Theoretische Physik 2: Elektrodynamik (Prof. A-S. Smith) Home assignment 9

WiSe 202 20.2.202 Prof. Dr. A-S. Smith Dipl.-Phys. Ellen Fischermeier Dipl.-Phys. Matthias Saba am Lehrstuhl für Theoretische Physik I Department für Physik Friedrich-Alexander-Universität Erlangen-Nürnberg Theoretische Physik 2: Elektrodynamik Prof. A-S. Smith) Home assignment 9 Problem 9. Dielectric tensor Anisotropic dielectric media are described by a dielectric tensor ε ik. So the following relation holds: D i = k=x,y,z ε ike k i = x, y, z). A flat, in x and y direction, infinite plate of thickness d with the dielectric tensor ε ik is placed in vacuum and in an external, homogeneous electric field E 0 = E 0x e x + e z. a) Calculate E r), D r) and P r) within the plate. b) Calculate the polarization charge density ρ p r) and for z = 0 and z = d) the surface polarization charge density σ p r). c) Calculate the angle of refraction θ of E at the interface z = 0 in terms of the incident angle θ 0. Problem 9.2 Surface-plasmon polaritons Consider an interface between a metal characterized by a dielectric function ε ω) = ω 2 p/ω 2 and an ideal dielectric, ε 2 ω) = ε = const., ε >. In each material the constitutive equations D i = ε iei, B i = H i, i =, 2 apply. The interface supports electromagnetic modes propagating along the interface surface-plasmon polaritons). Taking the interface as the z = 0 plane and choosing the propagation of the mode as the x-direction, choose as an ansatz for the fields E i r, t) = E i e iqx ωt) e κ i z, Bi r, t) = B i e iqx ωt) e κ i z for i =, 2),

with positive decay constants κ i > 0. a) Formulate appropriate continuity conditions for the amplitudes E i, B i across the interface, z = 0. b) Show that the magnetic fields are perpendicular to the interface and to the propagating direction, i.e. B = 0, By, 0). c) Sketch the dispersion ω = ωq). Show that for short wavelengths k ω p /c, the surface-plasmon polariton frequency approaches a constant ω s whereas for long wavelength the dispersion is linear ω = c s q to first order in q. Discuss the attenuation length l i = /κ i in the metal and the dielectric as a function of frequency. Problem 9.3 Superconductor The constitutive equation of a type-i superconductor relates the supercurrent density J s directly to the vector potential A via the second London equation, J s x) = n se 2 mc A x). Here m and e denote the mass and charge of the supercurrent carrier, and n s abbreviates their number density. a) Use Maxwell s equations to show that in the static case the magnetic field fulfills the field equation 2 B x) B x) λ 2 = 0, L and determine the London penetration depth λ L. Conclude that no homogeneous magnetic field can exist in the bulk of a superconductor. b) Consider the boundary z = 0 between a superconductor z > 0) and vacuum z < 0). A magnetic field B is applied parallel to the boundary z < 0). Solve for the magnetic field inside of the superconductor. c) Show that the field equation can be obtained by minimizing the total energy U = U matter +U field by varying with respect to the vector potential, A x) A x) + δa x). Here the variation of the matter and field energy follows from δu matter = c d 3 x J s x) δ A x) and δu field = 4π d 3 x B x) δ B x). The supercurrent J s x) and the magnetic field B x) have to be eliminated in favor of A x) to perform the variation. Problem 9.4 Nuclear Magnetic Resonance Nuclear magnetic resonance spectroscopy uses the magnetic moment of the nuclei of certain atoms to study physical, chemical, and biological properties of matter. The magnetization M due to the spin of the nuclei obeys the Bloch equations Mt) = γ Mt) Ht) T [ Mt) M0 ]. Here the gyromagnetic ratio γ determines the frequency of the Larmor precession. The second term is a phenomenological damping term introducing a characteristic energy) relaxation time T. Consider a strong d.c. field H 0 aligning the magnetization Mt) = M 0 H 0 in the static case. A small timedependent field δh t) is applied in addition to the d.c. field H 0. The probing field δh t) acts perpendicularly to H 0 at all times. a) Derive a constitutive equation for the induced magnetization δmt) = Mt) M 0 to linear order in δh t). Decompose the response into a component parallel and perpendicular to the static

external field, δ Mt) = δ M t) + δ M t), and show that they fulfill δ M t) + T δ M t) = 0, δ M t) γδ M t) H 0 + T δ M t) = γ M 0 δ H t). b) Discuss the free decay of the induced magnetization δ Mt) in the absence of external driving, i.e., δ H t) 0, for arbitrary initial condition δ Mt = 0). Hint: It is favorable to complexify the transverse magnetization δ M t). c) Derive the steady state response for a probing field rotating perpendicularly to the aligning field H 0 at constant angular frequency, δh t) = δh ω cos ωt, sin ωt, 0). Here the z-axis has been chosen parallel to H 0. Due date:

