Ashcroft Mermin eng · Problem 2

Failures of the Static Lattice Model

Physics — Statistical Mechanics · electron-surface-levels-weak-periodic-potential

Problem

Electron Surface Levels for a Weak Periodic Potential The method of Chapter 9 can be used to investigate electronic surface levels. Consider a semi-infinite crystal with a plane surface, perpendicular to a reciprocal lattice vector K (crystal surfaces are parallel to lattice planes). Taking the xx-axis to lie along K and the origin to be a Bravais lattice point, to a rough approximation we can represent the potential of the semi-infinite crystal by V(r)=U(r)V(\mathbf{r}) = U(\mathbf{r}), x<ax < a; V(r)=0V(\mathbf{r}) = 0, x>ax > a. Here U(r)U(\mathbf{r}) is the periodic potential of the infinite crystal. The distance aa lies between zero and the interplanar distance in the family of planes parallel to the surface, and should be chosen in any particular problem to give that U(r)U(\mathbf{r}) which most resembles the actual potential at the surface. We continue to assume that the Fourier components UKU_\mathbf{K} are real. However, if we wish the lowest level outside of the crystal to have energy 0, we can no longer neglect the zeroth Fourier component U0U_0 within the crystal, as we did in Chapter 9. The result of retaining U0U_0 is simply to shift the formulas of Chapter 9 for the levels within the crystal by that amount. Note that U0U_0 need not be small for the method of Chapter 9 to work, in contrast to the Fourier components UKU_\mathbf{K}, with K≠0K \neq 0. We examine a Bloch level for the infinite crystal with a wave vector k\mathbf{k} that is near the Bragg plane determined by K, but near no other Bragg plane, so that in a weak periodic potential the wave function of the level is a linear combination of plane waves with wave vectors k\mathbf{k} and k−K\mathbf{k} - \mathbf{K}. In Chapter 9 we required k\mathbf{k} to be real, to satisfy the Born-von Karman boundary condition. In a semi-infinite crystal, however, the component of k\mathbf{k} normal to the crystal surface need not be real, provided that it yields a decaying wave in the negative xx-direction (into the metal). Outside of the metal, the Bloch function must be joined onto a solution of the free-space Schr\"odinger equation that decays in the positive xx-direction (away from the metal). Thus, outside the metal we take ψ(r)=e−px+ik∥⋅r,x>a,(18.33)\psi(\mathbf{r}) = e^{-px + i\mathbf{k}_\parallel \cdot \mathbf{r}}, \qquad x > a, \tag{18.33} and inside, ψ(r)=eqx+ik0x+ik∥⋅r(ck+ck−Ke−iKx),x<a,(18.34)\psi(\mathbf{r}) = e^{qx + ik_0 x + i\mathbf{k}_\parallel \cdot \mathbf{r}}(c_\mathbf{k} + c_{\mathbf{k}-\mathbf{K}} e^{-iKx}), \qquad x < a, \tag{18.34} where k∥\mathbf{k}_\parallel is the part of k\mathbf{k} parallel to the surface, and the coefficients in (18.34) are determined by the secular equation (9.24) (with the energy E\mathcal{E} shifted by the constant U0U_0): (E−Ek0−U0)ck−UKck−K=0,−UKck+(E−Ek−K0−U0)ck−K=0.(18.35)\begin{aligned} (\mathcal{E} - \mathcal{E}^0_\mathbf{k} - U_0)c_\mathbf{k} - U_\mathbf{K} c_{\mathbf{k}-\mathbf{K}} &= 0, \notag \\ -U_\mathbf{K} c_\mathbf{k} + (\mathcal{E} - \mathcal{E}^0_{\mathbf{k}-\mathbf{K}} - U_0)c_{\mathbf{k}-\mathbf{K}} &= 0. \tag{18.35} \end{aligned} (a) Verify that for (18.35) to yield real energies for q≠0q \neq 0 it is necessary that k0=K/2k_0 = K/2. (b) Show that when k0=K/2k_0 = K/2, the resulting energies are E=ℏ22m(k∥2+14K2−q2)+U0±UK2−(ℏ22mqK)2.(18.36)\mathcal{E} = \frac{\hbar^2}{2m}\left(k_\parallel^2 + \frac{1}{4}K^2 - q^2\right) + U_0 \pm \sqrt{U_\mathbf{K}^2 - \left(\frac{\hbar^2}{2m}qK\right)^2}. \tag{18.36} (c) Show that continuity of ψ\psi and ∇ψ\nabla\psi at the surface leads to the condition p+q=12Ktan⁡(K2a+δ),(18.37)p + q = \frac{1}{2}K \tan\left(\frac{K}{2}a + \delta\right), \tag{18.37} where ckck−K=e2iδ.(18.38)\frac{c_\mathbf{k}}{c_{\mathbf{k}-\mathbf{K}}} = e^{2i\delta}. \tag{18.38} (d) Taking the case a=0a = 0, and using the fact that outside of the metal E=ℏ22m(k∥2−p2),(18.39)\mathcal{E} = \frac{\hbar^2}{2m}(k_\parallel^2 - p^2), \tag{18.39} show that (18.35) to (18.39) have a solution: q=−14KUKE0sin⁡2δ,q = -\frac{1}{4}K\frac{U_\mathbf{K}}{\mathcal{E}_0}\sin 2\delta, where sec⁡2δ=−(U0+UK)E0,E0=ℏ22m(K2)2.(18.41)\sec^2\delta = -\frac{(U_0 + U_\mathbf{K})}{\mathcal{E}_0}, \qquad \mathcal{E}_0 = \frac{\hbar^2}{2m}\left(\frac{K}{2}\right)^2. \tag{18.41} (Note that this solution exists only when U0U_0 and UKU_\mathbf{K} are negative, and ∣U0∣+∣UK∣>E0|U_0| + |U_\mathbf{K}| > \mathcal{E}_0.)

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