# Diapositiva 1

advertisement ```Magnetochimica
AA 2011-2012
Marco Ruzzi
Marina Brustolon
3. The exchange spin Hamiltonian
Spin Hamiltonians are Hamiltonians containing only spin operators.
The energy values of quantum particles should be obtained as
eigenvalues of an Hamiltonian containing spatial and spin operators. In
some cases it is very convenient to integrate over all spatial variables,
leaving only the spin operators multiplied by numerical parameters. By
studying the magnetic properties of the system (e.g. by magnetic
resonance methods) one can obtain these numerical parameters and
relate them to the property of the system.
For example, let us consider two electrons with repulsive coulombic
interaction, in two different orbitals. We know that an energy
contribution (exchange energy) depends on the relative orientations
of the two spins (parallel or antiparallel).
This is due to the Pauli principle, as two identical fermions (as two
electrons) must be described by an antisymmetric function, the
product of the spin and of the spatial functions must be
antisymmetric by exchanging the two electrons.
Therefore the spatial and spin functions are interdipendent as their
product must be antisymmetric.
(1, 2)   (r1, r2 )  (S1, S2 )
spatial function spin function
 '(1, 2)   A (r1, r2 ) S (S1, S2 )
 ''(1, 2)   S (r1, r2 ) A (S1, S2 )
The symmetric spin
function is a triplet
The antisymmetric spin
function is a singlet
as we will show.
e2
 '(1, 2)
 '(1, 2)  ETriplet
r12
e2
 ''(1, 2)
 ''(1, 2)  ESinglet
r12
The energies of the two
wavefunctions are different
because the two electrons have
a minor electrostatic repulsion
in the triplet state thanks to
the antisymmetric simmetry of
the spatial function.
The energies Etriplet and Esinglet can be referred to the two
triplet and singlet spin states A and S.
We will show that by defining a parameter:
J
1
ETriplet  E Singlet
2
we can build a spin Hamiltonian such that, acting on the spin
states, it gives the energy difference between the two states
due to their different spatial functions.
This convenient spin Hamiltonian is:
H  2JS1S2
where J is called the exchange integral.
This parameter measures therefore the extent of the
electrostatic repulsion between the two electrons depending
on the relative orientation of their spins.
Let obtain the eigenfunctions and eigenvalues of the spin
exchange Hamiltonian starting from the functions:
 ,  ,  and 
(uncoupled basis)
We will show that the eigenfunctions of this hamiltonian are the
singlet and triplet functions, and we will find that their energy
gap obtained from the eigenvalues of the spin Hamiltonian is in
fact 2J.
We will do the following:
1. Develop the spin Hamiltonian in its component operators;
2. Build the Hamiltonian matrix;
3. Diagonalize the Hamiltonian matrix finding eigenvalues;
4. Find the spin eigenfunction corresponding to each eigenvalue.
1. How to develop the spin hamiltonian H  2 JI1I 2
In the following slides we will use the symbol I for the angular momenta components.
2J (I1  I 2 )  2J (I x1I x 2  I y1I y 2  I z1I z 2 )
Let us express the spin components x and y as combinations of
raising and lowering operators:
I  I
Ix 
2
I  I
Iy 
2i
I+ I- are the so called “raising” and “lowering” operators (operatori scaletta).
I  I x  iI y
2 J ( I x1 I x 2 )  2 J
I   I x  iI y
1
J
( I 1  I 1 )( I 2  I 2 )   ( I 1 I 2  I 1 I 2  I 1 I 2  I 1I 2 )
4
2
2 J ( I y1 I y 2 ) 
J
( I 1 I 2  I 1I 2  I 1I 2  I 1I 2 )
2
2J ( I x1I x 2  I y1I y 2 )  J (I 1I 2  I 1I 2 )
Therefore:
2 J ( I1  I 2 )  2 J ( I x1I x 2  I y1I y 2  I z1I z 2 ) 
2 JI z1I z 2  J ( I 1I 2  I 1I 2 )
Now, let us apply this hamiltonian to the spin functions for
two electrons. We use the uncoupled basis:  ,  ,  and 
1
2 JI z1 I z 2    J 
2
1
2 JI z1 I z 2    J 
2
( I 1 I 2  I 1 I 2 )   0
( I 1 I 2  I 1 I 2 )   
1
J 
2
1
   J 
2
 2 JI z1 I z 2   
 2 JI z1 I z 2
( I 1I 2  I 1I 2 )   
( I 1I 2  I 1I 2 )   0
In conclusion:


2JI z1I z 2  J ( I 1I 2  I 1I 2 )


1
  J 
2
1
  J   J 
2
1
  J   J 
2
1
  J 
2
Whereas  and  are eigenfunctions of -2JI1I2, 
and  are not eigenfunctions of this spin hamiltonian.
Therefore we must find now the eigenvalues and
eigenfunctions of -2JI1I2 on the basis  and  .
2., 3. How to build and diagonalize the matrix of
the Hamiltonian
To find the eigenfunctions of a Hamiltonian on a given complete
functions basis we can build the matrix on that basis, and then
diagonalize it.




