For Gas of Electron in 3d the Density of States at the Fermi Surface Dn/de Can Be 3n/2e

Density of States

1. Last updated

2. Salvage as PDF

Page ID

312

i.ane Introduction

The density of states (DOS) is substantially the number of different states at a item energy level that electrons are allowed to occupy, i.e. the number of electron states per unit of measurement volume per unit energy. Bulk backdrop such as specific rut, paramagnetic susceptibility, and other transport phenomena of conductive solids depend on this function. DOS calculations let one to determine the full general distribution of states as a office of energy and can besides decide the spacing betwixt energy bands in semi-conductors\(^{[1]}\).

1.two Density of States for Waves

Earlier nosotros go involved in the derivation of the DOS of electrons in a material, it may be easier to first consider just an rubberband wave propagating through a solid. Elastic waves are in reference to the lattice vibrations of a solid comprised of detached atoms. Though, when the wavelength is very long, the atomic nature of the solid can be ignored and we can treat the material every bit a continuous medium\(^{[2]}\).

We begin with the i-D wave equation: \( \dfrac{\partial^2u}{\partial x^ii} - \dfrac{\rho}{Y} \dfrac{\fractional u}{\partial t^2} = 0\)

With which we then have a solution for a propagating plane moving ridge:

\[u = A e^{i(qx-\omega t)} \]

\(q\)= wave number: \(q=\dfrac{two\pi}{\lambda}\), \(A\)= amplitude, \(\omega\)= the frequency, \(v_s\)= the velocity of sound

Dispersion Relations:

\[ \omega = \nu_s q\nonumber\]

\[ \nu_s = \sqrt{\dfrac{Y}{\rho}}\nonumber\]

In equation(ane), the temporal factor, \(-\omega t\) can be omitted considering it is not relevant to the derivation of the DOS\(^{[two]}\). And then at present nosotros will use the solution:

\[ u =Ae^{i(qx)}\]

To begin, we must use some type of boundary atmospheric condition to the organisation. The easiest manner to do this is to consider a periodic boundary condition. With a periodic boundary status nosotros tin can imagine our system having two ends, ane being the origin, 0, and the other, \(L\). We now say that the origin end is constrained in a mode that it is always at the same state of oscillation as end L\(^{[2]}\).

This boundary condition is represented as: \( u(10=0)=u(ten=L)\)

Now we apply the boundary condition to equation (2) to become: \( e^{iqL} =1\)

At present, using Euler's identity; \( due east^{ix}= \cos(x) + i\sin(x)\) we can meet that there are certain values of \(qL\) which satisfy the above equation. Those values are \(n2\pi\) for whatever integer, \(n\). Leaving the relation: \( q =northward\dfrac{2\pi}{L}\)

If you cull integer values for \(n\) and plot them forth an axis \(q\) you get a 1-D line of points, known every bit modes, with a spacing of \({ii\pi}/{L}\) between each mode.

Equally \(L \rightarrow \infty , q \rightarrow \text{continuum}\).

We now have that the number of modes in an interval \(dq\) in \(q\)-infinite equals:

\[ \dfrac{dq}{\dfrac{two\pi}{L}} = \dfrac{Fifty}{2\pi} dq\nonumber\]

Using the dispersion relation we can find the number of modes within a frequency range \(d\omega\) that lies within\((\omega,\omega+d\omega)\). This number of modes in that range is represented by \(yard(\omega)d\omega\), where \(yard\omega\) is defined as the density of states.

So now nosotros run into that \(g(\omega) d\omega =\dfrac{L}{two\pi} dq\) which we turn into: \(one thousand(\omega)={(\frac{L}{ii\pi})}/{(\frac{d\omega}{dq})}\)

Nosotros do so in order to utilise the relation: \(\dfrac{d\omega}{dq}=\nu_s\)

and obtain: \(thousand(\omega) = \left(\dfrac{L}{2\pi}\correct)\dfrac{1}{\nu_s} \Rightarrow (g(\omega)=2 \left(\dfrac{L}{two\pi} \dfrac{1}{\nu_s} \right)\)

nosotros multiply past a cistron of two exist crusade there are modes in positive and negative \(q\)-space, and we become the density of states for a phonon in ane-D:

\[ thou(\omega) = \dfrac{L}{\pi} \dfrac{one}{\nu_s}\nonumber\]

ii-D

Nosotros tin can now derive the density of states for ii dimensions. Equation(2) becomes: \(u = A^{i(q_x x + q_y y)}\)

now use the same boundary conditions as in the 1-D example:

\[ eastward^{i[q_xL + q_yL]} = i \Rightarrow (q_x,q)_y) = \left( n\dfrac{2\pi}{Fifty}, grand\dfrac{two\pi}{L} \right)\nonumber\]

