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PHZ6426: Fall 2013 Problem set # 1: Solutions Instructor: D. L. Maslov [email protected] 392-0513 Rm. 2114 Office hours: TR 3:00 p.m.-4:00 p.m. Please help your instructor by doing your work neatly. “Units Rule”: Every (algebraic) final result must be supplemented by a check of units. Without such a check, no more than 75% of the credit will be given even for an otherwise correct solution. AM≡Ashcroft and Mermin ... 1. Estimates. [15 points] In the problems below, you need to obtain only an order-of-magnitude estimate without doing the actual calculations. “Long” solutions, even if correct, will not be accepted. (a) In certain materials, e.g., graphene, electrons behave as “Dirac fermions”, i.e., ultra-relativistic particles with dispersion E = h̄v0 k, where the “speed of light”, v0 , is much smaller in magnitude than the real speed of light (typically, v0 ∼ 106 m/s). Suppose that a gas of Dirac fermions is confined to a cubic container of side L. The total number of fermions is N . The temperature is equal to zero. Estimate the pressure that the gas exerts on the walls. Solution: Pressure is an intensive thermodynamic variable and, as such, it can depend on the number of particle and the system size only via the number density n = N/L3 . The units of pressure are [P ] = [force]/[L]2 = [energy]/[L]3 , therefore P must be of order of the energy per unit volume at T = 0. The relevant energy per fermion at T = 0 is the Fermi energy, EF ∼ h̄v0 n1/3 , and there are n fermions per unit volume. Therefore, P ∼ EF n ∼ h̄v0 n4/3 = h̄v0 (N/L)4/3 . (b) Estimate the Fermi energy of the Dirac gas in the previous example. Solution: See a): EF ∼ h̄v0 n1/3 = h̄v0 (N/L3 )1/3 . (c) A hydrogen atom is placed into a strong magnetic field B. The orbital effect of the field is to squeeze the electron orbit. Estimate the field at which the radius of the orbit is about a “factor of two” smaller than at B = 0. Solution: One can think in terms of relevant energies. The binding energy of an electron in the hydrogen atom is the Rydberg: ER ∼ me4 /h̄3 . If the electron were free, the magnetic field would quantize the energy level with the characteristic spacing determined by the cyclotron frequency: Ec = h̄ωc = h̄eB/mc (SGS). The field can be considered strong enough when Ec is comparable to Ry, from which one determines the characteristic field scale as Bch = m2 e3 c/h̄3 . Alternatively, one can compare the relevant spatial scales.The radius of the orbit in the absence of the field is the Bohr radius, aB = h̄2 /me2 . The characteristic scales of the wave function in p a magnetic field is the “magnetic length”, `B , related to h̄ωc via h̄ωc ∼ h̄2 /m`2B , which gives `B = h̄c/eB (the numerical coefficient happens to be equal to unity). Equating aB and `B , we arrive at the same result for the characteristic magnetic field as before. Substituting the numerical values of fundamental constants (stripped off the numerical prefactors), we obtain Bc ∼ 109 G= 105 T. (d) The typical binding energy in solids is on the order of a few eV, while the typical bond length is on the 3 order of one Å. Estimate the pressure it takes to crack a solid. Solution: As in a), P ∼ 1eV/Å ∼ 2 1011 N/m ∼ 106 atm. This is a “theoretical” value appropriate for ideal monocrystals. Cracking of real solids depends strongly on the number and type of defects they contain, the most important type of defects being so-called “dislocations”. Real “cracking pressures” are much smaller than the theoretical limit. (e) Estimate the temperature of the crossover between the Fermi-Dirac and Boltzmann statistics for electrons in a semiconductor. The number density is 1017 cm−3 and the effective electron mass is 0.1 of the electron mass in vacuum. 