### Condensed Matter IV

‘Nearly free’ electron approximation: we now add in the effects of the positive ion cores as a perturbing potential.

$V(x)=\sum_{G\neq 0}V_G\cos(Gx)$ where $G=\frac{2n\pi x}{a}$ are the reciprocal lattice vectors.

The spikier the potential, the more higher-frequency components are present (VG large for large G).

Physically, the magnitude of the VG coefficients are due to the number of electron shells, charges of ion cores, atomic spacing and so forth.

For our perturbation theory, we’ll just take the first term in the series.

$H'=V=V_0\cos\frac{2\pi x}{a}$

First order shift $\Delta E=<\psi|H'|\psi>=0$ as $\psi=Ae^{ikx}$

Second order shift $\Delta E_2=\sum_{\psi'}\frac{|<\psi|H'|\psi>|^2}{E_{\psi'}-E_{\psi}}$

This term is non-zero if $E_{\psi'}=E_{\psi}$ and $k-k'=\frac{2\pi n}{a}$

-> band gap opens up at $k=\pm\frac{n\pi}{a}$

The two energies at the band gap correspond to

Bloch waves are another way of looking at the above.

For any periodic potential, $V(\bold{r})=V(\bold{r}+\bold{R})$.

The Schrödinger equation has solutions of the form $\psi(\bold{r})=u(\bold{r})e^{i\bold{k}.\bold{r}}$

Substitute into the Schrödinger equation to energy eigenvalue equation; this will lead to the same result as above.

Reduced and extended zone schemes

Extended zone scheme

Reduced zone scheme- Umklapp everything back into 1st Brillouin Zone (translate everything back by $\frac{2n\pi}{a}$– lattice vector).

Fermi surface is no longer spherical.

Contours of constant energy in k-space (kx, ky plane). The edges of the k-square are at ±π/a

Effective mass: when an electron goes into a crystal, it is no longer a real particle. These quasi-particles appear to obey Newtonian mechanics, but with an effective mass m* that is not equal to the mass of a free electron. The effective mass contains various lattice forces which are not known explicitly- hence it can be infinite or negative.

Group velocity $v_g=\frac{\partial \omega}{\partial k}=\frac{1}{\bar{h}}\frac{\partial E}{\partial k}$

Acceleration $\frac{\partial v_g}{\partial t}\frac{1}{\bar{h}}\frac{\partial^2 E}{\partial k \partial t}$

$\frac{\partial v_g}{\partial t}\frac{1}{\bar{h}}\frac{\partial^2 E}{\partial k^2}\frac{\partial k}{\partial t}$

$\frac{\partial v_g}{\partial t}=\frac{1}{\bar{h}^2}\frac{\partial^2 E}{\partial k^2}(\bar{h}\frac{\partial k}{\partial t})$

A quick moment’s thought will show us that since $p=\bar{h}k$, then the bracketed term $(\bar{h}\frac{\partial k}{\partial t})$ must equal force. With that in mind, our acceleration equation looks like Newton II (acceleration = force/mass) but with an effective mass $\frac{1}{m*}=\frac{1}{\bar{h}^2}\frac{\partial^2 E}{\partial k^2}$

Metals, insulators and semiconductors

Metal: half-filled band means electrons can move into empty states.

Semiconductor: At T=0, the valence band is full and the conduction band is empty. Above T=0, we can excite electrons into the conduction band -> both electrons and empty states in the conduction and valence bands, therefore conduction can occur.

Insulator: large band gap so electrons cannot be excited out of the valence band (by the time you had enough energy to do so, you would have vaporised your sample anyway). The valence band is completely filled, so conduction cannot occur.

Notes

1. Sometimes bands overlap, so even if one band is filled, electrons may be able to move into the overlapping band (i.e. conduction can occur).
2. The conduction and valence bands don’t have to be the first bands- they just refer to the lowest unfilled and highest filled bands respectively.

Semiconductors

At T=0

This is the lowest unfilled band- the conduction band. At T=0 it is empty.

This is the highest filled band- the valence band. At T=0 it is completely filled.

At higher temperatures we start exciting electrons out of the valence band and into the conduction band. In the conduction band, we have a few electrons and mostly empty states, so conduction can occur here. Electrons have an effective mass me*.

In the valence band, we have mostly electrons and a few empty states- conduction can occur here also, but now electrons have a negative effective mass. It turns out to be easier to think of conduction in the valence band in terms of the movements of empty states rather than the electrons; these empty states behave like positively charged particles with positive effective mass, which we call holes.

Semiconducting materials: typical semiconductors include the Group IV elements Si and Ge and III-V compounds such as GaAs and InSb. In the primitive basis there are two atoms with 4+4 or 3+5 electrons; this total of eight electrons fills the four bands of s and p orbitals.

Direct gap semiconductor: in a direct gap semiconductor such as GaAs, the minimum of the conduction band occurs directly above the maximum of the valence band in k-space.

Optical absorption cannot occur until a photon has enough energy ($\bar{h}\omega \geq E_G$) to excite an electron from the valence band to the conduction band, thus creating an electron-hole pair.

Indirect gap semiconductor: in an indirect gap semiconductor such as Si or Ge, the minimum of the conduction band is not directly above the maximum of the valence band in k-space.

1. Photon has enough energy to excite an electron into the conduction band, but not enough momentum, so a phonon is also required to transfer momentum to the electron (phonon-assisted transition).
2. There is now sufficient energy for an electron to go from the valence band to the conduction band with δk=0 (no phonon required).

Experiment to determine band gap from optical absorption- some general notes.

1. EG is dependent on temperature; make sure temperature is controlled during experiment.
2. Radiation source of IR frequency is needed, and also a detector (obviously).
3. Plot absorption coefficient vs. photon energy as in the graphs above to find EG.

