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1
PHYS-505/551��� Τhe hydrogen atom
Lecture-1!!
2
Introduction-a n The study of the hydrogen atom is important
in quantum mechanics because it is the only atom where the Scrhoedinger equation can be exactly solved in the limit where all the interactions, except the electrostatic, between the proton and the electron can be ignored.!
3
Introduction-b!
n The Scrhoedinger equation takes the form:!
−!2
2m∇2ψ +V (r)ψ = Eψ (1.1)
∇2ψ + ε −U r( )( )ψ = 0 (1.2)
ε =
2mE!2
U r( ) = 2mV (r)
!2 V r( ) = − 1
4πε0
e2
r
4
Introduction-c!n Since the interaction Hamiltonian depends only on r, the proper
coordinate system for the study of this problem is the system of spherical coordinates, where:!
n 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
∂r2r + 1r2
1sinθ
∂∂θ
sinθ ∂∂θ
+1
sin2θ∂2
∂φ 2
#$%
&'(
(1.3)
5
Solution of Shroedinger equation-a !
n In solving the Schroedinger equation for the hydrogen atom we must take into account two important conservation principles:!
n The conservation of energy!n The conservation of angular momentum since the
Coulomb force between proton and electron is a central force. !
n The Schroedinger equation is solved with the method of separating variables!
!
6
Solution of Shroedinger equation-b!
n The wavefunctions for the hydrogen electron are given by:!
n As you may see they consist of a radial and an angular part!
ψ nlm r, θ , φ( ) = Rnl (r)radial part!Yl
m θ , φ( )angular part"#$ %$
(1.4)
7
The angular part-a
n The angular part of the wavefunction is given by the so-called spherical harmonics:!
Ylm θ , φ( ) = ε
2l +1( )4π
l − m( )!l + m( )!
Plm cosθ( )eimφ (1.5)
ε =−1( )m
, m ≥ 0
1 m < 0
'()
*)
Pl
m cosθ( ) associated Legendre function !
8
The angular part-b
n The associated Legendre functions polynomials are generated from the Legendre polynomials from the following relations:!
Pl
m x( ) ≡ 1− x2( )m / 2 ddx
#
$%&
'(
m
Pl (x)
Pl x( ) ≡ 1
2l l!ddx
"
#$%
&'
l
x2 −1( )l
9
The angular part-c���some associated Legendre polynomials
P00 = 1 P2
0 =12
3cos2θ − 1( )P1
1 = sinθ P33 = 15sinθ 1 − cos2θ( )
P10 = cosθ P3
2 = 15sin2θ cosθ
P22 = 3sin2θ P3
1 =32
sinθ 5cos2θ − 1( )
P21 = 3sinθ cosθ P3
0 =12
sinθ 5cos3θ − 3cosθ( )
10
The angular part-d
n The spherical harmonics are normalized and orthogonal to each other:!
n The spherical harmonics are eigenfunctions of the square angular momentum operator and of the angular momentum operator along the z-direction!
Yl
m θ , φ( )#$ %&*
Yl 'm'
θ , φ( )#$
%&
0
π
∫ sinθdθdφ0
2π
∫ = δll 'δmm' (1.6)
l2Yl
m = !2l(l +1)Ylm , lz Yl
m = !mYlm (1.7)
11
The angular part-e���the first few spherical harmonics
Y00 =
14π
"#$
%&'
1/2
Y2±2 =
1532π
"#$
%&'
1/2
sin2θe±2 iφ
Y10 =
34π
"#$
%&'
1/2
cosθ Y30 =
716π"#$
%&'
1/2
5cos3θ − 3cosθ( )
Y1±1 = ∓
38π
"#$
%&'
1/2
sinθe± iφ Y3±1 = ∓
2116π"#$
%&'
1/2
sinθ 5cos2θ − 1( )e± iφ
Y20 = 3
516π"#$
%&'
1/2
3cos2θ − 1( ) Y3±2 =
10532π
"#$
%&'
1/2
sin2θ cosθe±2 iφ
Y2±1 = ∓3
158π
"#$
%&'
1/2
sinθ cosθe± iφ Y3±3 = ∓
3564π
"#$
%&'
1/2
sin3θe±3 iφ
12
The angular part-e
n The integer number l is known as azimuthal quantum number and gets the values!
n The integer number m is known as magnetic quantum number and gets the values !
l = 0, 1, 2, ..., ∞
m = −l,............,+l
13
The radial part-a
n 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 ≡
4πε0!2
me2 = 0.529 ×10−10 m
Bohr radius!
Ln− l−1
2l+1 2r / na0( ) associated Laguerre polynomial !
14
The radial part-b n The associated Laguerre polynomials are
generated from the Laguerre polynomials from the following relations:
Lq− p
p x( ) ≡ −1( )p ddx
#
$%&
'(
p
Lq (x)
Lq x( ) ≡ ex d
dx"
#$%
&'
q
e− x xq( )
15
The radial part-c"Some 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
16
The radial part-c"Discussion!
n It can be shown that the radial part of the electrons wavefunction defines a function!
which satisfies the so-called radial equation ! u ≡ rRnl (r) (1.9)
−!2
2md 2udr 2
+ V +!2
2ml l +1( )
r 2
"
#$$
%
&''
effective potential" #$$$ %$$$
u = Eu (1.10)
17
The radial part-c"Discussion!
n The functions u satisfy the following boundary conditions:!
n Thus the radial equation describes an one-dimensional motion where at 0 we have a “wall” and at infinity the wavefunction becomes zero.!
n The radial equation contains the term!which is the so called centrifugal term.! !
u(0) = 0, u(∞) = 0, while 0 < r < ∞
!2l l +1( ) / 2mr 2( )
18
19
The total wavefunctions!n The total wavefunctions for the hydrogen
atom are given by:!
ψ nlm =2
na0
"
#$
%
&'
3 n − l −1( )!2n (n + l)!)* +,
3 e−r / na02rna0
"
#$
%
&'
l
Ln− l−12l+1 2r / na0( ))* +,Yl
m θ ,φ( )
ψ nlm
* ψn' l 'm' r 2
0
π
∫ sinθdrdθdφ0
2π
∫0
∞
∫ = δnn'δ ll 'δmm'
1.11( )
1.12( )
20
The energy spectrum of the hydrogen atom-a!
n The energies of the electron states are given by the following formula:!
n Where E1 is the ground state energy given by!
n The number n is called the principal quantum number. !
En = −
m2!2
e2
4πε0
$
%&'
()
2*
+
,,
-
.
//
1n2 =
E1
n2 , n = 1, 2, 3, ... (1.13)
En = −
m2!2
e2
4πε0
$
%&'
()
2*
+
,,
-
.
//= −13.6 eV
21
The energy spectrum of the hydrogen atom-b!
n One of the most impressive characteristic of the hydrogen atom energy spectrum is its degeneracy.!
n By degeneracy we mean that there can be more than one states with the same energy. This is obvious since the energy does not depend on the numbers l and m.!
n The principal quantum number n imposes the following restriction on the values of the azimuthal quantum number: !
n We can prove that the number of different states that have the same energy is given by !
dn = n2
l = 0, 1, 2, ..., n -1
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