3.23 Electrical, Optical, and Magnetic Properties of Materials, Homework #5 (Solutions)

Homework # 5, solution Homework is due on Wednesday October 17th, 5pm 1 Density of states We saw both in recitation and in class, that we can calculate the 3D density of state corresponding to a given dispersion formula E(�k), by using the following formula: � V dS g(E) = (2π) for the 3D case 3 � E(� E(� k)=E �� k)� � k where the integral is carried out over the ”isosurface” of energy E, i.e the sur­ face defined by the equation E(kx , ky , kz ) = E. The infinitesimal surface ele­ ment is dS and V is the volume of the 3D crystal. The 2D equivalent is the following: � A dl g(E) = (2π) for the 2D case 2 � E(� E(� k)=E �� k)� � k where the integral is carried out over the ”isosurface” of energy E, i.e the curve defined by the equation E(kx , ky ) = E. The infinitesimal element of length along the isosurface is dl and A is the area of the 2D crystal. Finally for the 1D case we have: � g(E) = (2Lπ) k0 dE(k) 1 for the 1D case | dk (k=k0 )| where k0 are all the k points such that E(k0 ) = E, and L is the length of the 1D crystal. Remark: in this exercise we are interested in the energy density of states without taking into account the spin degeneracy! If we were interested in the FULL density of states, we would have to multiply all our results by 2. 1.1 Free electron gas in 1D, 2D and 3D Now we will apply this to the 1D, 2D and 3D free electron gas. 1) Write down the dispersion relations E(�k) for a free electron gas in 1D, 2D and 3D. From those relations, describe the corresponding isosurfaces, i.e draw a typical isosurface for each case (1D, 2D, 3D). solution: The dispersion relations are the following: 1 E(kx ) = E(kx , ky ) = 2 h̄2 kx 2m in 1D 2 h̄2 (kx +ky2 ) 2m in 2D and E(kx , ky , kz ) = 2 h̄2 (kx +ky2 +kz2 ) 2m in 3D One can visualize the 1D and 2D dispersion by looking at the plots shown on figure 1. The isosurfaces are the regions of space where the energy is kept Figure 1: 1D and 2D energy dispersions for a free electron constant. As a consequence, we have the following results: h̄2 k2 1. In the 1D case, the isosurface is defined by the equation 2mx = E, so we see �that this isosurface reduces to two points, k0 and −k0 , where k0 = 2mE h ¯2 h ¯ 2 (k2 +k2 ) x y 2. In the 2D case, the isosurface is defined by = E. This equation 2m , k space of radius R = is nothing but the equation of a circle in k x y � 2mE . h ¯2 3. In the 3D case, the equation that defines the isosurface is the following: 2 h̄2 (kx +ky2 +kz2 ) 2m = E. So we see that we define a sphere in the kx , ky , kz � space with a radius R = 2mE . h̄2 2) For the 2D and 3D cases, find the appropriate set of coordinates to de­ scribe an isosurface. Using this set of coordinates, give a mathematical expres­ sion for the ”surface” element dS. Finally give an analytical form for the density of states. solution: Given the geometry of the isosurfaces in 2D and 3D, we see that the appro­ priate sets of coordinates are the following: • In the 2D case, we will use a set of planar polar coordinates r, θ (r2 = k kx2 + ky2 , tan(θ) = kxy ). The ”element of length” along the circle of radius r is dl = rdθ. 2 • In the 3D case, we will use the natural extension of the planar polar coordinates to 3 variables, which is nothing less than the set of spherical k coordinates r, θ, φ (r2 = kx2 + ky2 + kz2 , tan(φ) = kxy , sin(θ) = kkxz ). The ”element of surface” on a sphere of radius r is dS = r2 sin(θ)dθdφ. Let us now use the formulas given � at the beginning of this exercise. For the 2D � � E(�k)� = h̄2 k 2 + k 2 = h̄2 r using the polar coordinates. We case, we have: �� x y k m m � � � E(�k)� along the circle of radius r = 2mE need to integrate the inverse of �� . k h̄2 So the density of states in 2D is: � 2π rdθ A mA g2D (E) = (2π) = 2πh̄ 2 2 h ¯2 0 m r We see an interesting feature of the 2D density of states for the free electron gas: it is a constant! Let us move on to the � 3D case. This time the integral � is carried out over 2 h̄2 2mE � � a sphere of radius r = k 2 + k 2 + k 2 = h̄ r 2 . We have ��� E(k)� = k m � 2π � π r 2 sin(θ)dθdφ h ¯2 m r h̄ x y z m and so the density of states is: g3D (E) = V (2π)3 0 0 In this integral the remaining r term is independant of θ and φ, because r = � 2mE . So the integral is just 4π and the final answer is: h̄2 g3D (E) = V 2m 3/2 ) (2π)2 ( h ¯2 √ E 3) For the one dimensional case, use the third formula above to calculate the density of states. solution: For the 1D case, we use the formula given at the beginning of this exercise. We see that there are 2 k0 for which the dispersion � relation is verified in the . The direct expression for the density of states g(E) = (2Lπ) k0 dE(k) 1 | dk (k=k0 )| use of this formula gives us: g1D (E) = L π � m √1 2h̄2 E 4) Plot the 1D, 2D and 3D density of states as a function of the energy E. solution: g (E) g (E) g (E ) √ m , 2D and V 3D I give here the plots of L1D as a function of E mA ( 2m )3/2 π 2¯ h2 2π¯ h2 (2π)2 h ¯2 on figure 2. 1.2 The case of tightly bound electrons in 1D Let’s consider tightly bound electrons described by the following band dispersion (for a 1D linear chain of atoms): E(k) = �0 − 2γ cos(ka) 3 Figure 2: 1D, 2D and 3D normalized density of states for a free electron. The 1D curve blows up when E → 0, the 2D one is constant, and the 3D grows continuously from 0. 1) Define the interval of all possible values for E(k) when k is in the first Brillouin Zone. Plot the band dispersion for k in the first Brillouin Zone. solution: We see on the band dispersion that the periodicity in k is given by the cosine term. This term is periodic in k with the following period 2aπ . Moreover since the cosine is an even function in k, it is convenient to represent the band structure on a plot centered around k = 0, starting at k = − πa and ending at k = πa . The band dispersion is shown on figure 3. In order to plot this dispersion i considered γ to be positive. 2) Using the formula for the 1D case, calculate the density of states. Plot this density of states as a function of the energy in the appropriate interval (defined in question 1)). solution: Given the mathematical expression for the band dispersion, we see that: dE (k) dk = 2γa sin(ka). In the formula given for the 1D case, we see that we need to calculate this derivative at points k0 such that E(k0 ) = E. Before to calculate k0 , let’s have a look at what is the energy range for E. We see that because the cosine takes values only between −1 and 1, then the energy E must be located between �0 − 2γ and �0 + 2γ. In this interval of energy, k0 can be expressed as k0 = ± a1 arccos( �02−γE ). If we go back to the expression for the derivative of E(k), and if we use the � well-known trigonometric formula sin(x) = ± 1 − cos2 (x), then we see that: 1 | dE(k) dk |k=k0 = 1 |2γa sin(k0 a)| = 1 |2γa± � 1−( �0 −E 2 2γ ) | We can now express the density of states for a one dimensional nearest­ neighbour tight-binding as: g(E) = L 2πγa 1 � 4 1−( �0 −E 2 2γ ) Figure 3: Energy dispersion for a one dimensional nearest neighbour tightbinding model. The first Brillouin Zone runs from k = − πa to k = πa . On this plot �0 is 1 and so is γ. A plot of this density of state is shown on figure 4. Figure 4: Density of state of a 1D nearest-neighbour tight-binding model. 5 2 2.1 Tight-binding method geometrical description of the crystal Let us consider a simple cubic lattice (3D) of lattice parameter a, with one atom per unit cell centered at the origin of the coordinate system. Each atom has one valence electron and in the atomic limit (i.e in the limit where the atoms are far from each other) this electron is described by an s-type atomic orbital φs (�r). 