Properties of holes: the motion of electrons in a band with one empty state looks like the motion of one positively charge particle (hole). Since electron density is referred to as n, we’ll be using p for hole density.

Hole wavevector and momentum kh=-ke

Energy εh= -εe

Velocity $v_h=v_e=\frac{1}{\bar{h}}\frac{\partial \epsilon}{\partial k}$

Intrinsic semiconductors: an intrinsic semiconductor has no impurities, so n=p. We’ll now go ahead and work out these intrinsic carrier densities.

Put the bottom of the conduction band at EC and the top of the valence band at EV; EcEV=EG.

Electron density $n=\int_{E_C}^{\infty}g(\epsilon)f(\epsilon_d\epsilon$

$f(\epsilon)=\frac{1}{e^{\frac{\epsilon-\mu}{kT}}+1}\simeq e^{\frac{\mu-\epsilon}{kT}}$ for $\epsilon-\mu \gg kT$

$g(\epsilon)d\epsilon \propto \sqrt{\epsilon-E_C} d\epsilon$

Substituting all this into our expression for n gives

$n=A\int_{E_C}^{\infty}\sqrt{\epsilon-E_C} e^{\frac{\mu-\epsilon}{kT}} d\epsilon$

$n=A e^{\frac{\mu-E_C}{kT}}\int_0^{\infty}\sqrt{\epsilon-E_C} e^{\frac{E_C-\epsilon}{kT}} d\epsilon$

(We can let the lower limit go to zero as there aren’t going to be any electrons in the states from 0 to EC anyway).

Let $x=\frac{E_C-\epsilon}{kT}$

$n=A e^{\frac{\mu-E_C}{kT}}(kT)^{\frac{3}{2}}\int_0^{\infty}\sqrt{x}e^{-x}dx$

The integral over x will just come out to a constant, which we can absorb into A along with the factor of k3/2.

Hence $n=AT^{\frac{3}{2}}e^{\frac{\mu-E_C}{kT}}$

For holes, we start at $p=\int_{-\infty}^{E_V}(1-f(\epsilon)g(\epsilon)d\epsilon$

For most purposes, it should be sufficient to simply quote that, by symmetry

$p=A' T^{\frac{3}{2}}e^{\frac{E_V-\mu}{kT}}$

However, in an exam you might be expected to go through that derivation as well, in which case just proceed as above.

$np=N_CN_Ve^{\frac{E_V-E_C}{kT}}$

where

$N_C=A T^{\frac{3}{2}}=(\frac{2\pi m_e*kT}{\bar{h}^2})^{\frac{3}{2}}$

$N_V=A' T^{\frac{3}{2}}=(\frac{2\pi m_h*kT}{\bar{h}^2})^{\frac{3}{2}}$

$np=N_C N_Ve^{\frac{-E_G}{kT}}\simeq 10^{33}m^{-6}$

This is the Law of Mass action, and it always holds, independent of μ.

$n=p= \sqrt{N_C N_V}e^{\frac{-E_G}{2kT}}$

Extrinsic semiconductors have had impurities introduced to produce an excess of electrons or holes. This process is known as doping.

N-type semiconductors have been doped to produce an excess of electrons, for example by introducing Group V elements such as phosphorus into the crystal structure of a Group IV element such as Si or Ge. The Group V element atoms take the place of Group IV element atoms- four of its valence electrons are used in bonds, with an extra fifth electron left over.

These impurities are called donors because they donate extra electrons to the structure.

P-type semiconductors have been doped with electron acceptors to produce an excess of holes. In this case, the impurity atoms are from Group III, e.g. Ga or Al.

(One electron missing- equivalent to donating a hole.)

Acceptor/donor energy levels: model the ionised impurity atom and electron/hole as a hydrogenic system, with mass m* and permittivity ε0εr.

Ionisation energy $E=\frac{m*e^4}{2(4\pi \epsilon_0\epsilon_r \bar{h})^2}=\frac{m*}{m_e \epsilon_r^2}E_R$

where ER is the good old Rydberg energy (13.6 eV).

Substituting in typical values of εr~10, m*~0.1me gives us E~13.6 meV

At room temperature kT~25 meV, so we would expect all impurities to be ionised.

Now define an effective ‘Bohr radius’ for impurity wavefunctions.

$a*=\frac{4\pi\epsilon_0\epsilon_r\bar{h}^2}{m*e^2}=\frac{\epsilon_rm_e}{m*}a_0$

Where a0 is, of course the Bohr radius (0.53×10-10m).

If the impurity wavefunctions start to overlap, we get an impurity band which allows conduction to occur, leading to metallic behaviour.

Density of states and carrier densities

N-type

At room temperature, assume all donors are ionised, but few electrons are excited across the band gap.

nND+ND (actually, n=p+ ND+ for charge neutrality, but we’re saying that p is negligible).

From the law of mass action

$p=\frac{N_CN_V}{N_D}e^{\frac{-E_G}{2kT}}=\frac{n_i^2}{N_D}$

P-type

pNANA

$n=\frac{n_i^2}{N_A}$

Temperature dependence of extrinsic properties

1. Freeze-out range. The temperature is still low so not all impurities have been ionised yet. As the temperature increases, more impurities ionise.
2. Saturation range. All impurities are now ionised- carrier density is constant.
3. There is now sufficient energy to excite large numbers of electrons across the band gap, so we see a transition to intrinsic behaviour.

Conductivity of a semi-conductor

$\sigma=ne\mu_e+pe\mu_n$

To make things easier, it will often be assumed that one type of carrier dominates, or that μeh=constant.

Cyclotron resonance can be used to determined the effective mass and electron dispersion relation (more detail to be included at a later date).