H 
H
0 
0 
H
0 
H




0 
H
H 
H 
0 
H




H0 
H 
H 
H0 




0 
H
H
0 
0 
H
H 
In the present case  and  are eigenfunctions of the hamiltonian,
but the other two functions are not.
If a function of the basis is an eigenfunction of the Hamiltonian, all
its out diagonal integrals are equal to zero. We say that the function
is not “mixed” with other functions by the operator.
We obtain the Hamiltonian matrix elements:
1
 H    H    J
2
1
 H    H    J
2
 H    H   J
Therefore:
1
 J
2
0
0
1
J
2
J
0
0
J
1
J
2
0
0
0
0
1
 J
2
0
0
We must find the two linear
combinations of the functions 
and  which diagonalize this
block. To do this before we have
to find the eigenvalues of the
hamiltonian.
To diagonalize the 2x2 block we must find the energy
eigenvalues by solving the secular determinant:
1
J E
2
J
J
1
J E
2
3
J
2
1
E2   J
2
E1 
2
0
1

2
J

E

J
0


2

J  J 2  3J 2 J  2 J
E

2
2
Now we will find the linear combinations of
 and  corresponding to each of these
eigenvalues.
4. How to find the spin eigenfunctions
Then we can now solve the equations for the coefficients corresponding
to the two eigenstates:
 1

 i
i
J

E
c

Jc

0
i 1
2
 2






 Jci   1 J  E  ci  0 
i 2
 1  2


For E1 
3
J
2
For E2  
c12  c22
c11   c12
singlet
1
J
2
triplet
By normalization we obtain:
Singlet
1
(   )
2
Triplet
1
(   )
2
In conclusion
• The eigenfunctions of the spin exchange
operator are the coupled functions for two
spins &frac12;:
S, MS

