more Chapters on this topic:IntroductionTransport Eqs.Spin Proximity/ Spin InjectionSpin DetectionBoltzmann Eqs.Band currentScattering currentMeanfree pathCurrent near InterfaceOrdinary Hall effectAnomalous Hall effect, AMR effectSpinOrbit interactionSpin Hall effectNonlocal Spin DetectionLandau Lifshitz equationExchange interactionspd exchange interactionCoercive fieldPerpendicular magnetic anisotropy (PMA)Voltage controlled magnetism (VCMA effect)Allmetal transistorSpinorbit torque (SO torque)What is a hole?spin polarizationCharge accumulationMgObased MTJMagnetoopticsSpin vs Orbital momentWhat is the Spin?model comparisonQuestions & AnswersEB nanotechnologyReticle 11
more Chapters on this topic:IntroductionTransport Eqs.Spin Proximity/ Spin InjectionSpin DetectionBoltzmann Eqs.Band currentScattering currentMeanfree pathCurrent near InterfaceOrdinary Hall effectAnomalous Hall effect, AMR effectSpinOrbit interactionSpin Hall effectNonlocal Spin DetectionLandau Lifshitz equationExchange interactionspd exchange interactionCoercive fieldPerpendicular magnetic anisotropy (PMA)Voltage controlled magnetism (VCMA effect)Allmetal transistorSpinorbit torque (SO torque)What is a hole?spin polarizationCharge accumulationMgObased MTJMagnetoopticsSpin vs Orbital momentWhat is the Spin?model comparisonQuestions & AnswersEB nanotechnologyReticle 11

Perpendicular magnetic anisotropy (PMA) Spin and Charge TransportAbstract:The equilibrium magnetization of a ferromagnetic film is inplane, because of the demagnetization field. Due to the demagnetization field, the magnetic energy is smaller when the magnetization direction is inplane than perpendicular to plane. However, at an interface an electron may have an additional magnetic energy due to the spinorbit interaction. This additional energy may be substantial and it makes the total magnetic energy smaller in the perpendiculartoplane direction. As a result, the direction of the equilibrium magnetization becomes perpendiculartoplane. This effect is called the perpendicular magnetic anisotropy (PMA).The spinorbit interaction and consequently the PMA becomes strong only when the electron orbital becomes asymmetric (deformed) in one direction. When the orbital deformation is due to the breaking periodicity at an interface, the effect is called the interfacial PMA. When the orbital deformation is due to the crystal spacial asymmetry, the effect is called the bulk PMA.
Content1. Spinorbit interaction and PMA Perpendicular Magnetic Anisotropy describes the magnetic anisotropy, in which the direction of easy axes is perpendicular to the film surface and the direction of the hard axis is inplane of the filmEnergy or magnetic field. What is the PMA?What object or effect does the PMA describe?
The PMA describes the magnetic field H_{PMA}=H_{SO}H_{demag}, which is a sum of demagnetization magnetic field H_{demag} and the magnetic field of spin orbit interaction H_{SO}. In the case of a PMA sample, the direction of H_{SO} is along magnetization M, the H_{demag} is always opposite to the M and H_{SO}>H_{demag}. As a result, H_{PMA}>0 and M ( the total spin of localized electrons (magnetic moment)) is aligned along H_{PMA}. and perpendicularly to the film surface. The PMA energy E_{PMA }is the product of H_{PMA} and M (magnetic energy of localized electrons). In some cases it is preferable to use the PMA energy E_{PMA} instead of the magnetic field H_{PMA}. An example is the thermo activated magnetization reversal, when the magnetization reversal occurs when the energy of thermo fluctuation is larger than E_{PMA}. In some cases it is preferable to use the magnetic field H_{PMA} instead of the PMA energy E_{PMA} . An example is calculation of a torque in external magnetic field H_{ext}, when the magnetization procession M occurs around the field, which is the vector sum of H_{ext} and H_{PMA}. What is the PMA energy?The PMA energy E_{PMA} is defined as a difference between magnetic energies for the cases when when the magnetization is perpendiculartoplane and inplane (it is important) The primary parameter of the PMA effect is the magnetic field. Even the use of the PMA energy is convenient for some applications.Thickness dependence of H_{PMA} and E_{PMA}H_{PMA} and E_{PMA} are substantially different at the interface and in the bulk of a ferromagnetic material.The difference of H_{PMA} is because of difference of the orbital symmetry. The orbital symmetry at interface is substantially different that in the bulk because of a surface reconstruction at interface and the asymmetry of interaction with atoms from below and above for the interface atoms. Example: Fe film: at interface > H_{SO}>H_{demag}, but in the bulk H_{SO}< H_{demag}. As a result, equilibrium magnetization of a thinner Fe film is perpendicular to  film plane and of a thicker film is in plane (see here or here). FeTbB film: in the bulk H_{SO}< H_{demag}. Equilibrium magnetization of a thicker FeTbB film is