Lecture 30:The Hydrogen Atom

Phy851 Fall 2009

Example 2: Hydrogen Atom

• The Hamiltonian for a system consisting of anelectron and a proton is:

• In COM and relative coordinates, theHamiltonian is separable:

• The energy eigenvalue equations are:

pep

p

e

e

RR

e

m

P

m

PH rr

−−+=

0

222

422 πε

rCM HHH +=

R

ePHr

0

22

42 πεµ−=

M

PH CMCM 2

2

=

( ) )()()()( R

r

C

CMrCM

R

r

C

CM EEEEEEH ⊗+=⊗

)()()( C

CMCM

C

CMCCM EEEH =

)()()( R

rr

R

rRr EEEH =

Central Potential Ansatz

• For a spherically symmetric potential, we canchoose to form simultaneous eigenstates ofHr, L2, and Lz:

• Separate the radial and angular degrees offreedom:

• The Radial wave equation is then:

)(1

42

)1(

2)( ,

0

2

2

2

2

22

, rRr

e

rrrRE nnn ll

llhh

−

++

∂

∂−=

πεµµ

)()()(,,,,

Ω⊗= mnmn

rRlll

)(, ,1)(

rRrnr n

r

ll −=

),(,,)(

φθφθ mYm ll =Ω

)()(,,

RR

r mnE l=

)()(,,,,

R

n

R

r mnEmnH ll =

( ) )(2)(2 ,,1,,RR

mnmnL lllhl +=

)()(,,,,

RR

z mnmmnL lhl =

Natural Units

• Introduce dimensionless units:

• The radial equation becomes:

)(1

42

)1(

2)( ,

0

2

2

2

2

22

, rRr

e

rrrRE nnn ll

llhh

−

++

∂

∂−=

πεµµ

λρ=r

nnE εµλ22

2

h−=

)()(, ρλρ uRn =l

0)1(1

4

222

0

2

=

−

+−+′′ u

eu nερρπε

µλ ll

h

Consult the Sacred Text

• From p.781, Handbook of MathematicalFunctions:– IF

– THEN

– We just need:

0)1(1

4

222

0

2

=

−

+−+′′ u

eu nερρπε

µλ ll

h

04

1

4

1

2

122

2

=

−

−+

+++′′ u

nu r

ρα

ρα

20

2

4

2

2

12

hπεµλα enr =

++

Generalized Laguerre Polynomial

)1(4

1 2

+−=−

llα

nε=41

∞= ,,3,2,1,0 Krn # nodes in radialwavefunction

)()( )(21

2 ρρρ ααρ

rnLeCu

+−

=

Sort Out the Details

• Solve for εn:

• Solve for α:

• Solve for λ:

20

2

4

2

2

12

hπεµλα enr =

++)1(

4

1 2

+−=−

llα

nε=41

4

1=nε

144 22 ++= llα

12 += lα

( )12

42

20 ++= lh

rneµπε

λ

2

20

0

4

ea

µπε h

= `Bohr Radius’

( )120 ++= lrna

λ

Original Units

• The energy levels are given by:

• The principle quantum number is therefore:

• For a given n, the allowed l values are:

nnE εµλ22

2

h−=

( )220

2

1

1

2 ++−=

l

h

r

nna

Eµ

( )120 ++= lrna

λ4

1=nε

1++= lrnn

220

2 1

2 naEn µ

h−=

∞= ,,2,1,0 Krn

∞= ,,2,1,0 Kl

∞= K,3,2,1n

0min, =rn 11min,max −=−−= nnn rl

1,,0 −= nKl

J1018.2eV6.132

1820

2

1−×=−=−=

aE

µh

Radial Wavefunction:

€

ψn,l,m r,θ,φ( ) =8 n − l −1( )!

2n a0n( )3(n + l)!e−r / a0n 2r

a0n

l

Ln−l−1(2l+1) 2r

a0n

Yl

m θ,φ( )

)()( )(21

2 ρρρ ααρ

rnLeCu

+−

=

=

0,

2)(

an

rurRn l

,20 na

=λ

= +

−−

+−

0

121

1

0,

22)( 0

an

rL

an

reCrR n

anr

nll

l

l

r=λρ

)()(, ρλρ uRn =l

12 += lα

0

2

na

r=ρ

12

1+=

+l

α

Degeneracy of the nth level:

• We have:

lKll

Kl

K

,,1,

1,,1,0

,3,2,1

+−−=

−=

=

m

n

n

( )∑−

=

+=1

1

12n

ndl

l

2ndn =

)1(21

1

−+= ∑−

=

nn

l

l

( ) )1(12

12 −+−= nnn

2n=

2

eV6.13

nEn −=

Energy Level Diagram

• By taking spin into account (two spin statesper orbital), and using the Pauli principle, thedegeneracies of these levels explain much ofthe structure of the periodic table:– 2, 8, 18, 32,….

E

0

n=1

n=2

n=3n=4

ll=0 l=1 l=2 l=3

1s

2s 2p

3s4s

3p4p

3d4d 4f

(7)(5)(3)(1)

(1)

(4)

(9)(16)

(d)

Hydrogen Spectrum

• We know that a hydrogen atom absorbs andemits light only at specific frequencies

• This is due to the presence of quantumresonances

• There is a resonance associated with eachtransition between two energy levels– Resonance frequencies:

n=2

n=1

n=3n=4

Lymanseries(UV)

ω41ω31

ω21

Balmerseries(visible)

ω42 ω32

ω43

hnn

nn

EE ′′

−=ω