2 Solution: The crossover occurs at T ∼ TF = EF /kB ∼ h̄2 n2/3 /m∗ ∼ 10 K. 2. Dirac-Kronig-Penney model. [20 points] Consider the Dirac-Kronig-Penney model with an attractive potential: X U (x) = −U0 δ(x − na), n where U0 > 0. (a) Find numerically and plot as a function of the dimensionless parameter u ≡ mU0 a/h̄2 the widths of ? few first energy bands. (b) Find the analytic result for the energy spectrum in the tight-binding limit, when u 1. Solution (a) Unlike for the case of potential barriers where the energy was bounded from below by zero, now we can have both positive and negative energies. Consider first the case of E > 0. In this case, the equation for energy spectrum differs from Eq. (0.3) in the notes only in the sign of the second term on the right-hand side: cos ka = cos qa − u sin qa , qa (1) where u = mU0 a/h̄2 . The right-hand side of this equation is plotted as a function of qa/π in Fig. 1 for u = 2, 4, 10. For E < 0, the wavefunctions in intervals I and II change to ψI = Aeκx + Be−κx ψII = Ceκx + De−κx √ with κ = −2mE/h̄ being real. The equation for the energy spectrum can be obtained from the equation in the notes by replacing qa → iκa and u → −u. This gives cos ka = cosh κa − u sinh κa . κa (2) The right-hand side of this equation is plotted in Fig. 2 for u = 0.5, 1, 3, 4. The RHS is now a non-oscillatory function of κa, therefore, only one band is possible. (For u > 3, the RHS has a minium but the value of RHS at qa = 0 is already below −1, so only one solution is possible). (b) For E > 0, one simply needs to flip the sign of U0 in the result presented in the notes. This yields E = E0 + 2J (1 + cos ka) = E0 + 2J − 2J (1 − cos (ka)) , where E0 ≡ π 2 h̄2 π 2 h̄4 ; J ≡ . 2ma2 2m2 a3 U0 In contrast to the potential-barrier case, the band is hole-like near k = 0, where E ≈ E0 + 2J − Jk 2 a2 , 2 and electron-like near the edges of the Brillouin zone, where E0 + 2J − J (k ∓ π/a) a2 . The next band is electron-like at k = 0 and hole-like near the edges, etc. Now consider E < 0. If u 1, solution of (2) is only possible near the zero of RHS, defined by the equation (x ≡ κa, y ≡ ka) coth x0 = u . x0 (3) 3 1.0 2 1 0 0.5 0 1 2 3 4 5 6 7 8 x 1 2 3 4 x 5 6 7 8 K2 0 1 2 3 4 x 5 6 7 8 K1 K4 K2 K6 K 0.5 K K8 1.0 K3 8 6 4 2 0 0.5 1.0 1.5 x K 2 FIG. 1: Positive energies (E > 0). a) u = 2. Solutions are possible for almost all k except for narrow regions. b) u = 4. Forbidden gaps widen up. c) u = 10. Further widening of the gaps. d) Graphical solution of Eq. (1). Blue lines indicate the minimum and maximum values of cos ka. Since coth x0 ≈ 1 at x0 1, we have x0 ≈ u, which is already clear from the plots of RHS in Fig. 2 for u > 1. Define x = x0 + δ with δ x0 and expand the RHS to 1st order in δ : sinh (x0 + δ) x0 + δ u ≈ cosh x0 + δ sinh x0 − (sinh x0 + δ cosh x0 ) (1 − δ/x0 ) x0 u u uδ ≈ cosh x0 − sinh x0 + δ sinh x0 − cosh x0 + 2 sinh x0 + O δ 2 . x0 x0 x0 cos y = cosh (x0 + δ) − u The first term vanishes because of condition (3). The second term reduced with the help of (3) to