6
6.1
Atomic Structure
Variational principle
It’s a useful trick to compute ground state energy and wave function of
a system which is hard to solve exactly, by making a guess! Particularly
ground state energies are given quite accurately if your guess contains a
little intuition. . .
Let |ni, En be complete set of e’vectors and e’values fot H, and choose
any “trial state” |φi which is a guess for the ground state. Completeness
=⇒
|φi =
X
n
cn|ni
(1)
and now construct
hφ|H|φi
=
hφ|φi
≥
P
2
n |cn | En
P
2
n |cn |
P
2
n |cn | E0
P
2
n |cn |
(2)
= E0
(3)
since E0 is smaller than all other En by definition and |cn|2 ≥ 0. Thus
?
hφ|H|φi
hφ|φi
is a rigorous upper bound to the
ground state energy for any |φi
(4)
Game is then to choose for |φi a form which is reasonable and contains
a parameter (or more than one) which may be varied to minimize this
upper bound.
2
2
Example: ground state of 3D SHO. We know exact answer, ψ ∝ e−r /2x0
where x0 = (h̄/mω)1/2. Suppose we didn’t, might guess solution was
spherically symmetric and decayed exponentially for large distances. So
we’d choose trial wave function
|φi = e−αr
1
(5)
Now we need to calculate hφ|φi and hφ|H|φi, where H = p̂2/2m +
mω 2r2/2. First
hφ|φi =
Z
d3r|φ(r)|2
Z ∞
= 4π 0 r2dr e−2αr
π
4π Z ∞ 2
−x
=
x
dxe
=
.
{z
}
8α3 | 0
α3
2
(6)
For the next calculation we will need result ∇φ(r) = −α rr e−αr ,
giving
hφ|∇2|φi = −h∇φ|∇φi
Z
2
= α · 4π r2dr e−2αr
π
= −
α
(7)
so
p̂2
πh̄2
hφ| |φi =
2m
2mα
(8)
and finally,
2
hφ|r |φi = 4π
3π
= 5
α
Z ∞
0
r2dr r2e−2αr
(9)
So variational ground state energy for fixed α is
hφ|H|φi
E(α) =
=
hφ|φi
3πω 2
2α5
2
πh̄
+ 2mα
π
α3
h̄2 2 3 mω 2
=
α +
2m
2 α2
Extremum given by
2
(10)
∂E
h̄2
3 mω 2
=0 =
−
∂α2
2m 2 α4
3m2ω 2 1/4
=⇒ α = ( 2 ) = 31/4/x0
h̄
(11)
so
E(αmin) =
√
3h̄ω ' 1.732h̄ω
(12)
to be compared to exact answer of E0 = (3/2)h̄ω. Not bad, particularly
1/4
when you think that the wave fctn we chose, e−3 r/x0 , is exponentially
2
2
large compared to the true wave fctn, e−r /2x0 at large distances. This is
because for short distances r, where the expectation value is peaked, the
contributions to the energy are of order ²2 if error in wave function is ².
6.2
Helium ground state
Helium has atomic number (# protons) Z = 2, therefore 2 electrons, and 2
neutrons to boot. Assuming nucleus is nailed down at center of coordinate
system, Hamiltonian is
p22
2e2 2e2 e2
p21
+
−
−
+
(13)
H=
2m 2m
r1
r2
r12
where r1 and r2 are the distances of the 2 electrons from the nucleus, and
r12 is the distance between electrons. Take trial wave fctn.
φ = e−α(r1+r2 )
(14)
i.e. ind. of angles, so certainly an ` = 0 state, which we expect exact
ground state to be. From SHO problem in last section already know
normalizing factor
Z
Z
π2
3
−2αr2
3
−2αr1
= 6
· d r2 e
(15)
hφ|φi = d r1e
α
3
and
p̂21
p̂22
π πh̄2
hφ| |φi = hφ| |φi = 3 ·
2m
2m
α 2mα
Z
1
1
π
e−2αr
2
hφ| |φi = hφ| |φi = 3 · 4π r dr
r1
r2
α
r
π2
= 5
α
(16)
(17)
(18)
and
e−2α(r1+r2)
d r1d r2
r12
1 Z 3 3 e−x1−x2
=
d x1 d x2
32α5
x12
1
hφ| |φi =
r12
Z
3
3
(19)
(20)
where r = x/2α. Integrals look messy, six variables all coupled because of
R
x12 = |x1 − x2|. Use cute trick: note that d3x2 e−x2/x12 is potential of
charge distribution ρ(x2) = e−x2 , evaluated at point at radius x1. Electric
R
field is E(x1) = 0x1 d3x2 e−x2 /x1 by Gauss’s law. Potential is
dx0 Z x0 3
V (x1) = − ∞ 02 0 d x2 e−x2
x
Z ∞ dx0 Z x
Z ∞ dx0 Z x0
1 3
−x2
= x 02 0 d x2 e + x 02 x d3x2 e−x2
1 x
1 x
1
4π
=
[2 − e−x1 (2 + x1)]
x1
Z x
1
(21)
(22)
(23)
So
1
16π 2 Z ∞
hφ| |φi =
dx1 x1 e−x[2 − e−x1 (2 + x1)]
5
0
r12
32α
 
