Nearly Free Electron | Exercise 3, question 3

Physics Q&A
1

The answer claims that \bra{n} H_n \ket{n} = -\frac{\hbar^2 \kappa^2}{2m}. But when I calculate it the delta potential also introduces a factor.

See here my derivation:
\bra{n} H_n \ket{n} = \bra{n} -\frac{\hbar^2 }{2m} \frac{\partial^2}{\partial x^2}- \lambda \delta(x-na) \ket{n} = -\frac{\hbar^2 \kappa^2}{2m} - \lambda \int \kappa^2 e^{-2 \kappa |x-na|} \delta(x-na) =-\frac{\hbar^2 \kappa^2}{2m} - \lambda \kappa^2.

Did I do something that is not allowed?

I also had a unrelated question, in general \bra{\psi_i} H \ket{\psi_j} = \int_{-\infty}^{\infty} \psi_i^* H \psi_j. But in this tutorial session we’d use a single period as the boundaries. i.e: \bra{\psi_i} H \ket{\psi_j} = \frac{1}{a} \int_{0}^{a} \psi_i^* H \psi_j. I see that it allowes us to calculate stuf because otherwise the integral would blow up, but I don’t see how they’re one and the same, if they are. Or why we’re permitted to chose those bounds, or where they come from.

Thanks!

2

When taking the derivative, did you take into account that it is discontinuous at x=0? Then using \kappa=m\lambda/\hbar^2 you can rewrite the second term to eliminate \lambda.

Because the Hamiltonian and Bloch wave-functions repeat every lattice spacing, the integrand in
\langle \psi_i | H | \psi_j \rangle is identical in every unit cell (you just have to renormalize it properly). So all the other cells will have the same physics as one unit cell. Therefore we only need to integrate over one cell.

3

to me honest, for the kinetic term I just used what they used. Which is the free electron energy \epsilon_0 = \frac{\hbar^2 \kappa^2}{2m}. I did just try to calculate it given that \ket{n} = \kappa e^{-2 \kappa |k-na|}.

\bra{0} \frac{- \hbar^2}{2m} \frac{\partial^2}{\partial x^2} \ket{0} = \frac{1}{a} \frac{-\hbar^2}{2m} \int_{-a/2}^{a/2} \kappa^2 \frac{\partial^2}{\partial x^2} e^{-2 \kappa|x|} dx = \frac{1}{a} \frac{-\hbar^2}{2m} \kappa^2 \left( \int_0^{a/2} 4\kappa e^{-2 \kappa x} dx + \int_{-a/2}^0 4 \kappa e^{2 \kappa x} \right) = \frac{1}{a} \frac{-\hbar^2}{2m} 4 \kappa^3 \left( 1 - e^{-a \kappa} \right)

Which is something else entirely.

What I’m confused by though. The wavefunctions are defined over all space right? In the calculation above the effects of integrating over the BZ adds the complex exponential term. Which is peculiar to me.

Could you clarify me these things?

4

If you seek an explicit calculation for the kinetic energy of a particle with a delta potential, you can look at chapter 2.5 (I think) of Griffiths or check the wikipedia article. It is a well known result and we directly use it in the lecture notes.

BTW: The given |n\rangle has a wrong normalization factor; it should be \sqrt{\kappa}, fixed it now.

First part:

Calculation for derivative

(https://math.stackexchange.com/questions/2432987/how-do-i-calculate-derivative-of-sgnx):

\phi(x)=\sqrt{\kappa}\,e^{-\kappa |x|}.
\frac{d^{2}\phi}{dx^{2}} =\kappa^{2}\phi-2\kappa\sqrt{\kappa}\,\delta(x) =\kappa^{2}\phi-2\kappa\,\phi(0)\,\delta(x).

So

\begin{aligned} \langle T\rangle &= -\frac{\hbar^{2}}{2m}\int \phi(x)\,\phi''(x)\,dx \\[6pt] &= -\frac{\hbar^{2}}{2m} \left[ \kappa^{2}\!\!\int \phi^{2}\,dx -2\kappa\,\phi(0)^{2} \right] \\[6pt] &= -\frac{\hbar^{2}}{2m} \left[ \kappa^{2}\,\underbrace{\int \phi^{2}\,dx}_{\displaystyle =1} -2\kappa^{2} \right] \\[6pt] &= -\frac{\hbar^{2}}{2m}\,[\kappa^{2}-2\kappa^{2}] = \frac{\hbar^{2}\kappa^{2}}{2m}. \end{aligned}
\implies \langle n|H_n|n\rangle = \langle T\rangle + \langle V\rangle = \frac{\hbar^{2}\kappa^{2}}{2m} - \lambda\kappa = \frac{\hbar^{2}\kappa^{2}}{2m} - \frac{\hbar^{2}\kappa^{2}}{m} = -\frac{\hbar^{2}\kappa^{2}}{2m} \equiv \epsilon_0 \text{ (in the lecture notes)}.

Second part:
What we are doing is the following:

\int_{-\infty}^{\infty} (\cdots)\,dx = \sum_{\text{cells}}\int_{0}^{a} (\cdots)\,dx = N\underbrace{\int_{0}^{a} (\cdots)\,dx}_\text{cell} ,

where L=Na is the length of the system. Now what we are interested in is the integral over each individual cell, because (again) this quantity has all the information as this quantity is related to the energy bands and other observables. We are not actually interested in the total integral, as when we think of an infinite lattice, this integral over the crystal will diverge. So yes, the wavefunction is defined over the entire lattice, but we do not focus on the entire space when integrating.

(Now note that you can also drop the N factor and just consider the integral over one unit cell and renormalize the wavefunctions to get the same result).