I am reading this book:Solid State Electronic Devices by Ben G Streetman and Sanjay Kumar Banerjee.
I have some doubts in the article 3.2.2 Effective mass.
In this the aythors say that $E=\dfrac{1}{2}mv^2=\dfrac{1}{2}\dfrac{P^2}{m}=\dfrac{{\hbar} ^2}{2m}k^2$.

  • Electrons usually have thermal velocity of order $10^7$m/s. So Shouldn't we use $E=mc^2$ rather than $E=\frac{1}{2}mv^2$.

The author further says that electron mass is related to curvature of (E,k) relationship as $\dfrac{d^2E}{dk^2}=\dfrac{\hbar^2}{m}\tag{3.2(d)}$. then the author says that the effective mass is given by $m^{*}={{\hbar^2}/{{\dfrac{d^2E}{dk^2}}}}\tag{3.3}$

  • Is equation $3.3$ derived from equation $3.2(b)$. In equation $3.2(d)$ we have $m$ the original mass and in eqn $3.3$ we have $m^{*}$ the effective mass which is completely different from $m$.

In the end of the article the author says that the total force on the electron is given by $F_{tot}=ma$. After this the author says the external force applied on an electron is related to effective mass as: $F_{ext}=m^{*}a$.

  • From where this equation comes? Sometimes the effective mass $m^{*}$ is negative then the equation implies that if we apply certain external force on electrons in the crystal they will accelerate in the opposite direction, how this is possible? What's going on?
  • $\begingroup$ The effective mass is just a trick to avoid having to think about two positive and negative charge. $\endgroup$ – Danu Mar 21 '14 at 16:50
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    $\begingroup$ @Danu, does it indicate an instability in the system? $\endgroup$ – innisfree Mar 21 '14 at 17:02
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    $\begingroup$ Related: physics.stackexchange.com/q/101317 $\endgroup$ – user31782 Apr 30 '14 at 6:42

First of all, $E=mc^2$ is rest energy, not kinetic. Relativistic kinetic energy is given by $$E_K=(\gamma-1)mc^2,$$

where $$\gamma=\frac1{\sqrt{1-v^2/c^2}}.$$ Now, for a particle moving at speed of $10^7 m/s$, we have $\gamma\approx1.00504$. That's pretty close to 1, so non-relativistic approximation of $E_K=\frac12mv^2$ works quite nicely.

In $(3.2(d))$ the expression is given for free electron, so $m$ is its mass as of electron in vacuum.

In $(3.3)$ the expression defines the effective mass of crystal electron in terms of its dispersion relation in given band. In this case, of course, $m^*$ is indeed different from $m$ in general, though for free electron (i.e. crystal with constant potential everywhere) we'd get $m^*=m$.

Now to forces and negative effective mass. Electron in crystal doesn't have the usual parabolic dispersion relation as you know for vacuum electron. In some hypothetic 1D crystal with lattice constant $a=2$ dispersion relation for lowest band could look like


Effective mass in such a band will be $$m^*_{hyp}=1\left/\frac{d^2E}{dk^2}\right.=\frac12\sec(2k).\tag1$$

You can see that here effective mass at center of Brillouin zone is positive, at borders it's negative, and at $k=\pm\frac\pi4$ it has poles. This means that, in particular, linearly changing potential can't give electron too high energy no matter how long you wait.

Now suppose initially you have electron in the state at the border of Brillouin zone in this band. Its effective mass is negative. Let's do as you say, apply a force $F=m^*a$ to this electron and see what classical-mechanical calculations will lead to. As second Newton law says, $$\frac{dp}{dt}=F.$$

Electron quasimomentum is related to quasiwavenumber as $p=\hbar k$. Taking units where $\hbar=1$, we have $$\frac{dk}{dt}=F.$$ Now let's apply $F=m^*a$ to this equation:


Substituting $m^*_{hyp}$ from $(1)$, we get:


Now, $a$ could mean one of the following: group acceleration or phase acceleration, i.e. derivative of group velocity or phase velocity. What is important is that it depends on quasiwavevector: $a(k)$.

For group velocity substituting $a(k)=\frac{\text{d}}{\text{d}t}v_g(k)$ with $v_g(k)=E_{hyp}'(k)$ leads to tautology, meaning that the equation is correct for group velocity and any $F$, making our analogy with classical mechanics sensible.

Let's now apply a constant force $F$ at our particle, initially being in a state described by some wave packet. The solution to Newton's equation would be $$k(t)=Ft+k_0.$$

We can see that the particle is moving with ever increasing quasimomentum. But quasimomentum is defined up to translation by an arbitrary vector of reciprocal lattice. Thus, as $k$ crosses the boundary of Brillouin zone, we could subtract $\frac{2\pi}a$ from it, and see that it appears at the opposite side of Brillouin zone, and the group velocity changes sign.

Its energy $E(k)$ also depends on time periodically, and the same is for group velocity and expectation value of position. Thus what we have is oscillatory motion of the our wave packet.


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