The 1-dimensional Kronig-Penney Model: Trying to understand the relation between hopping energy and effective mass

The 1-dimensional Kronig-Penney Model predicts a relationship between energy, $$E$$ and wavenumber, $$k$$ of the form: $$\cos(ka)=\cos(qa) - \frac{m_e\,A\,t_0\,\sin(qa)}{\hbar^2\,qa}$$ where $$q=\sqrt{\frac{2m_e\,E}{\hbar^2}}$$ and $$m_e$$ is the electron mass, $$\alpha$$ is the lattice constant, $$A$$ [$$m^2$$] is a constant, and $$t_0$$ is the hopping energy. In the limit of small $$k$$ and small $$E$$, find an approximate dispersion relationship $$E(k)$$ for the model. Show that the effective mass $${m_e}^*$$ is related to the magnitude of the hopping energy, $$t_0$$ by: $${m_e}^*=m_e\left(1-\frac{m_e\,A\,t_0}{3\hbar^2}\right)$$

Using an approximation for working near the band edge: in this case, choosing $$ka \ll 1$$. Also, noting that $$qa \ll 1$$ for small values of $$k$$. I expand the trig functions to find terms to second order in $$k$$ and $$q$$, such that,

$$1-\frac{k^2\,a^2}{2}=1-\frac{m_e\,E\,a^2}{\hbar^2}-\frac{m_e\,A\,t_0}{q\,a\,\hbar^2}\left(qa-\frac{q^3\,a^3}{6}\right)$$ $$\implies \frac{k^2\,a^2}{2}=\frac{m_e\,E\,a^2}{\hbar^2}+\frac{m_e\,A\,t_0}{\hbar^2}\left(1-\frac{m_e\,E\,a^2}{3\hbar^2}\right)$$

$$\implies \frac{k^2\,a^2}{2}=\frac{m_e\,a^2}{\hbar^2}E\left(1-\frac{m_e\,A\,t_0}{3\hbar^2}\right)+\frac{m_e\,A\,t_0}{\hbar^2}$$

Rearranging this to obtain a dispersion relation:

$$E=\frac{\hbar^2\,k^2}{2\,m_e}\left(1-\frac{m_e\,A\,t_0}{3\hbar^2}\right)^{-1}+t_0\frac{A}{a^2}\left(1-\frac{m_e\,A\,t_0}{3\hbar^2}\right)^{-1}$$

I got everything right up to this point.....

....but then the solution says:

We can immediately identify the effective mass: $${m_e}^*=m_e\left(1-\frac{m_e\,A\,t_0}{3\hbar^2}\right)$$ as required. It is worth thinking about this result...It implies that the larger the hopping energy, the smaller the effective mass. Does this make sense to you? Note that this expression is valid for the lowest-lying energy levels. It is essentially the tight-binding model but in a different parameterisation.

How was the author able to

immediately identify the effective mass $$\large({\color{red}{\large{?}}}\large)$$

This is far from obvious to me. From a previous question asked by me: What does it mean to say that the Fermi energy is equal to the hopping energy?, I have seen energy expressions involving the hopping integral $$t$$, such as $$E_{\bf{ k}}=-2t\left[\cos(k_x\,a)+\cos(k_y\,a)+\cos(k_z\,a)\right],$$ $$E_F=+4t,$$ and with an offset, $$\epsilon$$:$$E_{\bf{ k}}=\epsilon-2t\left[\cos(k_x\,a)+\cos(k_y\,a)\right]$$

The author also writes

It implies that the larger the hopping energy, the smaller the effective mass. Does this make sense to you?

It doesn't make sense to me at all.

Could someone please explain what the author is saying, as I would really like to understand this?

Edit:

I still don't see how $${m_e}^*\propto \frac{1}{t_0}$$ It seems very much to me that since $$E=\frac{\hbar^2\,k^2}{2\,m_e}\left(1-\frac{m_e\,A\,t_0}{3\hbar^2}\right)^{-1}+t_0\frac{A}{a^2}\left(1-\frac{m_e\,A\,t_0}{3\hbar^2}\right)^{-1}$$ and $${m_e}^*=m_e\left(1-\frac{m_e\,A\,t_0}{3\hbar^2}\right)$$ then $$E=\frac{\hbar^2\,k^2}{2\,m_e}\frac{m_e}{{m_e}^*}+t_0\frac{A}{a^2}\frac{m_e}{{m_e}^*}\implies E\propto \frac{t_0}{{m_e}^*}\implies {m_e}^*\stackrel{\eqref{*}}\propto t_0$$