cosh2 x0 sinh2 x0 − cosh2 x0 =δ δ (sinh x0 − coth x0 cosh x0 ) = δ sinh x0 − sinh x0 sinh x0 δ . = − sinh x0 Since x0 ≈ u 1 and sinh x0 ≈ ex0 /2, this term is exponentially small and can be neglected. Finaly, the third term becomes uδ δ cosh x0 sinh x0 = coth x0 sinh x0 = δ . x20 x0 x0 This term is exponentially large. Keeping only the third term, the equation for δ reduces to cosh x0 x0 x0 δ = cos y cosh x0 x0 x = x0 + cos y cosh x0 cos y = δ 4 40 30 20 10 0 1 2 x 3 4 FIG. 2: Negative energies. Only one band is possible. or, replacing x0 by u and cosh u by eu /2 κa = u + 2ue−u cos y. Solving for energy and neglecting the square of the second term, we obtain for the energy dispersion within the allowed band h̄2 u2 + 4u2 e−u cos ka = −E00 − 2J 0 cos ka 2 2ma = −E00 + 2J 0 + 2J 0 (1 − cos ka) E = − h̄2 2 u 2ma2 h̄2 2 −u J0 = u e . ma2 E00 = − This is an electron-like band of an exponentially narrow width. 3. A bit of lattice dynamics: diatomic chain. [15 points] Show that the dispersion relation of a diatomic chain, consisting of atoms with masses M1 and M2 separated by distance a and connected by “springs” of constant K, is given by s 2 1 1 1 1 4 2 ω =K + ±K + − sin2 qa. M1 M2 M1 M2 M1 M2 What is the main difference between the dispersions of the monoatomic and diatomic chains? Solution: Equations of motion for two neighboring atoms of masses M1 and M2 read: M1 ün = K(un+1 + un−1 − 2un ), (4) M2 ün+1 = K(un+2 + un − 2un+1 ). (5) 5 Try a solution in the form of a propagating wave with two different amplitudes: A1 for atoms of mass M1 and A2 for atoms of mass M2 : un = A1 ei(qxn −ωt) , un+1 = A2 e i(qxn+1 −ωt) (6) , etc. (7) (8) where xn = na. Substituting this solution into eqs. of motion and canceling the common oscillatory factors leads to the system of equations for the amplitudes: −M1 ω 2 + 2K A1 − 2KA2 cos qa = 0 (9) 2 −2KA1 cos qa + −M2 ω + 2K A2 = 0. (10) This homogeneous system of linear equations has a non-trivial solution if and only if its determinant is equal to zero: −M1 ω 2 + 2K −2K cos qa (11) −2K cos qa −M2 ω 2 + 2K = 0, which is equivalent to the 4th order equation for ω: 1 4K 2 1 4 + ω2 + sin2 qa = 0. ω − 2K M1 M2 M 1 M2 (12) The solutions to the last equation are given by 2 ω± 1 1 = K + M1 M2 s 2 1 1 4 ±K + − sin2 qa. M1 M2 M1 M 2 (13) (14) The main difference compared to the monoatomic chains is in that there are now two branches of the spectrum, as shown in Fig. 3 The dispersion of the ω− branch becomes linear in the limit q → 0. This is an “acoustic” branch named so because the ratio of the amplitudes A1 2K cos qa = A2 2K − M1 ω 2 (15) in this case approaches unity for q = 0, i.e., the neighboring atoms oscillate in phase. The ω+ approaches a finite limit ω+ (0) = 2K(1/M1 + 1/M2 ) at q → 0. This is an “optical” branch, because A1 = −(M2 /M1 )A2 at q = 0, i.e., atoms oscillate out of phase around their common center of mass. If atoms carry opposite charges, oscillations of this type generate dipole moments and can be excited by electromagnetic waves. Notice that the wavenumber of electromagnetic radiation ql = ω/c is very small compared to the inverse lattice constant because the speed of light is so large, and thus light can excite only phonons with q → 0. 