1) Give a mathematical expression for the primitive basis vectors �a1 , �a2 , �a3 that describe this simple cubic lattice. Calculate the expression for the recipro­ cal primitive vectors �b1 , �b2 , �b3 . What is the volume of the unit cell in real space? What is the volume of the unit cell in reciprocal space, i.e the first Brillouin Zone? Relate those two volumes by an equation. solution: The simple cubic lattice is a Bravais lattice. As such all the lattice points are equivalent to each other. The most natural basis of primitive vectors are of course the cartesian vectors: �a1 = a�ex �a2 = a�ey �a3 = a�ez By definition, the primitive vectors of the reciprocal lattice �b1 , �b2 , �b3 are such that: �a1 · �b1 = 2π, �a2 · �b1 = 0 and �a3 · �b1 = 0 (and equivalent expressions for �b2 and �b3 ). For �b1 , we can easily conclude from the three relations above that this vector must be perpendicular to �a2 and �a3 . So in the case of the simple cubic lattice, this induces that this vector is perpendicular to �ey and �ez . So �b1 is a vector proportional to �ex . Now if we use the fact that �a1 · �b1 = 2π, we can conclude that �b1 = 2aπ �ex . The same kind of analysis for �b2 and �b3 leads us to the following final expressions for the primitive vectors in reciprocal space: �b1 = 2π ex a � �b2 = 2π ey a � �b3 = 2π ez a � The volume of the unit cell in real space is given by the volume of the parallepiped delimited by �a1 , �a2 and �a3 , which is nothing but Ω = �a1 ·(�a2 ×�a3 ) = a3 . The volume of the first Brillouin Zone is nothing but the volume of the unit cell in reciprocal space. Similarly to the real space case, this volume is given by 3 � Ω = �b1 · (�b2 × �b3 ) = (2aπ3) . We see that the relationship between the volume of the unit cell in real space Ω and the volume of the unit cell in reciprocal space � Ω is : � Ω = (2π)3 Ω This relationship is valid for ANY crystal structure. � for the atoms in this simple 2) Express the equilibrium position vectors R cubic crystal in terms of the �a1 , �a2 , �a3 basis. Sketch the first Brillouin Zone in reciprocal space. solution: � are given by: The direct space translation vectors R � = n1�a1 + n2�a2 + n3�a3 R where n1 , n2 , n3 are integer numbers. The first Brillouin Zone centered on the origin of the coordinate system in reciprocal space is defined by all the � possible � k vectors �k = k1�b1 + k2�b2 + k3�b3 , where k1 , k2 , k3 are in the interval − 21 , 21 . 6 2.2 calculation of the band dispersion The tight-binding method is an extension of the LCAO method, used in Chem­ istry, to periodic systems like solids. The spirit of the technique is the following: If one looks at the wavefunction for an electron in the atomic limit (when the atoms of the solid are far from each other), then this wavefunction looks very much like the atomic orbital φs (�r). Now if we bring the atoms closer and closer to each other, the wavefunction for the electron will be perturbed and at suffi­ ciently small distances between atoms, the electron will be able to ”hop” from atom to atom, and it will delocalize over the entire solid. Nevertheless when the electron is localized around one particular atom, let say the atom centered � , the wavefunction must still look very much like φs (�r − R � ). around position R So the idea of the tight-binding method is to write down the electronic wave� ), centered function as a linear combinaison of all the atomic orbitals φs (�r − R � in the crystal (the summation over R � below concerns all around each atom R atoms in the crystal): � �) ψs (�r) = R� cR� φs (�r − R But in order to be an ”acceptable” wavefunction, ψs (�r) must satisfy Bloch’s theorem. This is a big constraint for ψs (�r), and as a consequence, we can write down an expression for ψs (�r) for each wavevector �k: � �� �) ψs�k (�r) = √1N R� eikR φs (�r − R The √1 N is just a normalization factor (N being the number of unit cells in our �� crystal), and the coefficiants cR� are given by eikR . We are interested in the eigenenergy Es (�k) of the periodic one-electron effec­ tive hamiltonian Ĥ. To calculate this eigenenergy we use the following result: �ψs�k |Ĥ|ψs�k � �ψs�k |ψs�k � � � 1) Give the mathematical expression for ψs�k |ψs�k in terms of integrals over the volume of the crystal. In this expression you should obtain terms of the form: � ∗ � �� crystal φs (�r − R)φs (�r − R )d�r Es (�k) = Because of the Born-Von Karman boundary conditions, one can show that this integral is equal to the following one: � ∗ �� � crystal φs (�r)φs (�r − (R − R))d�r �� − R �. So you see that this integral only depends on� the relative distance R � Using this, �simplify the expression for ψsk� |ψsk� ? If