Primer on Quantum Chemistry

Please read my first year general chemistry (Chemical Principles) notes on the hydrogen atom and vibrations on a guitar string (1D) and on a drum head (2D), which you can down load here.

Distillation of what you should already know from a physics course:

1. Particles bounce: localized in space, momentum mv;
2. Waves diffract: delocalized throughout space, from one end of the Universe to the other. A diffraction pattern arises on a ripple tank or from a light source if (and only if) the two openings on a barrier are the same distance apart from each other as the wavelength of the ripples.

Constructive and destructive interference (above)

Diffraction pattern appearing at right

3. E=mc², c=l.n, E=hn,

Trivial: E=hc/l = mc² => mc = h/l waves

4. De Broglie said why not c => v ? Matter has associated wavelike properties mv = h/l. Proven at Bell Labs in New York (in a building that was demolished to make way for the WTC): electrons diffract off a Ni surface.
5. Schroedinger and Heisenberg applied mathematical methods for studying waves to deBroglie's idea (Schroedinger prodded along by Max Born to change doctoral thesis). This led to (beyond the ordinary general physics course):

Postulates of Quantum Mechanics concerning HY=EY

i)    This eigenvalue equation can be used to calculate physical properties of systems such as atoms, molecules, polymers, clusters… (cells?)

ii)    A physical property has an operator function. The Hamitonian, H(coord particles) is the operator for energy and is a sum of kinetic energy and potential energy terms. As it is a function, it depends on something; in this case, the simultaneous coordinates of all the particles in the system.

iii)    Systems exist in one particular "state" and each state has a mathematical function Y(coord particles) describing the wavelike behavior of the particles in the system.

iv)    There is one and only one answer for any state, E in the case of energy.

The wave function Y must correspond to physical reality, i.e. it must let the particle(s) it describes have only one probability of being found at each point in the Universe. Since Y is a function of coordinates, one only needs to plug in the coordinates for every particle and one can get the value for Y from the Schroedinger equation. According to Max Born's interpretation Y²dt is the probability of finding a particle at a point dt (otherwise known as a volume element; t can be a combination of the x,y,z coordinates).

H(coord. particles) Y (coord. particles) = E Y(coord. particles)

This is a differential equation that cannot be solved analytically (through the use of integral tables, etc.), but might possibly be solved numerically by making certain levels of approximation. A numerical solution will always involve a certain loss of accuracy and the loss depends on the crudeness of the approximation(s). It also gives what are called approximate energies and wavefunctions. Even the differential equation used for finding the orbits of the planets around the sun cannot be solved analytically but only approximately, as seen from the time of Newton and Leibnitz. The mathematical approximations used in Quantum Chemistry resemble the approximations used for the planets.

The Four Commonly Used Approximations of Quantum Chemistry

1st)    Separate one function which depends on many parameters into the product of a function that depends on some of the parameters and a function that depends on the rest of the parameters:

Fn(all particles) = Fn'(some particles) ∙ Fn"(other particles)

Y(all particles) =
Y_nuc(nuclei) ∙ Y_el(electrons in one nuclear framework)

Also called the Born-Oppenheimer Approximation, which gives the "reaction coordinate".

Think of the picture for the energy relative to two separate H atoms as they approach each other to form H₂. At any one internuclear distance there is one energy because there is only one Y_el(coord electrons) for that distance (the points on the graph below).

2nd Approx)    Fn(many particles) = Product of Fn(one particle each)

The Orbital Approximation: An orbital is a mathematical function that describes the wavelike behavior of one particle.

For single atoms these are "atomic orbitals" the product of which is called the electronic configuration, e.g. Li 1s²2s¹ ≡ 1s(a)∙1s(b)×2s(a)

(In this course, we will not consider antisymmetrizing electronic wavefunctions, which comes from the fact that electrons are indistinguishable.)

