Skip to article frontmatterSkip to article content
Site not loading correctly?

This may be due to an incorrect BASE_URL configuration. See the MyST Documentation for reference.

1D problem

Each atom has an atomic orbital ϕm(r)\phi_m(\vec{r}) with m=λm=\lambda for the adsorbate and m,=0,...,N1m,=0,...,N-1 the substrate. We can write the total system wavefunction (molecular orbital) ψk\psi_k as an LCAO, where kk indicates different energy levels of the molecular orbitals:

ψk=mcmkϕm\psi_k =\sum_m c_{mk} \phi_m

Or in bra-ket notation:

k=mmmk|k\rangle = \sum_m|m\rangle\langle m|k\rangle

where cmk=mkc_{mk}=\langle m|k \rangle.

Each molecular orbital satisfies the 1-electron Schrödinger equation

Hk=Ekk.H|k\rangle = E_k |k\rangle.

By inserting k|k\rangle and multiplying by n\langle n|, we get the Schrödinger equation in terms of matrix elements:

mcmknHm=Ekmcmknm.\sum_m c_{mk} \langle n|H|m \rangle = E_k \sum_m c_{mk} \langle n|m \rangle.

If we assume orthonormal atomic orbitals, nm=δnm\langle n|m \rangle= \delta_{nm}. We also write the Hamiltonian in the tight-binding approximation: $$ H =

(ββαββαββ)\begin{pmatrix} \ddots & \beta &&& \\ \beta & \alpha & \beta && \\ & \beta & \alpha & \beta & \\ && \beta & \ddots & \end{pmatrix}

which transforms our Schrödinger equation into

(Ekα)cnk=β(cn+1,k+cn1,k).(E_k-\alpha)c_{nk} = \beta (c_{n+1,k} + c_{n-1,k}).

This equation has three ‘boundary conditions’:

Now we take the Ansatz cn±,k=e±iθkcnkc_{n\pm \ell,k}=e^{\pm i\ell\theta_k} c_{nk}. Actually, this ‘plane wave Ansatz’ is a common Ansatz for a wave function in periodic systems, and corresponds to Bloch’s theorem. The book provides a derivation of this Ansatz from scratch. By inserting this Ansatz, the Schrödinger equation can then be written as

Xk:=Ekα2β=cosθkX_k := {E_k-\alpha \over 2\beta}=\cos \theta_k

This defines the energy band. Hence, β\beta is related to the band width.

We can also express the LCAO coefficient at any point in terms of the plane waves and the coefficient at n=0n=0 (actually not entirely clear to me how using positive and negative waves should be motivated here),

cnk=(aeinθk+beinθk)c0kc_{nk}= (ae^{in\theta_k} + be^{-in\theta_k})c_{0k}

or by re-expressing some parameters in terms of others

cnk=Acos(nθk)+Bsin(nθk)c_{nk}=A \cos(n\theta_k) + B\sin (n\theta_k)

Now we apply the boundary conditions. cNk=0c_{Nk}=0 gives

B=AcosNθksinNθkB=-{A\cos N\theta_k \over \sin N\theta_k}

and with the cosine/sine sum rule this yields

cnk=Asin(Nn)θksinNθkc_{nk} = A {\sin(N-n)\theta_k \over \sin N \theta_k}

which implies that A=c0kA=c_{0k}.

The other two boundary conditions can also be applied. Through some algebra (including some cosine/sine sum rules), one gets an ‘eigenvalue equation’:

(za+2cosθk)[zs+sin(N+1)θksinNθk]=η2(z_a+2\cos\theta_k)\left[z_s + {\sin(N+1)\theta_k \over \sin N\theta_k}\right] = \eta^2

where in the derivation process the following parameters were defined:

Solutions

Let’s consider solutions for η=0\eta=0. Then the solutions θk\theta_k to the eigenvalue problem satisfy either

za=2cosθk-z_a=2\cos\theta_k

(if 2za2-2\leq z_a\leq 2), or

zs=sin(N+1)θksinNθk.-z_s = {\sin(N+1)\theta_k \over \sin N \theta_k}.

The first equation gives one solution and the second one gives NN solutions.

The NN solutions are visible in the graph below, where the purple lines are two possible values of zs-z_s (1 and -1),

desmos-graph(1)

Play with the graph on Desmos

But when zs-z_s becomes larger than 1 or smaller than -1, a solution disappears -- at least it does on the real number line. We can consider solutions for complex θk=ξk+iμk\theta_k = \xi_k + i\mu_k, where μk>0\mu_k>0 (it does not make sense to include μk=0\mu_k=0, because then we get real solutions; I think the reason to not include negative μk\mu_k is because it will give duplicate solutions). We can then write Xk=cosξkcoshμkisinξksinhμkX_k=\cos\xi_k \cosh \mu_k - i\sin \xi_k \sinh \mu_k. However, the energy must be real, so sinξk=0\sin\xi_k=0, which gives ξk=jπ,j=0,1,2,...\xi_k=j\pi,j=0,1,2,.... Consequently cosξk\cos \xi_k is either 1 or -1, so ξk=0\xi_k=0 and ξk=π\xi_k=\pi already describe the complete space of solutions, higher jj just gives duplicates.

So overall we get two possible additional solutions by considering complex θk\theta_k,

Xk=coshμkorXk=coshμk.X_k=\cosh \mu_k \quad\text{or}\quad X_k=-\cosh \mu_k.

The positive one lies above the bulk band, and we call them P-states. The negative one lies below the bulk band, and we call them N-states.

In conclusion, there are N+1N+1 atoms, each with one electron, so there are N+1N+1 molecular orbital states, i.e. N+1N+1 solutions for θk\theta_k:

We now show that bulk band states are delocalized and P/N-states are localized. Find the corresponding graph on Desmos.

Bulk band states

cnk=Asin(Nn)θksinNθkc_{nk} = A {\sin(N-n)\theta_k \over \sin N \theta_k}

Oscillates in nn with constant period. So delocalized.

P-states

Large NN and if θk\theta_k has a positive imaginary part (μk>0)(\mu_k>0):

cnk=sin(N±n)θksinNθkeinθkc_{nk}={\sin(N\pm n)\theta_k \over \sin N\theta_k} \to e^{\mp in\theta_k}

If θk=iμk\theta_k=i\mu_k:

cnk=Aenμkc_{nk}=Ae^{-n\mu_k}

which is a decaying wavefunction.

Inserting the large-NN approximation for the wavefunction in the ‘eigenvalue equation’, we get

(za+2coshμk)(zs+eμk)=η2.(z_a + 2 \cosh \mu_k) (z_s + e^{\mu_k})=\eta^2.

By setting x=eμx=e^\mu we can solve this equation graphically; we get a function

y=(za+x+1x)(x+zs)y=(z_a + x + {1 \over x})(x + z_s)

and we find the intersections with

y=η2.y=\eta^2.

The graphical solution is shown below:

afbeelding

Find the Desmos graph here.

So for these particular parameters there are 2 localized P-states.

N-states

If θk=π+iμk\theta_k=\pi+i\mu_k:

cnk=A(1)nenμkc_{nk}=A(-1)^n e^{-n\mu_k}

Decaying but oscillatory wave-function.

By inserting the large-NN approximation for the N-state wavefunction in the eigenvalue equation, we get an algebraic equation

y=(zax1x)(zsx)y=(z_a - x - {1 \over x})(z_s - x)

which can be solved as before.

afbeelding

The NN-states appear at different parameter values than PP-states. They can even coexist. Check out Desmos!