Molecular SpectroscopyMolecular Spectroscopy

The hydrogen atom

NC State University

Electronic Structure of Hydrogen

The Schrdinger equation for hydrogen

Separation of variables: Radial and angular parts

Hydrogen atom wavefunctions Expectation valuesy g p

Spectroscopy of atomic hydrogenSpectroscopy of atomic hydrogen

Schrdinger equation for hydrogen:g y gThe form of the potential

The Coulomb potential between the electron and the proton isV = -Ze2/40r

The hamiltonian for both the proton and electron is: The hamiltonian for both the proton and electron is:

Separation of nuclear and electronic variables results in an electronic equation in the center-of-mass coordinates:

Separation of variablesSeparation of variables

It would be impossible to solve the equationwith all three variables simultaneously.yInstead a procedure known as separation of variablesis used.

The steps are:1 Substitute in (r ) = R(r)Y( )1. Substitute in (r,,) = R(r)Y(,)2. Divide both sides by R(r)Y(,)3. Multiply both sides by 2r2p y y

Schrdinger equation for hydrogen: The kinetic energy operator

The Schrdinger equation in three dimensions is:The Schrdinger equation in three dimensions is:

h2 2 V E h22 + V = E

Th t d lTh t d l d id iThe operator delThe operator del--squared is:squared is:

2 = 2

+ 2

+ 2

The procedure uses a spherical polar coordinate system.The procedure uses a spherical polar coordinate system.

x2

+y 2

+z2

p p p yp p p yInstead of x, y and z the coordiantes are Instead of x, y and z the coordiantes are , , and r.and r.

The wavefunctions of a rigid rotor are called spherical harmonics

The solutions to the and equation (angular part)Thesolutionstothe and equation(angularpart)arethesphericalharmonicsY(, )=()()Separationofvariablesusingthefunctions()and()allowssolutionoftherotationalwaveequation.

We can obtain a and equation from the aboveWecanobtaina and equationfromtheaboveequation.

Spherical Polar Coordinatesz

r

r cos()x

r sin() sin()

y r sin() cos()

The volume element in spherical polar coordinates

To solve the Schrodinger equation we need to integrate of all space. This is the same thing

f i l i t l Th las performing a volume integral. The volume element is:

dV = r 2dr sind d

This integrates to 4, which is the

dV = r 2dr sind d

g ,normalization constant. 4 stearadians also gives the solid angle of a sphere.

Separation of variablespThesphericalharmonicsarisefromtheproductof after substituting Y = ofaftersubstitutingY=

Multiplythroughbysin2/h2.p y g y /

Separation of variablespTheoperatorsinvariables and operateonfunctionand respectively so we can writeand,respectively,sowecanwrite

WhenwedividebyY=,weobtain

Now,theseequationscanbeseparatedusingseparationconstantm2.

Separation of variablespWehavealreadyseenthesolutiontothe equationfrom the example of rotation in two dimensionsfromtheexampleofrotationintwodimensions.

whichhassolutions

Nowthatwehavedefinedthevaluesofmaspositiveandnegativeintegers,the equationisalsodefined.g g , q

The solution of equation gives.

Legendre polynomialsSubstitute x = cosand the equation becomes:Substitutex cosandtheequationbecomes:

Thesolutionrequiresthat = with=0,1,2..Where istherotationalquantumnumber.Theazimuthal quantumnumberism.Th it d f Th l ti L dThemagnitudeof..ThesolutionsareLegendrepolynomialsP (x)=1 P (x)=1/2 (3x2 1)P0(x)=1 P2(x)=1/2(3x 1)P1(x)=x P3(x)=1/2(5x3 3x)

The spherical harmonics as solutions to the rotational hamiltonian

The spherical harmonics are the product of theThesphericalharmonicsaretheproductofthesolutionstothe and equations.Withnormalizationthesesolutionsare

Themquantumnumbercorrespondsthez component of angular momentumzcomponentofangularmomentum.Thenormalizationconstantis

The form of the spherical h iharmonics

Includingnormalizationthesphericalharmonicsare

Y00 = 1

4Y2

0 = 516 3cos2 1

4Y1

0 = 34 cos16

Y21 = 158 sincose

i

Y11 = 38 sine

i Y22 = 1532 sin

2e2i

Theformcommonlyusedtorepresentpanddorbitals are linear combinations of these functionsorbitalsarelinearcombinationsofthesefunctions