4. Density of states, Fermi energy. [15 points] As a reminder, the density of states (DoS), g(E), is defined in such a way that g(E)dE is equal to the number of levels (per unit volume of the sample) with energies in the interval from E to E + dE in the limit of dE → 0. One way to calculate the DoS is to equate g(E)dE to the number of states in the corresponding interval of the wavenumbers. In 1D, the interval dk contains 2 × dk/(2π/L) = dkL/π states (an extra factor of 2 accounts for two projections of the electron’s spin and L the length of the system). Another factor of 2 comes from two directions of the electron motion. Thus, the total number of states can be found in two equivalent ways: as Lg(E)dE and as 2dkL/π. Equating these two results, we obtain g(E) = 2 1 2dk = , πdE π |dE/dk| 6 FIG. 3: a) Acoustic (red) and optical (blue) branches of the spectrum for a diatomic chain as a function of qa/π. Parameters: M2 /M1 and K = 1. (the absolute value sign guarantees that the number of levels is a positive definite quantity). In class, we showed that the spectrum of electron in a 1D periodic potential and in the tight-binding limit can be written as E = 2t (1 − cos(ka)) (16) with −π/a ≤ k ≤ π/a. (a) Analyze and explain the behavior of the density of states at E → 0 and E → 4t. (b) Find the dependence of the Fermi energy on the electron number density. Solution (a) p |dE/dk| = 2ta |sin ka| = 2ta 1 − cos2 ka q 2 = 2ta 1 − (1 − E/2t) g(E) = (17) (18) 1 1 q . πta 2 1 − (1 − E/2t) (19) g(E) is plotted in Fig. 4 as a function of E/t. Notice that g(E) can be written as g(E) = 1 2 √ √ . πa E 4t − E (20) For E → 0 (energy at the bottom of the band) 1 1 g(E) ≈ √ √ , a tπ E which coincides with the density of states of a free electron system D0 (E) = cation m∗ = h̄2 /2ta2 . Near the top of the band, E ≈ 4t, and g(E) = (21) √ √ 2m∗ /πh̄ E upon identifi- 1 1 √√ . πa t 4t − E This again coincides with the DOS of free particles in 1D if the energy is measured from 4t. (22) 7 FIG. 4: DOS of a 1D tight-binding model (arbitrary units) as a function of E/2t. (b) The Fermi energy is determined from the condition Z EF 2 πa Z EF 2 1 = dE √ √ πa E 4t − E 0 0 √ with x ≡ E/4t. The integral is solved via a substitution y = x n= dEg(E) = Z 0 x0 √ 1 dx √ √ =2 x 1−x Z 0 x0 Z 0 EF /4t 1 dx √ √ x 1−x √ 1 dy p = 2 arcsin( x0 ). 2 1−y (23) (24) Solving for EF gives EF = 4t sin2 πan . 4 (25) In the limit of almost empty band, πna/4 1, this formula coincides with the result for free electrons EF = h̄2 kF2 /2m∗ upon identifying kF = πn/2 and m∗ = h̄2 /2ta2 . 5. Fermi-Dirac Statistics. [15 points] The chemical potential µ of a free electron gas is determined from the condition that the total number of occupied states is equal to the number of electrons Z dD k 1 n=2 , (26) (2π)D exp [(Ek − µ)/kB T ] + 1 where D is the dimensionality of space. Find the chemical potential for free electrons in D = 2. Analyze and sketch the dependence of µ on the temperature. Hint: Make a substitution x = exp((E − µ)/T ) to solve the integral Z ∞ 1 dE . exp(E/kB T ) + 1 0 Solution: A convenient feature of the 2D case is that one switch from integration over k to integration of Ek = h̄2 k 2 /2m: Z ∞ m 1 n= dEk . (27) 2 exp[(E − µ)/k πh̄ 0 k BT ] + 1 Note that the Fermi energy (the chemical potential at kB T = 0) in 2D is given by µ(0) = EF = h̄2 πn/m, (28) 8 so the last equation can be written as ∞ Z EF = dEk 0 1 . exp[(Ek − µ)/kB T ] + 1 (29) Introduce a new variable x ≡ eEk /kB T , (30) dEk = kB T dx/x; 1 ≤ x < ∞, (31) such that The integral then reduces to Z ∞ dx 1 x x exp(−µ/kB