For molecules, one can derive molecular orbitals,

e.g. The electronic configuration for N₂: 1s²∙2s²∙3s²∙p_x²∙p_y².

If 1 and 2 stand for the coordinates of electrons 1 and 2 then

Y(1,2)He » (1s(1)a(1)) ∙ (1s(2)b(2)) = 1s²

Y(1,2)H₂» (1s(1)a(1)) ∙ (1s(2)b(2)) = 1s²

3rd Approx.)   

Molecular Orbital Fn(one part.) solved by differential equations »

sum of Atomic Orbital Fns'(one part.) that are more easily solved by computer

This is called the LCAO-MO Approximation or linear combination of atomic orbitals - molecular orbitals method. The MO is approximated by a "weighted sum of AOs". The coefficient of an AO is the weighting factor for how much that AO (j) participates in the particular MO(i).

"MO" Y_i(1) » S_ij [c_ij “AO” F_j(1)]

E.g. for CO molecule p_i=(0.3)2p_xC + (0.5)2p_xO + (0.0)2p_yC + (0.0)2sC + etc.

4th) Replace atomic orbitals by other functions more easily handled by computer.

• Hydrogen-like AOs from the exact solutions for the H atom, aufbau principal,

2s = N (1-r/a) e^-b∙r where a,b,N constants; r distance of electron from nucleus.

  (See Chemical Principles chapter on H atom, picture of graphs of 1s, 2s, 3s orbitals.)

1s (no nodes) 2s (1 node) 3s (2 nodes)

Y_1s = e^-r Y_2s = (2-r)∙e^-r/2 Y_3s = (3-2r+r²)∙e^-r/3

• Modified H-like AOs Slater Type Orbitals (STO):

        2s = N e^{-z∙r ,}

N, z constants; r distance from nucleus, (This 2s function has no node.)

• Gaussian functions: 2s = N e^-a(r**2)

A single GF does not give a good approximation to the energy,
but an STO can be approximated by a weighted sum of gaussians,
STO (2s) ≈ GF (2s) c_i∙e^-a(i).(r**2) _,

ci , a constants; r distance from nucleus; (r**2) means r² (r squared).

The STO was developed because the (1-r) term in front of the exp function makes it a difficult mathematical function to work with. Slater found that one could drop it and not have the energy change drastically. Usually better results are obtained when STOs are used instead of a gaussian function. However, an STO can be fairly well approximated by combining a few gaussian functions, much like an MO can be approximated by combining AOs. This can be 'seen' in the next section.

Pictorial Quantum Chemistry

At this point let's go back upwards in the list of approximations and plot interesting functions. We will use the midpoint of a bond for the origin in a graph that plots the value of a wavefunction versus distance along the x-axis. This puts e^-|r| in the form exp(-abs(x-xo)) as input for computers. The absolute value is used because we are plotting the distance r to the right (+) or to the left (-) of the origin. One plugs in the value of xo as the position of a nucleus along the x-axis. So, the e^-|r| function for the right-hand, H_b nucleus would have the form:

_{exp(-abs(x-1.0))}

Approximation 4 pictorially: Atomic Orbitals

Exercise 1) Plot the 2s orbital centered at the origin (xo=0) as (a) a hydrogenic function, (b) an STO, (c) a gaussian function. Put the values of all constants equal to 1, and set the limits of x to between 0 and +5. Don't forget the (1-x) in the H-like function.

In this graph the H-like function is plotted in red, STO in green, and the Gaussian in blue. The Y axis gives the value of the orbital at various values of distance X.

Some introductory textbooks show the 2s function as a combination of an inner sphere with a positive sign covered by an outer sphere with a negative sign. Can you relate the graph in exercise 1a with this picture? The value of the H-like 2s function is positive at x<1 and negative at x>1. At x=1, there is a node, that is, the value of the 2s function (2s orbital) equals zero.