We now consider an surface area for each point in \(q\)-space =\({(ii\pi/50)}^2\) and find the number of modes that lie within a apartment ring with thickness \(dq\), a radius \(q\) and area: \(\pi q^2\)

Number of modes within interval: \(\frac{d}{dq}{(\frac{L}{2\pi})}^2\pi q^2 \Rightarrow {(\frac{L}{ii\pi})}^ii two\pi qdq\)

Now business relationship for transverse and longitudinal modes (multiply past a cistron of 2) and set equal to \(g(\omega)d\omega\) We get

\[thousand(\omega)d\omega=2{(\frac{L}{2\pi})}^2 2\pi qdq\nonumber\]

and apply dispersion relation to get \(ii{(\frac{50}{2\pi})}^2 2\pi(\frac{\omega}{\nu_s})\frac{d\omega}{\nu_s}\)

which simplifies to the 2-D result:

\[ 1000(\omega)= \dfrac{Fifty^two}{\pi} \dfrac{\omega}{{\nu_s}^2}\nonumber\]

three-D

We can now derive the density of states for 3 dimensions. Equation(2) becomes: \(u = A^{i(q_x x + q_y y+q_z z)}\)

now use the same boundary conditions as in the 1-D case to get:

\[e^{i[q_x x + q_y y+q_z z]}=1 \Rightarrow (q_x , q_y , q_z)=(north\frac{2\pi}{L},m\frac{ii\pi}{L}50\frac{2\pi}{L})\nonumber\]

We now consider a book for each indicate in \(q\)-space =\({(2\pi/L)}^3\) and find the number of modes that lie within a spherical crush, thickness \(dq\), with a radius \(q\) and volume: \(4/3\pi q ^iii\)

Number of modes inside shell:

\[\frac{d}{dq}{(\frac{Fifty}{2\pi})}^3\frac{4}{iii}\pi q^three \Rightarrow {(\frac{L}{2\pi})}^3 iv\pi q^2 dq\nonumber\]

Assuming a common velocity for transverse and longitudinal waves we tin can account for one longitudinal and two transverse modes for each value of \(q\) (multiply by a factor of 3) and set equal to \(g(\omega)d\omega\):

\[g(\omega)d\omega=3{(\frac{L}{ii\pi})}^three 4\pi q^2 dq\nonumber\]

Use dispersion relation and let \(L^iii = V\) to get \[3\frac{Five}{{2\pi}^3}4\pi{{(\frac{\omega}{nu_s})}^2}\frac{d\omega}{nu_s}\nonumber\]

and simplify for the three-D issue:

\[ m(\omega) = 3 \dfrac{V}{2\pi^2} \dfrac{\omega^two}{\nu_s^3}\nonumber\]

Density of States for Electrons

Now that we take seen the distribution of modes for waves in a continuous medium, we movement to electrons. The calculation of some electronic processes like absorption, emission, and the general distribution of electrons in a fabric require usa to know the number of bachelor states per unit book per unit free energy. The density of states is once again represented by a part \(k(E)\) which this time is a function of energy and has the relation \(g(Eastward)dE\) = the number of states per unit volume in the energy range: \((E, East+dE)\).

We begin past observing our system as a free electron gas confined to points \(thou\) contained within the surface. We do this so that the electrons in our system are gratis to travel around the crystal without being influenced past the potential of atomic nuclei\(^{[3]}\). Using the Schrödinger wave equation we can make up one's mind that the solution of electrons confined in a box with rigid walls, i.e. the Particle in a box problem, gives rise to standing waves for which the immune values of \(thousand\) are expressible in terms of three nonzero integers, \(n_x,n_y,n_z\)\(^{[ane]}\).

