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Chapter 2

Some Examples

2.1 Hydrogen Atom

Hydrogen atom is an exactly solvable problem in Quantum Mechanics. As the solu-

tions of H atoms are basis for us to solve many electron atoms and molecules, let us

look at them briefly. For detailed derivations, I recommend the book of Levine, Quan-

tum Chemistry or McQuarrie, Quantum Chemistry. These books are available in our

Library.

The electronic Hamiltonian of H atom is

H = −12∇2 − 1

r(2.1)

where the first term is the kinetic energy of the electron, and the second term is the

nucleus–electron repulsion. Note that we are using atomic units here. Now, the partial

differential equation,

HΨ(x, y, z) = EΨ(x, y, z) (2.2)

1

2 CHAPTER 2. SOME EXAMPLES

can not be solved in Cartesian coordinates analytically. This is because, the technique of

separation of variables cannot be used here due to the term 1/r, as r =√

x2 + y2 + z2.1

Fortunately, this problem can be solved if we change the coordinates from Cartesian to

spherical polar coordinates. In spherical polar coordinates the three coordinates used

are (r, θ, φ): see figure below.

x

y

θ

φ

zr

X

Y

Z

(x,y,z)

Figure 2.1: Geometric definition of spherical polar coordinates

Spherical polar coordinates and Cartesian coordinates are related through the fol-

lowing relationships:

x = r sin θ cos φ (2.3)

y = r sin θ sin φ (2.4)

z = r cos θ (2.5)

Thus,

r =√

x2 + y2 + z2 (2.6)

θ = arccos

(z√

x2 + y2 + z2

)(2.7)

φ = arctan(y

x

)(2.8)

1I recommend you to remember the separation of variables techniques from Mathematics text books

2.1. HYDROGEN ATOM 3

Since the partial differential equation (i.e. SE for H atom) is separable in spherical

polar coordinates,

Ψn,l,m(r, θ, φ) = Rn,l(r) Sl(θ) Tm(φ) = Rn,lYl,m(θ, φ) (2.9)

where R(r), S(θ), and T(φ) are solutions of partial differential equations separated in

r, θ, and φ coordinates respectively. For details of these solutions, I recommend you

to read the aforementioned textbooks on Quantum Chemistry. Here, n, l, m are the

quantum numbers.

n = 0, 1, 2, · · · (2.10)

l = 0, 1, · · · , n− 1 (2.11)

m = −l,−l + 1, · · · , 0, · · · , l − 1, l (2.12)

See the Table 1, and Table 2 for radial and spherical harmonic solutions of hydrogen

atoms.

n l Orbital Rn,l(r)

1 0 1s Z3/22 exp(−ρ/2)

2 0 2s Z3/2(1/8)1/2 (2− ρ) exp(−ρ/2)

1 2p Z3/2(1/24)1/2ρ exp(−ρ/2)

3 0 3s Z3/2(1/243)1/2(6− 6ρ + ρ2) exp(−ρ/2)

1 3p Z3/2(1/486)1/2(4− ρ)ρ exp(−ρ/2)

2 3d Z3/2(1/2430)1/2ρ2 exp(−ρ/2)

Table 1: Radial solutions of H atom in atomic units; Here ρ = (2Z/n)r.

Use Mathematica and visualize the atomic orbitals in Cartesian coordinates (by plot-

ting using the functions given the above two tables). Understand the sign of wavefunc-


l m Yl,m(θ, φ)

0 0(

14π

)1/2

1 0( 3

4π

)1/2cos θ

±1 ∓( 3

8π

)1/2sin θ exp(±iφ)

2 0( 5

16π

)1/2(3 cos2 θ − 1)

±1 ∓(

158π

)1/2cos θ sin θ exp(±iφ)

±2(

1532π

)1/2sin2 θ exp(±2iφ)

Table 2: Spherical harmonics part of H atom wavefunction

tions,and radial–angular nodes. Also, understand how the shapes of the orbitals are

formed by these functions.

Energy of a hydrogen atom is

En = −12

Z2

n2 a.u. (2.13)

Note that for m > 0, atomic orbitals are complex. However, we can create real

wavefunction from suitable linear combination of degenerate atomic orbitals. For e.g.

1√2(2p+1 + 2p−1) = constant · exp(−Zr/2) r sin θ cos φ

= constant · exp(−Zr/2) x ≡ 2px (2.14)1√2i

(2p+1 − 2p−1) = constant · exp(−Zr/2) r sin θ sin φ

= constant · exp(−Zr/2) y ≡ 2py (2.15)

Try yourself to construct the real 3d functions 3dxz, 3dyz, 3dxy, and 3dx2−y2 by your-

self. Plot all the 3d functions in Mathematica.