Euler relationEuler relationLinearcombinationsareformedusingtheEulerrelation

e+i =cos +isinei =cos isin

cossine+i +e+i e+i e+i 2 2i

Projectionalongthezaxisisusuallytakenusingz = rcos Projection in the x y plane is taken usingz=rcos.Projectioninthex,yplaneistakenusingx=rsincosand y=rsinsin

Solutions to the 3-D rotational hamiltonian

There are two quantum numbers is the total angular momentum quantum number

m is the z-component of the angular momentum The spherical harmonics called Y are functions whose

probability |Y |2 has the well known shape of the s, p and d orbitals etcd orbitals etc.= 0 is s , m = 0= 1 is p , m = -1, 0 , 1 1 is p , m 1, 0 , 1= 2 is d , m = -2 , -1, 0 , 1, 2= 3 is f , m = -3 , -2 , -1, 0 , 1, 2, 3, , , , , , ,

etc.

Space quantization in 3D Specification of the azimuthal

quantum number mz implies that the angular momentum M = +2

zthat the angular momentum about the z-axis is Jz = hm.

M = +2

M = +1

This implies a fixed orientation between the total angular momentum and the z

M = 0

component.

The x and y components

M = -1

M = 2 The x and y components cannot be known due to the Uncertainty principle.J = 2

M = -2

Standing waves on a sphereStanding waves on a sphere

These are the spherical harmonics Ylm, whichare solutions of the angular Schrodinger equation.

Orthogonality of g ywavefunctions

I i li ti h Ignoring normalization we have: s 1

p cos sincos sinsin p cos, sincos, sinsin d 1/2(3cos2 - 1), cos2cos2, cos2sin2,

cossincos cossinsincossincos , cossinsin

The differential angular element is sindd/4 The differential angular element is sindd/4 The limits = 0 to and = 0 to 2. The angular wavefunctions are orthogonalThe angular wavefunctions are orthogonal.

Orthogonality of g ywavefunctions

For the theta integrals we can use the substitution x = cos and dx = sind For example, for s and p-type rotational wave functions we

have

< s | p > cos sin d0

= x dx1

1

= x2

2 1

1

= 12 12 = 0

MAPLEMAPLE worksheet on spherical harmonicsTh f f h h i l h i Y ( ) i i The form of the spherical harmonics Ylm (,) is quite familiar. The shape of the s-orbital resembles the first spherical harmonic Y spherical harmonic Y00.

Attached to this lecture are three MAPLE worksheets that illustrate the s p and d orbitals respectively The that illustrate the s, p and d orbitals respectively. The idea is to obtain an interactive picture of the mathematical form and the plots of the functions.p

Disclaimer: The spherical harmonics have been simplified by formation of linear combinations to remove p yany complex numbers.

MAPLE

Th Y h i l h i h th f f

worksheet on spherical harmonics

The Y00 spherical harmonic has the form of ans-orbital.

There is only one angular function for There is only one angular function for

MAPLE worksheet on spherical harmonics

The Y10 Y11 , and Y1,-1 spherical harmonics have the form of p-orbitals.

There are three angular functions for There are three angular functions for

MAPLE

Th Y Y Y Y d Y h i l

worksheet on spherical harmonics

The Y20 , Y21 , Y2,-1 , Y2,2 and Y2,-2 spherical harmonics have the form of d-orbitals.

There are five angular functions for.g

The effective potential: result of the solution of the angular partthe solution of the angular part

The solutions for the angular part results in a g pterm in potential energy equal to:

This term contains the contributions to the This term contains the contributions to the energy from angular terms. Together with the Coulomb potential the Together with the Coulomb potential the

effective potential energy is:

The radial equation for hydrogenThe radial equation for hydrogen

Making the abo e appro imations e ha e anMaking the abo e appro imations e ha e anMaking the above approximations we have anMaking the above approximations we have anradial radial hamiltonianhamiltonian (energy operator)(energy operator)

The solutions have the form: The solutions have the form: RRn,n,ll = = NNn,n,ll ll ee--/2/2 LLn,n,ll(r)(r)wherewhere = 2Zr/na= 2Zr/na00 andand aa00 = 4= 400hh22/me/me22wherewhere 2Zr/na 2Zr/na00 andand aa00 4 400hh /me/meNNn,n,l l is the normalization constant.is the normalization constant.LLn,n,ll(r) (r) is an associated Laguerre polynomial.is an associated Laguerre polynomial.