perpendicular to  plane Averaging over film thickness The measured parameters of the PMA magnetic field H_{PMA} and PMA energy E_{PMA} where t is the film thickness (fact) In case when is substantially different at interface and bin the bulk, the magnetic properties of a thin film substantially depend on the film thickness (see here or here).Dependence of H_{PMA} and E_{PMA} on magnetization directionBoth the demagnetization magnetic field H_{demag} and magnetic field of spin orbit interaction substantially depend on the magnetization direction. It makes H_{PMA} substantially dependent on the magnetization direction(direction dependence of H_{demag}_{}) The demagnetization magnetic field H_{demag} is always directed perpendicularly to film surface and opposite to the magnetization direction. Its magnitude is proportional to the perpendicularto plane component of the magnetization. (direction dependence of H_{SO}_{}) The magnetic field os spin orbit interaction H_{SO} is directed along to the magnetization direction (along the magnetic field applied to the atomic orbital). Its magnitude depends on the orbital symmetry (or degree of orbital deformation). Usually atomic orbital is deformed along the film normal. As a result, the H_{SO} may be substantially larger when magnetization M is perpendicular to the film surface in comparison to the case when the magnetization is in the plane.
Why the magnetization direction depends on the magnetization direction with respect to the film normal?There is a discontinuity at the film interface. As a result, atomic orbitals are deformed at the interface. Since the effective magnetic field of spinorbit interaction H_{SO}. depends on the orbital deformation (See SO interaction), H_{SO} becomes larger when the magnetization is perpendiculartoplane and H_{SO} becomes smaller when the magnetization is inplane. Correspondingly, the magnetic energy becomes different for the perpendiculartoplane and inplane magnetization directions. Why do we care whether the equilibrium magnetization is inplane or perpendiculartoplane?The energy of the spinorbit interaction of only one interface layer may be huge and it may substantially exceed the total magnetic energy of all other electrons in a ferromagnetic film. This large energy of the spinorbit interaction is used to store a data in small volume of magnetic medium (e.g. a nanomagnet of a MRAM cell, a magnetic domain of a hard disk). Since the electron orbital at the top of a ferromagnetic metal are deformed in the direction of the interface, the energy of spinorbit interaction is largest for the film with the PMA ( the direction of the equilibrium magnetization is perpendiculartoplane).
Why do we need a high magnetic energy in order to reduce the volume of data storage media?In a magnetic media the data are stored by means of two opposite equilibrium magnetization directions. The energy of barrier between these two equilibrium states should be substantially (at least 2050 times) larger than thermo energy kT. Otherwise, the magnetization can be thermally switched and the date can be lost (See thermoactivated magnetization switching). The barrier energy is linearly proportional to the volume of the storage cell. In order to make the storage cell smaller, the barrier and consequently the magnetic energy should be larger. How the spinorbit interaction makes the equilibrium magnetization perpendiculartoplane?At the interface the electron orbital is substantially deformed towards the interface. Due to this deformation, the spinorbit interaction is significantly enhanced. This means that there is a large effective magnetic field of the spinorbit interaction H_{SO}, when the magnetization is along the deformation (perpendiculartointerface). However, there is no such field, when magnetization is inplane. The magnetic field H_{SO} may become larger than demagnetization field and the negative magnetic energy become smaller for perpendicular direction than for the inplane direction. As a result, the perpendiculartoplane direction of the magnetization becomes energetically favorable. How to measure the strength of the PMA?The anisotropy field is used to measure the strength of the PMA. The larger the anisotropy field is, the stronger the PMA is. What are the reasons why some ferromagnetic films have inplane magnetization and some films have perpendiculartoplane magnetizations?Two factors are factors are important to understand the equilibrium magnetization direction of a ferromagnetic film: (1) the directional dependence of the spinorbit interaction; (2) the orbital deformation at an interface Calculation of PMA(step 1) The spin‐orbit interaction and the orbital deformationThe Einstein theory of relativity states that an electron moving in a static electric field experiences an effective magnetic field, which is called the effective magnetic field of the spinorbit (SO) interaction H_{SO} .The electric field of atomic nucleus may induce a