The Aufbau Principle uses the H-like orbitals for all the atoms, but what is the difference between the 2s orbital for He or for Zn? The 2s orbital (indeed, any orbital) has the same mathematical form for all atoms; the only thing that changes from one element to another is the value of the constant in the exponential.

Exercise 2) Plot two gaussian 2s orbitals (between x = -5 and 5, both x_o = 0, i.e. centered at the origin) on the same graph, one with a=1and the other with a=2. (N=1). Suppose one is for Li and the other for Be. Remember the rules for electronegativity as one proceeds to the right in a row of the Periodic Chart and also the trend in atomic radius proceeding to the right. Look at the graph, assume equal area under each function, and assign one to Li and the other to Be.

The green curve (closer in to the origin) is for a=2, the red a=1.

Ordinarily, we want the areas under the curves to be the same: ∫Y²dx = 1, which means that if one sums up the probabilities of finding the particle at every point in space, one arrives at a value of one or a 100% probability of finding the particle somewhere in the Universe (in our case a one dimensional Universe along the x axis). The constants, N in front of the e function can be adjusted in the orbital so that ∫N²Y²dx = 1 for both atoms. Then Y_Li and Y_Be are normalized and the areas under each of the curves in Exercise 2 are equal to 1 (when x is taken from -infinity to +infinity). We want both areas to be equal because they are the areas of an orbital, which describes the wavelike behavior of one electron, no matter which element the orbital is on.

Exercise 3). Repeat exercise 2 but put in the value of the normalization constant N for each 2s orbital. This is done by solving the integral

∫(N²* exp(-a∙r∙r ))²dx = 1 for N. Answer: N = (2∙a/p)**(3/4) for a gaussian

This normalization constant, N for an STO is equal to ((z**3)/p)**1/2.

In this case, the areas under the curves are equal. That means an electron described by either orbital has the same probability of being found in all of space. The fact that the curves have different values at any one point, dx, in space reflects the chemical property of electronegativity. A 2s electron on Be would be closer to the nucleus than on Li. (Why? What is an orbital?)

Exercise 4(a). Plot an STO and one gaussian function on the same graph, where a and z = 1. (Don't forget to include the particular normalization constant N.)

Notice the difference between the two functions at the nucleus (x=0) and far out (x=5). Besides affecting the energy of an atom, the first difference can have consequences in NMR and second in bonding. The "chemical shift" in NMR is determined by the contact between the electrons and the nuclei. Bonding is affected by the amount orbitals on different atoms "overlap" each other. The distance beween nuclei for atoms such as C, N, O, and H is about 2 atomic units. So there is an appreciable difference between the results obtained for an STO or for a gaussian orbital. The STO turns out to give better results (closer to experiment), but there is a way to get an approximation of an STO by using a weighted sum of gaussians. Sound familiar?

Exercise 4(b). Plot 4 functions on the same graph: each of three gaussians with different values of the a coefficient, a1=0.1, a2=1, a3=3, and the fourth function is a sum of the three functions. You might want to amuse yourself by putting an STO on a graph and adjusting the values of the gaussian function as so that their sum resembles the STO as much as possible; or play around with the values of the c_i's for each function i. Why limit yourself to 3 functions? The more the merrier.

Merrier? What is meant by better results? Well, the Schroedinger equation will give the unique energy of a system in a given state (ground, first excited singlet, first triplet) if it could be solved. As more approximations are used to solve the equation, the energy obtained will be higher than the true value. If we stick to one level of approximation (some call this the level of theory), higher or lower energies can still be obtained, e.g. the use of H-like, STO, or gaussians. Generally, H-like give the lowest and gaussians the highest energy. At each level of theory, there is an asymptotic limit that is approached as more parameters can be optimized.

We can be assured of the best combination of parameters when the condition dE_i / dc_i = 0. This is called the Variation Principle. In the 1960’s and through to the present day, sets of parameters, i.e. the coefficients and exponents of gaussian functions, have been developed for most elements and with differing numbers of functions in the set. These are called basis sets.

