The Fermi Surface is the surface in reciprocal (`vec k`) space that represents the highest energy levels of Fermi-Dirac particles (FD particles) at a temperature of absolute zero Kelvin. FD particle momentum is `vec p=ℏ veck` where `vec k` is called the wave vector and has a very restricted set of values. `ℏ` is the reduced Planck constant, `h`, which equals `h/(2pi)`. The FD particles obey Fermi-Dirac statistics which means that there can be only two FD particles , spin up and spin down, in an allowed spatial quantum state and have the temperature dependence
`f(E,T)=1/(exp((E-E_F)/(k_BT_K))+1)\ \ \ \ \ (1)`
where `E` is particle kinetic energy, `T_K` is absolute temperature, `E_F` is the Fermi energy (to be discussed later), and `k_B` is the Boltzmann constant. Note that, at `T_K=0`, `f(E,T)=1` for `EltE_F` and `f(E,T)=0` for `EgtE_F`. Equation 1, `f(E,T)`, multiplied by the 3D density of states, `D(E)`, is plotted in green on Canvas 4.Most often the term Fermi Surface applies to the electrons in solids. But solids have both negative electrons and positive ions and the charges of both of these cause complications. So, in this document, we will discuss un-charged particles that obey Fermi-Dirac statistics and are in a container that has infinite potentials at the boundaries. Neutrons with spins of 1/2 would be an example of this type of particle but there we have the problem that neutrons cannot be confined inside a container. So, even though our FD particles are hypothetical, this will be a good graphical illutration of how the Fermi surface evolves. For those who want to look farther into shapes of Fermi surfaces the following link is suggested: Fermi Surface Viewer
Before people get all over my case about solids having energy bands, I need to remind them that energy bands are due to the Coulomb interactions of the negative electrons with the positive ions of the solid. With a large number, `N`, of positive ions, the electron energy levels split in to an equally large number `N` of very closely spaced values called bands. This splitting is due to the periodic potential imposed by the ions. But here there are no positive ions. So here the only relevant potential is that imposed by the boundaries of the solid. This potential, called the work function for solids, is large enough compared to the particle kinetic energy to be consideded infinite. For the 2 dimensional (2D) case, all the particles in a single ring have the same energy even though the `vec k` space directions of the different states are different. Similarly even though the `vec k` space directions of the different states are different for the 3 dimensional (3D) case, all the particles in a single spherical shell have the same energy. The containing spatial states are therefore called "degenerate". The only requirement here is that the Schrodinger wave functions of the FD particles have to be zero at the boundary of the solid. We will call any one wave function a spatial state. FD particles in different spatial states must not interfere with each other.
In one dimension, the Fermi surface is really the index of the occuppied state with the largest index. For one dimension the Schrodinger wave function has to be
`psi_n(x)=sin((npix)/L)\ \ \ \ (2)`
where n is an positive or negative integer and L is the length of the container and it is assumed that the container potential at `x=0` and `x=L` is infinite. Here the wave vector `k_n=(npi)/L` so we can write`psi_n(x)=sink_nx\ \ \ \ (2a)`
and the momentum is `p_n=ℏk_n`. We will name the quantity `deltak=pi/L` so that `k_n=ndeltak`. The function `psi_n(x)` will be called the `nth` momentum state. The energy associated with this wave function is its momentum squared divided by twice the FD particle mass, `m_e`, or `p_n^2/(2m_e)``E_n= ℏ^2p_n^2/(2m)\ \ \ \ \(3)`
where `ℏ` is Planck's constant and `m` is the particle mass. From now on, in this document, we will use the following expressions for `p_n` and `E_N`:
`p_n=ℏndeltak`
`E_n= (ℏndeltak)^2/(2m)\ \ \ \ \(4)`
`E_N= ℏ^2/(2m) N^2deltak^2\ \ \ \ \(5)`
Canvas 1 shows the FD particle pairs for an adjustable number of FD particle pairs. The widths of the `deltak` intervals in `vec k` space are labeled `dk`.`N=sqrt(E_N/(ℏ^2/(2m) ( deltak^2))) \ \ \ \ \(5a)`
Obviously the value of `(dN)/(dE_N)` is:`(dN)/(dE_N)=1/2 sqrt(1/(E_N))sqrt(1/(ℏ^2/(2m) deltak^2) \ \ \ \ \(5b)`
This expression for absolute zero temperature is the 1D basis of the Maxwell-Boltzmann energy distribution laws where `(dN)/(dE_N)` is proportional to `1/sqrt(E)`.`E_F= ℏ^2/(2m) N_p^2deltak^2\ \ \ \ \(6)`