2.2. HE ATOM 5

2.2 He atom

Electronic Hamiltonian:

H = −12∇2

1 −12∇2

2 −2r1− 2

r2+

1r12

(2.16)

where the first two terms are kinetic energies of the two electrons (1 and 2), the third

and the fourth terms are the nuclear (with Z = 2) – electron attraction, and the last

term is the electron electron repulsion. We can rewrite this as,

H = H1 + H2 +1

r12(2.17)

where H1 are H2 Hamiltonian operators for hydrogen–like system with Z = 2 corre-

sponding to electron 1 and 2.

SE for this problem is not solvable exactly, due to the 1/r12 term. Thus we employ

approximate techniques here. If we ignore the 1/r12 term, the problem is exactly solv-

able, and it is clear (without solving) that the solution is 1s(1)1s(2) for the ground state.

Let us approximate the actual (ground state) wave function as

Ψ(1, 2) ≈ 1s(1)1s(2)

This approximation is called the independent electron approximation. Without doing any

approximation to the Hamiltonian, we can compute the approximate energy of the

Helium ground state:

E =

⟨1s(1)1s(2)

∣∣∣∣H1 + H2 +1

r12

∣∣∣∣ 1s(1)1s(2)⟩

(2.18)

=⟨1s(1)

∣∣H1∣∣ 1s(1)

⟩〈1s(2)|1s(2)〉+ 〈1s(1)|1s(1)〉

⟨1s(2)

∣∣H2∣∣ 1s(2)

⟩+⟨

1s(1)1s(2)∣∣∣∣ 1r12

∣∣∣∣ 1s(1)1s(2)⟩

(2.19)

= E1s × 1 + 1× E1s + J1s 1s (2.20)

= 2E1s + J1s 1s (2.21)

Here

E1s = −12

22

1= −2 a.u. (2.22)


Figure 2.2: Pictorial representation of Columbic interactions within the J integral

The J1s 1s is called the Coulomb integral as it is the total Columbic interaction between

1s charge density of electron 1 and 2:

J1s1s =

⟨1s(1)1s(2)

∣∣∣∣ 1r12

∣∣∣∣ 1s(1)1s(2)⟩

(2.23)

=∫ ∫

dr1 dr21s(1)∗1s(2)∗1

r121s(1)1s(2) (2.24)

=∫ ∫

dr1 dr2 |1s(1)|2 1r12|1s(2)|2 (2.25)

See also the Figure 2.2 The J integral can be evaluated

J1s 1s =58

Z = 1.25 a.u. (2.26)

Thus, with the independent electron approximation, and using 1s functions, we get

the energy of He ground state as

E = 2×−2a.u. + 1.25a.u. = −2.75a.u.

Experimentally determined value is −2.9 a.u.. Thus, a serious error is present in our

approximation .

We can do an improvement by applying the variational principle. Let us define our

trial wave function as

Ψ(1, 2) = 1s(1)1s(2)

2.2. HE ATOM 7

with

1s(1) =1√π

ζ3/2 exp(−ζr)

The above equation is identical to the true 1s atomic orbital of hydrogen atom, but with

Z = ζ. If we use this, the “trial” energy becomes

E = ζ2 − 2ζZ +58

ζ

with Z = 2 for our case. We can now minimize2 E by varying ζ: minimum will have

∂E∂ζ

= 0

In this way we can see that we get the lowest energy as −2.8 a.u. for ζ = 1.7. This

estimate of energy is better than with ζ = 2.

Why is ζ < 2? This is because electrons screen the nuclear charge, and thus the

effective interaction of an electron with the nucleus is lower than for the one electron

case.

2.2.1 Electronic Spins

According to Pauli’s principle by interchanging the electron labels of two electrons,

wave function should change its sign. This is the property of a wave function describing

electrons.

P12Ψ(1, 2) = −Ψ(1, 2)

Here P12 exchanges the labels 1 and 2. This is called the antisymmetry property of a

wave function. Clearly, our approximate wave function for He atom 1s(1)1s(2) is not

antisymmetric.

We are now introducing the spin functions to the orbitals, which describes the elec-

tronic spin angular momentum. Electrons with “up” spin in a 1s orbital can be repre-2because of the variational principle


sented by 1s(1)α(1), and the one with “down” spin can be written as 1s(1)β(1) . Spin

functions are orthonormal:

〈α|α〉 = 〈β|β〉 = 1 (2.27)

〈α|β〉 = 〈β|α〉 = 0 (2.28)

For He ground state, we have 1s ��. Thus

Ψ(1, 2) = 1s(1)α(1) 1s(2)β(2)

However, the above equation cannot be correct as electrons are indistinguishable, and

it not correct to say that electron with label 1 is α spin and with label 2 is β spin.

