Download Lecture 4 (October 1, 2007): Quantum Statistical Mechanics

Ph235
Autumn 2008
F. Merritt
Jul 4, 2008
Lecture 5 (October 8, 2008):
Atomic systems and the periodic table. Part 1
Hydrogenic atoms:
We know that the energy levels of an electron in a coulomb field are given by
mc 2  Z 2 2
(1.1)
En  
2n 2
1
where Ze is the charge of the nucleus and   137.04  is the dimensionless “fine
structure constant”. Since me c 2  0.511 MeV , the ground state energy (equal to the
ionization energy) for hydrogen is
EGH  E1  13.6 eV
(1.2)
There is a simple “heuristic derivation” of equation (1.1). For circular orbits, we
know there is an integral number of de Broglie wavelengths in a closed orbit, so in the
nth subshell
h
2
pn 

 n
(1.3)
n 2 rn / n rn
so the kinetic energy is
2 2
p2
n
(1.4)
Tn  n 
2m 2mrn2
The potential energy is
Ze2
(1.5)
Vn  
rn
From the virial theorem (or from balancing centrifugal force against coulomb force)
1 2 n2 1 Ze2
1 m  Ze2
(1.6)
Tn   12 Vn 


 2 2
2m rn2
2 rn
rn
n
Then
1 Ze 2
1 m  Z 2e4
1
En  Tn  Vn   2 Vn  
 
2 rn
2 n2 2
(1.7)
2 2
E1
1
1 2 2 2 1
2  Ze  1
   mc  
 2   mc  Z  2 
2
2
n
n2
 c  n
Note from (1.1) that the velocity of the electron is equal to  / n , so the
nonrelativistic approximation is valid. Higher n states have larger radius and smaller
momentum -- but we shouldn’t take the orbit picture too seriously.
Note that the mass in (1.1) is really the “reduced mass”  . For hydrogen this is
almost the same as me , but for positronium it is half me .
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Ph235
Autumn 2008
F. Merritt
Jul 4, 2008
A striking feature of the hydrogen energy levels is the degree of degeneracy. The
energy of the state depends only on the principle quantum number n. It is independent of
the angular momentum l, and it is independent of the z-component m. The latter is not
hard to understand: a rotation of the atom will mix up the different m states, but will
have absolutely no effect on the total energy. So the degeneracy with respect to m is just
a consequence of rotational invariance.
The degeneracy for different values of l (with the same n) is more subtle. In fact,
this occurs only for the Coulomb potential (e/r). It can be understood as a symmetry,
which reflects the fact that the Runge-Lenz vecor is conserved (Shankar 15.4.6):
1
e2 R
(1.8)
N
P L  L P 
2m
R
One can define raising and lowering operators
N x  iN y
(1.9)
N11 
2
which raise the angular momentum, but which commute with the Hamiltonian.
It is interesting to explore this symmetry with the Coulomb and the Harmonic
Oscillator potential (Shankar section 15.4), but I will not pursue that. The important
point for our purposes is that this symmetry depends on the 1/R potential; for potentials
with different r-dependence, the energy of the angular momentum eigenstates can depend
on l as well as on n. We will see this clearly as we move up the periodic table to higher
Z. We do not have to go far.


Helium (Z=2)
The helium ion  He  has only one electron, but Z  2 . This means the ground
state energy is a factor of 4 times that of hydrogen, and in general
EnHe   4  EnH
(1.10)
We might expect the helium atom (with 2 electrons) to have energy of twice this, so a
ground state of E1He  8E1H  108.8 eV . But experimentally the ground state energy
has -79.0 eV.
It’s not hard to identify the problem. Equation (1.10) assumes that the
Hamiltonian for helium has a coulomb potential like hydrogen. But for all atoms with N
electrons, the inter-electron interactions are quite large. The Hamiltonian for atomic
systems of N electrons is
N 
Ze2 
e2
 2
 N N
(1.11)
H   
 j2 
 
rj 
j 1 
 2m
 j 1 k  j 1 rj  ri
It is best to explicitly include the effect of the second term – but it is a lot easier to
think of the inner electrons as providing “shielding” for the jth electron. In the case of
helium, the ground state electrons are clearly equivalent, so we might imagine that each
one provides 50% shielding for the other (since each one must be the “inner” electron
half the time). That would mean that the effective Z of the helium nucleus is about 1.5,
and that would give a ground state energy of -61.2 eV. This at least moves us in the right