π 2  5  5π 2
=
= 5
2α5 4
8α
So variational energy
4
(24)
(25)

E0var

α6  π 2h̄2 4π 2e2 5π 2e2 

= 2
−
+
π mα4
α5
8α5
h̄2α2 27 2
=
− eα
m
8
(26)
(27)
Minimize:
2h̄2α 27 2
27e2m
dE0var
=0=
− e,
α=
dα
m
8
16h̄2
so at minimum variational energy is
2

4
me
27
var
.
E0 (αmin) = −  
16
h̄2
(28)
(29)
Recall 1 Ryd = (me4)/(2h̄2), so estimate is −(27/16)2 · 2 · 1 Ryd '
−77.5 eV . Measured value is −79.0 eV–lower, of course.
6.3
He excited states
How can we use var. princ. to say anything about excited states? Theorem
explicitly says E0var ≥ E0, ground state energy. Point is if [Ô, H] = 0 for
some operator Ô, can choose eigenstates of H to be simul. e’states of Ô,
so IF we choose trial state |φi to be e’state of Ô as well, proof of var.
princ. can be run through again, this time note that sum over all e’states
|ni of H will run only over those n which have same Ô e’value! (i.e., other
cn’s are 0). So theorem will read
hφ|H|φi
≥ lowest energy corresponding to state with same value O as |φi
(30)
hφ|φi
For atomic problems, know angular momentum always good q.-no. in
absence of ext. fields, so we can take |φi to have, e.g. ` = 1, and find
following special case of theorem:
hφ|H|φi
≥ lowest energy corresponding to state with ` = 1
(31)
hφ|φi
5
So let’s pick trial state, but do it intelligently. First note that Hamiltonian has a new symmetry arising since there are 2 sets of coordinates,
ri, Si, pi...i = 1, 2 for 2 electrons. Note Hamiltonian invariant under exchange operator P12 defined such that P12ψ(1, 2) = ψ(2, 1), where (1, 2)
short for (r1, S1, ...r2, S2...). In other words, we have [P12, H] = 0 so can
classify eigenstates as even or odd under exchange (in complete analogy
with our discussion1 of parity Π).
Full exchange operator P12 exchanges all labels, but since current H
doesn’t depend on spin, also invariant under the more limited symmetry
Pr1r2 where we exchange only spatial coordinates. Pauli principle (see
below) says can only consider 2-particle states overall antisymmetric under P12, P12ψ(1, 2) = ψ(2, 1) = −ψ(1, 2). However can always consider
functions both symmetric and antisymmetric under Pr1r2 , which also commutes with H.
In absence of Coulomb interaction e2/r12 we would have two decoupled
Hydrogen problems, so the ` = 1 states would consist of one e− in a
hydrogenic ` = 0 and one in ` = 1. Since we want low-energy states, pick
a 1s state φ1s ∼ e−αr and a 2p state,2 φ2p ∼ ze−βr . Let’s use this as
guide to construct trial spatial3 wave fctns., but make them eigenfctns of
Pr1r2 :
φ±(r1, r2) = φ1s(r1)φ2p(r2) ± φ1s(r2)φ2p(r1)
= z2e−(αr1+βr2) ± z1e−(αr1+βr2)
(32)
(33)
Note the two functions φ± are even and odd respectively under exchange
of spatial coordinates. Note both4 have total ` = 1.
Have to compute same types of integrals as for ground state, except now
1 Note
2 = 1 and Π2 = 1.
crucial point is that eigenvalues of both P12 and Π must be ±1 since P12
the choice of the |10i state ∝ z is for convenience: by rotational symmetry could have taken any of the other two 2p states with
φ2p ∼ (x ± iy)e−αr also–we’re only interested in the energy right now. For counting purposes note it looks as though there will be 2×
3=6 spatial states, but Pauli principle will not allow any of them to be multiplied by any spin wave fctn. See below.
3 I’ll ignore the spin part of the wave fctns. for now since the energy doesn’t depend on them, but remember we can’t consider just
any spin wave fctn. χms m0s , because of Pauli.
4 You are adding two angular momenta ` = 0 and ` = 1 here. According to the a.m. addition rules, |` − ` | ≤ ` ≤ ` + ` So since
1
2
1
2
one of our single-particle wave functions has ` = 0 the sum has definite ` = 1. More complicated for higher `’s.
2 Note
6
some of them have angular parts. Only quote results5:
var
E`=1,+
= −4.245Ryd
var
E`=1,−
= −4.261Ryd
; E`=1,+ = −4.247Ryd
; E`=1,− = −4.265Ryd
(34)
(35)
Questions:
1. Q: Why are variational energies more accurate than for ground state?
A: Since 2 electrons are in relative ` = 1 state, the probability of
finding them near each other is smaller than in ground state, thereby
