Ph235 F. Merritt
Autumn 2008 Oct 16, 2008
ORMAT Lecture 7 (October 13, 2008): (version 1.1; 11-Oct-2008::21:12)
Atomic systems and the periodic table. Part 3
Higher Z atoms (Z>4) and Hund’s Rules
In the last lecture we covered notation for electron (nl) subshells, and for the standard spectroscopic term . We also reviewed the periodic table, including “Hund’s Rule Zero” which gives the ordering of subshells which governs the structure of the periodic table:
We derived the ground state configuration for the first four elements. But to go further than that, we need to use Hund’s Rules 1-3. That is what we will do today. The rules are:
Hund #1: The lowest energy state will have the maximum value of S (total spin) consistent with the electron configuration specified by Hund Zero.
Hund#2: The lowest state will have the maximum value of L (total orbital angular momentum) consistent with the previous rules Hund#0 and Hund#1.
Hund#3: For shells that are half-filled or less, the lowest state is the one with ; for shells that are more than half filled, the minimum energy state is the one with .
Note that the first rule tells us how to determine S, the second how to determine L, and the third how to determine J. The first element for which all 3 of these come into play is Carbon (Z=6), so I will use that as the primary example. But the rules will be derived and extended to all members of the multiplet.
Before going through the rules in detail, let’s look at the results obtained by applying these to first 10 elements, repeating the table of electron configurations and extending it to include the spectroscopic term. That way, we can immediately see what the rules predict and make sure that we understand how to apply the rules as we develop them.
Z Sym Name e configuration Spect. term
1 H Hydrogen (1s)
2 He Helium (1s)2
3 Li Lithium (1s)2 (2s)
4 Be Beryllium (1s)2 (2s)2
5 B Boron (1s)2 (2s)2 (2p)
6 C Carbon (1s)2 (2s)2 (2p)2
7 N Nitrogen (1s)2 (2s)2 (2p)3
8 O Oxygen (1s)2 (2s)2 (2p)4
9 F Flourine (1s)2 (2s)2 (2p)5
10 Ne Neon (1s)2 (2s)2 (2p)6
The GSU website can help with understanding all of this, both with alternate explanations and most of all with graphs and sketches to illustrate the physics. The Atomic Physics site is:
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/atomstructcon.html#c1
The most relevant items are “Many Electron Atoms”, “Term Symbols”, “Hund’s Rules”,
and also “Coulomb Interaction”, “Spin-spin interaction”, “Spin-orbit interaction”, and “LS Coupling”. Some of these cover the same material and/or link to each other.
Rationale for Hund#1
Carbon has a filled core, and the two valence electrons must be in the subshell. But the different m-states should have equal energies, so it’s not obvious whether the ground state should have L=0, 1, or 2. The H1 rule tells us to maximize total spin S, which seems strange because spin has no dynamical role in the Hamiltonian (yet); but if we recall the difference between orthohelium and parahelium, the reason is clear. We want the 2 p-electrons to be in a totally antisymmetric spatial wavefunction, since this will minimize the ee interaction. Since the total wavefunction (spacespin) must be antisymmetric under exchange, this means the spin wavefunction must be totally symmetric. The way to insure this is to put both electrons in the same spin state (clearly symmetric under exchange), and this automatically gives us the maximum value of S. In the (2p) subshell, the ground state must have S=1 for 2 electrons, and for 3 electrons (corresponding to a half-filled p-subshell).
When the subshell is half filled, all the electrons in the ground state must have different values of (=1, 0, and -1 for the p-subshell), since they have the same value of (). But if the shell is half filled, that means that all of the m-states are occupied, so when we add a fourth electron it must have the same m as one of the first
3 -- and that means it must have a different spin (spin-down). All of the additional electrons in this subshell must have spin down, because all the spin-up states were filled by the first 3. Therefore, after the subshell is half filled, each additional electron must decrease the total spin by , so that S=0 when the subshell is completely filled.
When we add a fourth electron, it must have ; but the spin state must still be totally symmetric, so we can write it as , which has a total spin of S=1. (???? Prove this. Do we need to include spatial part? I think so.
[It perhaps is conceptually simpler to think of the more-than half-filled subshell as a superposition of a filled subshell plus a subshell of “hole states”, where a hole state is defined to be a state with a positively charged electron (like a positron). To minimize the total energy, minimize the energy of the hole system – which means putting the hole states in the same spin state, so their spin wf is symmetric and their space wf is antisymmetric; this minimizes the hole-hole interaction energy, and so minimizes the energy of the atomic system. So for subshells that are half-filled or less, use the “electron” picture, but for subshells that are half-filled or more, use the “hole” picture.]
The generalization to higher -subshells is obvious. for the first electron, increases in half-integral steps with each new electron until the subshell is half filled, and then decreases in half-integral steps until it is zero and the subshell is filled. This means that the multiplicity in the spectroscopic term starts with the value 2, increases in unit steps until the subshell is half filled, and then decreases in unit steps until it equals 2 again and the subshell is filled.
Rationale for Hund#2
Returning to Carbon as an example, we now know that in the ground state the two electrons must be in the same spin state, and that the spatial part of the wf must be antisymmetric under exchange. It seems plausible that the lowest energy state would have the electrons orbiting in the same direction, so that they pass each other at a minimal frequency. But the two electrons cannot have the same quantum numbers, because the state must be antisymmetric (or equivalently, because of the Pauli Exclusion Principle).
