Quantum Mechanics and the hydrogen atom
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:
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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)
x = r sin θ cos φ
y = r sin θ sin φ
z = r cosθ
Rewrite all derivatives in (r,θ,φ), gives Schrödinger equation;
h2 ⎛ ∂ 2 ∂ ⎞
h 2 ⎛⎜ 1 ∂
∂
1 ∂ 2 ⎞⎟
Ψ + V ( r ) Ψ = EΨ
+
−
sin θ
⎜ r
⎟Ψ −
∂θ sin 2 θ ∂φ 2 ⎟⎠
2m ⎝ ∂r ∂r ⎠
2m ⎜⎝ sin θ ∂θ
This is a partial differential equation, with 3 coordinates (derivatives);
Use again the method of separation of variables:
Ψ (r ,θ ,φ ) = R(r )Y (θ ,φ )
Bring r-dependence to left and angular dependence to right (divide by Ψ):
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QM
Oθφ
Y (θ ,φ )
⎤
1 ⎡ d 2 dR 2mr 2
r
E
V
r
R
(
(
)
)
=
−
=λ
+
−
⎥
⎢
2
R ⎣ dr dr
Y (θ ,φ )
h
⎦
Separation of variables
Where does the quantisation in QM come from ?
⎤
1 ⎡ d 2 dR 2mr 2
(
(
)
)
r
+
E
−
V
r
R
⎥=λ
⎢
2
R ⎣ dr dr
h
⎦
Radial equation
Angular equation
2 ⎞
⎛ 1 ∂
∂
1
∂
⎟Y (θ ,φ )
− ⎜⎜
sin θ
+ 2
QM
2⎟
Oθφ
Y (θ ,φ )
sin
θ
∂
θ
∂
θ
sin θ ∂φ ⎠
=λ
−
= ⎝
Y (θ ,φ )
Y (θ ,φ )
−
Once more separation of variables:
Derive:
∂ 2Y
∂φ 2
= sin θ
Y (θ ,φ ) = Θ(θ )Φ(φ )
1 ∂ 2Φ 1 ⎛
∂
∂Θ
⎞
2
2
−
=
sin
θ
sin
θ
+
λ
sin
θ
Θ
=
m
⎜
⎟
Φ ∂φ 2 Θ ⎝
∂θ
∂θ
⎠
Simplest of the three: the azimuthal angle;
∂ 2Φ(φ )
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∂
∂Y
sin θ
+ λ sin 2 θY
∂θ
∂θ
∂φ
2
+ m 2Φ(φ ) = 0
(again arbitrary constant)
Where does the quantisation in QM come from ?
A differential equation with a boundary condition
∂ 2Φ(φ )
∂φ
2
+ m 2Φ(φ ) = 0
and
Φ(φ + 2π ) = Φ (φ )
Solutions:
Φ(φ ) = eimφ
Boundary condition;
Φ(φ + 2π ) = eim(φ +2π ) = Φ(φ ) = eimφ
e 2πim = 1
m is a positive or negative integer
this is a quantisation condition
General: differential equation plus a boundary condition gives a quantisation
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Where does the quantisation in QM come from ?
Φ(φ ) = e
First coordinate
Second coordinate
with integer m
(positive and negative)
imφ
m 2 ⎞⎟
∂Θ ⎛⎜
1 ∂
sin θ
+ λ − 2 ⎟Θ = 0
sin θ ∂θ
∂θ ⎜⎝
sin θ ⎠
λl = l(l + 1)
Results in
with
and
l = 0,1,2,K
angular
momentum
m = −l,−l + 1,K, l − 1, l
⎤
1 ⎡ d 2 dR 2mr 2
r
+
(
E
−
V
(
r
)
)
R
⎢
⎥ = l(l + 1)
2
R ⎣ dr dr
h
⎦
Third coordinate
Differential
equation
Results in quantisation of energy
2
Z 2 ⎛⎜ e 2 ⎞⎟ me
En = − 2 R∞ = − 2 ⎜
n ⎝ 4πε 0 ⎟⎠ 2h 2
n
Z2
with integer n (n>0)
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angular
part
radial
part
Angular wave functions
Operators:
h⎡ 1 ∂
∂
1 ∂2 ⎤
L = ⎢
+ 2
sin θ
⎥
i ⎣ sin θ ∂θ
∂θ sin θ ∂φ 2 ⎦
h ∂
Lz =
i ∂φ
2
Angular momentum
r
L = ( Lx , L y , Lz )
There is a class of functions that are simultaneous eigenfunctions
LzYlm (θ ,φ ) = mhYlm (θ ,φ )
L2Ylm (θ ,φ ) = l(l + 1)h 2Ylm (θ ,φ )
with
l = 0,1,2,K
Spherical harmonics (Bolfuncties)
Y00 =
1
4π
Ylm (θ ,φ )
m = −l,−l + 1,K, l − 1, l
Vector space of solutions
2
∫ Ylm (θ ,φ ) dΩ = 1
3
Y11 = −
sin θeiφ
8π
Y10 = −
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and
Y1,−1 =
3
cosθ
4π
3
sin θe −iφ
8π
Ω
∫ YlmYl 'm'dΩ = δ ll 'δ mm'
*
Ω
Parity
Pop Υlm (θ , φ ) = Υ (π − θ , φ + π ) = (− )l Υlm (θ , φ )
The radial part: finding the ground state
⎤
1 ⎡ d 2 dR 2mr 2
+ 2 (E − V (r ))R ⎥ = λ
⎢ r
R ⎣ dr dr
h
⎦
Find a solution for
l = 0, m = 0
h2 ⎛
2 ⎞ Ze 2
−
R = ER
⎜ R"+ R ' ⎟ −
2m ⎝
r ⎠ 4πε 0 r
Physical intuition; no density for r → ∞
h2
Ze 2
−
=0
ma 4πε 0
Solution for the
length scale paramater
a0 =
4πε 0 h 2
Ze 2 m
Bohr radius
R (r ) = Ae −r / a
trial:
A
R
R ' = − e −r / a = −
a
a
A
R
R" = 2 e −r / a = 2
a
a
−
2 ⎞ Ze 2
h2 ⎛ 1
=E
⎜ 2 − ⎟−
ar ⎠ 4πε 0 r
2m ⎝ a
must hold for all values of r
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Prefactor for 1/r:
Solutions for the energy
2
2 ⎞
e
h2
2⎛
⎟ me
E=−
= − Z ⎜⎜
2
⎟
2ma
⎝ 4πε 0 ⎠ 2h
Ground state in the
Bohr model (n=1)
Hydrogen Atom:
Schrödinger Equation and Quantum Numbers
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 l gives the angular
momentum; l can take on integer values from 0 to n - 1.
