Physics 211B : Solution Set #1 [1] Drude formula – Consider a hypothetical monovalent s-band metal with a simple cubic crystal structure. The valence band dispersion is given by the tight binding result, ε(k) = −2t cos(kx a) + cos(ky a) + cos(kz a) . Compute the DC conductivity tensor σαβ . Show that σαβ = σ δαβ is diagonal, and obtain an expression for σ. Numerically evaluate any integrals. The following result may prove useful: Zπ Zπ p du dv δ(cos u + cos v + 2λ) = 4 K( 1 − λ2 ) Θ(1 − λ2 ) , −π −π where K(x) is the complete elliptic integral of the second kind. Compare your result with the Drude value you would obtain by approximating the band as parabolic, based on its curvature at the zone center. Solution: We must evaluate 2 Z σαβ = 2e τ d3k α β v v (2π)3 ∂f − , ∂ε Ω̂ where the dispersion is ε(k) = −2t cos(kx a) + cos(ky a) + cos(kz a) . The velocity is v(k) = 1 ∂ε 2ta = sin(kx a), sin(ky a), sin(kz a) . ~ ∂k ~ As a consequence of the cubic symmetry, σαβ = σ δαβ is diagonal, and we may write 2ta 2 Z d3k ∂f 2 σ = 2e τ sin (kz a) − . ~ (2π)3 ∂ε 2 Ω̂ The monovalency condition means that there is one electron per unit cell, which in turn means the chemical potential lies at µ = 0 so the s-band is half-filled. Changing variables to u = kx a, etc., e2 τ t σ= 3 2 2π ~ a Zπ Zπ Zπ du dv dw sin2 (w) δ(cos u + cos v + cos w) . −π −π −π 1 We now use Zπ Zπ I(λ) = du dv δ(cos u + cos v + 2λ) −π −π 1−2|λ| Z =4 −1 (1) √ 1 1 q 2 2 1−x 1 − x + 2|λ| (2) Z1 ds 2 2 1 + |λ| − 2 1 + |λ|2 s2 + 1 − |λ| s4 −1 p 1 − |λ| 8 K = 4 K 1 − λ2 . = 1 + |λ| 1 + |λ| = 4 1 − |λ| q (3) (4) Thus, σ= where 8C e2 τ t · . π 3 ~2 a (5) Z1 p q C = dx 1 − x2 K 1 − 14 x2 . 0 Numerical integration gives C ' 2.59011. Expanding the dispersion about the zone center, we find ε(k) = −6t + ta2 k2 + O(k 4 ) , (6) hence the effective mass is given by ta2 ≡ ~2 2m∗ =⇒ m∗ = ~2 . 2ta2 (7) The Drude conductivity is then σ= ne2 τ 2ta2 e2 τ t 2 = ne τ · = 2x · , m∗ ~2 ~2 a (8) where x ≡ na3 is the dimensionless density, equal to the average number of electrons per unit cell (0 ≤ x ≤ 2). For a monovalent metal, the band is half filled, and x = 1, and the prefactor 2x is 2. The exact value of the prefactor is 8C/π 3 ' 0.668. [2] Thermal transport in a magnetic field – Consider a metal with a parabolic band ε(k) = ~2 k2 /2m∗ in the presence of a uniform magnetic field B. Use the Boltzmann equation* to compute (a) the resistivity tensor ρ, (b) the thermal conductivity tensor κ, (c) the thermopower tensor Q, and (d) the Peltier tensor u. Assume T is small, and work to lowest nontrivial order in the temperature T . Also assume a constant relaxation time τ . Does the Wiedemann-Franz law hold for the matrices κ and ρ ? 