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The Hydrogen Atom
• This is of course the most important model system for
understanding electronic structure.
• New wrinkle: we’re dealing with a 3D system, instead of
the 1D model systems from previous lecture.
−h ∂
e
ˆ
H≠
−
2
2m ∂r
r
2
2
2
• The strategy is to exploit the spherical symmetry of the
problem, which allows us to decouple the radial and
angular parts of the problem. The angular parts lead to
angular momentum eigenstates.
The Hamiltonian
Recall from our previous lecture that the 1D Hamiltonian operator is
h2 ∂2
−
+ U (x )
2
2m ∂x
By analogy, we can imagine the form of the Hamiltonian for an
electron moving around a proton (i.e., H atom):
h2 ⎛ ∂2
∂2
∂ 2 ⎞ e2
⎜⎜ 2 + 2 + 2 ⎟⎟ −
−
2me ⎝ ∂x
∂y
∂z ⎠ r
Kinetic energy term, now in 3D
Potential energy
where r is the distance between the proton and electron. Actually, this
expression would be correct if the proton were infinitely heavy relative to the
electron, but that’s not quite the case; the correct expression is ...
∂2
∂ 2 ⎞ e2
h2 ⎛ ∂2
h 2 2 e2
⎜⎜ 2 + 2 + 2 ⎟⎟ − = −
∇ −
−
2 µ ⎝ ∂x
∂y
∂z ⎠ r
2µ
r
where µ is the “reduced mass” of the system
1
1
1
=
+
µ m p me
which is close to, but not exactly equal to, the electron mass. The origin of our
coordinate system is now at the center of mass between the electron and
proton, which of course will stay quite close to the proton.
The key to solving the hydrogen atom is to take advantage of the spherical
symmetry, i.e., convert to radial coordinates (r,θ,φ). The potential part of the
Hamiltonian is already in radial form, so it’s just a matter of getting the kinetic
energy operator into the radial coordinates. This is a standard exercise in
beginning QM classes, but it is really quite tedious.
The final result is
2
2
2
⎡
⎤
∂
∂
∂
∂
2
1
1
∂
2
∇ = 2+
+ 2 ⎢ 2 + cot θ
+
⎥
∂r
∂θ sin 2 θ ∂φ 2 ⎦
r ∂r r ⎣ ∂θ
Plugging this into the Hamiltonian yields:
2
2
2
2
⎛ ∂2
⎛
⎞
∂
∂
∂
2
e
1
h
h
⎜
⎜⎜ 2 +
⎟⎟ − −
+
+
cot
Hˆ = −
θ
∂θ sin 2 θ
2 µ ⎝ ∂r
r ∂r ⎠ r 2 µr 2 ⎜⎝ ∂θ 2
Purely radial term
⎛ ∂2 ⎞⎞
⎜⎜ 2 ⎟⎟ ⎟
⎟
⎝ ∂φ ⎠ ⎠
Purely angular term
The angular part is often referred to as the “total angular momentum”
operator, and gets a special symbol:
2
2
2
ˆ2
⎞
⎛
2
e
L
h
∂
∂
⎟⎟ − +
⎜⎜ 2 +
Hˆ = −
2 µ ⎝ ∂r
r ∂r ⎠ r 2 µr 2
This separation of variables implies that we can solve independently for the
radial and angular eigenfunctions. Neither one of these are trivial derivations,
but they are both analytical.
Angular Part
The potential does not depend on the angles; i.e., there is “free rotation”.
These eigenfunctions are the so-called spherical harmonics (maybe think of
them as the normal modes of a spherical drum, like a balloon). They are
described by 2 quantum numbers:
l=0,1,2,3,..., for the θ angle
m=0,±1,±2,...,±l for the φ angle
and the eigenfunctions take the form
Yl m (θ , φ ) = f l (θ )e imϕ
with eigenenergies
E = h 2l (l + 1)
Now we’re dealing with
complex-valued
eigenfunctions ...