T ) + 1 Z ∞ dx 1 = kB T exp(µ/kB T ) x x + exp(µ/kB T ) 1 Z ∞ 1 1 − = kB T , x x+a 1 EF = kB T (32) 1 (33) (34) where a ≡ exp(µ/kB T ). Each of the two integrals in the last line is divergent, but the difference of them is convergent. To solve the integral in the last line, replace the upper limit by an arbitrarily large number, to be taken to infinity at the end: Z X Z ∞ 1 1 1 1 − = lim − X→∞ 1 x x+a x x+a 1 X +a = lim (ln X − ln ) X→∞ 1+a X(1 + a) = lim ln X→∞ X +a = ln(1 + a). (35) Substituting this result into the equation for µ, we get EF = ln(1 + exp(µ/kB T )), kB T (36) µ = kB T ln [exp(EF /kB T ) − 1] . (37) or For kB T EF , exp(EF /kB T ) 1, and µ is very close to its zero-temperature value (EF ) µ = EF − kB T exp(−EF /kB T ) + . . . This is the limit of quantum, Fermi-Dirac, statistics. In the opposite limit, when kB T EF , exp(EF /kB T ) ≈ 1 + EF /kB T , and µ = kB T ln(EF /kB T ) < 0. (38) This is the limit of classical, Maxwell-Boltzmann, statistics.Because µ is positive at low temperatures and negative at high temperatures, there must be a temperature at which it vanishes. Equation (37) predicts that this temperature is T0 = EF /kB ln 2. A 2D free electron gas is a rare case of a many-body problem when the dependence of the chemical potential on the temperature can be found in a closed form. 9 1 0 1 T 2 3 K1 K2 FIG. 5: The chemical potential in units of EF as a function of the temperature in units of EF /kB . 6. “Two-band Drude model”. [20 points] Suppose that a 2D electron gas contains charge carriers of two types, characterized by different charges qi , densities ni , effective masses mi , and relaxation times τi (i = 1, 2). The electric field is in the plane and the magnetic field is perpendicular to the plane. Using semiclassical equations of motion for each of the carriers pi ṗi = qi E + qi vi ×B− τi with i = 1, 2, show that the longitudinal resistivity and the Hall coefficient are given by ρ1 ρ2 (ρ1 + ρ2 ) + (ρ1 R22 + ρ2 R12 )B 2 ; (ρ1 + ρ2 )2 + (R1 + R2 )2 B 2 R1 ρ22 + R2 ρ21 + R1 R2 (R1 + R2 )B 2 ρxy = = , B (ρ1 + ρ2 )2 + (R1 + R2 )2 B 2 ρxx = (39) RH (40) where ρi ≡ mi /qi2 ni τi and Ri ≡ 1/ni qi . (41) Analyze the special case q1 = −q2 and the limits B → 0, B → ∞. Show that in the limit of ρ1 → ρ2 and R1 → R2 , ρxx and RH reduce to corresponding expressions for a single type of carriers with the number density 2n1 = 2n2 . Note that, contrary to the single-carrier case, the resistivity (53) does depend on B. These results apply to 3D as well. Solution: As two subsystems of carriers are independent, they can be considered as two resistors in parallel. The currents are divided between the bands while the voltage drops are the same. There is no current and, therefore, no electric field along the magnetic field, so the problem is effectively 2D. Then Ex jx1 = ρ̂1 (42) Ey jy1 Ex jx2 = ρ̂2 , (43) Ey jy2 where the in-band resistivities are obtained by inversion of the in-band conductivies. For each of the bands, σi 1 ωci τi , i = 1, 2, σ̂i= 2 −ωci τi 1 1 + (ωci τi ) therefore, −1 ρ̂i= = (σ̂i ) = ρi 1 −ωci τi ωci τi 1 , i = 1, 2, (44) 10 where σi = e2 ni τi /mi , ρi = 1/σi , and ωci ≡ qi B/mi . Eqs. (42,43) represent a system of 4 equations for 6 variables. To make the system complete, we need two more equations. These are jx = jx1 + jx2 , (45) jy = jy1 + jy2 = 0. (46) The last line means that there is no current in the direction transverse to the applied electric field (Hall geometry). Each of the subsystems can carry the y-component of the current, but their sum must be equal to zero. To solve the system, exclude jx2 in favor of jx1 , and jy2 in favor of jy1 . Equating the like components of E in Eqs.