At absolute zero Kelvin, `T_K=0`, all of the states with energy less than `E_F` will be occupied. For real solids, the value of `E_F` is much, much greater than thermal energy at room temperature. If the Kelvin temperature is more than 0 then some of the states with energy below `E_F` will be unoccupied and some of the states above `E_F` will be occupied. Temperature causes a particle to be perturbed from its momentum state given by Equation 2. But its final state has to be one of the states like that of equation 2 and, in addiiton, the final state must not aleady be paired. So, the change of energy is always a discrete integer amount rather than infinitesimally variable:`deltaE_(n->p)=ℏ^2/(2m)(p^2-n^2)deltak^2\ \ \ \ \(7)`
where n and p are the initial and final state indices.The first requirement is that the wave function be zero at the `x=0` and `x=L` boundaries so a 1D solution must be `psi(x)=sin kx` for any allowed `k`. Therefore `k` would be `k=pin/L=ndeltak` where n is an integer. The second requirement is that all wave functions be orthogonal on the interval `x=0->L`. Orthogonal means that the interference product oftwo states integrated over `x=0->L` must be zero. That results in the following expression
`int_0^Lsin npix/L sin mpix/L dx=0\ \ m ne n\ \ \ \(8)`
where m and n are integers. Expanding the integrand into the cosines of the sums and difference of the arguments we have:`sin npix/L Lsin mpix/L=1/2 {cos[(m-n)pix/L]-cos[(m+n)pix/L]}\ \ \ \ \(9)`
which obviously integrates to zero on the interval `x=0->L`. Totally analogous results apply for the `y` and `z` coordinates.Suppose the container is square with side `L`. The basic unit of wave vector is then
`delta k=pi/L`
Then we can write that we may have two components of wave vector
`k_r=delta k hat r`
`k_phi=delta k hat phi`
`hat r=cos phi hat x + sin phi hat y`
`hat phi=-sin phi hat x + cos phi hat y`
`hat r * hat r=cos^2phi+sin^2phi=1`
`hat phi * hat phi=cos^2phi+sin^2phi=1`
`hat r * hat phi=-cos phi * sin phi+sin phi cos phi=0`
Then the `vec k` states of lowest magnitude form in rings of radius `k_n=ndeltak` in `k` space. If there are `m` states in the `nth` `k_r` ring then the locations in `k` space can be written
`(ndeltak)sum_(i=1)^m [cos(2pii/m)hatx+sin(2pii/m)haty]`
and the `k` vector can be written:`vec k_n=(n delta k)/sqrt(2) (hat r+hat phi)`
The states in `k_x` and `k_y` space are shown in Canvas 2.The quadrilaterals are of the area of the state (`deltak^2`) The red and blue dots are symbols for spin up and spin down FD particles. In this display, each state contains a pair of FD particles so the occupation of the states shown is 100%. However, if there are not enough FD particles to fill the outer ring, then its occupation may not be 100% and some states may contain just one FD particle. Generally the number of spin up and spin down FD particles in a given container will be equal. Otherwise the container will have a net magnetic moment. Simple solids do not usually have a magnetic moment.
`E_n= ℏ^2/(2m) n^2deltak^2\ \ \ \ \(10)`
and the number of states out through the Nth ring is just the area of a circle of radius `n` and is computed by geometry to be`N_n= pin^2\ \ \ \ \(11)`
Now we solve equation 10 for n :`n=1/(deltak^2) E_n/(2mℏ) \ \ \ \ \(10a)`
`N_n=(pi)E_n/(2mℏdeltak^2)\ \ \ \ \(11a)`
A plot of this function as well as an actual state count (small circle) is shown in Canvas 2a. Then the density of energy states, (number of states per unit energy increment), `D(E)`, is just the derivative of `N_n` with respect to `E_n` and is:`D(E)=(dN_n)/(dE_n)=(pi)/(2mℏdeltak^2)\ \ \ \ \(12)`
which is a constant with units `1/("Energy")`. Again, for 2D, this is the basis of Maxwell-Boltzmann's energy distribution law.We will assume that our solid is a cube of side `L`. For 3 dimensions (3D) the simple scalar wave vector `k_n=npi/L` becomes its 3D analog
`vec k=deltak[n_x hat x+n_y hat y+n_z hat z]\ \ \ \ (13)`
where [`hat x,hat y, hat z`] on the are unit vectors in the (x,y,z) directions. so the `k` is really a vector specified by three integers, [`n_x,n_y,n_z`]. The energy associated with this state is:`E=ℏ^2/(2m) (vec k*vec k)=ℏ^2/(2m)deltak^2 (n_x^2+n_y^2+n_z^2)\ \ \ \ (14)`
At absolute zero all of the lowest energy states are occupied. Then the `k` states of lowest magnitude form in spherical shells of radius `k_n=ndeltak` in `k` space. These energy states have equal `k` magnitude described by the vector:`vec k_n=n delta k(cos theta cos phihat x+costheta sin phi hat y+sin theta hat z)\ \ \ \ (15)`
where `theta` is the latitude angle and `phi` is the longitude angle. If there are `m` points in a longitudinal ring then we can express their position by the equation`(ndeltak)sum_(i=1)^m [cos(2pii/m)hatx+sin(2pii/m)haty]`
just as for the 2D case. If there are `l` points in a latitudinal ring we can express their locations by the equation`(ndeltak)sum_(i=1)^l [cos(2pii/l)haty+sin(2pii/l)hatz]`