Solutions of the radial equation

The normalization constant depends on n The normalization constant depends on n anandd

The associated Laguerre polynomials are:The associated Laguerre polynomials are:

Normalization of the radialNormalization of the radial functions

Each of the radial equation solutions is a polynomialEach of the radial equation solutions is a polynomialmultiplying an exponential. The normalization ismultiplying an exponential. The normalization isobtained from the integral:obtained from the integral:obtained from the integral:obtained from the integral:

R l* R l r

2dr = 1

Rnl Rnl r dr 10

The volume element here isThe volume element here is rr22drdr which is the rwhich is the rThe volume element here isThe volume element here is rr22dr dr which is the rwhich is the rpart of the spherical coordinate volume elementpart of the spherical coordinate volume elementrr22sinsindrddrddd..

MAPLE worksheetMAPLE worksheetNormalization of the radial functions

A MAPLE worksheet attached to this lecture illustratesth li ti f th fi t th di l f ti Ththe normalization of the first three radial functions. Theworksheet includes plots of the functions.

When examining a plot keep in mind that you can plotthe wave function or the square of the wave function.qWe often plot the square of the wave function, becausethe integral of the square of the wave function gives the

b bilitprobability.

Hydrogen 1 s radial f tiwavefunction

2.0 The 1s orbital has no

nodes and decays exponentially

1.5exponentially.

R1s = 2(1/a0)3/2e-/2

F 1 1/1.0

R1,

0

1s For n = 1, = 1/a0 n = 1 and = 0 are

th t b

0.5

the quantum numbers for this orbital. 1086420

The Radial Distribution in Hydrogen 2s and 2p orbitals

0.6

2p

0.4Rn,

l

2sp

0.2

0.0

1086420

Radial wave functions for HRadial wave functions for H

The Rydberg Constanty g The energy levels calculated using the

Schrdinger equation permit calculation of the Rydberg constant.

One major issue is units. Spectroscopists often use units of wavenumber or cm-1. At first this seems odd, but h = hc/ = hc where is the value of the transition in wavenumbers.

~~

R = 1hce4

32 2 2h2 in cm-1

hc 32202h2

The simple form for H energy levels

Using the Rydberg constant the energy of the Using the Rydberg constant the energy of the hydrogen atom can be written as:

E = Rn2

~

nwhere R = 107 690 cm-1 where R = 107,690 cm .In units of eV R = 13.6 eV.

Shells and subshellsShells and subshells

All of the orbitals of a given value of n for a shell All of the orbitals of a given value of n for a shell. n = 1, 2, 3, 4 .. correspond to shells K, L, M, N Orbitals with the same value of n and different

values of form subshells. = 0, 1, 2, ... correspond to subshells s, p, d Using the quantum numbers that emerge from Using the quantum numbers that emerge from

solution of the Schrdinger equation the subshells can be described as orbitalssubshells can be described as orbitals.

Hydrogenic orbitals

s orbitals are spherically symmetrical. The 1s s orbitals are spherically symmetrical. The 1s wavefunction decays exponentially from a maximum value of (1/a03)1/2 at the nucleus. maximum value of (1/a0 ) at the nucleus.

p orbitals have zero amplitude at r =0, and the electron possesses an angular momentum of electron possesses an angular momentum of hl(l +1). The orbital with m = 0 has zero angular momentum about the z axis The angular momentum about the z axis. The angular variation is cos which can be written as z/r leading to the name p orbitalleading to the name pz orbital.

Spectroscopy of atomic hydrogenSpectroscopy of atomic hydrogen

S t t d i b 1 Spectra reported in wavenumbers, cm-1

Rydberg fit all of the series of hydrogen spectra with a single equation,

Absorption or emission of a photon of frequency = R 1n12

1n22

p p q y occurs in resonance with an energy change, E = h (Bohr frequency condition).( q y )

Solutions of Schrdinger equation result in further selection rules.further selection rules.

