# A Effective Mass lecture notes

1. Oct 4, 2017

### SchroedingersLion

Greetings,

I have come across the effective mass in my solid state physics lectures, and I have troubles understanding it. I have uploaded the part of the script (handwritten 4,5 pages) that I don't understand.

At page 3, it is shown that the effective mass in the vicinity is negative. What does this mean?

What does the divergence mean?

And in general, what do you think from my short explanation of the effective mass:
"The effective mass is the mass a bloch electron within a crystal potential would have if it were to be described as a free particle, on which classical physics (e.g Lorentz-force for outer fields) can be applied.
In general, it is written as a tensor yielding different elements depending on the direction of the applied forces"

It would be nice if someone could give an explanation.
Thanks!

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2. Oct 5, 2017

### DrDu

I just want to add some physical intuition: When you increase the crystal-momentum k of a particle in a lattice near the Brillouin zone boundary, you not only increase the (absolute value of its) velocity but you also change the reflection coefficient from the Bloch planes formed by the lattice. If the increase of reflection outweights the increase in velocity, than the particle will on average slow down rather than accelerate. Hence its effective mass is negative.

3. Oct 12, 2017

### Henryk

Let me try to explain and I hope I won't confuse you more. Your statement
is correct.
Now, how do we get to the effective mass formula?
The basis is the "semi-classical model of electron dynamics". What it means is to use classical-like equations for electron dynamics and take into account band structure (which is a quantum effect).
The classical equation of motion for a charged particle is $m \frac {d\vec v}{dt} = F = e(\vec E + \vec v \times \vec B)$
The quantum effect come in the following way:
A time dependent Bloch functions are of the form $\psi(r,t) = exp^{-i \frac {E(k)} \hbar t} exp^{ikr} \phi(r)$, i.e. plane waves multiplied by a periodic functions $\phi(r)$. Note that the period of the functions $\phi(r)$ is the interatomic distance.
If you are looking on a scale of thousands of atoms, the periodic functions can be averaged out and a localized electron can be represented by a wave packet of angular frequency $\omega = \frac {E(k)} \hbar$. from wave theory we know that a wave packet propagates with a group velocity $v = \frac {d\omega}{dk} = \frac 1 \hbar \frac {dE(k)} {dk}$
Now, you understand that if we differentiated the velocity of the wave packet with respect to time (i.e. classical acceleration) and note that the energy is only the function if the k- vector we get the formula $a = \frac {dv}{dt} = \frac 1 \hbar \frac {d^2E(k)} {dk^2} \frac {dk} {dt} = \frac 1 {\hbar^2} \frac {d^2E(k)} {dk^2} \frac {d\hbar k} {dt}$. The last term is just a force. Now, compare that to the Newton's second law and you have an equation or motion for a body of mass given by
$\frac 1 m = \frac 1 {\hbar^2} \frac {d^2E(k)}{dk^2}$
I hope that this does make sense to you.
For a real free electron in vacuum, the equations reduce to the effective mass equal to the mass of an electron (non-relativistic, of course). However in a crystal, the energy is not a quadratic function of k vector and its curvature can be (should I say must be) negative sometimes, implying a negative effective mass.
What it means is that under the influence of en external static field, the momentum of an electron increases. But, at the same time. it's interaction with the periodic potential can increase as well resulting in decrease of it's propagation velocity , hence the negative effective mass.
The points of diverging effective mass correspond to the points in band structure where an increase of electron momentum does not change it's velocity.