1
   
2
1
   
2

1,1
1
2
  2 JS1S 2    J
0,0
1
1
3
(   )  2 JS1S2
(   )  J
2
2
2
1,0
1
1
1
(   )  2 JS1S2
(   )   J
2
2
2
1, 1
1
2
  2 JS1S 2    J
Singlet or Triplet?
The energy of a system of two electrons, described by different
spatial wavefunctions, and near enough to feel a substantial
electrostatic inter-electronic repulsion, depends on the relative
orientation of their spins.
The spins can be parallel (triplet, three states) or antiparallel (a
singlet). The energy gap is equal to 2J, where J is called
exchange integral. The energy order of the spins state depends
on the sign of J.
E
E
0
2J&gt; 0
0
2J&lt; 0
E
E
2J&gt; 0
0
When J&gt;0 the ground
state is a triplet
0
2J&lt; 0
When J&lt;0 the ground
state is a singlet
The spin coupling for two interacting electrons is said to be:
J 0
Ferromagnetic
J 0
Antiferromagnetic
Warning!
• Some authors use a spin exchange
Hamiltonian with a J’=-2J
H  J ' S1  S 2
• Sometimes the triplet-singlet energy
separation is called 2K
When the exchange integral is relevant in
determining the electronic properties of a pair
of electron spins?
1. Two electrons in same molecule
2. Biradicals
3. Complexes of pairs of paramagnetic transition
ions
4. Near paramagnetic centers S=1/2 in a solid
1. Two electrons in same molecule
Usually in molecules all the molecular orbitals are doubly occupied.
Therefore in the ground state each pair of electrons is in singlet
state, and the molecule is diamagnetic.
Molecules in an excited state with the two electrons occupying
two different orbitals can be in a singlet or in a triplet state.
Usually for the states with the same spatial configuration of
orbitals the triplet is at lower energy than the singlet, i.e. J&gt;0.
Photoexcitation takes a molecule from its ground singlet state
to an excited singlet, but spin-orbit coupling can produce the
intersystem-crossing with the two spins passing from a singlet to
a photoexcited triplet state at lower energy.
Photoexcited triplet states
Energy
difference
for antracene
2 eV
Se gli stati di singoletto e di
tripletto hanno dei livelli
vibrazionali molto vicini, in
presenza di perturbazioni dovute
all’accoppiamento spin-orbita (pi&ugrave;
grande per atomi pesanti) uno
degli spin si gira e si passa dallo
stato di singoletto a quello di
tripletto che &egrave; pi&ugrave; basso in
energia. Questo fenomeno si
chiama “Intersystem Crossing” o
ISC.
Dallo stato di tripletto la
molecola pu&ograve; subire una
emissione spontanea di un fotone
(fosforescenza) passando allo
stato fondamentale.
Diagramma di Jablonski
Triplet ground states
In carbenes and nitrenes the two unpaired electrons are
accommodated in a  orbital and in a  orbital respectively
on a single C or N atom. These species can be obtained by
photolysis of a suitable precursor in a glassy matrix or in a
crystal, since in solution they would be non persistent.
N
H
P h e n y lc a r b e n e
P h e n y ln itr e n e
Energy
difference 2J
for
phenylnitrene
0.35 eV
CH3
H3C
H3C
CH3
N
CH3
H3C
H3C
CH3
CH3
N
Radicali nitrossido
CH3
O
O
CH3
H3C
O
O
N
O
CH3
CH3
H3C
CH3
O
Energy
difference 2J
10-6-10-8 eV
O
N
O
CH3
CH3
Biradicali nitrossido
The two Cu(II) ions (S=1/2), are coupled antiferromagnetically.
J depends on the bridge ion.
0.1 eV
Inorg. Chem., 2012, 51 (15), pp 7966–7968
A radical
nitronyl
nitroxide with
its spin density
distribution
The structure
of the radical
crystal
Polyhedron 24 (2005) 2368–2376
The pattern of
exchange
interactions
between pairs of
radicals at
different positions
The triplet states are degenerate?
E
E
2J&gt; 0
0
0
2J&lt; 0
The three triplet states,
corresponding to Ms= +1, 0, -1
(as far as the spin exchange
Hamiltonian H= -2J S1S2 is
concerned), should have the
same energy -1/2 J.
BUT:
The energies of the three states are different, and depend on the Ms
quantum number, thanks to magnetic energy contributions:
a. in presence of a magnetic field B, for the Zeeman interaction
between the two spins magnetic moments and the field;
b. for the dipolar interaction between the magnetic moments of the
two electrons. As this interaction is always present, also without an
external magnetic field, it is called Zero Field Splitting (ZFS). This
will be treated in the following lectures.
Spin Hamiltonian for two interacting
electron spins in a magnetic field
Let us consider first the spin Hamiltonian given by the sum of
the Zeeman terms and of the exchange term. In an magnetic
field B0, is the following:
H  g1 B S1 B0  g 2  B S 2 B0  2JS1S 2
Consider B0//z. Then:
H  g1 B S z1 B0  g 2  B S z 2 B0  2JS1S 2
If g1g2 the eigenfunctions of the Zeeman term and of the exchange term are
respectively those of the uncoupled and of the coupled basis. This means that neither
of the two basis diagonalizes the Hamiltonian matrix, and the eigenfunctions of H will be
a specific linear combination depending on the relative values of the difference in
Zeeman energies and the exchange parameter J. This combinations will be found by
diagonalizing the Hamiltonian matrix.
In conclusion
•
•
•
•
•
•
We have seen that an energy contribution due to electrostatic
electron-electron repulsion can be represented by a spin hamiltonian,
-2JS1S2
The eigenfunctions of the exchange Hamiltonian are the triplet and
singlet functions.
If J&gt;0 the triplet lies lower in energy than the singlet; the absolute
value of energy difference is 2J.
The value of J depends on the extent of electrostatic interaction
between the two electrons; approximately it decreases linearly on
increasing the distance r between the electrons.
J is “large” for electrons on the same molecule, as in photoexcited
triplets for molecules with a ground singlet state, or in molecules with
ground triplet states, as carbene and nitrene. In these cases all the
magnetic interactions are smaller than the spin exchange term (the spin
coupled basis is a good basis).
J is “small” for biradicals, which are usually two stable radicals linked
together by a non-conjugated chain. The energy difference 2J can be
of the same order as kT, and the populations of the triplet and singlet
in this case depend on the temperature. The spin eigenfunctions depend
on the relative values of the various terms of the spin Hamiltonian.
```