Spherical Harmonics
l=1, m=-1,0,+1 (“p”)
l=0, m=0 (“s”)
1
0
Y0 =
4π
3
Y =
cos θ
4π
0
1
3
pz =
cos θ
4π
+1
1
3
=
sin θeiϕ
8π
3
px =
sin θ cos ϕ
8π
−1
1
3
=
sin θe −iϕ
8π
3
py =
sin θ sin ϕ
8π
Y
Y
d orbitals
l=2, m=-2,-1,0,+1,+2
(
)
5
Y =
3 cos 2 θ − 1
16π
0
2
+1
2
Y
15
=
sin θ cos θeiϕ
8π
Y2−1 =
15
sin θ cos θe −iϕ
8π
+2
2
15
=
sin 2 θe 2iϕ
32π
−2
2
15
=
sin 2 θe − 2iϕ
32π
Y
Y
Knowing the Angular Eigenfunctions,
We Can Now Do Radial Part
2
2
2
ˆ2
⎞
⎛
2
e
L
h
∂
∂
⎟⎟ − +
⎜⎜ 2 +
Hˆ = −
2 µ ⎝ ∂r
r ∂r ⎠ r 2 µr 2
ψ (r , θ , ϕ ) = ψ r (r )Yl m (θ , ϕ )
Lˆ2ψ (r , θ , ϕ ) = ψ r (r )Lˆ2Yl m (θ , ϕ )
= ψ r (r )h 2l (l + 1)Yl m (θ , ϕ )
= h 2l (l + 1)ψ (r , θ , ϕ )
⎡ h 2 ⎛ ∂ 2 2 ∂ ⎞ e 2 h 2l (l + 1) ⎤
⎜⎜ 2 +
⎟⎟ − +
Hˆ ψ (r , θ , ϕ ) = ⎢−
⎥ψ r (r )
2
r ∂r ⎠ r
2 µr ⎦
⎣ 2 µ ⎝ ∂r
“Effective potential”, incorporating
a “centrifugal force”. Solve for
each value of l.
purely
radial!
Centrifugal Force in H-atom
l > 0 (p, d, f, etc.)
l = 0 (s)
• For l=0, wavefunction is always a maximum at r = 0.
• For l>0, wavefunction is always zero at r = 0.
Radial eigenfunctions of H atom
The eigenenergies are simple, and do not depend on l ...
C
E=− 2
n
where C is a constant that involves the reduced mass, electron
charge, and Planck’s constant.
The eigenfunctions are pretty awful, and do depend on l ...
Rnl (r ) = −Cnl r e
l − r / na0
2 l +1
n +l
L
⎛ 2r ⎞
⎜⎜
⎟⎟
⎝ na0 ⎠
where Cnl is a complicated “normalization constant”, and L are the
“associated Laguerre functions”, and a0 is the “Bohr radius”, a
constant.
Ground state of H atom
3
2
1 ⎛ 1 ⎞ − r / a0
⎜⎜ ⎟⎟ e
R00 (r ) =
π ⎝ a0 ⎠
ψ (r ) [amplitude]
r 2ψ 2 (r ) [probability]
Amplitude at r=0 is large, but probability of finding system there is zero!
Some more H atom
wavefunctions
1. For constant l, number of
nodes along r increases
with increasing n.
2. For l=n, all nodes are in the
angular coordinates (2p,
3d, etc.).
Only a very few, very simple QM systems
can be solved analytically
The ones we’ve just looked at are analytical, or at least quasi-analytical, as is
the hydrogen atom, which we will review in the next lecture. But that’s about it
for analytical solutions. Everything else involves some numerical computation.
The indispensable tools for modern quantum mechanics are perturbation
theory and the variational principle.
Perturbation theory
• Relates the solutions of one QM problem to another that has already been
solved (“reference system”).
• Works well when the 2 QM systems are similar to each other, i.e., the one
you are trying to solve represents a relatively small “perturbation” relative
to the reference system.
• For example, the eigenstates of the anharmonic (Morse) oscillator can be
solved using the harmonic oscillator as the reference system. Works well
for the first few eigenstates, but not near dissociation.
Basics of Perturbation Theory
Hˆ = Hˆ ° + Hˆ ′
Ĥ °ψ ° = E °ψ °
ψ = ψ ° +ψ ′
Divide the Hamiltonian into two parts: A “zeroorder” Hamiltonian whose eigenstates you know,
and a “perturbation”.
E = E° + E′
Write the eigensolutions of the full system
as sum of the zero-order solutions and a
perturbation.