2 Solution: We begin with the Boltzmann equation, i ∂f h e ∂δf δf ε−µ − = ∇T v×B· − . v · eE + T ∂ε ~c ∂k τ We take ε(k) = ~2 k2 /2m∗ , and we write δf (k) = k · A(ε) ⇒ vx ∂δf ∂δf ~ − vy = ∗ (kx Ay − ky Ax ) , ∂ky ∂kx m from which we obtain the linear relations eEx + 1 +ωc τ 0 Ax 0 ~τ ∂f −ωc τ eEy + 1 0 Ay = ∗ m ∂ε Az eEz + 0 0 1 The solution is easily obtained: 1 Ax ∗ ∂f 0 ~τ /m +ωc τ Ay = 1 + ωc2 τ 2 ∂ε Az 0 −ωc τ 1 0 ε−µ T ε−µ T ε−µ T ∂x T ∂y T . ∂z T eEx + 0 eEy + 0 2 2 eEz + 1 + ωc τ ε−µ T ε−µ T ε−µ T ∂x T ∂y T . ∂z T The electrical and thermal currents are given by Z 2e dε g(ε) ε Aα jα = − 3~ Z 2 jαq = dε (ε − µ) g(ε) ε Aα . 3~ We now read off the transport coefficients from the relations j = ρ−1 E − ρ−1 Q∇ T jq = uρ−1 E − (κ + uρ−1 Q)∇ T . We find −ωc τ 0 ρ−1 1 0 2 2 0 1 + ωc τ Z 1 −ωc τ 0 2e ε(ε − µ) ∂f τ +ωc τ ρ−1 Q = − ∗ dε g(ε) − 1 0 3m T ∂ε 1 + ωc2 τ 2 2 2 0 0 1 + ωc τ Z 1 −ωc τ 0 2e ∂f τ +ωc τ uρ−1 = − ∗ dε (ε − µ) g(ε) ε − 1 0 3m ∂ε 1 + ωc2 τ 2 0 0 1 + ωc2 τ 2 Z 1 −ωc τ 0 2 τ 2 ε (ε − µ) ∂f +ωc τ κ + uρ−1 Q = dε g(ε) − 1 0 ∗ 3m T ∂ε 1 + ωc2 τ 2 2 2 0 0 1 + ωc τ . 2e = 3m Z 1 ∂f τ +ωc τ dε g(ε) ε − ∂ε 1 + ωc2 τ 2 0 3 We evaluate the integrals using the Sommerfeld expansion, and invoking the density of states √ 2 (m∗ )3/2 √ ε. g(ε) = π 2 ~3 With an energy-independent relaxation time τ , we obtain 1 −ωc τ 0 2τ ne 1 +ωc τ %−1 = σ = 1 0 m∗ 1 + ωc2 τ 2 2 2 0 0 1 + ωc τ The quantity ρ−1 Q is proportional to the same matrix as is ρ−1 , and one readily finds that Q is a multiple of the unit matrix, Q=− π 2 kB2 T 2eεF ·I . Here we have used the results Z ∂f dε g(ε) ε (ε − µ) − = 12 π 2 (kB T )2 g(µ) + O(T 4 ) ∂ε g(εF ) = m∗ kF 3n = . 2 2 π ~ 2εF We also find u = T Qt (−B) = − π 2 (kB T )2 Finally, the thermal conductivity tensor is 1 ∗ nτ /m 1 2 2 κ = 3 π kB T +ωc τ 1 + ωc2 τ 2 0 2eεF ·I . −ωc τ 1 0 0 0 2 2 1 + ωc τ . Thus, the Wiedemann-Franz law, κ= π 2 2 −1 k Tρ 3e2 B holds at the matrix level as well. [3] Two bands – Calculate the frequency-dependent conductivity tensor for a direct gap semiconductor in the presence of a magnetic field B = B ẑ. You should begin with the Boltzmann equation in the relaxation time approximation (f 0 → f¯0 , δf → δf¯ for holes), ∂δf ∂f 0 e ∂δf δf − ev · E − v×B· =− , ∂t ∂ε ~c ∂k τ and the conduction and valence band dispersions, εv (k) = εv0 − 12 ~2 mvαβ −1 k α k β εc (k) = εc0 + 12 ~2 mcαβ −1 k α k β . 4 Assume the two bands behave independently, and solve the two Boltzmann equations for the conduction electrons and valence holes. In each case, try a solution of the form δf (k, t) = k µ Aµ (ε(k)) e−iωt . The currents are jc jv d3k = 2e (2π)3 Z − vc δfc + vv δf¯v . Ω̂ Compute σαβ along principal axes of the effective mass tensors. You may assume that mv and mc commute, i.e. they have the same eigenvectors. You should further assume that B lies along a principal axis. Solution: This problem is essentially solved in the notes in section 1.7. All that is left to do is to sum the contributions from the valence and conduction bands: σxx (ω) = pv e2 τv nc e2 τc 1 − iωτc 1 − iωτv + 2 2 τ2 ∗ ∗ 2 2 mc,x (1 − iωτc ) + ωc,⊥ τc mv,x (1 − iωτv )2 + ωc,⊥ v ωc,⊥ τc ωv,⊥ τv pv e2 τv nc e2 τc p + σxy (ω) = − p ∗ 2 2 τ2 ∗ ∗ ∗ 2 2 mc,x mc,y (1 − iωτc ) + ωc,⊥ τc mv,x mv,y (1 − iωτv )2 + ωc,⊥ v σzz (ω) = nc e2 τc 1 1 pv e2 τv + . ∗ ∗ mc,z 1 − iωτc mv,z 1 − iωτv [4] Spin disorder resistivity (for the brave only!) – Consider an isolated trivalent Tb impurity ion in a crystal field. Application of Hund’s rules gives a total angular momentum J = 6. A cubic crystal field splits this 13-fold degenerate multiplet into six levels: two singlets, one doublet, and three triplets. The ground state is a singlet. Using the first Born approximation, calculate the temperature-dependent resistivity in a free electron model with a scattering Hamiltonian Nimp Himp = −A (g − 1) X δ(r − Rj ) S · Jj /~2 , j=1 where r and S are the conduction electron position and spin operators Rj and Jj are the impurity position and angular momentum of the j th Tb impurity. A is the strength of the exchange interaction, and g = 23 is the gyromagnetic factor. (a) In general the relaxation time is energy-dependent: τ = τ (ε). Show that the resistivity is given by ρ = m/ne2 hτ i, where the average is with respect to the weighting function ε g(ε) (−∂f 0 /∂ε). Show also that 1 ≤ hτ −1 i, hτ i which provides an upper bound for ρ which can often be computed. 5 (b) Use the results of (a) to derive the approximate expression for the resistivity ρ ' ρ0 pij Qji , where (Ei − Ei )/kB T e−Ei /kB T pij = P −E /k T · k B 1 − e−(Ei −Ej )/kB T ke 2 2 2 Qij = 12 i J + j + 12 i J − j + i J z j , where the ionic energy levels are denoted by Ei and where the summations run over the (2J + 1) crystal field states. Show that ρ0 = 3πm (g − 1)2 A2 nimp 8e2 ~3 εF . (c) Show that the high temperature limiting value of ρ is J(J + 1) ρ0 . This is often called the spin-disorder resistivity. Solution: The collision term in Boltzmann equation is, from Fermi’s golden rule, 2 2π X X Ikσ [f ] = Pi jkσ Himp ik0 σ 0 δ Ej + εk − Ei − εk0 fk0 σ0 (1 − fkσ ) ~ ij k0 σ 0 2 2π X X 0 Pi f k σ Himp ikσ 0 δ Ej + εk0 − Ei − εk fkσ (1 − fk0 σ0 ) , − ~ 0 0 ij k σ where Pi is the Boltzmann weight for the ion in state i: exp(−Ei /kB T ) . Pi = P n exp(−En /kB T ) Using plane wave states ψk (r) = V −1/2 exp(ik · r), the matrix element is obtained: X 1 0 jkσ Himp ik0 σ 0 = − A (g − 1) ~−2 ei(k−k )·R jσ 0 S · J iσ . V R We assume the impurity positions are uncorrelated, so that 2 2 X jkσ Himp ik0 σ 0 2 = A (g − 1) jσ 0 S · J iσ 2 ei(k−k0 )·(R−R0 ) ~2 V 2 0 R,R = Nimp V2 A2 (g − ~4 1)2 0 jσ S · J iσ 2 1 + (Nimp − 1)δkk0 . The term proportional to δkk0 cancels when inserted into the collision integral. We are left with X Z d3k 0 2 0 2π 2 2 0 − Ei − εk jσ S · J iσ δ E + ε Ikσ [f ] = 5 A (g − 1) nimp j k ~ (2π)3 0 ijσ Ω̂ n o × Pj fk0 σ0 (1 − fkσ ) − Pi fkσ (1 − fk0 σ0 ) . 