Spectroscopic transitionsSpectroscopic transitions

A t iti i t f f t t A transition requires a transfer from one state with its quantum numbers (n1, 1, m1) to another state (n2, 2, m2).

Not all transitions are possible: there are Not all transitions are possible: there are selection rules, = 1, m = 0, 1

These rules demand conservation of angular momentum. Since a photon carries an intrinsic

fangular momentum of 1.

Expectation values

The expectation (or average) value of an observable is given The expectation (or average) value of an observable is given b th l f lb th l f l

< r > = *rd

by the general formula.by the general formula.

0

As written the above integral describes the expectation valueAs written the above integral describes the expectation valueof the mean value of the radius, r. Integration over the angular of the mean value of the radius, r. Integration over the angular of the mean value of the radius, r. Integration over the angular of the mean value of the radius, r. Integration over the angular part gives 1 because the spherical harmonics are normalized.part gives 1 because the spherical harmonics are normalized.The volume element can be written The volume element can be written dd = r= r22dr. The mean value is:dr. The mean value is:

< r > = *r r2dr = * r3dr0

0

MAPLE worksheet onExpectation (Average) values

There is a MAPLE worksheet attached to this lecture There is a MAPLE worksheet attached to this lecture that illustrates the use of expectation values. The that illustrates the use of expectation values. The example of the position is given for the 1s, 2s and example of the position is given for the 1s, 2s and 3s radial wave functions. The expectation value or 3s radial wave functions. The expectation value or average value of r gives the average distance of anaverage value of r gives the average distance of anaverage value of r gives the average distance of anaverage value of r gives the average distance of anelectron from the nucleus in a particular orbital. Sinceelectron from the nucleus in a particular orbital. Sincethe 2s orbital has one radial node and the 3s orbital hasthe 2s orbital has one radial node and the 3s orbital hasthe 2s orbital has one radial node and the 3s orbital hasthe 2s orbital has one radial node and the 3s orbital hastwo radial nodes the average distance of the electrontwo radial nodes the average distance of the electronfrom the nucleus is shown to increase as:from the nucleus is shown to increase as:

for 3s > for 2s > for 1s for 3s > for 2s > for 1s

The average energy

The expectation (or average) value of the energy is given The expectation (or average) value of the energy is given b th l f lb th l f l

< E > = *Hd

by the general formula.by the general formula.

0

since the since the hamiltonianhamiltonian H is the energy operator. If the wave functionH is the energy operator. If the wave functionis not normalized this is often written as:is not normalized this is often written as:is not normalized this is often written as:is not normalized this is often written as:

*H d

< E > = H d

0

* d

d0

The Variation MethodThe Variation Method

The variational theoremThe He atom

The variational theoremThe variation method allows us to obtain an approximation to the ground state energy of the system without solving the Schrdinger equationthe Schrdinger equation.The variation method is based on the following theorem:

Given a system with hamiltonian operator H, then if is any normalized well-behaved function that satisfies the boundary conditions it is true thaty

*Hd E0where E0 is the true value of the lowest energy eigenvalueof H. This important theorem allows us to calculate an

b d f th d t tupper bound for the ground state energy.

Practical significanceThe variation method serves as the basis for all methods that use combinations of hydrogen-like orbitals to solve for the eigenfunctions (wave functions) and eigenvalues (energies)the eigenfunctions (wave functions) and eigenvalues (energies) of atoms and molecules.

The radial part of the hydrogen like wave functions is modifiedThe radial part of the hydrogen-like wave functions is modified by a variational parameter, which is minimized. The theorem allows us to set the derivative with respect to any parameter equal to zero to find the value of that parameter that minimizes the energy:

*Hd = 0

We can be sure that the energy calculated in this way will be

Hd = 0

We can be sure that the energy calculated in this way will be greater than the true energy (an upper bound).