Thus we have groups of gaussian functions to approximate an STO, but we do not have to limit our basis set to only one STO for one type of orbital. We can use a 2s and a 3s STO on the same C atom, or even a 4s also, and the same for the p orbitals. The difference between the 2s and 3s are the two different values of z_i and a_i used for each. Basis sets with more than one STO for the different H-like orbitals are called double zeta sets. Those including higher orbitals than those usually associated with an atom, e.g. 2p on H, 3d on C, are called polarized sets.

The group (headed by John Pople, co-winner of the 2000 Nobel Prize in Chemistry) that devised the Gaussian series of programs arrived at a nomenclature for basis sets:

1. Minimal basis sets: STO-nG or a set of n gaussians for each type of atomic orbitals usually associated with an atom, e.g. N 1s, 2s, and 2p_xyz^.
2. Double zeta: n-m1G or n gaussians in the core orbital(s) and two STO for the valence orbitals (m gaussians in one and 1 gaussian in the other STO). The ones most often used are the 3-21G and 6-31G sets. For more involved reasons these are more properly called extended basis sets. They include sets such as 6-311G and 6-311+G, the first having three valence STOs, the other having 4 with the + meaning a diffuse gaussian (very small a). Diffuse functions are needed to describe anions well.
3. Polarized: 6-31G* 6-31G**,where the first * is for d functions on group III, IV, V, VI, VII, VIII elements and the second * is for a set of 2p_x, 2p_y, and 2p_z orbitals on each H atom. The * designation is being replaced by one specifying the type, e.g. 6-31G(d).

This designation, while in wide spread use, is not the only one.

Approximation 3 pictorially: Molecules, MOs, and AOs

Let us now consider molecules and molecular ions, specifically the H₂ and HHe⁺ cases. Both are 2 valence electron systems.

The bonding and antibonding s and s* molecular orbitals in each system can be considered as combinations of the two atomic orbitals, each centered on one nucleus. The bonding MO adds the two AOs and the antibonding subtracts.

Exercise 5. Plot s and s*, where s = 1s_A + 1s_B and s* = 1s_A - 1s_B. Make separate plots for bond length = 1, 2, 4, and 5 units and the limits of the plots are -10 and 10.

Exercise 6. Electron density can be defined as Y²dx. Plot the squares of the functions obtained in exercise 5. You might like to plot Y and Y²on the same graph for bond length = 2.

Now let us consider the HHe+ ion, using 1s_H = e^-r and 1s_He = e^-2r. As seen in exercise 2, the more electronegative He should have the greater value of z. This reflects the smaller atomic radius for He that would appear when one electron (definition of orbital?) is attracted to the He nucleus with +2 charge compared to the H atom with +1 nuclear charge.

Exercise 7). Plot s and s*, where s = 1s_H + 1s_He and s* = 1s_H - 1s_He and the internuclear distance = 2.

If you did exercise 7 correctly, there is something clearly wrong. The area underneath the curve in the vicinity of the He nucleus should be greater than that for the H nucleus. That is, we interpret “area under curve” = “electron density” = “probability of finding one electron between point d(x₁) and d(x₂)”. This wave function is wrong only in the sense that it leads to higher energy. The LCAO-MO approximation, however, uses a weighted sum of AOs, i.e. each AOj in MOi is multiplied by a coefficient c_ij. The best set of c_ij gives the lowest energy. (The computer program solves for the best set of c_ij when dE/d(c_ij) = 0.)

Exercise 8). Plot s and s*, where s = (0.2)1s_H + (0.5)1s_He and s* = (0.7)1s_H - (0.4)1s_He and the internuclear distance = 2.

Given particular values for the parameters in the AOs for H and He and given the bond length, there can be found a set of coefficients c_ij which lead to the best energy given this level of theory (approximations).