substantial H_{SO}, because the electron moves with a very high speed on its atomic orbital in close proximity to the atomic nucleus. However, the value of the H_{SO }substantially depends on the orbital symmetry. It is because the electron experiences different directions of the H_{SO} _{ }on different parts of the orbit that may compensate each other. For example, in the case of a spherical orbital the contributions to the H_{SO} are equal and opposite in sign and the resulting H_{SO}=0. In the case of a deformed orbital, when the orbital is elliptical or/and the orbital center is shifted from the nuclear position, the H_{SO}_{ }becomes substantial and proportional to the degree of the orbital deformation. Also, the H_{SO} is linearly proportional to an external magnetic field H_{ext}. Figure 1 demonstrates how the H_{SO} changes depending on the direction of the orbital deformation and the direction of the external magnetic field. When the orbital is spherical (Fig.1a), the H_{SO} equals to zero independently on the direction of the H_{ext}. When the orbital is deformed and the H_{ext} is applied perpendicularly the orbital deformation (see Fig.1b), the H_{SO} is also zero, because the orbital is symmetrical along the direction of the H_{ext}. For the cases of Figs. 1a and 1b, the magnetic energy of the electron E_{mag} is equal to: where S is the electron spin and is m_{B} the Bohr magneton. When the orbital is deformed and the H_{ext} applied along the orbital deformation (Fig.1c), the H_{SO} becomes a non‐zero and proportional to the H_{ext}. In this case the E_{mag} equals to: The absolute value of the electron magnetic energy is larger in the case shown in the Fig. 1c and smaller in the cases shown in the Figs. 1a and 1b. Therefore, the orbital deformation substantially changes the electron magnetic energy and therefore magnetic properties of the ferromagnetic film.
(step 2) PMA of a thin filmThe equilibrium magnetization of an isotropic ferromagnetic thin film can be either in‐plane or perpendicular‐to‐plane depending on the deformation of the electron orbitals at the film interface and the thickness of the film. The interaction of analyte molecules with interface electrons of the ferromagnetic film leads to the orbital deformation of the interface electrons and consequently to a change of the PMA. Consequently, the change of magnetization direction of the ferromagnetic film due to the change of the PMA is used as the molecular detection mechanism in the disclosed invention. The physical phenomenon of the PMA and the reason, why the orbital deformation defines the strength of the PMA, are explained as follows. Figure 2 shows a cross‐section of a nanomagnet as an array of its electronic orbitals. The magnetization of a thicker film (Fig. 2a) is inplane, while the magnetization of a thinner film (Fig.2b) is perpendiculartoplane. The PMA is the reason, why the magnetization changes its direction depending on the film thickness. The equilibrium magnetization direction is the direction of the smallest magnetic energy of the whole film. For the bulk electrons the magnetic energy is smallest when the magnetization is inplane. For the interface electrons the magnetic energy is smallest when the magnetization is perpendiculartoplane. In the case of a thicker film, the number of bulk electrons is larger and the total magnetic energy of the film is smaller when the magnetization is inplane. In the case of a thinner film, the number of bulk electrons is smaller while the number of interface electrons remains the same. As a result, the magnetization becomes perpendiculartoplane for the substantially thin film. The reason why the dependence of the magnetic energy on the magnetization direction is different for the bulk and interface electrons, is their orbital shape. The orbital of the bulk electrons is spherical. They do not experience any H_{SO}. The magnetic energy E_{,b} and E_{⊥,b} of a bulk electron for the case of inplane and perpendiculartoplane magnetization can be calculated as where S is electron spin, μ_{B} is the Bohr magneton, H_{M} is the intrinsic magnetic field induced by the magnetization, H_{D} is the demagnetization field, which is directed perpendiculartoplane and proportional to the perpendicular component of the H_{M}. DE_{B} is the difference of the magnetic energy for two magnetization directions. For the bulk electron, the DE_{B} is negative. The interface electrons experience the spinorbit magnetic field H_{SO}, additionally to H_{M} and H_{D}, because their orbitals are deformed. The H_{SO} is a nonzero only when the magnetization direction is along the deformation and therefore perpendiculartoplane. For an interface electron, the magnetic energies E_{,i} and E_{⊥,i } for the inplane and perpendiculartoplane magnetizations, respectively, and their difference DE_{i} can be calculated as The H_{SO} is proportional to the degree of the orbital deformation. Even when the deformation is small, H_{SO} > H_{D} and the ΔE_{i} is positive. The difference of the magnetic energy for the  and ⊥ magnetization directions is called the PMA energy (E_{PMA}) and can be calculated as where N_{B} and N_{i} are the numbers of the bulk and surface electrons, correspondingly. Since ΔE_{B} is negative, Eq. (1.5) can be simplified as