3. The magnetic quantum number, m , gives the l
direction of the electron’s angular momentum, and can
take on integer values from –l to +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 dV
around a given point is then |ψ|2 dV.
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Radial Probability Distributions
Spherical shell of thickness dr, inner radius r
and 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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Hydrogen Atom Wave Functions
This figure shows the three probability distributions for n = 2 and l
= 1 (the distributions for m = +1 and m = -1 are the same), as well
as the radial distribution for all n = 2 states.
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Hydrogen “Orbitals” (electron clouds)
r 2
Ψ (r ,t )
Represents the probability to find a particle
At a location r at a time t
The probability density
The probability distribution
Max Born
The Nobel Prize in Physics 1954
"for his fundamental research in
quantum mechanics, especially for his
statistical interpretation of the wavefunction"
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Atomic Hydrogen Radial part
Analysis of radial equation yields:
Enlm = −
with
R∞ =
Z2
n
2
R∞
me e 4
8ε 0 h 3c
Wave functions:
r
Ψnlm (r , t ) = Rnl (r )Υlm (θ , φ )e −iEn t / h
For numerical use:
u (r )
R = nl
r
ρ / r = 2 Z / na
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unl (ρ ) =
2Z
na0
a = 4πε 0h 2 / μe 2
(n − l − 1)!e − Zρ / n ⎛ 2Zρ ⎞l +1 L2l +1 ⎛ 2Zρ ⎞
⎜
⎟
⎟
n − l −1⎜
n
n
2n(n + 1)!
⎝
⎠
⎝
⎠
Quantum analog of electromagnetic radiation
Classical electric dipole radiation
Transition dipole moment
Quantum jump
Classical oscillator
I rad
r& 2
&
∝ er
I rad ∝
2
* r
∫ψ 1 erψ 2 dτ
The atom does not radiate when it is in a stationary state !
The atom has no dipole moment
μii = ∫
*r
ψ 1 r ψ 1dτ
=0
Intensity of spectral lines linked
to Einstein coefficient for absorption:
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Bif =
μ fi
2
6ε 0h 2
Selection rules
Mathematical background: function of odd parity gives 0 when integrated over space
∞
Ψ f x Ψi =
In one dimension:
∞
∫
Ψ *f xΨi dx
−∞
=
∫ f ( x)dx
f ( x) = Ψ *f xΨi
with
−∞
∞
0
∞
0
∞
∞
∞
−∞
−∞
0
∞
0
0
0
∫ f ( x)dx = ∫ f ( x)dx + ∫ f ( x)dx = ∫ f (− x)d (− x ) + ∫ f ( x)dx = ∫ f (− x)dx + ∫ f ( x)dx
∞
= 2 ∫ f ( x)dx ≠ 0
if
Ψi
f ( − x) = f ( x)
and Ψ f
opposite parity
0
=0
if
f (− x) = − f ( x)
Ψi
and Ψ f
Electric dipole radiation connects states of opposite parity !
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same parity
Selection rules
depend on angular behavior of the wave functions
Parity operator
r
r
All quantum mechanical wave functions
have a definite parity
r
r
Ψ (− r ) = ± Ψ (r )
θ
φ
r
−r
r
r
Pr = −r
( x , y , z ) → ( − x, − y , − z )
(r ,θ , φ ) → (r , π − θ , φ + π )
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r
Ψ f r Ψi ≠ 0
If
Ψf
and
Rule about the
Ψi
have opposite parity
Ylm
functions
PYlm (θ , φ ) = (− )l Ylm (θ , φ )
Hydrogen Atom:
Schrödinger Equation and Quantum Numbers
“Allowed” transitions between energy levels occur between states
whose value of l differ by one:
Other, “forbidden,” transitions also occur but with much lower
probability.
“selection rules, related to symmetry”
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Selection rules in Hydrogen atom
Intensity of spectral lines given by
r
r
μ fi = ∫ Ψ*f μΨi = Ψ f − er Ψi
n
1) Quantum number
no restrictions
2) Parity rule for
Balmer series
l
Δl = odd
3) Laporte rule for
l
Angular momentum rule:
r
r r
l f = li + 1
From 2. and 3.
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so
Δl ≤ 1
Δl = ±1
Lyman series