To include electron indistinguishability to the wavefunction, we can write the above

equation as

Ψ(1, 2) = 1s(1)1s(2)1√2[α(1)β(2)− β(1)α(2)] (2.29)

The above equation is clearly antisymmetric. Here 1s(1)1s(2) is the spatial part of the

wavefunction, and the rest is the spin dependent part.

I have not considered 1√2[α(1)β(2) + β(1)α(2)] as this will make the wavefunction

symmetric, and will not follow Pauli’s principle.

We can also rewrite Eqn. 2.29 as a determinant

Ψ(1, 2) =1√2

∣∣∣∣∣∣∣∣∣1s(1)α(1) 1s(1)β(1)

1s(2)α(2) 1s(2)β(2)

∣∣∣∣∣∣∣∣∣ (2.30)

This determinant is called the Slater determinant. This determinant clearly introduces

the antisymmetric property (exchange of any two rows will change the sign of wave

function by -1). Also Pauli’s exclusion principle is satisfied (any two columns with

same entries will make the determinant zero). For an n electron system, we can write

2.2. HE ATOM 9

the n–electron wave function as

Ψ(1, . . . , n) =1√n!

∣∣∣∣∣∣∣∣∣∣∣∣∣∣

χ1(1) · · · χn(1)

... . . . ...

χ1(n) · · · χn(n)

∣∣∣∣∣∣∣∣∣∣∣∣∣∣(2.31)

where χ1, · · · , χn are the spin orbitals.

χi(r, ω) = φi(r)g(ω)

where φ(r) is the spatial part, and g(ω) is the spin function (either α or β) and ω is the

spin–coordinate (which is just arbitrary). Let us now try for the Be ground state,

Ψ(1, 2, 3, 4) =1√4!

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

1s(1)α(1) 1s(1)β(1) 2s(1)α(1) 2s(1)β(1)

1s(2)α(2) 1s(2)β(2) 2s(2)α(2) 2s(2)β(2)

1s(3)α(3) 1s(3)β(3) 2s(3)α(3) 2s(3)β(3)

1s(4)α(4) 1s(4)β(4) 2s(4)α(4) 2s(4)β(4)

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

(2.32)

Note: here I have used 1s and 2s functions as the spatial part. But this has to be

substituted with other functions to get better energy, as we have seen for He ground

state. For brevity, we can slater determinant as (there are many other short forms used

in various text books)

Ψ(1, · · · , n) = |χ1 · · · χn| (2.33)

For Be, we can write a short form

Ψ(1, · · · , n) =∣∣1s 1s 2s 2s

∣∣ (2.34)

Here 1s means spatial part is 1s, and spin part is the β function; otherwise it can be the

α function.

Note that the Hamiltonian operators are not having any spin terms, and thus the


energy is influenced only by the spatial part of the wave function. Let us check it:⟨1s(1)1s(2)

1√2[α(1)β(2)− β(1)α(2)]

∣∣∣∣H1 + H2 +1

r12

∣∣∣∣ 1s(1)1s(2)

× 1√2[α(1)β(2)− β(1)α(2)]

⟩=

⟨1s(1)1s(2)

∣∣∣∣H1 + H2 +1

r12

∣∣∣∣ 1s(1)1s(2)⟩×⟨

1√2[α(1)β(2)− β(1)α(2)]

∣∣∣∣ 1√2[α(1)β(2)− β(1)α(2)

]⟩=

⟨1s(1)1s(2)

∣∣∣∣H1 + H2 +1

r12

∣∣∣∣ 1s(1)1s(2)⟩×

12(< α(1)|α(1) >< β(2)|β(2) > − < α(1)|β(1) >< α(2)|β(2) >

− < α(1)|β(1) >< α(2)|β(2) > + < β(1)|β(1) >< α(2)|α(2) >)

=

⟨1s(1)1s(2)

∣∣∣∣H1 + H2 +1

r12

∣∣∣∣ 1s(1)1s(2)⟩×

12(1− 0− 0 + 1)

=

⟨1s(1)1s(2)

∣∣∣∣H1 + H2 +1

r12

∣∣∣∣ 1s(1)1s(2)⟩

(2.35)

Eqn. ?? is equal to Eqn. ??, proving this.

Then what is the role of these spin functions? Should we ever worry about it?

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