(42, 43) , we get a closed system for jx1 , jy1 : (ρ1 + ρ2 )jx1 − (ρ1 ωc1 τ1 + ρ2 ωc2 τ2 )jy1 = ρ2 jx (47) (ρ1 ωc1 τ1 + ρ2 ωc2 τ2 )jx1 − (ρ1 + ρ2 )jy1 = ρ2 ωc2 τ2 jx . (48) The solutions are given by ρ2 (ρ1 + ρ2 ) + ρ2 ωc2 τ2 (ρ1 ωc1 τ1 + ρ2 ωc2 τ2 ) jx , (ρ1 + ρ2 )2 + (ρ1 ωc1 τ1 + ρ2 ωc2 τ2 )2 ρ2 (ρ1 + ρ2 )ωc2 τ2 − ρ2 (ρ1 ωc1 τ1 + ρ2 ωc2 τ2 ) = jx . (ρ1 + ρ2 )2 + (ρ1 ωc1 τ1 + ρ2 ωc2 τ2 )2 jx1 = (49) jy1 (50) Plugging these solutions into Eq.(42), we get Ex = 2 2 2 2 ρ1 ρ2 (ρ1 + ρ2 ) + ρ21 ρ2 ωc1 τ1 + ρ1 ρ22 ωc2 τ2 ) jx . 2 2 (ρ1 + ρ2 ) + (ρ1 ωc1 τ1 + ρ2 ωc2 τ2 ) (51) The proportionality coefficient between Ex and jx is ρxx . It reduces to the form given in the formulation of the problem by re-writing ρi ωci τi = Ri B. (52) Relation between Ey and jx gives the Hall coefficient. In the special case q1 = −q2 > 0 (the last inequality is for the sake of convenience only), R1 = −|R2 | and ρ1 ρ2 (ρ1 + ρ2 ) + (ρ1 R22 + ρ2 R12 )B 2 ; (ρ1 + ρ2 )2 + (R1 − |R2 |)2 B 2 R1 ρ22 − |R2 |ρ21 + R1 R2 (R1 − |R2 |)B 2 . = (ρ1 + ρ2 )2 + (R1 + R2 )2 B 2 ρxx = (53) RH (54) Suppose that n1 = n2 . (Such a system is called “compensated”: the charge densities are equal in magnitude and opposite in sign. Examples: semiconductors doped with donors and acceptors simultaneously; semimetals (bismuth, graphite, antimony, etc.) Then the Hall coefficient ceases to depend on the magnetic field. Furthermore, if ρ1 = ρ2 , then RH = 0. This is what one should expect for a compensated system because RH = 1 1 + =0 q1 n1 q 2 n2 for q1 n1 = −q2 n2 . For identical subsystems, ρ1 = ρ2 ≡ ρ and R1 = R2 ≡ R 2ρ3 + 2ρR2 B 2 = ρ/2; 4ρ2 + 4R2 B 2 2Rρ2 + 2R3 B 2 = = R/2, 4ρ2 + 4R2 B 2 ρxx = (55) RH (56) which is simply the effect of doubling the charge density. 11 The same results can be obtained in a more formal way by noticing that, since two bands are connected in parallel, the total conductivity σ̂ = σ̂1 + σ̂2 and the total resistivity is −1 = (σ̂1 + σ̂2 ) −1 α1 σ1 σ1 σ2 2 + 1+α2 2 1+α 2 1+α1 1 α 2 σ2 σ1 α1 σ1 − 1+α2 + 1+α2 1+α2 1 2 1 + ρ̂ = (σ̂) . For brevity, denote αi ≡ ωci τi . Then σ̂ = + α2 σ2 1+α22 σ2 1+α22 ! . Now, ρxx = σ1 1+α21 + σ2 1+α22 Det , where Det = σ2 σ1 + 1 + α12 1 + α22 2 + α1 σ1 α 2 σ2 + 1 + α12 1 + α22 2 . The rest is just algebra: ρxx = σ1 1+α21 + σ1 1+α21 2 σ2 1+α22 1 = (1 + α22 ) (1 + α12 ) = 1 (1 + α22 ) (1 + α12 ) + + σ2 1+α22 α1 σ1 1+α21 + α 2 σ2 1+α22 σ1 1 + α22 + σ2 1 + α12 2 2 σ12 σ22 σ1 σ2 α1 σ1 α2 σ2 σ1 σ2 α1 α2 + + + 2 1+α 2 + 2 + 2 2 2 2 2 2 2 2 2 1+α 1+α (1+α2 )(1+α1 ) ( 1 21 (1+α1 ) (1+α2 ) 2 )(1+α1 ) σ1 1 + α22 + σ2 1 + α12 σ12 1+α21 + σ22 1+α22 2 = = = 2 + 2σ1 σ2 (1+α22 )(1+α21 ) 2 (1 + α1 α2 ) σ1 1 + α2 + σ2 1 + α1 2 (1 + α2 ) + σ22 (1 + α12 ) + 2σ1 σ2 (1 + α1 α2 ) 1 + α22 /ρ1 + 1 + α12 /ρ2 (1 + α22 ) /ρ21 + (1 + α12 ) /ρ22 + 2 (1 + α1 α2 ) /ρ1 ρ2 ρ1 ρ22 1 + α22 + ρ2 ρ21 1 + α12 ρ21 (1 + α12 ) + ρ22 (1 + α22 ) + +2 (1 + α1 α2 ) ρ1 ρ2 σ12 Recalling that ρi αi = Ri B, we obtain ρxx ρ1 ρ2 (ρ1 + ρ2 ) + ρ1 R22 + ρ2 R12 B 2 = 2 ρ1 + ρ22 + 2ρ1 ρ2 + ρ21 α12 + ρ22 α22 + 2α1 α2 ρ1 ρ2 ρ1 ρ2 (ρ1 + ρ2 ) + ρ1 R22 + ρ2 R12 B 2 = , 2 (ρ1 + ρ2 ) + (R1 + R2 )2 B 2 which is the desired result. Similar for ρxy .