the set of atomic orbitals � �) φs (�r − R form an orthonormal basis, what is the mathematical expression � �R� for ψs�k |ψs�k ? solution: � � Let’s write down the scalar product of ψs�k (�r) with itself: ψs�k |ψs�k . � � � ψs�k |ψs�k = crystal ψs∗�k (�r)ψs�k (�r)d�r 7 By plugging in the mathematical expression for ψs�k (�r), this expression becomes: � � � �� −R �) � i� k(R ∗ � �� ψsk� |ψsk� = N1 R, �� e � R crystal φs (�r − R)φs (�r − R )d�r At this stage, � we will use the hint given in the text above. Indeed if one considers � )φs (�r −R � � )d�r, it is quite easy to see that translating the integral crystal φ∗s (�r −R � does not change the value of the integral because �r by a direct space vector R of the Born-Von Karman boundary conditions on the solid. So we can simplify the expression above to the following one: � � � �� −R �) � i� k(R ∗ �� � ψsk� |ψsk� = N1 R, � R �� e crystal φs (�r)φs (�r − (R − R))d�r It doesn’t seem that we have achieved a lot here, but actually it is not true! � and R � � to R � and R��� = Indeed if we change the summation variable from R � � −R � , we see that the argument of the double sum does NOT depend on R �. R � just gives us the total number of unit cell This means that the sum over R 1 in the crystal, N , which cancels out � with the � N term. So in the end, we end up with the following expression for ψs�k |ψs�k : � � � � ��� � ψs�k |ψs�k = R��� eikR crystal φ∗s (�r)φs (�r − R��� )d�r Moreover if the basis of atomic orbitals is orthonormal then those integrals are all zero, exept the one for R��� = �0. And the final result is: � � ψs�k |ψs�k = 1 if orthonormal atomic orbitals We will now consider that the set of atomic orbitals form an or­ thonormal basis � � 2)�Now that � we have expressed ψsk� |ψsk� , we need to focus on the numer­ ator ψs�k |Ĥ|ψs�k . Because Ĥ is a periodic hamiltonian we can show that the numerator can be written as: � � � �� � ˆ s (�r − R � )d�r ψs�k |Ĥ|ψs�k = R� eikR crystal φs (�r)Hφ � ˆ s (�r)d�r Let us denote by �s and γ the following integrals: �s = crystal φs (�r)Hφ � ˆ s (�r− R)d� � r (when R � is a nearest-neighbour to the atom and γ = crystal φs (�r)Hφ centered around� the origin��0). Using those parameters, give a mathematical � only up to expression for ψ � |Ĥ|ψ � when one considers the sum over R sk sk nearest neighbours. solution: � � Using the expression for ψs�k |Ĥ|ψs�k given above and the definition for the nearest-neighbour transfert integral, wee find the following result: � � � �� ψs�k |Ĥ|ψs�k = �s + R� eikR γ where the sum runs over the nearest neighbours of the atom centered at the � = ±�a1 , origin. Those nearest neighbours have the following coordinates R � � R = ±�a2 and R = ±�a3 . 3) Using the results of question 1) and 2), plot the function Es (�k) along the following lines in the first Brillouin Zone: from Γ ( 2aπ (0, 0, 0)) to X ( 2aπ (1, 0, 0)), then from X to L ( 2aπ (1/2, 1/2, 1/2)), and finally form L to Γ. 8 solution: From questions 1) and 2), we see that the energy dispersion can be written as: E(�k) = �s + 2γ(cos(kx a) + cos(ky a) + cos(kz a)) We can then plot this dispersion along high-symmetry lines. This plot is shown on figure 5. Figure 5: Band structure of our 3D nearest-neighbour tight-binding model in a simple cubic structure. 4) Using a Taylor expansion for Es (�k) around �k ≈ �0, show that the band dispersion can be written as: Es (�k) ≈ α − h̄2� k2 2m∗ and relate α and m∗ to the parameters �s , γ and a. From that expression what can you infer about the behavior of electrons with a wavevector close to the Γ point? How does the effective mass m∗ vary with the ”hopping integral” γ? solution: Using a Taylor expansion for the cosines around �k ≈ 0, we find: E(�k) ≈ �s + 2γ ∗ (1 − (kx a)2 2 +1− (ky a)2 2 +1− (kz a)2 ) 2 = (�s + 6γ) − a2 γ�k 2 By identification, we can extract the values for m∗ and α: α = �s + 6γ and m∗ = h ¯2 2a2 γ Remark: it is interesting and physically appealing that the effective mass is proportional to the INVERSE of the transfert integral γ. Indeed this coefficiant is a measure of how ”easy” it is for an electron to ”hop” from an atom to its nearest neighbour. It is then logical that when γ is big, meaning that the electron easily transfer to a neighbouring atom, then the effective mass is small, such that this 9 particle is very ”light”. On the contrary, when γ is very small, then the particle hardly moves away corresponding to the physical picture of a very ”heavy” particle. 