D r) = ε 0 ε E inside r) = ε 0 = ε 0 ε xx ε xy ε xz ε yx ε yy ε yz ε zx ε zy ) ε xx εxzεzx E 0x + εxz ) ε yx εyzεzx E 0x + εyz E 0x 0 ε zx E 0x ) inside = P 0 r) = D r) ε 0E r) = 0 outside ) ε xx εxzεzx E 0x + εxz ) P r) = ε 0 ε yx εyzεzx E 0x + εyz ) ε zx E 0x + inside b) ρ p r) = P z) = 0 σ p z = 0) = P 0z P z = P z = ε 0 εzx E 0x + σ p z = d) = P z 0 = P z = σ p z = 0) ) ) c) It holds tan θ = E x E z, tan θ 0 = E 0x tan θ = E 0x ε zx E 0x = tan θ 0 ε zx tan θ 0 The special case of an isotropic medium: ε ik = εδ ik tan θ = ε tan θ 0 ) Solution of Problem 9.2 Surface-plasmon polaritons Surface plasmons are waves localized at an interface. The half spaces are characterized by complex dielectric functions ε,2 ω) = ε,2 ω) + iε,2 ω). The imaginary parts correspond to absorption, so we assume that the frequency ω is such that both media are almost transparent, i.e. ε,2 ω) is taken as real. To simplify notation, the dependence on the frequency is suppressed in the following. Maxwell s equations hold in each half space Ampère: Bi r, t) = ε i c t E i r, t) Faraday: Ei r, t) = c t B i r, t) Coulomb: [εi Ei r, t)] = 0 Gauß: Bi r, t) = 0 As a consequence, the fields obey a wave equation in each region { } εi c 2 2 t 2) Ei r, t) = 0 B i r, t) 2

As an ansatz for the surface plasmons we write E i r, t) = E i e iqx ωt) e κ i z, Bi r, t) = B i e iqx ωt) e κ i z, for i = z > 0)and i = 2z < 0), thus the wave is propagating along the x-direction and decays exponentially away from the interface. The boundary conditions are as usual, the tangential compontents of E and H are continuous, and the normal components of D and B are also continuous. Since here B = H is this problem, B is continuous across the interface. The wave equation gives the dispersion relation magnetic Gauß law implies ε i ω 2 c 2 q 2 + κ 2 i = 0 κ i = q 2 ε iω 2 c 2 iqb x κ B z = 0 iqb x + κ 2 B z = 0 Since κ κ 2 one first concludes B z = 0 and then B x = 0. Thus B = 0, B y, 0). >From Ampère s law κ B y = ε iω c E,x, κ 2 B y = ε 2iω c E 2,x However, since E x is continuous, surface plasmons can only exist if κ ε = κ 2 ε 2 Hence we expect solutions only if ε and ε 2 are of opposite sign. For the metal-dielectric interface, ε ω) = ωp/ω 2 2, ε 2 ω) = ε = const. this implies that surface-polaritons have a frequency below the plasma frequency. Together with the dispersion relation q ε 2 ε ω 2 c 2 = q ε 2 ε 2ω 2 2 c 2 q = ω ε ε 2 c ε + ε 2 For the metal-dielectric interface this can be inverted to ω 2 = { } c 2 q 2 + ε) + εωp 2 [c 2ε 2 q 2 + ε)] 2 εωp) 2 2 + 2c 2 q 2 ε )εωp 2 This is the representation of ω in terms of q. For long wavelengths q ω p /c the dispersion becomes linear ω = cq cq c 2 q 2 ε ε 2ωp 2 + Oq 4 ), q 0 with a velocity corresponding to the speed of light in the dielectric medium. For short wavelength q the frequency approaches ω ω p / + ε. For the case of a metal-vacuum surface, ε =, the relations simplify ω 2 = ωp/2 2 + c 2 q 2 ωp 4 + c 4 q 4 ωq) = cq + Oq 2 ), ωq ) ω p / 2 The attenuation length l i can be found by using the relation l i = /κ i and substituting the expression for κ i in terms of the frequency. 3