where t is the film thickness, β is the constant, which depends on the symmetry of crystal lattice. In the simplest case of a cubic lattice, it can be calculated as where a is the lattice constant. Figure 3 shows the energy of the perpendicular magnetic anisotropy E_{PMA} of a FeB thin film as function of its thickness for two cases of different orbital deformation at the interface with the interface magnetization ΔE_{i} of 1.15 and 0.85 mJ/m2. The magnetization of the thinner film is perpendiculartoplane. The magnetization of the thicker film is inplane. The thickness of the film at which the magnetization changes direction depends on the degree of orbital deformation at the interface and therefore on the value of interface magnetization energy ΔE_{i}. For example, at the thickness of 1 nm, the magnetization can be switched between the inplane and perpendiculartoplane directions by the modulation of the orbital deformation. The orbital deformation at interface can be of two types. The orbital can be elongated or shortened along the interface normal. Both types of the orbital deformation lead to the increase of the H_{SO} and the PMA energy. How and why the electron orbital is deformed?The spinorbit interaction is stronger in the case of a nonsymmetrical electron orbital. The less symmetrical orbital is, the stronger H_{SO} the electron experiences. The spinorbit interaction is weaker when induced by centrosymmetric electrical field. Even though the p and dorbital of a hydrogen atom is nonsymmetrical, the H_{SO} is small for the p and d electrons in this fully centrosymmetric electrical field of one nuclear. The case is different, when many atoms interacts. Such interaction makes their orbits nonsymmetrical. The case of an atom at an interface is prominent. At two side of the interface the atom experience different electrostatic force, which makes the atom orbital well nonsymmetric. As a result, the electron of this orbital experience the strong H_{SO}.
How the polarity of the orbital influence the sensor output signal?It does not matter whether orbital is elongated or squeezed. Both deformations induce the same polarity of the change of H_{SO}. A breaking the spacial symmetry is important, but not the polarity of the breaking. The most effective enhancement of H_{SO} is when the center of electron orbital is shift of the position of the nuclear. (See more details in the spinorbit interaction) Does the interface roughness influence the PMA?Yes, very much. The interface PMA exists only at a very smooth interface. Even a moderate roughness of the interface causes the reduction and disappearance of the PMA. Does the PMA increases when the film thickness decreases?Yes, in the case of the interfacetype PMA, the PMA increases when the ferromagnetic layer becomes thinner. See Fig.3 Is it possible to get film with a large PMA by decrease the film thickness?Yes, it is possible. See Fig.3. However, often when the ferromagnetic film or layer becomes very thin, the film roughness sharply increase and even the film may becomes discontinuous. It causes reduction and disappearance of the PMA. However, there sever growth techniques (tricks), which allow to grow a very thin or thick layers with a high PMA. For example, using a conventional sputtering of amorphous FeB on amorphous SiO2 it is only possible to grow film with perpendicular magnetization in the range of thicknesses between 0.8 nm and 1.5 nm. I have made FeB with strong PMA and perpendicular magnetization as thin as 0.1 nm and as thick as 2.5 nm.
Can an electron deep in bulk experience the strong spinorbit interaction?Yes. It is the case of a single crystal metal with anisotropy axes along the film normal. Another case is the asymmetrical rearrangement of atoms along growth direction. For example, FeTbB has a strong bulk type PMA. Also, it has a weak the the interface PMA. As a result, The equilibrium magnetization of a thin FeTbB is inplane and a thick FeTbB is perpendicularto plane. The slope of Fig.3 becomes positive instead of negative. Which interface induce the strongest PMA?There are many possible interfaces with a strong PMA. The famous interfaces are: (1) Co(111)/Pt(111); (2) Fe(001)/ MgO (001); (3) Fe/Ta; (4) Fe/W The strength of the PMA depends very much on growth technique. It means how thin film and smooth the interface can be obtained. The model, which is described below, is wellmatched to all experimental fact. For example, the existence of H_{SO} well explains the linear dependence of inplane component of magnetization as function of applied inplane magnetic field (See anisotropy field) .