reducing Coulomb energy hφ|e2/r12|φi, which is what keeps our trial
wave fctn. from being exact!
2. Q: Why is E`=1,+ higher than E`=1,− in (34-35)? A: Electrons in
spatially antisymmetric wave fctn φ− spend less time near one another
(φ−(r1 = r2) = 0!)
` = 0 excited states
There are also ` = 0 excited states, variational wave fctns. for which one
can construct by combining φ1s and φ2s. Again we can construct even and
odd eigenstates of Pr1r2 :
φ`=0,± = φ1s(r1)φ2s(r2) ± φ1s(r2)φ2s(r1)
(36)
The odd combination φ`=0,−is about 1eV lower in energy than φ`=1,−
(why?), and the even one φ`=0,+ ∼ degenerate with it.
6.4
Pauli principle revisited
So far we have neglected spin. Pauli princ. states that of all the states
with spin and spatial degrees of freedom one could construct, only those
which are eigenstates of P12 with eigenvalue -1, i.e. completely antisymmetric,6 occur in nature. The ground state of He was spatially symmetric,
5 Note for odd spatial state φ
− there will be three levels (` = 1,S = 1) which are split by spin-orbit coupling in the real He atom–see
var
Figure. This splitting is however small, so exact result quoted in (35) is accurate reflection of error in calculation of E`=1,−
6 More generally, a many-fermion wave fctn. must satisfy P ψ(1, 2, 3...) = −ψ(1, 2, 3...) for any i, j.
ij
7
Pr1r2 ψ0(r1, r2) = ψ(r1, r2). So the spin state
√ must be antisymmetric, i.e.
it must be spin singlet, χs = (↑↓ − ↓↑)/ 2. Excited ` = 1 states we’ve
just looked at are only a bit more complicated. Note φ+ from (33) must
go along with a spin singlet wave fctn. as well, but the spatially antisymmetric fctn. φ− must be multiplied by a spin triplet wave fctn. to satisfy
Dr. Pauli:
ψ`=1,S=0 = φ+χs
ψ`=1,S=1 = φ−χt
(3×1=3 states)
(3 × 3=9 states)
(37)
(38)
These sets of states are given special notations in atomic physics, e.g.,
since we are combining 1s and 2p electrons, a state might be referred
to as 1s2p. This would be suff. to specify the energy in the absence
of the electron-electron interaction. Or you might find the states (37)
and (38) referred to as 1P and 3P multiplets7, following the general form
2S+1
L. Upon application of the spin-orbit interaction, Sz and Lz are no
longer good quantum numbers, so we use total angular momentum j, mj
to label states. The rules of combining a.m.8 show that the 1P orbital
does not split: only allowed j is j = 1 since S = 0, so in this scheme the
states are called 3P1. The 3P multiplet does split however, into 3P0,3P1,
and 3P2, where the term notation is 2S+1Lj . The splittings starting from
the hydrogenic levels are shown in the figure. Note the order of the fine
structure splittings is reversed from that of hydrogen.9
Make sure you can account for 12 states in every column of the Figure!
More jargon: S = 1 states of helium referred to as ortho-, S = 0 as
para-helium.
We can now summarize what we have learned about the low-lying states
of the He atom so far, plus I’ll add some states we haven’t considered:
7 Generalization
of singlet and triplet
allowed values of j are |` − S|...` + S.
9 See e.g. H. Bethe and E. Salpeter,Quantum Mechanics of One- and Two-Electron Atoms, New York:Plenum, 1977, p. 183 et. seq.
8 Recall
8
Figure 1: Splitting of He excited states derived from 1s2p hydrogenic states. First splitting due to e− −e− interaction,
second to spin-orbit coupling.
Lifetimes
Won’t try to calculate decay rates for any of the transitions between lowlying states in He, but note that in figure above the 1s2p 1S excited state
is metastable (very long lifetime!) i.e. transition a won’t occur as normal
dipole transition since it involves ∆` = 0. Transition b from the 1s2p
para state is ok (goes fast), but the transition c from the 1s2p ortho state
is ”forbidden” since it involves a ∆S = 0, i.e. a transition due to the
Zeeman coupling of the magnetic field, much smaller. The transitions e
and f take place quickly (electric-dipole allowed), but the electrons then
get stuck in metastable states.
9
Figure 2: Low-lying He levels and transitions.
10