Phys 446 Solid State Physics Lecture 11 Nov 29(Ch 11)(Ch. 11)Magnetic properties of materialsLecture 11The interaction of radiation with matterˆ()DEGGMaterials equationsEM wave()()DEjEεωσω==GGFor your references: The interaction of radiation with matterDecoupled form of MaxEq. in vacuum: Same in material medium(solid state) [SI] 2GG[CGS] 22HHH∂∂∇GGG222EEEttεµ σµ∂∂∇= +∂∂GGG22Httεµσµ∂∂∇=+∂∂Plane EM wave (light)Solution:Definition of fundamental quantitiesIn vacuum, B = µ0H; µ0= 4π ×107(SI units: N⋅A-2); B − magnetic induction; H − magnetic field intensityWhen a material medium is placed in a magnetic field, the medium is magnetized This is described by themagnetization vectorM-the magneticmagnetized. This is described by the magnetization vector M -the magnetic dipole moment per unit volume.B = µ0H + µ0MMagnetization is induced by the field ⇒ assume that M is proportional to H:M = χH → B = µ0(1 + χ)H or B = µH; µ = µ0(1 + χ); µr= 1 + χχµ0(χ)µµµ0(χ)µrχχ - magnetic susceptibility of the medium (no physical relationship to the electric susceptibility). Real crystals are anisotropic, and the susceptibility is represented by a second-rank tensor. For simplicity, we shall ignore anisotropic effects.We assumed that M is proportional to H, the external field - ignored such things as demagnetization field, which were included in the electric case. This is justifiable in the case of paramagnetic and diamagnetic materials because M is very small compared to H (typically χ = M/H ~ 10−5), unlike the electric case in whichχ ~1unlike the electric case, in which χ 1. But when we deal with ferromagnetic materials, where M is quite large, the above effects must be included. Note also that χ can be dependent on the applied magnetic field. I thi d fi th ti tibilit f llIn this case, we can define the magnetic susceptibility as follows:HM∂∂=χThe magnetization can be defined asH∂HM∂∂−=Ewhere E is the total energy of the system. These definitions are more general.Classification of materialsAll magnetic materials may be grouped into three magnetic classes, depending on the magnetic ordering and the sign and magnitude of thedepending on the magnetic ordering and the sign and magnitude of the magnetic susceptibility: • diamagnetics: the magnetic susceptibility is negative -ggpygthe magnetization is opposite to the applied magnetic field. Usually its magnitude is ~ -10-6to -10-5. In diamagnetic materials the susceptibility nearly has a constant value gpyyindependent of temperature. Example: Ionic crystals and inert gases. •i iiiiMilllHTh ibili i•paramagnetics: χ is positive, i.e. M is parallel to H. The susceptibility is also very small: 10-4to 10-5. The best-known examples of paramagnetic materials are the ions of transition and rare-earth ions.• ferromagnetics: very large positive χ (e.g. 105), spontaneous magnetization below a certain temperaturemagnetization below a certain temperature. Will discuss later, as well as antiferromagnetics and ferrimagneticsMagnetism of a free atom1. spins of electrons S2. orbital momentum of electrons around the nucleus L3. A change of the orbital momentum caused by the external magnetic fieldi1d2( ii ibi)paramagnetism: 1 and 2 (positive contribution)diamagnetics: 3 (negative contribution)Can obtain the same formula classically: Consider an electron rotating about the nucleus inClassical consideration: diamagnetismConsider an electron rotating about the nucleus in a circular orbit; let a magnetic field be applied.Before this field is applied, we have, according to pp , , gNewton's second law,rmF200ω=Fith tt ti C l bf bt th lF0is the attractive Coulomb force between the nucleus and the electron, and ω0is the angular velocity.()Applied field → an additional force: the Lorentz force()Bv×−= eFLFL= -eBωr ⇒ F0–eBωr = mω2r eB20−=ωωm2Reduction in frequency → corresponding change in the magnetic moment.Th h ithf f tti i i l ttth h ithThe change in the frequency of rotation is equivalent to the change in the current around the nucleus: ∆I = (charge) × (revolutions per unit time) meBZeI221π−=∆m22πThe magnetic moment of a circular current is given by the product (current) x (area of orbit)reeBxy122BmrermeBexyxy42212−=−=∆ππµH〈2〉〈2〉+〈2〉Th di t f th l t f thHere 〈rxy2〉= 〈x2〉+ 〈y2〉.The mean square distance of the electrons from the nucleus is 〈r2〉 = 〈x2〉 + 〈y2〉+ 〈z2〉. hill ilh di ibi〈2〉〈2〉〈2〉〈2〉/3For a spherically symmetrical charge distribution 〈x2〉= 〈y2〉= 〈z2〉= 〈r2〉/3222rr=Bre22∆rNZe220µ3rrxy=⇒⇒Bm6−=∆µDiamagnetism in ionic crystals and crystals composed of inert gas atoms: m6χ−=⇒gyypgthey have atoms or ions with complete electronic shells. Another class of diamagnetics is noble metals, which will be discussed later.Quantum-mechanical calculation of atomic susceptibilitiesIn the presence of a uniform magnetic field the Hamiltonian of an ion 1) In the total kinetic energy term the electron momentum is replaced:AhAih il i dpg(atom) is modified in the two major ways:p → p + eA, where A is the vector potential associated with the magnetic field: B = ∇×A .1We assume that the applied field is uniform so that2) The interaction energy of the field with each electron spin must be addedBrA ×−=21to the Hamiltonian: where µBis the Bohr magneton S – spin momentum02meB==µAs the result the total energy of electrons will have a form:0HaveLet T0- the kinetic energy in the absence of the applied field:The cross term is the brackets can beThe cross term is the brackets can be rewritten taking into account thatNote that although r and p are quantum-mechanical operators, here we can gpqp,work with them as with classical variables because only non-diagonal components enter this product (i.e. there no terms which contain, e.g., x components of bothrandpwhich do not commute).components of both r and p which do not commute).Assume that B is along z direction → can rewriteFinally we find for the field-dependent correction to the total Hamiltonian:where L is the total orbital momentum:This equation is the basis for theories of the magnetic susceptibility of individual atoms, ions, or molecules.The energy correction due to the applied field is small compared to electron energies;µ=58×10-5eV/T→forB=1TµB=58×10-5eVFirst