes are pushed together, one obtains a larger box – thereby lowering the energy of the lowest eigenstate – which is known as the bonding orbital. The two electrons can take opposite spin states and can thereby both fit in the bonding orbital. The first excited state is known as the antibonding orbital “Born-Oppenheimer” approximation8,9_{. We hope to calculate the eigenenergies of the system as}

a function of the distance between the positively charged nuclei.

For simplicity, let us consider a single electron and two identical positive nuclei. We write the Hamiltonian as H = K + V1+ V2 with K = p 2 2m being the kinetic energy of the electron and

Vi=

e2

4π0|r − Ri|

is the Coulomb interaction energy between the electron position r and the position of nuclei Ri.

Generally this type of Schroedinger equation is hard to solve exactly. (In fact it can be solved exactly in this case, but it is not particularly enlightening to do so). Instead, we will attempt a variational solution. Let us write a trial wavefunction as

|ψi = φ1|1i + φ2|2i (5.1)

8_{Max Born (also the same guy from Born-Von Karmen boundary conditions) was one of the founders of quantum}

physics, winning a Nobel Prize in 1954. His daughter, and biographer, Irene, married into the Newton-John family, and had a daughter named Olivia, who became a pop icon and film star in the 1970s. Her most famous role was in the movie of Grease playing Sandra-Dee opposite John Travolta. When I was a kid, she was every teenage guy’s dream-girl (her, or Farrah Fawcett).

9_{J. Robert Oppenheimer later became the head scientific manager of the American atomic bomb project during}

the second world war. After this giant scientific and military triumph, he pushed for control of nuclear weapons leading to his being accused of being a communist sympathizer during the “Red” scares of the 1950s and he ended up having his security clearance revoked.

5.3. COVALENT BOND 49 01 01 0 0 02 3456758 47  57456758 47  01 01 012 7  7

Figure 5.5: Molecular Orbital Picture of Bonding. In this type of picture, on the far left and far right are the orbital energies of the individual atoms well separated from each other. In the middle are the orbital energies when the atoms come together to form a molecule. Top: Two hydrogen atoms come together to form a H2 molecule.

As mentioned above in the particle-in-a-box picture, the lowest energy eigenstate is reduced in energy when the atoms come together and both electrons go into this bonding orbital. Middle: In the case of helium, since there are two electrons per atom, the bonding orbitals are filled, and the antibonding orbitals must be filled as well. The total energy is not reduced by the two Helium atoms coming together (thus helium does not form He2). Bottom: In the case of LiF, the energies of the

lithium and the fluorine orbitals are different. As a result, the bonding orbital is mostly composed of the orbital on the Li atom – meaning that the bonding electrons are mostly transferred from Li to F — forming a more ionic bond.

where φiare complex coefficients, and the kets |1i and |2i are known as “atomic orbitals” or “tight

binding” orbitals10_{. The form of Eq. 5.1 is frequently known as a “linear combination of atomic}

orbitals” or LCAO11_{. The orbitals which we use here can be taken as the ground state solution of}

the Schroedinger equation when there is only one nucleus present. I.e. (K + V1)|1i = 0|1i

(K + V2)|2i = 0|2i (5.2)

where 0 is the ground state energy of the single atom12. I.e., |1i is a ground state orbital on

10_{The term “tight binding” is from the idea that an atomic orbital is tightly bound to its nucleus.}

11_{The LCAO approach can be improved systematically by using more orbitals and more variational coefficients}

— which then can be optimized with the help of a computer. This general idea formed the basis of the quantum chemistry work of John Pople. See footnote 3 above in this section.

12_{Here }

0is not a dielectric constant or the permittivity of free space, but rather the energy of an electron in an

50 CHAPTER 5. CHEMICAL BONDING nucleus 1 and |2i is a ground state orbital on nucleus 2.

For simplicity, we will now make a rough approximation that |1i and |2i are orthogonal so we can then choose a normalization such that

hi|ji = δij (5.3)

When the two nuclei get very close together, this orthogonality is clearly no longer even close to correct. We then have to decide: either we keep our definition of the atomic orbitals being the solution to the Schroedinger equation for a single nucleus, but we give up on the two atomic orbitals being orthogonal; or we can give up on the orbitals being solutions to the Schroedinger equation for a single nucleus, but we keep orthonormality. It is a good exercise to consider what happens when we give up orthonormality, but fortunately most of what we learn does not depend too much on whether the orbitals are orthogonal or not, so for simplicity we will assume orthonormal orbitals. An effective Schroedinger equation can be written down for our variational wavefunction which (unsuprisingly) takes the form of an eigenvalue problem13

X

j

Hijφj = Eφi

where

Hij = hi|H|ji

is a two by two matrix in this case. (The equation generalizes in the obvious way to the case where there are more than 2 orbitals).

Recalling our definition of |1i as being the ground state energy of K + V1, we can write14

H11 = h1|H|1i = h1|K + V1|1i + h1|V2|1i = 0+ Vcross (5.4)

H22 = h2|H|2i = h2|K + V2|2i + h2|V1|2i = 0+ Vcross (5.5)

H12 = h1|H|2i = h1|K + V2|2i + h1|V1|2i = 0 − t (5.6)

H21 = h2|H|1i = h2|K + V2|1i + h2|V1|1i = 0 − t∗ (5.7)

In the first two lines

Vcross= h1|V2|1i = h2|V1|2i

is the Coulomb potential felt by orbital |1i due to nucleus 2, or equivalently the Coulomb potential felt by orbital |2i due to nucleus 1. In the second two lines (Eqs. 5.6 and 5.7) we have also defined the so-called hopping term15,16

t = −h1|V2|2i = −h1|V1|2i

13_{To derive this eigenvalue equation we start with an expression for the energy}

E = hψ|H|ψi hψ|ψi

then with ψ written in the variational form of Eq. 5.1, we minimize the energy by setting ∂E/∂φi= ∂E/∂φ∗i = 0. 14_{In atomic physics courses, the quantities V}

crossand t are often called a direct and exchange terms and are

sometimes denoted J and K. We avoid this terminology because the same words are almost always used to describe 2-electron interactions in condensed matter.

