Derivation of the Heisenberg Hamiltonian
for The Hydrogen Molecule
As we have seen that dipolar interaction and spin orbit coupling can't be the main factor underlying the model of interacting atomic moments. The appropriate leading factor of the interaction must be of the order of 0.1eV to 1eV. Quantum mechanics under the consideration of the Pauli exclusion principle provides the source of interaction which is surprisingly electrostatic rather than spin dependent.
We'll consider the very simple case of solid comprising N=2 Hydrogen atoms, which in fact is the hydrogen molecule, in order to illustrate how the Pauli principle can lead to magnetic effects even when the Hamiltonian doesn't contain spin-dependent terms.
We consider the hydrogen molecule as consisting of two protons at fixed distance from each other, and two electrons. We include in the Hamiltonian all the electrostatic interactions between the protons and electrons (and neglecting spin depended terms). Hence the Hamiltonian is of the form:
(1)
and the corresponding Schrodinger equation is:
(2)
where P_{1}, P_{2} are the electron momentum operators, m and e are the electron mass and charge respectively, and r = r_{αβ} is the distance between particles α and β (1,2 refer to electrons, a,b refer to protons and r = r_{ab}). Since there are no magnetic (spin) interaction terms in the Hamiltonian, the wave function for the electrons can be written as a product of the spatial and spin wave functions:
Ψ(r_{1}σ_{1}, r_{2}σ_{2}) =
φ( r_{1}, r_{2})
Χ(σ_{1},σ_{2}) .
Where σ_{1} and σ_{2} are the projection of the
electron spins along a given axis usually denoted as the z axis.
One may regard Χ( σ_{1}) as |↑> and Χ(
σ_{2}) as |↓>.
According to the Pauli exclusion principle this wave function must be antisymmetric
under electron interchange, namely it must change sign under the simultaneous
interchange of both space and spin coordinates:
ψ(r_{1}σ_{1}, r_{2}σ_{2}) = -
ψ(r_{2}σ_{2}, r_{1}σ_{1})
Hence a symmetric spatial function must be associated with an antisymmetric
spin function and an antisymmetric spatial function must be associated with a
symmetric spin function. Note that all solutions to (2) must be symmetric or
antisymmetric only, because the potential energy is symmetric and hence the
Hamiltonian commutes with the parity operator.
Recalling spin addition for s_{1} = 1/2 and s_{2} =1/2 we get
four spin wave functions describing the total spin of the two electron system:
the Singlet state which is antisymmetric under electron interchange-
Χ_{s} = 1/√2(|↑↓> - |↓↑>) with
total spin S=0 and hence projection on the z axis S_{z} = 0,
and three Triplet states which are symmetric-
Χ_{t} = |↑↑> with S=1, S_{z} =1
Χ_{t} = 1/√2( |↑↓> + |↓↑> ) with S=1, S_{z} =0
Χ_{t} = |↓↓> with S=1, S_{z} =-1
Since the total wave function is the product of its spin and space functions the spatial function associated with the singlet state must be symmetric and the one associated with the triplet states must be antisymmetric. We note that the symmetric solution of the Schrodinger equation (2) corresponds to antiparallel alignment of electronic spins (which in this case are considered as the atomic spins) and the antisymmetric solution to parallel alignment of the spins. Thus
even though the Hamiltonian doesn't consist of magnetic (spin) terms, the energy of the system depends on the relative orientation of the spins, since the different spatial functions, symmetrically speaking, solving the Schrodinger equation correspond to different energy values.
We denote the spatial functions solving the Schrodinger equation (2) as φ_{symmetric} = φ _{s} and φ_{antisymmetric} = φ_{a}.
Assuming that the two hydrogen atoms constituting the molecule are in their
ground state and the two protons are well separated so that the interaction
between the two atoms is weak, we can use the Heitler-London approximation
which expresses φ_{s} and φ_{a} by the hydrogen atom ground state wave function:
(3)
Where φ(r_{αi} ) is the normalized ground state wave function of the hydrogen atom consisting of the ith electron (i=1,2) and the αth nucleus (α=a,b) and γ = ∫ φ(r_{a1})φ(r_{b1})dr_{1} = (1 + r/a_{0} + r^{2}/3a_{0}^{2})e^{-r/a0}.
The energies of the molecule for the above states are given by:
(4)
The symbol indicating a complex conjugate is omitted since φ _{s} and φ _{a} are real functions. Substituting the Heitler-London expressions for φ _{s} and φ _{a} we find that
(5)
Where
(6)
and
In the expressions for E(r)_{↑↓} and
E(r)_{↑↑}, terms which are independent of r are omitted since
they represent the energy of two hydrogen atoms at an infinite distance apart.
