The presence of a second electron induces a term of repulsion between electrons.
This term is positive so it increases the energy of the orbitals. The rest of the equation is similar to the Hamiltonian of the hydrogen. We can compare the energy of the orbitals with those two models.
The repulsion is large in the 1s orbital wherein the electrons are close to each other (the radius being small). For instance, the energy of the orbital 1s2 of the Helium is at -2.9au (atomic units). If we calculate the energy without the repulsion term, i.e. as a monoelectronic atom (Ĥ= Ĥ1 + Ĥ2), we find -4au (E1=E2=-Z2/2n2=2au). Another consequence is that the orbitals s and p don’t have the same energy anymore. The configurations 1s12s1 has an energy of -2.17au and the configuration 1s12p1 an energy of -2.13au.
We can separate the Hamiltonian into one monoelectronic Hamiltonian term and the term of repulsion between the electrons.
ĥ(i) is the monoelectronic Hamiltonian of one electron. We sum up the terms for all the electrons to obtain the Hamiltonian. To resume, the repulsion spreads the degenerated orbitals from the monoatomic model.
The repulsion can be large enough to have an orbital of level n+1 of lower energy than an orbital of n. It is the case for orbitals d and f and it is why we use the following technique to fill the orbitals with electrons:
The 4s orbital is thus filled before the 3d orbital.
The operators that commute are simply the sum of the operators applied to the electrons independently (but perturbed by the repulsions). We write these operators with a lower case to denote them from the global operators (Ĥ→ĥ, Ŝ→ŝ, …).
The electrons are indistinguishable. It means that if we exchange an electron of an atom with another electron of the atom on the same orbital or on a different orbital, the atom remains unchanged. We can thus introduce a permutation operator Pij that commutes with the Hamiltonian Ĥ and the proper value of which is ±1. This last affirmation is true because we get back to the initial conditions if we apply the permutation operator Pij twice. The identity operator Ê is the operator with which nothing changes: ÊΨ=Ψ.
The value of p is imposed by the type of particle. Bosons have a integer spin and pboson=1. Fermions have half-integer spins and pfermion=-1. Electrons are thus fermions. Some nuclei are bosons and some are fermions. In an atom with N electrons, N! permutations of electrons are possible. For instance, the lithium has 3 electrons with spherical coordinates (r1, σ 1), (r2, σ2) and (r3, σ 3) (considering the orbitals 1s and 2s) that can be permuted between 3 spots a, b and c. There are 6 possible permutations 6=3! = 3x2x1): P12, P13, P23, (P13,P12), (P13,P23) and Ê.
We don’t know which electron is in which spot (a, b or c). The wave function Ψ is one of the functions
To write the wave function we use the determinant of Slater which is a combination of the possible states.
with a, b, c, …, z are the spin orbitals.
In mathematics, a determinant is read in diagonal. The elements of the diagonals (from top left to bottom right) are multiplied and the diagonals are summed up. Then we deduce the diagonals in the other way (from top right to bottom left). In the case of the lithium we have
The determinant of Slater has a few properties:
- if we exchange two lines or two columns, the sign of the determinant changes.
- if two columns are identical, the determinant equals 0.
It can be translated by the law of Fermi: 2 electrons must differ by at least one quantic number.
Let’s apply this to the helium and a few of its excited states. He has 2 electrons that are distributed on the 1s orbital in its ground state. There can be two electrons on this orbitals if they have different spins (1/2 and -1/2). In the notation of Slater, the spin is indicated by the presence of a line above the orbital if the spin=-1/2.
For the excited state wherein one electrons jumped to the orbital 2s, there are 4 possible states depending on the spin of the electrons:
If instead the electron jumped to the 2p orbital, there are 12 possible states:
Coupling of Russel-Saunders
This coupling is also called the L-S coupling. We consider it for multi-electron atoms with weak spin-orbit couplings. In this case, the orbital angular moments of the individual electrons add to form a resultant orbital angular momentum L and the same is true for the spin moments to form a spin angular moment S. L and S combine to form the total angular moment J.
The resultant of two vectors is their sum. If we consider two operators Ĵ1 and Ĵ2 (which can be L and S), their resultant Ĵ = Ĵ1+ Ĵ2.
Ĵz is simply equal to the sum of Ĵ1z and Ĵ2z and Ĵ2 is the development of the square of the sum
We can look at the coupling between the electrons in He. The electrons have a spin s=1/2 and ms=±1/2. Let’s say that ms=1/2 is the state α and that ms=-1/2 is the state β. The coupling can be αα, αβ, βα or ββ coupling. The coupled Ms=ms1+ms2 can thus be equal to 1, 0, 0 or -1. We write these values in a table. I insist on the fact that Ms=0 is degenerated.