6 0 is annihilated by I: Note that the Fermi-Dirac distribution fkσ e−βEj 1 eβ(εk −µ) 0 Pj fk00 σ0 (1 − fkσ ) = P −βEn β(ε 0 −µ) δ Ej + εk0 − Ei − εk β(ε −µ) +1 e k +1 e k ne −βE β(ε −µ) 0 i 1 e e k = P −βEn β(ε 0 −µ) δ Ej + εk0 − Ei − εk β(ε −µ) +1 e k +1 e k ne 0 = Pi fkσ (1 − fk00 σ0 ) . We therefore write 0 fkσ = fkσ + δfkσ 0 fkσ = fk0 = , 1 eβ(εk −µ) +1 and find 0 0 + δfkσ ) (1 − fk00 σ0 − δfk0 σ0 ) − δfkσ ) − Pi (fkσ Pj (fk00 σ0 + δfk0 σ0 ) (1 − fkσ 0 0 δfk0 σ0 − Pj fk00 σ0 + Pi 1 − fk00 σ0 δfkσ . ) + Pi fkσ = Pj (1 − fkσ The Boltzmann equation then takes the form X ∂f 2π jσ 0 S · J iσ 2 evk · E − = 5 A2 (g − 1)2 nimp ∂ε ~ ijσ 0 Z 3 0 dk 0 0 0 − Ei − εk δfk0 σ0 δ E + ε P (1 − f ) + P f × j j i k kσ kσ (2π)3 Ω̂ − Pj fk00 σ0 + Pi (1 − fk00 σ0 ) δfkσ . We now sum both sides on σ and divide by two. Using X X jσ 0 S · J iσ 2 = j J α i i J β j × σ0 S α σ σ S β σ0 σ = = σ αβ 1 2 3 ~ S(S + 1)δ 2 1 2 j J i 4~ α β j J i iJ j , we obtain X 2 Z d3k 0 ∂f 2π 2 2 j J i 0 − Ei − εk evk · E − = 5 A (g − 1) nimp δ E + ε j k ∂ε ~ (2π)3 ij Ω̂ 0 0 ) + Pi fkσ δfk0 σ0 − Pj fk00 σ0 + Pi 1 − fk00 σ0 δfkσ × Pj (1 − fkσ It is convenient to define the structure factor X 2 S(u) ≡ Pi j J i δ(u − Ej + Ei ) . ij 7 . (9) Note that S(−u) = e−βu S(u). We assume a solution of the form ∂f 0 ∂ε and plug this into our Boltzmann equation. Further assuming an isotropic Fermi surface, the δfk0 term integrates to zero since it is proportional to vk0 , aside from energy-dependent (hence rotationally isotropic) factors. We then have δfk = eτ (εk ) · E 1 π = 3 A2 (g − 1)2 nimp τ (εk ) 2~ Z∞ Z 3 0 dk du S(u) δ(u + εk0 − εk ) (2π)3 −∞ Ω̂ h i × 1 − fk00 + e−βu fk0 . Furthermore, since u = εk − εk0 , we have 1 − fk00 + e−βu fk0 = 1 − fk00 , 1 − fk0 giving 1 π = 3 A2 (g − 1)2 nimp τ (εk ) 2~ Z∞ Z 3 0 1 − fk00 dk 0 − εk ) δ(u + ε du S(u) k (2π)3 1 − fk0 −∞ Ω̂ (a) The conductivity is determined from Z 3 ∂f dk eτ (ε ) v · E v − , j = 2e k k k (2π)3 ∂ε Ω̂ which, for free electrons, gives ne2 hτ i m ,Z Z ∂f ∂f hτ i = dε ε g(ε) τ (ε) − dε ε g(ε) − . ∂ε ∂ε σ= Now the triangle inequality requires for any real functions f (ε) and g(ε) Z 2 Z Z dε f (ε) h(ε) ≤ dε f (ε) dε h(ε) , so taking 1/2 ∂f f (ε) ≡ ε g(ε) − τ (ε) ∂ε ∂f 1 1/2 h(ε) ≡ ε g(ε) − ∂ε τ (ε) 8 we conclude D 1 E τ (ε) · ≥1 τ (ε) hence ρ= m m D1E ≤ . ne2 hτ i ne2 τ (b) Let us compute τ −1 . Noting that ∂f = β f0 1 − f0 , − ∂ε we have D1E τ Z 3 Z 3 0 2m π 2 dk dk 2 = · 3 A (g − 1) βnimp 3 3n 2~ (2π) (2π)3 Ω̂ Ω̂ Z × du S(u) δ(u + εk0 − εk ) fk0 1 − fk00 vk2 Z Z Z πm A2 (g − 1)2 = β nimp du S(u) dε dε0 δ(u + ε0 − ε) 3~n (2π)6 ~4 Z Z × f 0 (ε) 1 − f 0 (ε0 ) dSε |v| dSε0 |v 0 |−1 . The factor f 0 (ε) 1 − f 0 (ε − u) varies on a scale kB T . If kB T µ and if u µ then we can approximate Z Z 2 dSε |v| dSε0 |v 0 |−1 ≈ 4πkF2 , and the energy integral becomes Z dε f 0 (ε) 1 − f 0 (ε) = ω . −1 eβω Thus, D1E τ and ρ≤ m = A2 (g − 1)2 (kF /~)4 nimp 12π 3 ~n Z m D 1 E 3π mA2 (g − 1)2 = nimp ne2 τ 8 e2 ~εF Z du du βu S(u) , −1 eβu βu S(u) eβu − 1 since n = kF3 /3π 2 and εF = ~kF2 /2m. Note that Z du X β(Ej − Ei ) e−βEi 2 βu P S(u) = j J i eβu − 1 eβ(Ej −Ei ) − 1 n e−βEn ij X ≡ pij Qji , ij 9 where β(Ej − Ei ) e−βEi P −βE n eβ(Ej −Ei ) − 1 n e 2 1 + 2 1 i 2 z 2 Qji = j J i = 2 j J i + 2 j J i + j J i . pij = Therefore, the upper bound on the resistivity is ρ+ = ρ0 pij Qji , with ρ0 = 3πm A2 (g − 1)2 nimp . 82 ~3 εF (c) At high temperatures, βu/(eβu − 1) → 1 and therefore pij → Pi , independent of j. In this limit, then, X 2 X X X −βEi j J i e−βEn pij Qji = e ij i n j X X = i J2 i e−βEn n i = J(J + 1) . Hence, ρ+ = J(J + 1)ρ0 at high temperatures. [5] Cyclotron resonance in Si and Ge – Both Si and Ge are indirect gap semiconductors with anisotropic conduction band minima and doubly degenerate valence band maxima. In Si, the conduction band minima occur along the h100i (hΓXi) directions, and are six-fold degenerate. The equal energy surfaces are cigar-shaped, and the effective mass along the hΓXi principal axes (the ‘longitudinal’ effective mass) is m∗l ' 1.0 me , while the effective mass in the plane perpendicular to this axis (the ‘transverse’ effective mass) is m∗t ' 0.20 me . The valence band maximum occurs at the unique Γ point, and there are two isotropic hole branches: a ‘heavy’ hole with m∗hh ' 0.49 me , and a ‘light’ hole with m∗lh ' 0.16 me . In Ge, the conduction band minima occur at the fourfold degenerate L point (along the eight h111i directions) with effective masses m∗l ' 1.6 me and m∗t ' 0.08 me . The valence band maximum again occurs at the Γ point, where the hole masses are m∗hh ' 0.34 me and m∗lh ' 0.044 me . Use the following figures to interpret the cyclotron resonance data shown below. Verify whether the data corroborate the quoted values of the effective masses in Si and Ge. Solution: We found that σαβ = ne2 Γ−1 αβ , with e Γαβ ≡ (τ −1 − iω) mαβ ± αβγ B γ c −1 ∗ (τ − iωτ )mx ±eBz /c ∓eBy /c ∓eBz /c (τ −1 − iω)m∗y ±eBx /c . = ±eBy /c ∓eBx /c (τ −1 − iω)m∗z 10 Figure 1: Constant energy surfaces near the conduction band minima in silicon. There are six symmetry-related ellipsoidal pockets whose long axes run along the h100i directions. The valence band maxima are isotropic in both cases, with m∗hh (Si) ' 0.49 me m∗hh (Ge) ' 0.34 me m∗lh (Si) ' 0.16 me m∗lh (Ge) ' 0.044 me . With isotropic bands, the absorption is peaked at ω = ωc = eB/m∗ c, assuming ωc τ 1. Writing ω = 2πf , the resonance occurs at a field m∗ c e hc m∗ 1 hf = · · · 2 e me 2πaB (e2 /aB ) m∗ = 3.58 × 10−7 G · · f [Hz] me m∗ = 8590 G · , me B(f ) = 2πf · 11 Figure 2: Cyclotron resonance data in Si (G. Dresselhaus et al., Phys, Rev, 98, 368 (1955).) The field lies in a (110) plane and makes an angle of 30◦ with the [001] axis. where we have used hc = 4.137 × 10−7 G · cm2 e ~2 aB = = 0.529 Å me e2 