Spinorbit (SO) interaction is the origin of Perpendicular magnetic anisotropy (PMA)
for more details see here Fact 1. Relativistic origin of spinorbit interaction An object, which moves in an electrical field, experience an effective magnetic field. This effective magnetic field is 100% magnetic field and it is indistinguishable from any other magnetic field. This magnetic field is called the SO magnetic field H_{SO}. The origin of the SO magnetic field is the relativistic nature of the electromagnetic field. Fact 2. The electrical field of atomic nuclear induces the SO magnetic field, which originates the perpendicular magnetic anisotropy (PMA) Because of its relativistic nature, the SO effect is weak. Only a very strong electrical field may induce sizable SO magnetic field. Only the electrical field of atomic nuclear is sufficient strong to create sufficiently strong SO magnetic field (130 kGauss), which induces the PMA Fact 3. Only an electron in a nonsymmetrical orbital experiences SO interaction
An electron in spherical orbital (s orbital) does not experience any SO interaction (explanation is here). Only when orbital symmetry is broken, the electron experience SO interactions. There are several possibilities to break the orbital symmetry. Each of them induces H_{SO}. The magnitude H_{SO} is proportional to the degree of the breaking symmetry. Fact 4. The electrical field of nuclear cannot break time inverse symmetry. The SO magnetic field exists only when there is external magnetic field. The H_{SO} exists only when there is some external magnetic field. The H_{SO} is always is zero in absence of the external magnetic field. The H_{SO} is always in the same direction as the external magnetic field. The H_{SO} may be significantly larger than the external magnetic field
Fact 5. The SO is direction dependent. It is the source of the PMA When orbital is deformed only in one direction (for example, only along the z direction), the H_{SO} exists only when the external magnetic field is directed along this direction (along the zdirection), but there is no H_{SO} when the external magnetic field is directed in a different direction (along the x or y direction) When the orbital is deformed perpendicularly to the film interface, the H_{SO} is induced only in this direction. The directional dependence of H_{SO} originates the PMA (See below)
Fact 6. The orbital symmetry in close proximity to nuclear determine the strength of the SO magnetic field
A substantial SO is induced only by a very strong magnetic field, which exists only in very close proximity to nuclear. Only this region makes substantial contribution to the H_{SO} and E_{PMA}. For this reason, the orbital symmetry in this region is critically important for the PMA. As a result: H_{SO} is larger when the center of orbital slightly shifted from the position of the nuclear comparing to a slight deformation of Fact 7. Mainly localized d or f electrons contribute to PMA Spins of the localized d or f electrons mainly contribute to the magnetization of a ferromagnetic metal. Therefore, the deformation and symmetry of these orbitals mainly determines the PMA. The contribution of the conduction electrons to the metal magnetization and the PMA exists, but it is very small. It is because the distribution of the spin directions of the conduction electrons is very different from that of localized d or f electrons (See here) The conduction electrons influence the PMA and the magnetization mainly due to the spd exchange interaction. Q. The orbital of d and f electrons are already not spherical. Do they experience the SO interaction and the magnetic anisotropy? A. It is correct. They do. There is magnetic anisotropy along some crystal orientation. Often it is not large. However, at the interface the orbital deformation might be much larger, which induces much larger H_{SO} and E_{PMA}. See also VCMA effect and SO torque Fact 8. The parity symmetry and spinorbit interaction
The SO interaction and the PMA depend only the direction of orbital deformation, but not its polarity. For example, the orbital deformation due to a shift of the center of electron orbit from position of atomic nuclear in + x direction induces absolutely equal H_{SO} and E_{PMA} as the shift in  x direction. Fact 8. Neither covalent nor ionic bonding is good for PMA and SO. The optimum bonding should be something between. In both case of the covalent bonding (E.g. Si, Fe) or the ionic bonding (E.g. NaCl, ZnO), the electron orbital is rather symmetric to induce any H_{SO} and E_{PMA}. In order to induce a large magnetic anisotropy, the orbital should be deformed asymmetrically. It is the case when the bonding is neither fully covalent nor fully ionic. A bonding across an interface (the interfacial PMA) or a bonding along some specific crystal direction in a compound crystal (E.g. CoPt, SmCo, GaAs, InP) (bulktype PMA) make optimum orbital deformation and induce a strong H_{SO} and E_{PMA}