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Autumn 2008
F. Merritt
Jul 4, 2008
direction (but overcorrects substantially). The effective Z that gives the correct groundstate energy in the absence of ee interactions is Z eff  1.70 . [We will derive this result
next week using the variational methods (Shankar 16.1)].
But this picture ignores the variation of effective Z with r, and also treats the two
electrons as independent, uncorrelated states. Recall that a) the overall electron wf must
be antisymmetric under exchange of electrons (since they are fermions), which means
that the product of space wf and spin wf for the 2-electron system must be odd. We
know that the spin wf can be either symmetric (triplet) or antisymmetric (singlet). The
triplet spin function must be associated with an antisymmetric space wf for the the 2e
system, and the singlet must be associated with a symmetric space function.
But we saw in Ph234 (Griffiths Section 5.1.2, or Eisberg) that the rms separation
between two particles in a symmetric wf is less than for antisymmetric. This is what we
would expect, since the exclusion principle prevents 2 electrons from occupying the same
spatial state (if they are in the same spin state). But this means the ee electrostatic term in
the Hamiltonian should be smaller for the triplet spin state (antisymmetric space) than for
the singlet spin state (symmetric space). The difference can be reasonably large
( 1 eV for the 2S state ).
[Suppose 2 electrons occupy states spatial a and b (orthogonal). Then the triplet
wf with Sz=0 is
 abtrip  12 (ab  ba)  12     
(1.12)
while the same state with singlet spin (S=Sz=0) is
 absing  12 (ab  ba)  12     
(1.13)
Note that both states are antisymmetric under total exchange. The triplet has S=1, since
applying S  will give a new state with S z  1 , whereas the singlet state has S=0, and
applying S  gives zero (kills the state). Also note that the two states have identical
expectation values for x1 and x2 , but different expectation values for  x1  x2  . There is
no classical analogue for this.
[Note also that when the 2 spatial states are the same, then the spatial part is simply aa,
which is obviously symmetrical. Then overall symmetry of the wavefunction requires
that the spin function be antisymmetric].
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But we know that correlations are very important. We have already seen that the
expectation value of the separation of two particles in states a and b is greater if the
particles are fermions than if they are bosons. This would mean that the ee interaction is
smaller, or equivalently that the effective shielding is smaller, for particles that are in a
spatially antisymmetric state. This means that the total energy of the state should be
larger in magnitude, and consequently lower in absolute magnitude, for identical
electrons that are in a spatially antisymmetric state.
Indeed, there are different energy levels for helium eigenstates that have S=1
(triplet) and S=0 (singlet). The former are called orthohelium, and have antisymmetric
spatial wavefunctions. The latter are called parahelium, and have symmetric spatial
wavefunctions. Because of the exclusion principle, orthohelium states (spatially
antisymmetric) have lower energy than parahelium states (spatially symmetric). This is
shown in Griffiths Figure 5.2 (page 213).
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Ph235
Autumn 2008
F. Merritt
Jul 4, 2008
The parahelium and orthohelium states are also shown on the physics website of
Georgia State University:
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/helium.html .
In fact, this website seems quite good to me, and is worth exploring. It covers most of
the topics that we’ll be addressing in the next two lectures:
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/atomstructcon.html#c1
The website covers other areas as well. See:
http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html
http://hyperphysics.phy-astr.gsu.edu/hbase/quacon.html#quacon
We will return to helium later in the course and look at different ways of dealing
with the Hamiltonian in a much more quantitative manner.
Spectroscopic notation for atomic systems
There are two ways of representing the eigenstates of an atomic system with
nuclear charge Z. The subshell occupancy is given by writing the state as a product of
terms (nl ) factors, where n is the principal quantum number and l is the orbital angular
momentum of the ith electron, with an exponent giving the number of electrons given as
a letter: s (sharp) for l=0, p (principal) for l=1, d(diffuse) for l=2, then f, g, h for l=3,4,5
in that subshell. (see Griffiths Table 5.1). This corresponds to using the basis in which
H i , Li 2 , Si 2 , Lzi , S zi are mutually diagonal (where the subscript refers to the individual
electrons).
But this does not completely specify the atomic state. One can also choose the
mutually commuting operators H , L2 , S 2 , J 2 , J z where these variables now refer to the
sums over all electrons. The quantum numbers L, S, and J are incorporated into the
standard spectroscopic notation of atomic states, as
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Autumn 2008
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Jul 4, 2008
2 S 1
LJ
(1.14)
The total angular momentum L is written using the SPDFG notation described above. So
the ground state of hydrogen is written as (S= 12 , L=0, J= 12 )
Hydrogen ground state: 2 S1/ 2
(1.15)
while that of helium is written as (S=0, L=0, J=0)
Helium ground state: 1S0
(1.16)
Lithium (Z=3)
Here the ground state must have a filled n=1 shell with a single electron in the
n=2 shell. We would expect the Lithium ion L (with 2 electrons) to be just like the He
2
ground state, but with energies higher by a factor of  32  . Then we would expect adding
the third electron to be like the n=2 state in hydrogen, which has an ionization energy of
about 3 eV. But do we expect the ground state to have this electron in the L=0 state or
the L=1 state?
Since we want the lowest energy state (i.e., the ground state), we want the 3rd
electron to see the full nuclear change – we want to minimize the shielding from the n=1
shell. This means the ground state wants l3 to be as small as possible within the current
shell (recall that the S-wave wf peaks at the origin). For the ground state of lithium,
l3  0 (s-wave):
Lithium ground state:
2
S1/2
(1.17)
[We should be a little careful with this explanation. All the n-level hydrogenic
eigenstates have the same energy, and all satisfy the virial theorem. That means that all
have the same 1/ r . So why should low l give lower energy for helium? The answer
is that the low-l states have a larger value of  (1/ r ) . Because the width of the 1/r
distribution is greater, they sample more of the unshielded low-r potential. The visible Z
is a highly non-linear function of r, so the low-r part of the wavefunction has a much
stronger weight than the high-r part. [It would be good to try to be more quantitative
about this.] The GSU webpage on lithium is a good example:
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/orbdep.html#c1
Beryllium (Z=4):
The 4th electron completes the (2s) subshell, so
Beryllium ground state: 1S0
(1.18)
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Ph235
Autumn 2008
F. Merritt
Jul 4, 2008
Boron (Z=5) through Neon (Z=10):
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