h =4.136 × 10−15 eV · s e2 = 27.2 eV = 2 Ry aB f = 2.40 × 1010 Hz . Thus, we predict Bhh (Si) ' 4210 G Bhh (Ge) ' 2920 G Blh (Si) ' 1370 G Blh (Ge) ' 378 G . All of these look pretty good. 12 Figure 3: Constant energy surfaces near the conduction band minima in germanium. There are eight symmetry-related half-ellipsoids whose long axes run along the h111i directions, and are centered on the midpoints of the hexagonal zone faces. With a suitable choice of primitive cell in k-space, these can be represented as four ellipsoids, the half-ellipsoids on opposite faces being joined together by translations through suitable reciprocal lattice vectors. Now let us review the situation with electrons near the conduction band minima: Si : 6-fold degenerate minima along h100i Ge : 4-fold degenerate minima along h111i (at L point) m∗l (Si) ' 1.0 me m∗l (Ge) ' 1.6 me m∗t (Si) ' 0.20 me m∗t (Ge) ' 0.08 me . The resonance condition is that σαβ = ∞, which for τ > 0 occurs only at complex frequencies, i.e. for real frequencies there are no true divergences, only resonances. The location of the resonance is determined by det Γ = 0. Taking the determinant, one finds e2 m∗ e2 det Γ = (τ −1 − iω) m∗l · (τ −1 − iω)2 m∗t 2 + 2 Bz2 + ∗t 2 Bx2 + By2 . c ml c 13 Figure 4: Cyclotron resonance data in Ge (G. Dresselhaus et al., Phys, Rev, 98, 368 (1955).) The field lies in a (110) plane and makes an angle of 60◦ with the [001] axis. Assuming ωτ 1, the location of the resonance is given by 2 ω = eBk m∗t c 2 m∗ + t∗ ml eB⊥ m∗t c 2 , where Bk ≡ Bz and B⊥ ≡ Bx x̂ + By ŷ. Let the polar angle of B be θ, so Bk = B cos θ and B⊥ = B sin θ. We then have 2 n o m∗t 2 ω = cos θ + ∗ sin θ m s l m∗ m∗ t B(f ) = 8600 G · cos2 θ + t∗ sin2 θ , me ml 2 eB m∗t c 2 where again we take f = ω/2π = 2.4 × 1010 Hz. 14 According to the diagrams, the field lies in the (110) plane, which means we can write q q B̂ = 12 sin χ ê1 − 12 sin χ ê2 + cos χ ê3 , where χ is the angle B̂ makes with ê3 = [001]. Ge We have m∗t = 0.051 , m∗l m∗t = 0.082 me and we are told χ = 60◦ , so B̂ = q 3 8 ê1 − q 3 8 ê2 + 12 ê3 . The conduction band minima lie along h111i, which denotes a set of directions in real space: ±[111] : n̂ = ± √13 (ê1 + ê2 + ê3 ) ⇒ cos2 θ = (B̂ · n̂)2 = 1 12 ⇒ B = 1950 G ±[111̄] : n̂ = ± √13 (ê1 + ê2 − ê3 ) ⇒ cos2 θ = (B̂ · n̂)2 = 1 12 ⇒ B = 1950 G ±[1̄11] : n̂ = ± √13 (−ê1 + ê2 + ê3 ) ⇒ cos2 θ = (B̂ · n̂)2 = ⇒ cos2 θ = (B̂ · n̂)2 = ±[11̄1] : n̂ = ± √13 (ê1 − ê2 + ê3 ) √ 7−2 6 12 √ 7+2 6 12 ⇒ B = 1510 G ⇒ B = 710 G . All OK! Si Again, B lies in the (110) plane, this time with χ = 30◦ , so B̂ = q 1 8 ê1 − q 1 8 ê2 + q 3 4 ê3 . The conduction band minima lie along h100i, so ±[001] : n̂ = ±ê3 ⇒ cos2 θ = (B̂ · n̂)2 = 2 2 ±[010] : n̂ = ±ê2 ⇒ cos θ = (B̂ · n̂) = ±[100] : n̂ = ±ê1 ⇒ cos2 θ = (B̂ · n̂)2 = These also look pretty good. 15 3 4 1 8 1 8 ⇒ B = 1820 G ⇒ B = 2980 G ⇒ B = 2980 G .