quantum mechanics and the hydrogen atom

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Since we cannot say exactly where an electron is, the Bohr picture of the atom, with electrons in neat orbits, cannot be correct.

Quantum theory describes electron probability distributions:

Quantum Mechanics and the hydrogen atom

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Hydrogen Atom: Schrödinger Equation and Quantum Numbers

Potential energy for the hydrogen atom:

The time-independent Schrödinger equation in three dimensions is then:

Equation 39-1 goes here.

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Where does the quantisation in QM come from ?

The atomic problem is spherical so rewrite the equation in (r,θ,φ)

φθ cossinrx = φθ sinsinry = θcosrz =

Rewrite all derivatives in (r,θ,φ), gives Schrödinger equation;

Ψ=Ψ+Ψ⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

+∂∂

∂∂

−Ψ⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

− ErVmr

rrm

)(sin

1sinsin

122 2

2

2

22

2

φθθθ

θθhh

This is a partial differential equation, with 3 coordinates (derivatives);Use again the method of separation of variables:

( ) ( ) ( )φθφθ ,,, YrRr =Ψ

Bring r-dependence to left and angular dependence to right (divide by Ψ):

( )( )( )

( ) λφθφθθφ =−=⎥

⎦

⎤⎢⎣

⎡−+

,,21

2

22

YYO

RrVEmrdrdRr

drd

R

QM

h Separation of variables

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( )( ) λ=⎥⎦

⎤⎢⎣

⎡−+ RrVEmr

drdRr

drd

R 2

22 21

hRadial equation

Angular equation ( )( )

( )

( ) λφθ

φθφθθ

θθθ

φθφθθφ =

⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂+

∂∂

∂∂−

=−,

,sin

1sinsin

1

,, 2

2

2

Y

Y

YYOQM

YYY θλθ

θθ

θφ

22

2sinsinsin +

∂∂

∂∂

=∂∂

−

Once more separation of variables: ( ) ( ) ( )φθφθ ΦΘ=,Y

Derive: 222

2sinsinsin11 m=⎟

⎠⎞

⎜⎝⎛ Θ+

∂Θ∂

∂∂

Θ=

∂Φ∂

Φ− θλ

θθ

θθ

φ(again arbitrary constant)

Simplest of the three: the azimuthal angle;

( ) ( ) 022

2=Φ+

∂Φ∂ φφφ m

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A differential equation with a boundary condition

( ) ( ) 022

2=Φ+

∂Φ∂ φφφ m and ( ) )(2 φπφ Φ=+Φ

Solutions:

( ) φφ ime=Φ

Boundary condition; ( ) ( ) ( ) φπφ φπφ imim ee =Φ==+Φ +22

12 =ime π

m is a positive or negative integerthis is a quantisation condition

General: differential equation plus a boundary condition gives a quantisation

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( ) φφ ime=ΦFirst coordinatewith integer m (positive and negative)

Second coordinate 0sin

sinsin

12

2=Θ⎟

⎟⎠

⎞⎜⎜⎝

⎛−+

∂Θ∂

∂∂

θλ

θθ

θθm

Results in ( )1+= lllλ with Kl ,2,1,0=

and llKll ,1,,1, −+−−=m

angularpart

angularmomentum

Third coordinate ( )( ) ( )1212

22 +=⎥

⎦

⎤⎢⎣

⎡−+ ll

hRrVEmr

drdRr

drd

R

Differentialequation Results in quantisation of energy

2

2

0

2

2

2

2

2

24 h

en

menZR

nZE ⎟

⎟⎠

⎞⎜⎜⎝

⎛−=−= ∞ πε

with integer n (n>0)

radialpart

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Angular wave functions

⎥⎦

⎤⎢⎣

⎡

∂∂

+∂∂

∂∂

= 2

2

22

sin1sin

sin1

φθθθ

θθiL h

with Kl ,2,1,0= and llKll ,1,,1, −+−−=m

Operators:

φ∂∂

=i

Lzh

There is a class of functions that are simultaneous eigenfunctions

),,( zyx LLLL =r

Angular momentum

( ) ( ) ( )φθφθ ,1, 22lmlm YYL hll += ( ) ( )φθφθ ,, lmlmz YmYL h=

Spherical harmonics (Bolfuncties) ( )φθ ,lmY

π41

00 =Y φθπ

ieY sin83

11 −=

θπ

cos43

10 −=Y

φθπ

ieY −− = sin

83

1,1

Vector space of solutions

( ) 1, 2 =Ω∫Ω

dYlm φθ

''''*

mmllmllm dYY δδ=Ω∫Ω

( ) ( ) ( ) ( )φθπφθπφθ ,,, lmlmopP Υ−=+−Υ=Υ l

Parity

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The radial part: finding the ground state

Find a solution for 0,0 == ml

ERRr

ZeRr

Rm

=−⎟⎠⎞

⎜⎝⎛ +−

0

22

4'2"