15_{The minus sign is a convention for the definition of t. For many cases of interest, this definition makes t positive,}

although it can actually have either sign depending on the structure of the orbitals in question and the details of the potential.

16_{The second equality here can be obtained by rewriting H}

5.3. COVALENT BOND 51 The reason for the name “hopping” will become clear below. Note that in the second two lines (Eqs. 5.6 and 5.7) the first term vanishes because of orthogonality of |1i and |2i. Thus our Schroedinger equation is reduced to a two by two matrix equation of the form

 0+ Vcross −t −t∗ _ 0+ Vcross   φ1 φ2  = E  φ1 φ2  (5.8) The interpretation of this equation is roughly that orbitals |1i and |2i both have energies 0 which

is shifted by Vcross due to the presence of the other nucleus. In addition the electron can “hop”

from one orbital to the other by the off-diagonal t term. To understand this interpretation more fully, we realize that in the time dependent Schroedinger equation, if the matrix were diagonal a wavefunction that started completely in orbital |1i would stay on that orbital for all time. However, with the off-diagonal term, the time dependent wavefunction can oscillate between the two orbitals.

Diagonalizing this two-by-two matrix we obtain eigenenergies E± = 0+ Vcross± |t|

the lower energy orbital is the bonding orbital whereas the higher energy orbital is the anti-bonding. The corresponding wavefunctions are then

ψbonding = 1 √ 2(φ1± φ2) (5.9) ψanti−bonding = 1 √ 2(φ1∓ φ2) (5.10)

I.e., these are the symmetric and antisymmetric superposition of orbitals. The signs ± and ∓ depend on the sign of t, where the lower energy one is always called the bonding orbital and the higher energy one is called antibonding. To be precise t > 0 makes (φ1+ φ2)/

√

2 the lower energy bonding orbital. Roughly one can think of these two wavefunctions as being the lowest two “particle-in-a-box” orbitals — the lowest energy wavefunction does not change sign as a function of position, whereas the first excited state changes sign once, i.e., it has a single node (for the case of t > 0 the analogy is precise).

It is worth briefly considering what happens if the two nuclei being bonded together are not identical. In this case the energy 0 for an electron to sit on orbital 1 would be different from

that of orbital 2. (See bottom of Fig. 5.5) The matrix equation 5.8 would no longer have equal entries along the diagonal, and the magnitude of φ1and φ2would no longer be equal in the ground

state as they are in Eq. 5.9. Instead, the lower energy orbital would be more greatly filled in the ground state. As the energies of the two orbitals become increasingly different, the electron is more completely transferred entirely onto the lower energy orbital, essentially reducing to an ionic bond.

Aside: In section 22.4 below, we will consider a more general tight binding model with more than one electron in the system and with Coulomb interactions between electrons as well. That calculation is more complicated, but shows very similar results. That calculation is also much more advanced, but might be fun to read for the adventurous.

Note again that Vcross is the energy that the electron on orbital 1 feels from nucleus 2. How-

ever, we have not included the fact that the two nuclei also interact, and to a first approximation, this Coulomb repulsion between the two nuclei will cancel17 _{the attractive energy between the}

17_{If you think of a positively charged nucleus and a negatively charged electron surrounding the nucleus, from far}

outside of that electron’s orbital radius the atom looks neutral. Thus a second nucleus will neither be attracted nor repelled from the atom so long as it remains outside of the electron cloud of the atom.

52 CHAPTER 5. CHEMICAL BONDING nucleus and the electron on the opposite orbital. Thus, including this energy we will obtain

˜

E± ≈ 0± |t|

As the nuclei get closer together, the hopping term |t| increases, giving an energy level diagram as shown in Fig. 5.3.2. This picture is obviously unrealistic, as it suggests that two atoms should bind together at zero distance between the nuclei. The problem here is that our assumptions and approximations begin to break down as the nuclei get closer together (for example, our orbitals are no longer orthogonal, Vcrossdoes not exactly cancel the Coulomb energy between nuclei, etc.).

0

123

456789575

89575 89575

Figure 5.6: Model Tight Binding Energy Levels as a Function of Distance Between the Nuclei of the Atoms.

A more realistic energy level diagram for the bonding and antibonding states is given in Fig. 5.7. Note that the energy diverges as the nuclei get pushed together (this is from the Coulomb repulsion between nuclei). As such there is a minimum energy of the system when the nuclei are at some nonzero distance apart from each other, which then becomes the ground state distance of the nuclei in the resulting molecule.

Aside: In Fig. 5.7 there is a minimum of the bonding energy when the nuclei are some particular distance apart. This optimal distance will be the distance of the bond between two atoms. However, at finite temperature, the distance will fluctuate around this minimum (think of a particle in a potential well at finite temperature). Since the potential well is steeper on one side than on the other, at finite temperature, the “particle” in this well will be able to fluctuate to larger distances a bit more than it is able to fluctuate to smaller distances. As a result, the average bond distance will increase at finite temperature. This thermal expansion will be explored again in the next chapter.

Covalently bonded materials tend to be strong and tend to be electrical semiconductors or insulators (since electrons are tied up in the local bonds). The directionality of the orbitals makes these materials retain their shape well (non-ductile) so they are brittle. They do not dissolve in polar solvents such as water in the same way that ionic materials do.