5) If one considers that this crystal has only one valence electron per unit cell, will it be a metal or an insulator (explain your answer)? If this is a metal, what is the value of the Fermi energy? Show a 3D plot of the Fermi surface in the first Brillouin Zone (kx , ky and kz all vary between − πa and πa ). solution: We saw in recitation, that a band can accomodate 2N electrons, where N is the number of unit cell in the crystal and the coefficiant 2 takes nto account the spin degeneracy. In our case, we have one electron per unit cell, so N electrons in total and the band is half-filled. Our system is then a metal because empty energy states are arbitrarily close to the highest occupied energy, the Fermi energy. The Fermi energy is exactly the middle energy of our band because this band is symmetric with respect to the middle energy. So EF = �s . The Fermi surface is defined to be the location of all the �k vectors in the first Brillouin Zone, such that the energy E(�k) is equal to the Fermi energy. The equation that defines the Fermi surface is then: E(�k) = �s + 2γ(cos(kx a) + cos(ky a) + cos(kz a)) = EF = �s A 3D plot of the Fermi surface in the first Brillouin Zone is shown on figure 6. Figure 6: Fermi surface corresponding to a half-filled s band in a 3D simple cubic crystal in a nearest neighbour tight-binding model. 10 3 Si and GaAs Silicon and Gallium Arsenide crystals can be described by an fcc lattice with two atoms per unit cell. The primitive vectors �a1 , �a2 , �a3 for the fcc lattice expressed in a cartesian basis are given by: �a1 = a2 (0, 1, 1), �a2 = a2 (1, 0, 1), and �a3 = a2 (1, 1, 0) a is the lattice parameter. The basis vectors, that describe the positions of atoms inside a unit cell, are given by: �τ1 = a4 (1, 1, 1) and �τ2 = − a4 (1, 1, 1) 1) Find an expression for the reciprocal space primitive vectors �b1 , �b2 , �b3 . Give the expression for the Fourier coefficiant VG � in terms of the periodic po∗ tential V (�r). Establish a relationship between V−G � and VG �. solution: In general, the primitive basis vectors in reciprocal space are given by the following formulas: a2 ×� a3 • �b1 = 2π �a1�·(� a2 ×� a3 ) a1 • �b2 = 2π �a1�·a(�3a×� a3 ) 2 ×� a1 ×� a2 • �b3 = 2π �a1�·(� a2 ×� a3 ) Calculating those cross products gives us: �b1 = 2π a (−1, 1, 1), �b2 = 2π a (1, −1, 1), �b3 = 2π a (1, 1, −1) We have seen in the lectures on Bloch theorem (yes we saw it!), that the r) can be written as: fourier coefficiants VG � of the effective periodic potential V (� � � 1 VG r)eiG·�r d�r � = Ω unit cell V (� where Ω is the volume of the unit cell. Now let us take the complex conjuguate of VG � . We end up with: � � � � 1 ∗ (VG r)eiG·�r d�r)∗ = Ω1 unit cell (V (�r)eiG·�r )∗ d�r � ) = ( Ω unit cell V (� But since the potential is a real valued function, we see that V (�r)∗ = V (�r), and so we find: � � � � 1 ∗ (VG r)eiG·r� )∗ d�r = Ω1 unit cell V (�r)(eiG·r� )∗ d�r = V−G � � ) = Ω unit cell (V (� 2) Does Silicon have an inversion symmetry, i.e V (−�r) = V (�r)? Explain your answer. What about GaAs? solution: The origin of the coordinate system being in the middle of two nearest neighbour silicon atoms, we see that Silicon has an inversion symmetry. And so the potential function must have this symmetry too, so V (�r) = V (−�r). GaAs does not have that symmetry because in the process of changing �r to −�r, a Gallium atom becomes an Arsenide atom. ∗ 3) If a crystal have an inversion symmetry, show that VG � = V−G � = VG �. solution: If a crystal have an inversion symmetry, then because of question 1) and 2), we see that: 11 ∗ VG � = VG � � = V−G Indeed the first equality comes from question 1), and the second comes from � �r iG� the identification of like terms in the two sums V (�r) = G = V (−�r) = � VG �e � � r) iG(−� . � VG �e G 12