Approximation 2 pictorially: Configuration Interaction

Three ways of going back up beyond the orbital approximation are configuration interaction (CI), Density Functional Theory (DFT), and perturbation theory. These give what is called electron correlation since the electrons are less restricted to being described by one-electron wavefunctions, i.e. orbitals and a better description of the simultaneous motion of the electrons is given. We will only consider CI pictorially here since it lends itself more easily to visual interpretation. Most Quantum Chemistry calculations use DFT routinely but its mathematical description is beyond the scope of this course.

For a picture of CI, we must first go back to the results for the H₂ wavefunction at r = 2 in exercise 5 to see the effects without CI. We can consider the graph as a measure of two valence bond terms in H₂, those being the covalent bonding H-H term and the resonance ionic bonding H:^-H⁺ + H⁺H:^- (meaning a resonance picture of a lone pair on one H and no electrons on the other H atom and vice versa) term. The height of the curve at the midpoint of the bond corresponds to the amount of covalent character and the heights over the nuclei to ionic character.

Now, what exactly have we plotted? It was the sigma molecular orbital for H₂, but not the wavefunction for the H₂ molecule. This latter is just the electronic configuration, 1s² or 1s(1)a(1)∙1s(2)b(2). If we separate the spatial 1s(1)∙1s(2) part from the spin part a(1)∙b(2), we can concentrate on the spatial part which has two electrons described by the same 1s function. This also avoids the question how do we know electron 1 has up spin and 2 down spin and not vice versa.

Exercise 9). Plot the 1s(1)∙1s(2) wavefunctions for H₂ and HHe⁺, using an internuclear distance of 2 and the 1s MOs from exercises 5 and 8. You can save yourself some work if you notice you are plotting 1s squared.

Exercise 10). Do the same for the doubly excited 1s*(1)∙1s*(2) configuration. If you are really adventuresome, try plotting the singly excited wavefunction, but here you must take the average of the two ways you can assign two electrons to two different orbitals: [s*(1)∙1s(2)+s*(2)∙1s(1)].

During the plotting in exercise 8 you would notice that 1s(1)∙1s(2) = (1s_A(1)+1s_B(1)) ∙ (1s_A(2)+1s_B(2)). Multiplying this quadratic:

1s(1)∙1s(2) = [1s_A(1) ∙1s_A(2)+1s_B(1) ∙1s_B(2)+ 1s_B(1) ∙1s_A(2)+1s_A(1) ∙1s_B(2)]

The first two terms correspond to the ionic terms where two electrons are on the same atom in a diatomic molecule. The latter two are covalent in the sense that the two eleectrons are shared between two atoms. The doubly excited configuration leads to:

1s*(1)∙1s*(2) = [1s_A(1) ∙1s_A(2)+1s_B(1) ∙1s_B(2)-1s_B(1) ∙1s_A(2)-1s_A(1) ∙1s_B(2)]

The configuration interaction method uses weighted sums of electronic configurations to go beyond the orbital approximation, back up to the Grand Approximation 1.

Y(1,2) » c₁∙1s(1)∙1s(2) + c₂∙1s*(1).1s(2) +c₃∙1s*(1)∙1s*(2) +...

Here again the Variation Principle is used to get the best set of coefficients and once again the values of these coefficients change with the positions of the nuclei, i.e. bond lengths, angles.

As we get better wavefunctions, the calculated energy is lowered towards the true Schroedinger value. We can see this from the above graphs. Assume that at one particular internuclear distance, the bonding obtained from just the 1s(1).1s(2) electron configuration is not enough and the electrons reside too much on the nuclei (covalent terms too small and the ionic too large). We can combine configurations to form a more accurate wavefunction which leads to lower energy.

Exercise 11). Replot the 1s(1).1s(2) function for H2 from exercise 8 on the same graph as the correlated wavefunction Y(1,2) » c₁∙1s(1)∙1s(2) + c₂∙1s*(1)∙1s*(2), where c₁ = 0.9 and c₂ = -0.4.