The hamiltonian for H

The electronic hamiltonian for the hydrogen atom consists of a kinetic energy term for the electron and the Coulomb attractionkinetic energy term for the electron and the Coulomb attractionof the electron and proton (nucleus).

h2 2 h2m2

Z 2+

-

Of course the nuclear charge of hydrogen is Z = 1 so the

Ze2

r1Of course, the nuclear charge of hydrogen is Z 1 so theZ is included for completeness. We know that the solutionsOf the Schrdinger equation (H = E) gives energy levels:

n is the principal quantum numbern is the principal quantum number.a0 is the Bohr radius.En =

e22a0

1n2

The hamiltonian for He

For helium the same kinetic energy and Coulomb attractionterms are present, but there is also a Coulomb repulsion between the two electrons that must be included.

h22 2

- e2

r12 2m1 h2

2m22

Z 2++

-

Ze2

r

12

Ze2

r2r1

Because of the Coulomb repulsion there is no exact solution for He. To solve the problem we use two 1s orbitals from the solution for hydrogen and then apply the variational method.y g pp y

The He wave function

The hydrogen 1s wave functions for electrons 1 and 2 are: 3/2 3/2

Th fb h f t th t th t t l

f1 = 1Za0

3/2e Zr1/a 0, f2 = 1

Za0

3/2e Zr2/a 0

The aufbau approach for atoms assumes that the total wavefunction for a many-electron atom is just a product of oneelectron wave functions. In the present case:

Note that the hydrogen wave functions are normalized so:

= f1 f2Note that the hydrogen wave functions are normalized so:

f1* f1d = f2

* f2d = 1

Variational approach for the He atom

The He wave function used for the variation method is a product of two hydrogen 1s orbitals However instead ofproduct of two hydrogen 1s orbitals. However, instead of the nuclear charge Z we use a variational parameter .

1 3

= 1a0 e

r1/a0e r2/a0

has a physical interpretation. Since one electron tends toscreen the other from the nucleus, each electron is subject toa nuclear charge that is less than Za nuclear charge that is less than Z.The hamiltonian is:

h2 2 e2 h2 2 e2 e2 e2 e2H = h2m12 er1

h2m2

2 er2 + Zer1 + Z

er2 +

er12

Evaluation of the integrals

If we consider only the part of the hamiltonian in parentheses We have the solution to a hydrogen atom with two electrons in the 1s orbital.

h2

2 12 e

2

r h22 2

2 e2

r = 2 e2a

where the right hand side is twice the energy of a 1s electron. Using this result we have:

2m1 r1 2m2 r2 a0

Using this result we have:

*Hd = 2e2

a0 *d + Z e2 *d

r1 + Z e2 *d

r2 + e2 *d

r12

The integrals have the following values:*d *d 2 *d 5 e2*d = 1 , dr1 =

dr2 =

a0 , e

2 dr12 =

5 e8a0

Evaluation of the variational parameter We have:

2 5 e2

We now vary to minimize the variational integral:

*Hd = 2 2Z + 58ea0

We now vary to minimize the variational integral:

*Hd =

2 2Z + 58e2a0 = 0

2 2Z + 58

516The variational energy is:

16

*H d Z2 5Z 25 e2

Z 52 e2*Hd = Z2 + 58Z

25256

ea0 = Z

516

ea0

The variational energy: comparison with experiment

The experimental ionization energy of He is 24 5 eVThe experimental ionization energy of He is 24.5 eV.

Our first guess would be to calculate the energy of the 1s El t i H i th h d l l ith lElectron in He using the hydrogen energy level with a nuclear charge Z = 2, E = -Ze2/a0. This gives - 2(13.6) eV = -27.2 eV.

Using the value obtained by the variational method we have,E = -(27/16)e2/a0 = -(27/16)(13 6) eV = -22 95 eVE (27/16)e /a0 (27/16)(13.6) eV 22.95 eV.

The value is much closer to the true value. In accord with th i ti l th th t d t t i lthe variational theorem, the true ground state energy is less than that given by variational method.

Summaryy

The hydrogen atom is the only atom with an exact solution.y g y

Hydrogen wave functions are used as the approximationfor atomic wave functions in multielectron atomsfor atomic wave functions in multielectron atoms.

The variational principle states that any wave function we choose that satisfies the Schrdinger equation will give

an energy greater than the true energy of the system.

The variation method provides a general prescription forimproving on any wave function with a parameter by minimizing that function with respect to the parameterminimizing that function with respect to the parameter.