We can assign A(r) and B(r) a physical meaning. A(r) determines the electrostatic interaction energy of the two atoms on the assumption that each of the electrons is rigidly coupled to one of the nuclei (electron 1 to nucleus a and electron 2 to nucleus b). B(r) determines the energy of the electron exchange between the two atoms, since B(r) is the matrix element of the
electrostatic electron interaction energy U, taken between the state in which
electron 1 is coupled to nucleus a and electron 2 to b, and the state in which
electron 1 is coupled to nucleus b and electron 2 to nucleus a.
The exchange energy falls off rapidly (comparing with the coulomb interaction) with increasing the distance r = r_{ab} = | r_{a} - r_{b} |. One can see that the integrand of B(r) consists of the factors φ(r_{a1})φ(r_{b2}) and φ(r_{a2})φ(r_{b1}) that correspond to states in which each electron is attached to different nucleus. Thus B(r) depends on the overlap of the wave functions φ(r_{ai}) and φ(r_{ai}). Since the atomic function φ(r_{αβ}) is strongly localized in the neighborhood of each nucleus this overlap falls off rapidly. Hence the exchange interaction is considered as a short range interaction.
The lowest of E(r)_{↑↓} and E(r)_{↑↑} will correspond to the ground state energy of the hydrogen molecule and hence will determine the spin alignment of the two atoms. But a more restrictive rule determines which state of E(r)_{↑↓} and
E(r)_{↑↑} is favorable. The next figure shows schematically the variation of E(r)_{↑↓} and E(r)_{↑↑} with increasing r - the distance between the two nuclei.
Click to view larger image
It is evident from the figure that only the curve E(r)_{↑↓} can provide a bound
state i.e. the hydrogen molecule (when the two atoms are in ground state). Hence the hydrogen molecule can be formed only if the two electron spins are antiparallel.
This example illustrates clearly that the purely quantum-mechanical effect of
electron exchange leads to an effective interaction between the atoms which is
strongly dependent on the relative orientations of the atomic spins, although
the Hamiltonian doesn't take into consideration any spin variable. Hence it
explains the nature of magnetically ordered crystals, and the N=2 hydrogen
"solid" is considered as antiferromagnetic. This example also shows that the exchange effect
underlies the covalent bond.
For the hydrogen molecule example one may also introduce the energy splitting
E(r)_{↑↓}- E(r)_{↑↑}, which in the limit of
large spatial separation of the nuclei can be reduced to E(r)_{↑↓}- E(r)_{↑↑} = 2B(r) (since in the limit of large r we have γ = 1). For the hydrogen molecule the exchange energy is negative ( B(r)<0 ) so the state when the resultant
spin is S=0 has the lower energy than the state when S=1, and again, it's regarded as
antiferromagnet. For solids with B(r)>0 the lower state is with S=1 and hence
they are regarded as ferromagnets.
The next step is to introduce a convenient operator, which though in the above case is unnecessary, is of great importance in the case of real solids (crystals). Since the energy splitting E(r)_{↑↓}-E(r)_{↑↑} is small compared with all other excitation energies of the hydrogen molecule one may regard it as a simple four states (one singlet and three triplets) system and ignore higher energy states (k_{B}T is comparable to E(r)_{↑↓}- E(r)_{↑↑} ,but small enough that no states other than the four are thermally excited). If we represent a general state of the molecule as a linear combination of the four states, it is convenient to have an operator, known as the spin Hamiltonian, whose eigenvalues are the same as those of the original Hamiltonian within the four state manifold, and whose eigenfunctions give the total spin of the corresponding states. To construct the spin Hamiltonian for two electron system, we note that for each electron S_{i}^{2} = S_{i}(S_{i}+1) = 1/2(1/2+1) = 3/4 so that the total spin
satisfies
S^{2} = (S_{1} + S_{2})^{2} = 3/2 +
2S_{1}·S_{2}.
Since the eigenvalue of S^{2} is S(S+1), it follows that the operator
S_{1}·S_{2} has eigenvalue -3/4 for the
singlet state (S=0) and +1/4 for the triplet states (S=1). Consequently the
operator
H = 1/4[E(r)_{↑↓} + 3E(r)_{↑↑}] - [E(r)_{↑↓}-E(r)_{↑↑}] S_{1}·S_{2}
has the eigenvalue (E(r)_{↑↓} in the singlet state and
E(r)_{↑↑} in each of the triplet states, as desired. By
redefining the zero of energy, the term 1/4[E(r)_{↑↓} +
3E(r)_{↑↑}], which determines the mean energy of the atoms
and is common to all four states, may be omitted. Now we can write down the
spin Hamiltonian as
H_{e} = -J(r)
S_{1}·S_{2}
whereas J(r) = E(r)_{↑↓}- E(r)_{↑↑}
determines the exchange energy. Hence this operator is also called the
exchange Hamiltonian.
Since H_{e} is the scalar product of S_{1} and
S_{2}, it will favor parallel spins if J is positive and
antiparallel if J is negative.