Department of physics, NTNU TFY4340 Mesoscopic Physics Spring 2010

Solution to Exercise 4

a) With the given gaussians, the inner product ⟨i|j⟩ is

⟨i|j⟩ = C^{2} ∫ dx e^{−κ(x−ia)^{2}−κ(x−ja)^{2}}
= C^{2} ∫ dx e^{−κ(2x^{2}−2(i+j)ax+(i^{2}+j^{2})a^{2})}
= C^{2}e^{−κa^{2}(i−j)^{2}/2} ∫ e^{−2κ(x−(i+j)a/2)^{2}}
= C^{2}e^{−κa^{2}(i−j)^{2}/2} · \sqrt{π/2κ}

where we have completed the square in the exponent, as suggested in the exercise. In this exercise, all integrals extend from −∞ to +∞. As long as κa^{2}/2 ≫ 1, and choosing C = (2κ/π)^{1/4}, we have

⟨i|j⟩ ≃ δ_{ij}.

b) From the lectures, we have

γ = −⟨1| ∑_{m≠1} V_{m}|2⟩
ε_{0} = α − α′ = −⟨2| ∑_{m≠2} V_{m}|2⟩ + ⟨1| ∑_{m≠1} V_{m}|1⟩

In the notation of the lectures (p 29), we have here chosen j = l = 1 in the matrix element α′, j = l = 2 in the matrix element α, and j = l − 1 = 1 in the matrix element γ. Each of these matrix elements is a sum over the index m, and all terms are, at the outset, assumed to be ”small”. The largest of these small terms correspond to m = 2 in γ, m = 2 in α′, and m = 1 and m = 3 in α. All these terms in α and α′ are equal. Hence, if we keep only the largest term(s) in the sums over m, we end up with

γ = −⟨1|V_{2}|2⟩
ε_{0} = −⟨1|V_{2}|1⟩

Here, both matrix elements are integrals of products of gaussians. Just like in a), we may solve the integral by completing the square in the exponent. This yields

γ/C^{2}V_{0} = \sqrt{\frac{π}{2κ + β}} exp (−a^{2} · \frac{κ^{2} + κβ}{2κ + β} )
ε_{0}/C^{2}V_{0} = \sqrt{\frac{π}{2κ + β}} exp (−a^{2} · \frac{2κβ}{2κ + β} )

and finally

y = \frac{ε_{0}}{γ} = exp (a^{2} · \frac{κ(κ − β)}{2κ + β}) ,

which is larger than 1 if κ > β, i.e., if |j⟩ decays faster than V_{j}.