Physical Origin of perpendicular magnetic anisotropy (PMA)
The PMA exists, because the electron orbitals in a ferromagnetic metal are deformed in the direction perpendicular to the film interface. When magnetization is along this direction, the intrinsic magnetic field induced a substantial effective magnetic field of the spinorbit interaction H_{SO} , because orbital deformation in this direction. When the magnetization is inplane, there is no H_{SO}, because the orbital is not deformed in this direction. As a result, the absolute value of the magnetic energy increases, when the magnetization is perpendicular to the film, and decreases when the magnetization is in plane. Magnetization is in plane: Total magnetic field= H_{intristic} Magnetic energy= H_{intristic} · M Magnetization is perpendicular to plane: Total magnetic field= H_{intristic}+H_{SO} Magnetic energy= (H_{intristic} +H_{SO} )· M where M is the magnetization or the spin of localized electrons. Since the absolute value of the negative magnetic energy is larger in the case when the magnetization is perpendiculartoplane. As a result, the easy magnetization direction becomes perpendiculartoplane, because of the SO interaction
note: See also about demagnetization field
Anisotropy field H_{anis}
Anisotropy field and the dependence inplane magnetization vs applied inplane magnetic field reveals key features of the spinorbit interaction and the PMA (For example, see VCMA effect or SOT effect)
Important feature of PMA: Linear dependence of inplane component of magnetization on in plane magnetic field. See below the math to prove it. Since the fitting of a linear dependence is simpler, is resisted against the noise and other unwanted disturbing factors (like magnetic domain) and can be done with a high precision, measuring of H_{anisotropy} with a high precision is an important step to almost any magneto transport measurement.
As was explained here, Important facts about the spinorbit interaction and PMA are: fact 1: Due to the spinorbit interaction, there is a strong magnetic along the film normal (the zaxis). This magnetic field is called the magnetic field of the SO interaction. It is absolutely real magnetic field, which is generated relativistically due to electron movement in electrical field of atomic nuclear fact 2: The SO magnetic field is generated due to the orbital deformation along the zaxis. In the case of the uniaxial anisotropy it can be simplified that there is a SO magnetic field only along zaxis, but there is no inplane SO magnetic field. There is no field along the xaxis and yaxis. The magnitude of the SO magnetic field is proportional to degree of deformation of electron orbital in close proximity to atomic nuclear. fact 3: The magnitude of the SO magnetic field is linearly proportional to the total magnetic field (excluding SO field) along the z direction (along orbital deformation).
From the fact 3, the SO magnetic field H_{SO,z} can be described as
where G_{SO }is the proportionality constant; H is the external magnetic field and M is the magnetization. From Eq.(2.1) the magnetic energy can be calculated as: where Substituting Eqs. (2.1ab) into (2.2), the magnetic energy is calculated as In the case when external magnetic field is applied only inplane (along the xdirection) The magnetic energy is calculated as The Eq.(2.5) can be rewritten as where The equilibrium magnetization direction can be found from condition of minimum magnetic energy. The energy minimum can be found from Eq. (2.6) as The solution of Eq.(2.8) gives in plane magnetization M_{x} as where the anisotropy field is defined as The inplane magnetization M_{x} is linearly proportional to applied inplane magnetic field and it is calculated as Eq.(2.10) is wellmatched to the experimental measurements. (See Fig.3) Using Eq.(2.9a), the magnetic energy can be calculated as Without external magnetic field the magnetic energy is The PMA energy is defined as an energy difference between cases when magnetization is inplane and perpendicularly to the plane. From Eq.(2.3b) it can be calculated The ratio of the PMA energy to thermo energy is called delta. Measurements of the anisotropy field H_{anisotropy}3 following experimental method are most often used to measure H_{anisotropy} Using a Vibratingsample magnetometer (VCM) or a SQUID magnetometer.The magnetometer measures the magnetization. From measurement of the inplane magnetization as a function of inplane magnetic field, the linear fitting by Eq.