Taking a footstep back, nosotros expect at the free electron, which has a momentum,\(p\) and velocity,\(5\), related by \(p=mv\). Since the energy of a free electron is entirely kinetic we tin can disregard the potential energy term and country that the energy, \(Due east = \dfrac{one}{2} mv^2\)

Using De-Broglie'southward particle-moving ridge duality theory nosotros can presume that the electron has wave-like backdrop and assign the electron a wave number \(k\): \(thou=\frac{p}{\hbar}\)

\(\hbar\) is the reduced Planck's constant: \(\hbar=\dfrac{h}{ii\pi}\)

Now we substitute terms:

\[k=\frac{p}{\hbar} \Rightarrow g=\frac{mv}{\hbar} \Rightarrow five=\frac{\hbar k}{one thousand}\nonumber\]

Substitute \(5\) term into the equation for energy:

\[E=\frac{1}{2}m{(\frac{\hbar grand}{1000})}^ii\nonumber\]

We are now left with the dispersion relation for electron energy: \(E =\dfrac{\hbar^two k^2}{ii m^{\ast}}\)

where \(yard ^{\ast}\) is the effective mass of an electron.

As for the case of a phonon which we discussed earlier, the equation for allowed values of \(k\) is institute past solving the Schrödinger wave equation with the same boundary weather that nosotros used earlier.

We are left with the solution: \(u=Ae^{i(k_xx+k_yy+k_zz)}\)

and after applying the same purlieus conditions used earlier:

\[e^{i[k_xx+k_yy+k_zz]}=1 \Rightarrow (k_x,k_y,k_z)=(n_x \frac{two\pi}{L}, n_y \frac{ii\pi}{L}), n_z \frac{2\pi}{L})\nonumber\]

Nosotros can consider each position in \(grand\)-space being filled with a cubic unit jail cell volume of:

\(V={(2\pi/ L)}^iii\) making the number of allowed \(thousand\) values per unit volume of \(m\)-infinite:\(ane/(ii\pi)^three\)

We can picture the allowed values from \(East =\dfrac{\hbar^ii k^2}{2 chiliad^{\ast}}\) as a sphere almost the origin with a radius \(1000\) and thickness \(dk\). The allowed states are now found within the volume contained between \(yard\) and \(chiliad+dk\), meet Figure \(\PageIndex{ane}\).

Effigy \(\PageIndex{1}\)\(^{[1]}\). Spherical vanquish showing values of \(1000\) as points. The points contained within the shell \(k\) and \(k+dk\) are the allowed values.

The volume of the shell with radius \(k\) and thickness \(dk\) can exist calculated by simply multiplying the surface area of the sphere, \(iv\pi k^two\), by the thickness, \(dk\):

\[V_{shell}=4\pi k^2 dk\nonumber\]

Now we can form an expression for the number of states in the shell by combining the number of allowed \(1000\) states per unit of measurement volume of \(k\)-infinite with the volume of the spherical shell seen in Effigy \(\PageIndex{1}\).

Number of states: \(\frac{i}{{(2\pi)}^3}4\pi m^2 dk\)

Substitute in the dispersion relation for electron energy:

\(E =\dfrac{\hbar^2 k^ii}{2 yard^{\ast}} \Rightarrow k=\sqrt{\dfrac{2 one thousand^{\ast}East}{\hbar^2}}\)

which leads to \(\dfrac{dk}{dE}={(\dfrac{2 m^{\ast}E}{\hbar^2})}^{-1/2}\dfrac{m^{\ast}}{\hbar^2}\) at present substitute the expressions obtained for \(dk\) and \(k^2\) in terms of \(E\) back into the expression for the number of states:

\(\Rightarrow\frac{one}

ParseError: ")" expected (click for details)

Callstack:     at                  (Bookshelves/Materials_Science/Supplemental_Modules_(Materials_Science)/Electronic_Properties/Density_of_States), /content/body/div[iii]/p[27]/span, line one, column three                

{\hbar^two})dE\)

\(\Rightarrow\frac{ane}{{(2\pi)}^iii}4\pi{(\dfrac{2 m^{\ast}}{\hbar^2})}^ii{(\dfrac{2 m^{\ast}}{\hbar^two})}^{-i/ii})East(E^{-1/2})dE\)

\(\Rightarrow\frac{i}{{(ii\pi)}^iii}4\pi{(\dfrac{2 k^{\ast}E}{\hbar^ii})}^{3/2})E^{1/2}dE\)