Spherical shell states are represented as dots on the surface of a sphere as shown in Canvas 3. Each dot on the shell shown on Canvas 3 is surrounded by a quadrilateral that has area `delta k^2`. The radial shell separation is `delta k` so the volume in `k` space of each `k` cube is `deltak^3`. Then the total number of states (cubes) out to a `k` space radius of `Ndeltak` can be either counted as they are drawn or they can be computed by the equation`N_n=(4pi)/(3deltak^3)(ndeltak)^3\ \ \ \ (16)`
This equation is plotted in Canvas 3a. Surfaces like the sphere shown in Canvas 3 are often called the Fermi Surface. The Fermi surface is the shape of the surface in `k` space that has the highest energy level. All other energy levels are on shells lying under the outer shell. The Fermi surface does not have to be spherical in shape. In fact only the very simple metal elements of valence 1 with a single FD particle in the outer orbital like Lithium, Sodium, and Potassium have Fermi surfaces that closely resemble a sphere. More complex valence 2 metals like Cadmium, and Zinc are more more strongly affected by the lattice of the crystal. In some k space directions this results in reduced wave vector phase change, `(dk)/(ds)`, with position, `s`. For these `k` space regions, the spacing between the energy shells (layers) is reduced. This has the same effect as reducing the mass of the FD particle.Since the total number of FD particles, `N_e` in the cubic container is not related to the number of states in an integer number of spherical shells, the outer shell will generally only be partially filled. In fact the number of unfilled states will be
`N_("unfilled")="Integer"((4pi)/(3deltak^3)(N_(outer)deltak)^3)-2N_e\ \ \ \ \(17)`
where the function "Integer" takes the integer value of the enclosed expression. The unfilled shell number is computed by solving equation 17 for `N_(outer)`:`N_(outer)="Integer"(3N_e/(2pi))^(1/3)`
and the Fermi energy, `E_F`, of FD particles in that shell is:`E_F=(ℏN_(outer)deltak)^2/(2m_e)`
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If we have FD particle-FD particle collisions at least one of the FD particles will have its energy reduced and there is usually not a lower energy unfilled state for that FD particle to fill. Therefore we must conclude that valid FD particle-FD particle collisions cannot occur at absolute zero temperatures.
At temperatures above absolute zero, the `Nth` energy shell will continue to fill and even some of the higher energy shells will start to fill. Some FD particles have vacated lower energy states so the results of collisions can sometimes descend to these. The energy of the FD particles in the `Nth` energy shell will now be called the "chemical potential", `mu`, which is the same as the Fermi energy, `mu=E_F`, that we discussed before. For 3D the energy of an FD particle in the `nth` shell is
`E_n=(ℏndeltak)^2/(2m_e)`
so we can solve for `n` in terms of `E_N` and obtain:`n=sqrt(2m_eE_n)/(ℏdeltak)^2`
Then since `N_n=(4pi)/3n^3` we have the expression:`N_n=(4pi)/3((2m_eE_N)/(ℏdeltak)^2)^(3/2)\ \ \ \ \ (18)`
The density of states per unit energy, `D(E)`, is obtained by taking the derivative of `N_n` with respect to `E_n`:`D(E)=(dN_n)/(dE_n)=3/2(4pi)/3((2m_e)/(ℏdeltak)^2)^(3/2)sqrt(E_n)\ \ \ \ \(19)`
and this expression is again the basis for the Maxwell-Boltzmann law of 3D energy distribution. In terms of absolute temperature, `T_K`, the 3D energy distribution will exhibit the Fermi-Dirac temperature dependence and become:`(dN_n)/(dE_n)=3/2(4pi)/3((2m_e)/(ℏdeltak)^2)^(3/2)sqrt(E_n)/(exp((E_n-E_F)/(k_BT_K))+1)\ \ \ \ \ (20)`
A plot of equation 20 is shown in green in Canvas 4.
Using equation 11 we will now exhibit the unfilled outer shell or `Nth` shell. First we have to calculate the excess FD particles beyond the '(N-1)th' shell. Then, we select states for these FD particles. Remember, each state can contain both a spin up and a spin down FD particle. The FD particles receive their values of `vec k` and `vec v` from the outer shell state into which they are randomly placed. The random placement also includes a routine where the average number of spin up and spin down FD particles is made approximately the same so as to avoid a net magnetic moment. The Start button initiates incrementing of the Number of FD Particles slider with a step size of 100. A bar graph display of the values of `(dN_n)/(dE)` is given in Canvas 4. The shell number to which the bar height applies is given inside the bar. The bar heights are scaled to the `(dN_n)/(dE)` value of the outermost filled shell. The motion of the FD particles of the outer unfilled shell is shown in Canvas 5. The learner should note that the outer shell proceeds from vacant to filled as the number of FD particles is increased. After the current outer shell is filled, filling of the new (almost empty) outer shell starts.