There are usually two kind of complementary approximations which one use in band theory. One is assuming parabolic dispersion $$E(k)=\frac{\hbar^2k^2}{2m_e^*} + E_0$$ Neglecting the offset energy $$E_0$$, this is equivalent to the dispersion of a free electron (for example in the vacuum) which can be described (at low speed $$) by the Hamiltonian $$H=\frac{p^2}{2m_e}$$ The first equation can be derived from the second by using $$p=-i\hbar\partial_x$$ and by assuming plane waves wacefunctions $$\psi(r,x)\propto e^{i k x}$$. In this sense, the mass $$m_e^*$$ is called effective, because it is in general different from the "real" mass of the electron. Now if you compare the first equation with your dispersion relation that you wrote: $$E=\frac{\hbar^2k^2}{2m_e}\times \text{(something)}^{-1} + \text{(something else)}$$ you recognize that $$m_e^*=m_e \times \text{(something)}$$ In general, you can simply define the effective mass in terms of the second derivative of the dispersion, that is $$m_e^*=[(\partial_k^2 E)/\hbar^2]^{-1}$$ which again gives the same result.

Now regarding the second part of your question, I mentioned that there are two main kind of approximations in band theory. The second is the tight binding: $$E=-2t\cos{k}$$ where I just consider the 1-dimensional case. Now close to $$k\approx0$$ one recovers the parabolic dispersion because one has $$E=-2t\cos{k}\approx -2t + t k^2$$ Therefore by comparing with the parabolic dispersion you can identify $$t=\frac{\hbar^2}{2m_e^*}$$ which is the relation between the hopping parameter and the effective mass. Alternatively, if you want, you can again define the effective mass in terms of the 2nd derivative of the dispersions and obtain $$m_e^*=[(\partial_k^2 E)/\hbar^2]^{-1}=\frac{\hbar^2}{2t}$$ Now, you can clearly see that the larger the hopping $$t$$, the smaller the effective mass, as the author says.

Long story short, the effective mass and the hopping are defined as $$t=\frac{\hbar^2}{2m_e^*}=\frac12\partial_k^2 E(k)$$

Edit:

To answer the last edit of the question, which has the equation $$E=\frac{\hbar^2\,k^2}{2\,m_e}\frac{m_e}{{m_e}^*}+t_0\frac{A}{a^2}\frac{m_e}{{m_e}^*}\implies E\propto \frac{t_0}{{m_e}^*}\implies {m_e}^*\stackrel{\eqref{*}}\propto t_0$$

I notice that these implications are false. The energy is proportional to the inverse of the effective mass. The energy is written as a sum of two terms. The first term is proportional to $$k^2$$ and can be written as $$tk^2$$ where $$t=\hbar^2/(2m_e^*)$$ is the hopping (as I wrote above). The second term do not depend on the momentum. The constant $$t_0$$ is not the hopping. Also, you cannot write that the energy is proportional to $$t_0$$ because you cannot just ignore the first term. Up to a constant term, the energy is proportional to $$k^2$$ and the proportional constant is $$t\neq t_0$$.

• Sorry for the delayed response.I'm just a bit confused about "Now, you can clearly see that the larger the hopping $t$, the smaller the effective mass, as the author says." Since $E=\frac{\hbar^2\,k^2}{2\,m_e}(1-\frac{m_e\,A\,t_0}{3\hbar^2})^{-1}+t_0\frac{A}{a^2}(1-\frac{m_e\,A\,t_0}{3\hbar^2})^{-1}$ and ${m_e}^*=m_e(1-\frac{m_e\,A\,t_0}{3\hbar^2})$ then $E=\frac{\hbar^2\,k^2}{2\,m_e}\frac{m_e}{{m_e}^*}+t_0\frac{A}{a^2}\frac{m_e}{{m_e}^*}\implies E\propto \frac{t_0}{{m_e}^*}\implies {m_e}^*\propto t_0$. So it seems the hopping is directly proportional to effective mass? Thanks. Jun 9, 2020 at 6:49
• No. The energy is proportional to the inverse of the effective mass, and I think we can agree on this. The energy is written as a sum of two terms. The first term is proportional to $k^2$. This term can be written as $t k^2$ where $t$ is the hopping. The second term do not depend on the momentum. The constant $t_0$ is not the hopping, it is just part of the constant term. I think that the notation of the book are maybe confusing but they are correct. Jun 9, 2020 at 15:02
• Also, you cannot write that the energy is proportional to $t_0/m_e^*$ because you cannot just cancel the first term. The second term which is proportional to $t_0$ is just a constant energy shift. Jun 9, 2020 at 15:04
• I generally agree with this answer. However, I wouldn't oppose effective mass approximation and tight-binding approximation. First of all, you demonstrate that the former is possible in the context of the latter. In addition, as a method of approximating the band structure, tight-binding is usually opposed to the nearly free electrons approach. Jun 10, 2020 at 8:59
• @Vadim I agree, the two approximations are not "opposed". Actually, they become equivalent in the low energy description, because one can bee seen as the approximation of the other, and vice versa, at the bottom of the band. Jun 10, 2020 at 9:04