To know the resulting states of the coupling between the electrons, we take the largest S=s1+s2 and look for its Ms in the results of the coupling. It is Smax=1 with Ms=-1, 0 and 1. We discard these values of Ms from the 4 couplings that we obtained earlier. If there are one or more remaining Ms in the table, we do the same for S=Smax-1, etc. Here, there is one last Ms=0 that comes from S=0. The 4 couplings correspond thus to a singlet state (S=0 and Ms=0) and 1 triplet state (S=1 and Ms=1, 0 or -1). S goes from ∣s1+s2∣ to ∣s1-s2∣ by step of 1. The multiplicity is equal to 2S+1. The difference between the states is always by step of ΔS=1 and ΔMs=1. The states can be represented this way:
Note that the vectors for Ms=1 or Ms=-1 are not two times the length of the vectors for ms1 and ms2. The vertical component is. Note too that the resulting vector for Ms=0 may have a nonzero length (S=1).
If 3 electrons are coupled, we proceed step by step. First we do the coupling between two electrons. As previously we obtain two different states, S12=0 (singlet) and S12=1 (triplet). There is no need yet to determine Ms. The last electron is coupled next to the two possible states S12. The coupling with the singlet S12=0⊗s3=1/2 gives S123=1/2. It is a doublet (Ms=1/2 or Ms=-1/2, remember that ΔMs=1 between ∣S12+s3∣ to ∣S12-s3∣). The coupling S12=1⊗s3=1/2 gives a total of S123=3/2 (1+1/2), leading to two states S123=3/2 and S123=1/2 (ΔS=1). It means that there is a quadruplet (2(3/2)+1=2) and a doublet (2(1/2)+1=4). In total, there are 2n=8 possible states, n being the number of electrons.
In the case of degenerated orbitals as the orbitals p or d, the amount of configurations is larger (see the part about the determinant of Slater) but the method is identical and is also applied to the quantic number l. Let’s take a few examples: the electronic configurations (1s)1(2s)1, (1s)1(2p)1 and (2p)1(3p)1.
The degeneration is 4: the two electron can have a positive or a negative spin (α or β). We have thus 2×2=4 degenerated configurations:
There is only one value for the orbital number L=0: l1s=0⊗l2s=0. As a result, ML=0. As we have seen above, the coupling between two electrons s1S=1/2⊗s2s=1/2→S=1, 0 gives one singlet (S=0, degeneration =1) and one triplet (S=1, deg=3).
We write the electronic term depending on L and S. L gives a capital letter S, P, D, F the same way n gives s, p, d, f. The electronic degeneration is written as an exponent before the orbital letter. The electronic terms for the configuration (1s)1(2s)1 are thus 1S, 3S. The degeneration of the terms is the multiplication of the degeneration of L and S and we sum up the degeneration of the electronic terms: 1.1 (for 1S) +1.3 (for 3S) = 4. We have thus a concordance between the degeneration of the terms and the degeneration of configuration.
The degeneration is 12 in this case: both electrons can have a positive or negative spin (α or β) but the 2p electron can be on 3 different orbitals (2p–, 2p0 or 2p+). We have thus (1.2).(3.2)=12 degenerated configurations (written earlier in the chapter).
The coupling l1s=0⊗l2p=1 gives L=1 (ML=-1, 0, 1). There is no L=0 here: if we look at the details of the coupling.
We find the ML’s of L=1 but once they are discarded, there is no other ML. There is thus no L=0. An easier way to know it is that L goes from ∣l1=l2∣ (=1) to ∣l1-l2∣ (=1-0=1).
As L=1, the electronic terms will be xP with x=1 and 3 from the spin. As always, the coupling between two electrons gives a triplet and a singlet. The electronic terms are thus 1P, 3P with a degeneration (3.1)+(3.3)=12.
The degeneration is 36 ((3.2).(3.2)): 3 orbitals with 2 possible spins for each electron. The coupling l2p=1⊗l3p=1 gives L=2 (ML=-2, -1, 0, 1, 2), L=1 (ML=-1, 0, 1) and L=0 (ML=0). The coupling of spins gives one singlet and one triplet as usual. We have thus the electronic terms
1S, 1P, 1D, 3S, 3P, 3D that can also be written as 1(S, P, D), 3(S, P, D) or even 1,3(S, P, D)
The degeneration are respectively (1.1), (3.1), (5.1), (1.3), (3.3) and (5.3) for a total of 36.
If we want to couple a third electron, we obtain the electronic terms by coupling separately the electronic terms if the electrons are not equivalent. For instance, if we add a 3d electron (2D) to the 2p3p electrons, we proceed this way:
One can see that some electronic terms are obtained several times (2F for instance). Those terms are not identical and have not the same energy because they come from different couplings.
