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PHYS-505/551 Τhe hydrogen atom

Lecture-1

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Introduction-a

The study of the hydrogen atom is importantin quantum mechanics because it is the onlyatom where the Scrhoedinger equation can beexactly solved in the limit where all theinteractions, except the electrostatic, betweenthe proton and the electron can be ignored.

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Introduction-b

The Scrhoedinger equation takes the form:

!

h2

2m"2# +V (r)# = E# (1.1)

!2" + # $U r( )( )" = 0 (1.2)

! =

2mEh2

U r( ) = 2mV (r)

h2 V r( ) = !

14"#0

e2

r

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Introduction-c Since the interaction Hamiltonian depends only on r, the proper

coordinate system for the study of this problem is the system ofspherical coordinates, where:

Since the mass of the proton is much larger than the electron’s,the proton has been considered as a heavy motionless particle.

!2 =

1r"2

"r 2 +1r 2

1sin#

""#

sin# ""#

+1

sin2#"2

"$ 2

%&'

('

)*'

+' (1.3)

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Solution of Shroedinger equation-a

In solving the Schroedinger equation for thehydrogen atom we must take into account twoimportant conservation principles:

The conservation of energy The conservation of angular momentum since the

Coulomb force between proton and electron is acentral force.

The Schroedinger equation is solved with the methodof separating variables

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Solution of Shroedinger equation-b

The wavefunctions for the hydrogenelectron are given by:

As you may see they consist of a radialand an angular part

! nlm r, " , #( ) = Rnl (r)radial part{

Ylm " , #( )

angular part1 24 34

(1.4)

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The angular part-a

The angular part of the wavefunction is given bythe so-called spherical harmonics:

Ylm ! , "( ) = #

2l +1( )4$

l % m( )!l + m( )!Pl

m cos!( )eim" (1.5)

# = %1( )m, m & 0

1 m < 0

'()

*)

Pl

m cos!( ) associated Legendre function

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The angular part-b

The associated Legendre functionspolynomials are generated from theLegendre polynomials from the followingrelations:

Pl

m x( ) ! 1" x2( )m / 2 ddx

#$%

&'(

m

Pl (x)

Pl x( ) ! 1

2l l!ddx

"#$

%&'

l

x2 (1( )l

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The angular part-csome associated Legendre polynomials

P00 = 1 P2

0 =12

3cos2! " 1( )P1

1 = sin! P33 = 15sin! 1 " cos2!( )

P10 = cos! P3

2 = 15sin2! cos!

P22 = 3sin2! P3

1 =32

sin! 5cos2! " 1( )P2

1 = 3sin! cos! P30 =

12

sin! 5cos3! " 3cos!( )

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The angular part-d

The spherical harmonics are normalized andorthogonal to each other:

The spherical harmonics are eigenfunctions of thesquare angular momentum operator and of theangular momentum operator along the z-direction

Yl

m ! , "( )#$ %&*

Yl 'm'

! , "( )#$

%&

0

'

( sin!d!d"0

2'

( = )ll ')mm' (1.6)

l2Yl

m = h2l(l +1)Ylm , lz Yl

m = hmYlm (1.7)

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The angular part-ethe first few spherical harmonics

Y00 =

14!

"#$

%&'

1/2

Y2±2 =

1532!

"#$

%&'

1/2

sin2(e±2 i)

Y10 =

34!

"#$

%&'

1/2

cos( Y30 =

716!

"#$

%&'

1/2

5cos3( * 3cos(( )

Y1±1 = m

38!

"#$

%&'

1/2

sin(e± i) Y3±1 = m

2116!

"#$

%&'

1/2

sin( 5cos2( * 1( )e± i)

Y20 = 3

516!

"#$

%&'

1/2

3cos2( * 1( ) Y3±2 =

10532!

"#$

%&'

1/2

sin2( cos(e±2 i)

Y2±1 = m3

158!

"#$

%&'

1/2

sin( cos(e± i) Y3±3 = m

3564!

"#$

%&'

1/2

sin3(e±3 i)

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The angular part-e

The integer number l is known as azimuthalquantum number and gets the values

The integer number m is known as magneticquantum number and gets the values

l = 0, 1, 2, ..., !

m = !l,............,+l

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The radial part-a

The radial part of the solution is given by:

Rnl (r) = Ne!r / na0

2rna0

"

#$%

&'

l

Ln! l!12l+1 2r / na0( )() *+ (1.8)

N =2

na0

!

"#$

%&

3 n ' l '1( )!2n (n + l)!() *+

3 a0 !

4"#0h2

me2 = 0.529 $10%10 m

Bohr radius

Ln! l!1

2l+1 2r / na0( ) associated Laguerre polynomial

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The radial part-b The associated Laguerre polynomials are

generated from the Laguerre polynomialsfrom the following relations:

Lq! p

p x( ) " !1( )p ddx

#$%

&'(

p

Lq (x)

Lq x( ) ! ex d

dx"#$

%&'

q

e( x xq( )

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The radial part-cSome associated Laguerre polynomials

L00 = 1 L0

2 = 2L1

0 = !x + 1 L12 = !6x + 18

L20 = x2 ! 4x + 2 L2

2 = 12x2 ! 96x + 144L0

1 = 1 L03 = 6

L11 = !2x + 4 L1

3 = !24x + 96L2

1 = 3x2 ! 18x + 18 L23 = 60x2 ! 600x + 1200

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The radial part-cDiscussion

It can be shown that the radial part of theelectrons wavefunction defines a function

which satisfies the so-called radial equation u ! rRnl (r) (1.9)

!h2

2md 2udr 2

+ V +h2

2ml l +1( )

r 2

"

#$$

%

&''

effective potential1 2444 3444

u = Eu (1.10)

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The radial part-cDiscussion

The functions u satisfy the following boundaryconditions:

Thus the radial equation describes an one-dimensional motion where at 0 we have a “wall”and at infinity the wavefunction becomes zero.

The radial equation contains the termwhich is the so called centrifugal term.

u(0) = 0, u(!) = 0, while 0 < r < !

h2l l +1( ) / 2mr 2( )

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The total wavefunctions The total wavefunctions for the hydrogen

atom are given by:

! nlm =2

na0

"

#$%

&'

3 n ( l (1( )!2n (n + l)!)* +,

3 e(r / na02rna0

"

#$%

&'

l

Ln( l(12l+1 2r / na0( ))* +,Yl

m - ,.( )

! nlm

* !n' l 'm' r 2

0

"

# sin$drd$d%0

2"

#0

&

# = 'nn'' ll ''mm'

1.11( )

1.12( )

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The energy spectrum of thehydrogen atom-a

The energies of the electron states are given by thefollowing formula:

Where E1 is the ground state energy given by

The number n is called the principal quantumnumber.

En = !

m2h2

e2

4"#0

$%&

'()

2*

+,,

-

.//

1n2 =

E1

n2 , n = 1, 2, 3, ... (1.13)

En = !

m2h2

e2

4"#0

$%&

'()

2*

+,,

-

.//= !13.6 eV

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The energy spectrum of thehydrogen atom-b

One of the most impressive characteristic of thehydrogen atom energy spectrum is its degeneracy.

By degeneracy we mean that there can be more thanone states with the same energy. This is obvious sincethe energy does not depend on the numbers l and m.

The principal quantum number n imposes thefollowing restriction on the values of the azimuthalquantum number:

We can prove that the number of different states thathave the same energy is given by

dn = n2

l = 0, 1, 2, ..., n -1