Hydrogen Atom: Schrödinger Equation and Quantum Numbers. Potential energy
for the hydrogen atom: The time-independent Schrödinger equation in three.
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. MNW-L2
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 MNW-L2
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 MNW-L2
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: MNW-L2
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 ) MNW-L2
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” MNW-L2
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