2 πεh

Physical intuition; no density for ∞→r

trial: ( ) arAerR /−=

aRe

aAR ar −=−= − /'

2/

2"aRe

aAR ar == −

Er

Zearam

=−⎟⎠⎞

⎜⎝⎛ −−

0

2

2

2

421

2 πεh

must hold for all values of r

04 0

22=−

πεZe

mahPrefactor for 1/r:

mZea 2

20

04 hπε

=

Solution for thelength scale paramater

Bohr radius

Solutions for the energy

2

2

0

22

2

242 h

h emeZma

E ⎟⎟⎠

⎞⎜⎜⎝

⎛−=−=

πε

Ground state in the Bohr model (n=1)

( )( ) λ=⎥⎦

⎤⎢⎣

⎡−+ RrVEmr

drdRr

drd

R 2

22 21

h

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There are four different quantum numbers needed to specify the state of an electron in an atom.

1. The principal quantum number n gives the total energy.

2. The orbital quantum number gives the angular momentum; can take on integer values from 0 to n - 1.


ll

3. The magnetic quantum number, m , gives the direction of the electron’s angular momentum, and can take on integer values from – to + .l l

l

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Hydrogen Atom Wave Functions

The wave function of the ground state of hydrogen has the form:

The probability of finding the electron in a volume dVaround a given point is then |ψ|2 dV.

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Radial Probability Distributions

Spherical shell of thickness dr, inner radius rand outer radius r+dr.

Its volume is dV=4πr2dr

Density: |ψ|2dV = |ψ|24πr2dr

The radial probablity distribution is then:

Pr =4πr2|ψ|2

Ground state

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This figure shows the three probability distributions for n = 2 and = 1 (the distributions for m = +1 and m = -1 are the same), as well as the radial distribution for all n = 2 states.

Hydrogen Atom Wave Functions

l

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Hydrogen “Orbitals” (electron clouds)

( )2,trrΨRepresents the probability to find a particleAt a location r at a time t

The probability densityThe probability distribution

The Nobel Prize in Physics 1954"for his fundamental research in

quantum mechanics, especially for his statistical interpretation of the wavefunction"

Max Born

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Atomic Hydrogen Radial part

Analysis of radial equation yields:

∞−= RnZEnlm 2

2

chemR e

30

4

8ε=∞with

Wave functions:

( ) ( ) ( ) hr /,, tniElmnlnlm erRtr −Υ=Ψ φθ

( ) ( )( ) ⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

+−−

= +−−

+−

nZL

nZe

nnn

naZu n

nZn

ρρρ ρ 22!12!12 12

1

1/

0

ll

l

ll

naZr /2/ =ρ 220 /4 ea μπε h=

For numerical use:

( )rruR nl=

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Quantum analog of electromagnetic radiation

Classical electric dipole radiation Transition dipole moment

Classical oscillator Quantum jump

2reIrad&&r∝ 2

2*1 τψψ dreIrad ∫∝ r

The atom does not radiate when it is in a stationary state !The atom has no dipole moment

01*1 == ∫ τψψμ driir

20

2

6 hε

μ fiifB =Intensity of spectral lines linked

to Einstein coefficient for absorption:

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Selection rules

Mathematical background: function of odd parity gives 0 when integrated over space

In one dimension: dxxfdxxx ifif ∫∫∞

∞−

∞

∞−

=ΨΨ=ΨΨ )(* with if xxf ΨΨ= *)(

( ) dxxfdxxfdxxfxdxfdxxfdxxfdxxf ∫∫∫∫∫∫∫∞∞∞

∞

∞

∞−

∞

∞−

+−=+−−=+=000

0

0

0)()()()()()()(

0)(20

≠= ∫∞

dxxf )()( xfxf =−if

0= if )()( xfxf −=−

iΨ and fΨ opposite parity

iΨ and fΨ same parity

Electric dipole radiation connects states of opposite parity !

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Selection rules

depend on angular behavior of the wave functions

rr

rr−

θ

φ

Parity operator

( ) ( )πφθπφθ +−→−−−→

−=

,,,,),,(),,(

rrzyxzyx

rrP rr

All quantum mechanical wave functionshave a definite parity

( ) ( )rr rrΨ±=−Ψ

0≠ΨΨ if rr

If and have opposite parityfΨ iΨ

Rule about the mYl functions

( ) ),(),( φθφθ mm YPY ll

l −=

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“Allowed” transitions between energy levels occur between states whose value of differ by one:

Other, “forbidden,” transitions also occur but with much lower probability.


l

“selection rules, related to symmetry”

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Selection rules in Hydrogen atom

Intensity of spectral lines given by

ififfi re Ψ−Ψ=ΨΨ= ∫rrμμ *

1) Quantum number nno restrictions

2) Parity rule for l

odd=Δl

3) Laporte rule for l

1r

lr

lr

+= if

Angular momentum rule:

so 1≤Δl

From 2. and 3. 1±=Δl

Lyman series

Balmer series

quantum mechanics and the hydrogen atom

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