In the case of equivalent electrons, the amount of states is more limited. For instance we had 36 possibilities for the 2p3p configuration. This amount drops to 15 for the 2p2 configuration. This amount is equal to X!/(X-N)!N! where X is the degeneration of the state and N the number of electrons. In our case, 2p2, N=2 and X=6 for the orbital 2p: there are 3 orbitals (2p+, 2p0 and 2p–) and 2 spins. There are thus 6.5/2=15 possibilities. To determine the electronic terms of this coupling, we will make a list of the 15 possibilities and identify each one by the couple (ML,MS). To do so we draw a table with a column for each one of the 6 states and we put pairs of electrons together. In a seventh column, we write down the (ML, MS) of the couple of electrons.
Once it is done, we seek for the emergents. An emergent is one couple of projections that is unique. We take the one with the largest values, i.e. (2,0) in this case and we determine the electronic term from which the pair comes. ML=2 means that L=2 and MS=0 means that S=0. The corresponding electronic term is 1D that has a degeneration of 5. As a consequence we will find 5 pairs of projections in the table that come from the electronic term 1D.
We can remove them from the list and look for another emergent: (1,1). The corresponding electronic term is 3P with a degeneration of 9. From the 15 possibilities, 14 correspond to 2 electronic terms (3P and 1D). The last emergent is (0,0), what corresponds to the electronic term 1S. The electronic terms of the configuration 2p2 are thus 1D 3P 1S. In comparison with inequivalent electrons (2p3p), the 3D, 1P and 3S terms disappeared. Also note that several pairs were several times in the list and we randomly selected which one corresponds to which electronic term. For instance, we don’t know which (1,0) comes from 1D or from 3P.
In terms of energy, the rules of Hund tell us that the term with the highest MS is the most stable. If there are identical MS projections, then the highest ML is the most stable. For instance, the 3P coupling of 2p2 has a lower energy than 1D itself lower than 1S.
Here come two properties of the L-S coupling:
- Complete orbitals are all 1S:
As a result, they do not modify the other orbitals. Note that even if they do not perturb the other orbitals, they do contribute to the energy of the electronic states.
- A “hole” (the absence of electron) in one orbital gives the opposite signs of ML and MS than the presence of this electron. For instance
But the sign doesn’t change the state. As a result, the carbon (2p2) and the oxygen (2p4) have the same state 2P.
The magnetic moment induced by the spin of the electron interacts with the magnetic field induced by the electric current resulting from its orbital movement. This relativist effect induces a further separation of the states and an additional term in the Hamiltonian.
The coupling introduces a new observable J=L+S
J goes from L+S to ∣L-S∣ by step of 1. Considering this, [Ĥ,L2] and [Ĥ,Ŝ2]≠0 but [Ĥ,Ĵ2]=0.
The relativist effects grow with the atomic number of the atom: the larger the atomic number, the heavier the nucleus. Its charge increases as well meaning that the electrons have to revolve faster to compensate.
Energy of the states
The energy of the states can be found from the equation of Schrödinger
From that, we can write
The energy is thus the average value of Ĥ. To determine it, we start from a linear combination of determinants of Slater (ML,MS). An emergent determinant of Slater always gives a proper function. From this point, we can apply operators L+, L–, Ŝ+ or Ŝ– to increase/decrease the value of ML or MS and find their energy.
We do it for each state that has at least an emergent. If there is no emergent, we can apply the rules of orthogonality: 2 wave functions that differ by at least one proper value of operators that commute together are orthogonal. Eventually, the proper functions are normalised.
With this method, the energies of the states can be found. One simple example says more than a thousand words. The coupling of two electrons gives a singlet (S=0, MS=0) and a triplet (S=1, MS=1, 0, -1). We will place the values of MS in boxes for S=1 and S=0, as shown below.
An emergent is found for MS=1 (S=1) so we can find the wave function ΨS=1,Ms=1 (the state αα). We also have the wave function of Ψ1,-1 (the state ββ) because it is an emergent too.
To obtain the wave function of the state MS=0, we apply the operator of descent Ŝ– on Ψ1,1 (we could have applied Ŝ+ on Ψ1,-1 instead).
One electron in the α state is thus now in the β state. We need to norm the function. To do so, we must have the norm N equal to
Applied here, we have
The orbitals are orthonormal, meaning that the integrals equal 1 if the two terms of the integrant are identical and zero if they are different: ∫αα=1, ∫ββ=1, ∫αβ=0, ∫βα=0. The cross terms of the square (in blue) are thus equal to zero and the others equal 1.
As a result,
To obtain the wave function of the singlet, we can use the property of orthogonality between Ψ1,0 and Ψ0,0. Posing β(1)α(2)=x and α(1)β(2)=y,
Meaning that, once normed,
The same method can be applied on more complex cases. For instance with two 2p electrons: