Anderson-Newns-Grimley model

The ANG Hamiltonian describes an adatom-substrate system with a single adatom state $a$ and substrate states $k$ . The electrons can (de)localize over any of these states. Each state can be occupied by an up-spin electron ( $\sigma=+$ ) and a down-spin electron ( $\sigma=-$ ).

On the adatom, this can optionally lead to Coulomb repulsion between the electrons. This repulsion is a two-body interaction, described by an operator

W = {1\over 2} \sum_{ijkl} w_{ijkl} c_i^\dag c_j^\dag c_k c_l

(1)

By the Pauli principle, the two electrons occupying the adatom state must have opposite spins. Thus we only have the matrix elements

w_{+-+-}=w_{-+-+}=U

(2)

W={1 \over2} U \left( c_+^\dag c_-^\dag c_-c_+ + c_-^\dag c_+^\dag c_+ c_- \right)

(3)

From the anticommutation relations of the creation and annihilation operators for different spins, it turns out that all these creation/annihilation operator pairs can be combined into number operators, so that

W = U n_+n_-.

(4)

This operator simply counts whether the adatom state is doubly occupied, and if so, assigns an energy penalty $U$ .

By adding up all the energy contributions, the Hamiltonian reads

H_\mathrm{ANG} = \sum_\sigma \left\{ \left[ \varepsilon_a n_{a\sigma} + \sum_k \left( \varepsilon_k n_{k\sigma} + V_{ak} c_{a\sigma}^\dag c_{k\sigma} + h.c. \right) \right] \right\} + U n_{a+}n_{a,-} \tag{4.33}

(5)

where $h.c.$ implies the Hermitian conjugate of the term before it. In summary, we have:

the self-energy of electrons on the adatom state, $\varepsilon_a$ , which is counted for each electron on the adatom;
the self-energy on the substrate, $\varepsilon_k$ ;
the interaction or hybridization between the substrate and adsorbate $V_{ak}$ . Note that hybridization is unrelated to the spin: a spin-up electron can delocalize between adatom and substrate, and the spin-down electron can also do so, independently of one another;
the electron-electron repulsion $U$ on the adatom, which is counted only if it is doubly occupied.

Hartree-Fock approximation¶

In the Hartree-Fock approximation, the electron-electron repulsion is written as the average interaction with the opposite-spin electron. The e-e repulsion part of the Hamiltonian is written

U n_{a\sigma}n_{a,-\sigma}=U n_{a\sigma}\langle n_{a,-\sigma}\rangle + U n_{a,-\sigma} \langle n_{a,\sigma}\rangle - U \langle n_{a\sigma}\rangle\langle n_{a,-\sigma}\rangle

(6)

This approximation allows us to include the repulsion energy into the self-energy, as a spin-dependent self-energy:

\varepsilon_{a\sigma} = \varepsilon_a + U\langle n_{a,-\sigma} \rangle

(7)

Now, in the Hartree-Fock approximation, we can write the Hamiltonian as the sum of two decoupled Hamiltonians:

H_\mathrm{ANG}=\sum_\sigma H^\sigma

(8)

with

H^\sigma=\varepsilon_{a\sigma} n_{a\sigma}+ \sum_{k} \left[\varepsilon_k n_{k\sigma}+ \left(V_{ak} c_{a\sigma}^{\dagger} c_{k\sigma}+V_{ka}^{*} c_{k\sigma}^{\dagger} c_{a\sigma} \right) \right]

(9)

Important notes:

There are only two possible values for $\sigma$ , up (+) and down (-). Consider that the entire system is completely noninteracting, i.e., each state is occupied by two opposite-spin electrons but they do not interact. Then the total energy of the system can be obtained by solving the system where every state is occupied once, and then multiplying times two. Essentially, $H=H^++H^-=2H^+$ . Now we do say that there is an interaction, but only on the adatom. Because Hartree-Fock decouples $H^+$ and $H^-$ , we can still solve $H^+$ and $H^-$ separately, but $H^+$ depends on the solution of $H^-$ and the other way around. The Hartree-Fock approximation will therefore eventually lead us to a self-consistent procedure.
The total Hartree-Fock-approximated Hamiltonian now leaves out the term $- U \langle n_{a\sigma}\rangle\langle n_{a,-\sigma}\rangle$ . Why? That’s how people do Hartree-Fock: they want to consider the electrons separately. An up-spin electron sees the entire mean-field of the down-spin electron, and this entire energy -- not just half of it -- affects its energy level. Later on, we’ll act like we’re very surprised that the total energy and sum of eigenvalues differ by exactly that value.

Eigenvalues of $H^\sigma$ ¶

The Schrödinger equation in terms of $H^\sigma$ and the ground state $\Phi_0^\sigma$ is

H^\sigma \Phi_0^\sigma = \varepsilon_0 \Phi_0^\sigma \tag{4.36}

(10)

where $\varepsilon_0$ is the ground-state energy. Recall that $\Phi_0^\sigma$ in the occupation representation is something like $[111\cdots00]$ .

By the Hartree-Fock approximation, we can now solve this Schrödinger equation for a single spin state. The Hamiltonian $H^\sigma$ describes all energy levels in the system, but each state is occupied by at most one electron. All electrons have spin $\sigma$ . Or, equivalently, because we do not consider electron-electron repulsion anywhere other than the adatom, we can say that this is the Hamiltonian where the electron on the adatom has spin $\sigma$ .

In the discussion below, we drop the sub/superscript $\sigma$ , and solve for this non-interacting electron system with Hamiltonian

H=\varepsilon_{a} n_{a}+ \sum_{k} \left[\varepsilon_k n_{k}+ \left(V_{ak} c_{a}^{\dagger} c_{k}+V_{ka}^{*} c_{k}^{\dagger} c_{a} \right) \right]

(11)

We can always add $\sigma$ back in by remembering that $\varepsilon_a$ is actually $\varepsilon_{a\sigma}$ , and all states in the system can have $\sigma=+,-$ .

If we add one more electron in a state $m$ , we get an additional energy $\varepsilon_{m\sigma}$ :

H c^\dag_{m} \Phi_0 = (\varepsilon_0 + \varepsilon_{m}) c_m^\dag \Phi_0.

(12)

By subtracting Eq. (4.36) $\times c^\dag_{m}$ we get

[H, c^\dag_{m}] \Phi_0 = \varepsilon_{m} c_m^\dag \Phi_0 \tag{4.39}

(13)

where $[.,.]$ is a commutator bracket.

The energy levels $m$ are some unspecified energy levels of the perturbed system (adatom plus substrate). They could be the set of substrate states plus the adatom state ( $m=k$ or $a$ ), or a set of new molecular orbitals of the combined system. Either way, the change of basis can be written in terms of creation operators as

c_m^\dag = \langle m | a \rangle c^\dag_a + \sum_k \langle m|k \rangle c^\dag_k \tag{4.41}

(14)

By commuting this operator with $H$ and doing some algebra with the (anti)commutator properties of creation/annihilation operators, one gets

[H, c_k^\dag] = \varepsilon_k c_k^\dag + V_{ak} c_a^\dag \quad\text{for $m=k$}

(15)

[H, c_a^\dag] = \varepsilon_a c_a^\dag + \sum_k V^*_{ka} c_k^\dag \quad\text{for $m=a$}

(16)

On the other hand, one can put (4.41) directly into (4.39), and using the above commutators one gets

\begin{aligned} &\left( \langle m|a \rangle \varepsilon_{a} + \sum_k \langle m|k \rangle V_{ak} \right) c_{a}^\dagger \\ &+ \sum_k \left( \langle m|a \rangle V_{ka}^* + \langle m|k \rangle \varepsilon_k \right) c_{k}^\dagger \\ &= \left( \langle m|a \rangle c_{a}^\dagger + \sum_k \langle m|k \rangle c_{k}^\dagger \right) \varepsilon_{m} \end{aligned} \tag{4.46}

(17)

The coefficients in front of $c_a^\dag$ and $c_k^\dag$ should be equal, so

\varepsilon_m \langle m|a \rangle = \varepsilon_a \langle m|a \rangle + \sum_k V_{ak} \langle m|k\rangle

(18)

\varepsilon_m \langle m|k \rangle = \varepsilon_k \langle m|k \rangle + V_{ka}^* \langle m|a\rangle

(19)

These equations are also referred to as “equations of motion” (?).

\varepsilon_m = \varepsilon_a |\langle m|a\rangle|^2 + \sum_k \left[ \varepsilon_k |\langle m|k\rangle|^2 + (V_{ak} \langle m|k\rangle \langle a|m\rangle + h.c.) \right]

(20)

Now, we’re going to bring back the spin. We take the noninteracting system with energy levels $\varepsilon_m$ , fill each level with two electrons, and set the energy on the adatom to $\varepsilon_a\to\varepsilon_a + U\langle n_{a,-\sigma}\rangle=\varepsilon_{a\sigma}$ . Bringing back this electron-electron interaction also makes the (still unspecified) energy levels change as $\varepsilon_m\to\varepsilon_{m\sigma}$ since they may depend on the adatom state.

Summing over all the energy states, and including electrons with both spins,

\begin{aligned} \sum_{m,\sigma} \varepsilon_{m\sigma} &= \sum_{m,\sigma} \bigg[ \sum_q \varepsilon_q |\langle m\sigma|q\sigma\rangle|^2 \\ &+ \sum_k(V_{ak}\langle m\sigma|k\sigma\rangle\langle a\sigma|m\sigma\rangle + h.c.) \bigg] \\ &+ U\sum_\sigma \langle n_{a,-\sigma}\rangle \sum_m |\langle m\sigma|a\sigma\rangle|^2 \end{aligned}

(21)

where $q$ runs over all $k$ and $a$ ; $\varepsilon_q$ are the state ‘self-energies’ i.e. plane-wave energies (also for the adatom (?)). Now, $\sum_m |\langle m\sigma|a\sigma\rangle|^2$ is the sum of probabilities over all states $m$ that an electron localizes on the $a$ -state. Hence it is simply equal to $\langle n_{a\sigma} \rangle$ .

So, surprise: writing out the two terms for $\sigma=-,+$ , we now get two times the mean e-e repulsion term:

\begin{aligned} \sum_{m,\sigma} \varepsilon_{m\sigma} &= \sum_{m,\sigma} \bigg[ \sum_q \varepsilon_q |\langle m\sigma|q\sigma\rangle|^2 \\ &+ \sum_k(V_{ak}\langle m\sigma|k\sigma\rangle\langle a\sigma|m\sigma\rangle + h.c.) \bigg] \\ &+ 2 U\langle n_{a,-}\rangle \langle n_{a+} \rangle \end{aligned}

(22)

The sum of energies of occupied states should give us the total system energy, right?

To check this we take the expectation of $H_\mathrm{ANG}$ , which also yields the total energy of the system:

E =\langle \Psi_0 | H_\mathrm{ANG} | \Psi_0 \rangle

(23)

Here $\Psi_0$ is the Slater determinant of single-particle states $|n\sigma\rangle$ . To take the expectation of $H_\mathrm{ANG}$ in second-quantization notation, we write

\begin{align} \langle \Psi_0 | H_\mathrm{ANG} | \Psi_0 \rangle &= \sum_{n,\sigma} \langle n\sigma | H_\mathrm{ANG} | n\sigma \rangle \\ &= \sum_{n,\sigma} \sum_{q,q'} \langle n\sigma |q\sigma\rangle\langle q\sigma| H_\mathrm{ANG} |q'\sigma\rangle \langle q'\sigma | n\sigma \rangle \end{align}

(24)

For example, there are terms like $\langle q\sigma | c^\dag_{k\sigma} c_{a\sigma} |q'\sigma\rangle=\delta_{q'a} \delta_{qk}$ . To see this, one may write $c_{a\sigma}=|0\rangle\langle a|$ , and $c^\dag_{k\sigma}=|k\rangle\langle0|$ . In this way, the term $\varepsilon_a n_{a\sigma}$ for example has expectation

\sum_{n} \varepsilon_a |\langle n\sigma|a \sigma\rangle|^2

(25)

Continuing like this, we can write

\begin{align} E &=\sum_{n,\sigma} \bigg[ \sum_q \varepsilon_q |\langle n\sigma | q\sigma\rangle|^2 \\ &+ \sum_k(V_{ak}\langle n\sigma | k\sigma\rangle \langle a\sigma | n\sigma\rangle + h.c.)\bigg] \\ &+U\langle n_{a-}\rangle \langle n_{a+} \rangle \end{align}

(26)

which (wow!) differs from $\sum_{m\sigma}\varepsilon_{m\sigma}$ by a term $U\langle n_{a,-\sigma}\rangle \langle n_{a\sigma} \rangle$ .

In conclusion,

E = \sum_{m\sigma}\varepsilon_{m\sigma} - U\langle n_{a,-}\rangle \langle n_{a+} \rangle.

(27)

Anderson-Newns-Grimley model

Hartree-Fock approximation¶

Eigenvalues of HσH^\sigmaHσ¶

Eigenvalues of $H^\sigma$ ¶