We have now represented the electrons in a 3 dimensional \(yard\)-space, like to our representation of the elastic waves in \(q\)-space, except this time the trounce in \(k\)-infinite has its surfaces defined by the energy contours \(E(k)=Due east\) and \(East(yard)=Eastward+dE\), thus the number of allowed \(grand\) values inside this shell gives the number of available states and when divided by the beat thickness, \(dE\), nosotros obtain the function \(thou(E)\)\(^{[two]}\).

\[one thousand(E)=\frac{1}{{4\pi}^2}{(\dfrac{2 m^{\ast}E}{\hbar^2})}^{three/ii})E^{1/ii}\nonumber\]

we must at present account for the fact that whatever \(yard\) state can contain two electrons, spin-up and spin-downwards, so we multiply by a gene of two to get:

\[g(Due east)=\frac{i}{{ii\pi}^2}{(\dfrac{ii m^{\ast}Eastward}{\hbar^ii})}^{3/two})E^{1/ii}\nonumber\]

The above expression for the DOS is valid only for the region in \(grand\)-infinite where the dispersion relation \(Due east =\dfrac{\hbar^two thousand^ii}{2 1000^{\ast}}\) applies. As the energy increases the contours described by \(E(1000)\) become non-spherical, and when the energies are large enough the beat out will intersect the boundaries of the kickoff Brillouin zone, causing the shell volume to subtract which leads to a decrease in the number of states. If the volume continues to subtract, \(m(Due east)\) goes to naught and the shell no longer lies inside the zone. The free energy at which \(chiliad(E)\) becomes zippo is the location of the tiptop of the valance band and the range from where \(g(E)\) remains cipher is the ring gap\(^{[2]}\).

Figure \(\PageIndex{2}\)\(^{[1]}\) The left hand side shows a two-band diagram and a DOS vs.\(Due east\) plot for no band overlap. The correct hand side shows a two-band diagram and a DOS vs. \(Due east\) plot for the example when at that place is a band overlap.

In simple metals the DOS tin can be calculated for nearly of the energy ring, using:

\[ yard(E) = \dfrac{1}{2\pi^2}\left( \dfrac{2m^*}{\hbar^ii} \right)^{3/2} E^{1/2}\nonumber\]

even so when we reach energies most the height of the band we must use a slightly dissimilar equation. To derive this equation we can consider that the next band is \(Eg\) ev below the minimum of the first band\(^{[ane]}\). This is illustrated in the upper left plot in Figure \(\PageIndex{ii}\). The energy of this second band is: \(E_2(g) =E_g-\dfrac{\hbar^2k^2}{2m^{\ast}}\). Now nosotros tin derive the density of states in this region in the same way that we did for the rest of the band and go the result:

\[ one thousand(Due east) = \dfrac{1}{two\pi^two}\left( \dfrac{2|chiliad^{\ast}|}{\hbar^2} \right)^{iii/2} (E_g-Eastward)^{1/two}\nonumber\]

Solving for the DOS in the other dimensions will exist similar to what we did for the waves. i.east. for two-D we would consider an expanse chemical element in \(k\)-space \((k_x, k_y)\), and for ane-D a line element in \(one thousand\)-infinite \((k_x)\). Figure \(\PageIndex{three}\) lists the equations for the density of states in iv dimensions, (a breakthrough dot would be considered 0-D), along with respective plots of DOS vs. free energy. Figure \(\PageIndex{4}\) plots DOS vs. energy over a range of values for each dimension and super-imposes the curves over each other to further visualize the unlike behavior between dimensions.

Figure \(\PageIndex{iii}\)\(^{[four]}\)

Figure \(\PageIndex{4}\)\(^{[three]}\)

References

1. Sachs, M., Solid State Theory, (New York, McGraw-Colina Book Company, 1963),pp159-160;238-242.
2. Omar, Ali M., Elementary Solid State Physics, (Pearson Education, 1999), pp68- 75;213-215.
3. Semiconductor Physics (online) http://britneyspears.ac/physics/dos/dos.htm
4. Density of States (online) www.ecse.rpi.edu/~schubert/Course-ECSE-6968%20Quantum%20mechanics/Ch12%20Density%20of%20states.pdf

Contributors and Attributions

• ContribMSE5317

orozcoliew1990.blogspot.com

Source: https://eng.libretexts.org/Bookshelves/Materials_Science/Supplemental_Modules_(Materials_Science)/Electronic_Properties/Density_of_States