Solution of Problem 9.3 Superconductor a) Use Ampère s law in the stationary case E/ t = 0, B = 4π J c s = 4πn se 2 A taking the curl) mc 2 B) = 4πn se 2 4πn s e 2 A = B mc 2 mc 2 B) 2 B 4πn s e 2 = B. mc 2 With Gauß s law of magnetism, B = 0, and identifying λ 2 L 4πn se 2 /mc 2, the field equation follows, 2 B B λ 2 = 0. L For a homogeneous field B x) = B 0 = const, the field equation gives λ 2 L B 0 = 0. Actually this holds only for type I superconductors discussed in this problem. For type II superconductors and in particular for high-t c superconductors, magnetic fields can penetrate and build small flux tubes. In conventional type II superconductors, the flux tubes are arranged in a two-dimensional hexagonal lattice, the so-called Abrikosov lattice. b) The magnetic field at the boundary Bz 0) = B 0 ê x. By symmetry the field in the superconductor can depend only on the distance to the surface z. Then the field equation yields therefore we get d 2 dz 2 B zz) λ 2 B z z) = 0, z > 0. L B z z) = B 0 exp z ), z > 0. λ L This solution is compatible with the requirement that the bulk superconductor does not support a magnetic field. Thus, the field decays exponentially on the characteristic length scale λ L, which justifies the name penetration length. London s constitutive equation thus successfully describes the Meißner effect, i.e., a magnetic field is expelled from a superconductor. There is an interesting analogy to the Debye-Hückel theory of an electrolyte or a plasma): there the electric fields are screened, and the constitutive equation relates the charge density to the scalar potential. c) The total energy is given by U = U matter + U field, U matter = c Note: The electrostatic analogue of U matter would read δu matter = d 3 x ρ x)δϕ x) d 3 x J s x) A x), U field = 8π d 3 x B x) 2. 4

Consider A x) A x) + δ A x) and linearize, δu = δu matter + δu field = d 3 x J c s x) δa x) + d 3 x B x) 4π δb x) = { d x A x) 4π λ 2 δa x) + L = { d x 4π λ 2 L = { } d x δa x) 4π A x) + [ A x)] [ A x) ] [ δ A x) ] } partial integration A x) δa x) [ + δa x) ] [ ] } A x) a b) c = a b c) λ 2 L = 0 for U to be a minimum. Since δ A x) is arbitrary, Taking the curl once more, λ 2 L A x) + [ A x)] = 0. λ 2 L λ 2 L B x) + [ B x)] = 0 B x) + [ B x)] 2 B x) = 0 since B = A = 0) λ 2 L B x) 2 B x) = 0. Solution of Problem 9.4 Nuclear Magnetic Resonance Nuclear magnetic resonance is an important tool in solid state physics. Usually one observes the nuclear spin of the protons since they possess a large gyromagnetic ratio. a) Insert Mt) = M 0 + δ Mt), Ht) = H 0 + δ H t) in the equation of motion, Mt) = γ Mt) Ht) T [ Mt) M 0 ]. and keep only terms linear in the small perturbations δ Mt) = γ[ M0 + δ Mt)] [ H 0 + δ H t)] T δ Mt) = γ M 0 δ H t) + γδ Mt) H 0 T δ Mt) Decomposing δ Mt) = δ M t) + δ M t) such that δ M t) H 0 = 0 and δ M t) H 0 = 0 yields δ M t) + T δ M t) = 0 δ M t) γ δ M t) H 0 + T δ M t) = γ M 0 δ H t). 5

b) No driving δ H t) 0. Then the parallel component relaxes exponentially δ M t) = δ M 0) exp t/t ). Since the work necessary to change the magnetization is given by δu = δw = d 3 x H δm = d 3 x H δm, only the parallel component is concerned with energy relaxation. T quantifies the time scale where energy from the spins is equilibrated to the other degrees of freedom of the sample. In NMR jargon it is referred to as spin-lattice relaxation time or longitudinal relaxation time. For the transverse magnetization, we introduce cartesian components H 0 = H 0 ê z ), δṁxt) ω L δm y t) + T δm x t) = 0 δṁyt) + ω L δm x t) + T δm y t) = 0 with the Larmor frequency ω L = γh 0. It is favorable to use the complex variable δmt) = δm x t) + i δm y t). Then both equations can be combined to a single complex first order differential equation, with the obvious solution For the initial condition δṁt) + i ω LδMt) + T δmt) = 0, δmt) = δm0) exp t/t ) exp i ω L t). δm0) = δm e iϕ = δm x 0) + i δm y 0), i.e., δm x 0) = δm cos ϕ, and δm y 0) = δm sin ϕ, we have δm x t) = ReδMt) = δm e t/t cosω L t ϕ) δm y t) = ImδMt) = δm e t/t sinω L t ϕ), and one recovers the damped harmonic oscillator rotating clockwise. So far we find that the transverse relaxation time equals the longitudinal relaxation time. c) Complexify the rotating field, δht) = δh x t) + i δh y t). Then δht) = δh ω e iωt ; for counter-clockwise rotation, merely take the complex conjugate. The equations of motion in complex notation read δṁt) + i ω LδMt) + δmt) = i γm 0 δh ω i ωt e T It is clear that δmt) can rotate only with the same angular frequency, δmt) = δm ω e i ωt. The complex amplitude is determined by the algebraic equation i ω + i ω L + T ) δm ω = i γm 0 δh ω. 6

The linear response is characterized by the complex susceptibility χω) with δm ω = χω)δh ω, χω) = iγm 0 iω ω L ) + /T ; decomposing into real and imaginary parts, χω) = χ ω) + i χ ω), χ ω) = ω ω L)T 2 γm 0 + T 2 ω ω L) 2, χ ω) = T γm 0 + T 2 ω ω L) 2. 7