In general, we can write the antisymmetric spatial part of the state as
1)
Since we want to maximize L, let’s make as great as possible, which means . The , so this state must have or To find out which, apply . If the state has , this will raise it from up to ; but if the state had , then this operator will give zero. That is what happens (make sure you understand why), so the state had , and since for Carbon (p-subshell), this means the ground state has
But how do we handle 3 valence electrons? Or more? How do we make a wavefunction which is totally antisymmetric under exchange (and that means exchange of any two electrons)? The way to do this is to write the state as a “Slater determinant” (see Griffiths page 210, problem 5.7).
Let’s consider a general state in -subshell with 3 electrons. Then if the subshell is half filled or less, all of the electrons must have spin-up, so all of the m’s must be different. Since we want to maximize L, choose them to be
2)
Then the state which is totally antisymmetric under exchange is uniquely defined as
3)
where the first state in each term refers to particle 1, the second to 2, and the third to 3. Check that this indeed swithches sign if you exchange any two particles. Now the total of this state is clearly . And if we apply we necessarily kill the state, because we will make 2 elements the same. Therefore, the total L of the system is just the value of , so . For the case of Nitrogen (Z=7), this gives .
This procedure can get a little complicated for more electrons, where we would need to use (which you have probably never encountered). A nice way of making the wavefunction totally antisymmetric is to use the Slater determinant. For N particles in N different states, draw a matrix, with columns corresponding to states (=a,b,c,… and rows corresponding to particles (=1,2,3,…). The the antisymmetric state is just the normalized determinant of this matrix. For N=3, this becomes
4)
which is the same as . This works because the determinant changes sign under the interchange of any two rows, so the resulting wf is necessarily antisymmetric under exchange.
But what happens when a subshell becomes more than half filled? Then all of the electrons can’t be in the same spin state. That means that we can no longer factor out the common spin wavefunction and deal with the spatial part alone. There are 3 ways to handle this:
a) Note that when a subshell is exactly half filled, L=0 necessarily, since all of the (2l+1) m-states are occupied. For the second half of the subshell, we just start over (so far as L is concerned) are repeat the (2l+1) numbers we have written down for the first half.
b) Note that a filled subshell necessarily has L=0 and S=0. Argue that a subshell of P states and Q electrons is equivalent to a filled subshell with P-Q “hole” states. Think of the holes as positively charged electrons with negative mass. The hole states also carry angular momentum, so we simply maximize L for the hole states. For a subshelll that is more than half filled, the hole states will all have spin up, so can proceed exactly as we did for the first half. This gives the same (2l+1) values of L that we found for the first half. (Note that the order is reversed this time, logically speaking, but you will see that doesn’t matter much.)
c) Follow exactly the same formal procedure that we used for the first half, using Slater determinants, but now include the correlation between spin and space wavefunctions.
For example, in the case of Nitrogen, with 4 electrons in the subshell, a possible state would be , with a,b,c defined as above in equation , with . This state has and ; but it is not a valid solution because it is not totally antisymmetric. We make the totally antisymmetric state by constructing a Slater determinant:
5)
Multiplying this out gives 24 terms, and the state is of course totally antisymmetric under exchange. It is also an eigenstate of and with eigenvalues . We can prove this easily by applying or to the state, both of which destroy the state because they make 2 of the states the same (and because of the antisymmetry under exchange, this gives an eigenvalue of zero). So we conclude that this state has , which is the same result our other methods gave.
Rationale for Hund#3
We now have determined both L and S for the ground state, and the wavefunction is necessarily totally antisymmetric. But we are still free to adjust the relative orientation of L and S. In general, we could have any value of J ranging from a maximum of down to a minimum of . The ground state will be the one of these with lowest total energy. Using the Hamiltonian that we have assumed up until now, all of these would be degenerate. But we have left out a dynamical term involving electron spin.
Because the electrons are orbiting in a radial Coulomb field, in the instantaneous rest frame of an electron there will be a magnetic field given by (Shankar , p.468)
6)
The -field that an electron sees in its instantaneous rest frame is consequently proportional to the angular momentum of its orbit!
Equation is not exactly correct for 2 reasons. First, the instantaneous rest frame of the electron is not an inertial frame, and correcting for this reduces the apparent
-field by a factor of 2. Moreover, in a multielectron atom the -field is not a pure Coulomb field because of shielding effects from the other electrons. But it is nevertheless a radial field, so equation gives the right direction of the -field and the right order of magnitude.
We have already seen (in Ph234) that the interaction energy of an electron in a magnetic field is given by (Shankar equation 14.4.19, page 389)
7)
where g=2 for the electron. Combining this with equation gives the approximate interaction energy of
8)
That is, the energy of the electron is minimized when its spin is antiparallel to its orbital angular momentum.
Note that the S and L of equation are those of the individual electron, rather than the summed S and L that we have been dealing with. When we are dealing with a subshell that is less than half filled, following the conventions we have used above with each electron in a spin-up state, we always have and also for each electron that we add. The total interaction energy from is consequently
9)
where and now refer to the total quantum numbers of the system. The lowest energy state (the ground state) is consequently the one with S and L antiparallel, so.