Hˆ ψ = Eψ
these Hˆ °ψ ° + Hˆ ′ψ ° + Hˆ °ψ ′ + Hˆ ′ψ ′ =
terms
E °ψ ° + E ′ψ ° + E °ψ ′ + E ′ψ ′
equal
Hˆ ′ψ ° + Hˆ °ψ ′ ≈ E ′ψ ° + E °ψ ′
2nd
order
terms
Substitute into
Schrodinger eqn.
and simplify.
this term is
exactly zero
∫ψ °Hˆ ′ψ ° + ∫ψ °Hˆ °ψ ′ ≈ ∫ψ °E ′ψ ° + ∫ψ °E°ψ ′
ˆ ′ψ ° + ψ °(Hˆ ° − E °)ψ ′ ≈ E ′ ψ °ψ ° integrated probability is 1
°
ψ
H
∫
∫
∫
ˆ ′ψ ° ≈ E ′
ψ
°
H
∫
More Advanced
Perturbation Theory
0
ˆ
′
∆En = ∫ψ H ψ n + ∑
0
n
E −E
0
n
j≠n
ψ n =ψ + ∑
0
n
ˆ ′ψ
ψ
H
∫
0
j
j ≠n
0
ˆ
′
∫ψ H ψ n
0
j
E −E
0
n
0
j
+ ...
0
n
2
2nd order term
0
j
perturbation theory
can be applied to
eigenfunctions as well
as eigenenergies
Variational principle
• A very important theorem in modern quantum mechanics.
• Provides an upper bound to the eigenenergies.
• The basic strategy is to “guess” a wavefunction and obtain its energy
using your Hamiltonian operator. This energy will always be larger than
the true eigenenergy. Moreover, the energy will approach the true energy
as the trial wavefunction becomes more like the true wavefunction.
E0 =
ˆψ
ψ
H
0
0
∫
∫ψ ψ
0
•
Eφ =
ˆφ
φ
H
∫
0
∫ φφ
Eφ ≥ E0
E
The usual approach is to define the trial wavefunction as a linear
superposition of multiple “basis functions” with linear “variational
parameters”, i.e.,
N
ϕ = ∑ cn f n
n =1
•
It is fairly easy to show that the energy will reach a minimum when the
determinant of this matrix is zero:
Simple Example: Trial Wavefunction
with 2 Basis Functions
ϕ = c1 f1 + c2 f 2
Eφ =
ˆφ
H
φ
∫
∫ φφ
(c f + c f )Hˆ (c f + c f )
∫
=
∫ (c f + c f )(c f + c f )
c ∫ f Hˆ f +c c ∫ f Hˆ f +c c ∫ f Hˆ f +c ∫ f Hˆ f
=
c ∫ f f +c c ∫ f f +c c ∫ f f +c ∫ f f
1 1
2 2
1 1
2
1
1
2
1
1 1
2 2
1
1 1
1 2
1 2
2 2
1 1
1
2 2
2
1 2
1 2
1 2
2
2 1
c12 H11 + c1c2 H12 + c1c2 H 21 + c22 H 22
= 2
c1 S11 + c1c2 S12 + c1c2 S 21 + c22 S 22
2
2
1
2
2
2
2 2
2
c12 H11 + c1c2 H12 + c1c2 H 21 + c22 H 22
Eφ = 2
c1 S11 + c1c2 S12 + c1c2 S 21 + c22 S 22
Taking advantage of the variational principle, we aim to minimize this expression
with respect to the “variational parameters” c1 and c2, i.e.,
∂Eφ
∂c1
∂Eφ
∂c2
=0
=0
These two expressions lead to the following linear equations:
c1 (H11 − ES11 ) + c2 (H 21 − ES 21 ) = 0
c1 (H12 − ES12 ) + c2 (H 22 − ES 22 ) = 0
This now becomes a standard problem in linear algebra, which can be
expressed as a matrix problem.
Matrix Formulation of Quantum
Eigenvalue Problem
H11 − ES11
H12 − ES12
H 21 − ES 21
H 22 − ES 22 L = 0
M
M
L
O
If basis set is chosen to be “orthonormal”, i.e.,
H11 − E
H 21
M
H12
L
H 22 − E L = 0
M
O
H ij = ∫ f i Hˆ f j
Sij = ∫ f i f j
Sij = δ ij
Except for very simple
problems, solving these
equations (essentially root
finding of polynomial equation)
is handled numerically (e.g.,
Householder diagonalization).