(2.10) gives H_{anisotropy}. merit: Highreliability direct measurements weakpoints: Due to sensitivity limitations, only a relatively large samples can be measured. Using the Anomalous Hall effectThe Hall angle is linearly proportional to perpendicular component of the magnetization (See here). The measurement of the Hall angle as a function of inplane magnetic field gives dependence M_{x}/M. The linear fitting by Eq.(2.10) gives H_{anisotropy}. merits: (1) Nanosized object can be measured (2) It can be combined with other magnetotransport measurements weakpoints: (1) Its measurement precision is rather sensitive to existence of magnetic domains. (2) It takes a relative a long time for a measurement. Using the Tunnel magnetoResistance (TMR)The method uses a magnetic tunnel junction (MTJ), in which the magnetization of the “reference” layer is inplane and the magnetization of the “free” layer is perpendiculartoplane. When a magnetic field is applied in the inplane direction, the magnetization of the “free” layer turns toward the magnetic field. From the measurement of the tunnel resistance, the angle between “free” and “reference” layers is calculated. It gives inplane component of magnetization of "free" layer vs the inplane magnetic field. From this data, the linear fitting by Eq.(2.10) gives H_{anisotropy}. merits: (1) Nanosized object can be measured (2) It can be combined with other magnetotransport measurements (3) It is fast measurements weakpoints: (1) Comparing with previous two methods, it is more indirect measurement. It is easy to get a systematical error with this method. (2) there is an undesirable influence of the dipole magnetic field from the reference electrode (3) The MTJ configuration is limited to a specific ferromagnetic metal, which has to provide a sufficient magneto resistance.
Anisotropy influenced by different effects
Demagnetization field
Interfacetype PMA
Bulktype PMA
Nonlinear effect for spinorbit interaction and PMAUnder a strong magnetic field, the dependence the spinorbit interaction and consequently PMA from perpendicular magnetic field deviates from a linear dependence. There are several reasons for that. The first reason is that the magnetic field may deform the atomic orbital, which enhances the SO and consequently PMA. As a result, the strength G_{SO} of the spinorbit interaction (See Eq. 2.1) becomes dependent on the magnetic field. As was explained here, the breaking the symmetry only along the zdirection (perpendicularly to the interface) affects the PMA, the G_{SO} may be changed only by H_{z }.Than, Eq. 2.1 can be re written as where H_{SO} is the effective magnetic field of SO interactions; G_{SO} is the proportionality constant; H_{intristic,z} is the zcomponent of the total magnetic field; H_{nl} is the magnetic field, at which the strength of SO interaction increases in two times. It means the proportionality coefficient between H_{SO } and H_{intristic,z} increases in two times. H_{nl} is usually substantially larger than magnetization M. Usually it is about a few Teslas. Which parameters are influenced by the nonlinear SO interaction? The dependence of the inplane component of magnetization vs inplane magnetic field deviates from linear at the magnetic field close to anisotropy field H_{anis} (the "nonlinear tail") It changes the value of anisotropy field H_{anis}
the equilibrium magnetization direction can be calculated from The effective anisotropy field H_{anis } can be calculated as where is the anisotropy field in case without any nonlinear SO component click here to expand and see how to obtain Eqs.(5.12a) and (5.14)
Nonlinear spinorbit interaction The total intrinsic magnetic field is the sum of the extrinsic magnetic field H and the magnetization field M. Therefore, Eq. (5.1) becomes The magnetic energy can be calculated as where component proportional to M_{z} is and component proportional to M_{x} is The total energy is the sum of Eqs. (5.3) and (5.4): In the case when the magnetic field is applied inplane (H_{x} =0), Eq.(5.7) is simplified to or where The minimum of the energy corresponds to the equilibrium magnetization direction, which can be found as Therefore, the equilibrium magnetization direction can be calculated from or where is the anisotropy field in case without any nonlinear SO component. In the case field H is smaller and is not close to H_{anis}, the following condition is satisfied From (5.13), (5.12a) and (2.10) the effective anisotropy field can be calculated as
Content of this page represents my personal view and it is reflected my own finding. It may slightly different from the "classical" view on PMA, which is described in following references M. T.Johnson et. al. Reports on Progress in Physics(1996) ; P.Bruno PRB (1989);
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