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 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 (feature 1) The spinorbit interaction and the orbital deformation The Einstein theory of relativity states that an electron, which moves in a static electrical field, experiences an effective magnetic field. This effective magnetic field is called the effective magnetic field of the spinorbit interaction H_{SO} (See here for details).
The electrical field of atomic nuclear may induce a substantial H_{SO}, because in an atomic orbital the electron is moving very close to the atomic nuclear at a high speed. However, the value of H_{SO} substantially depends on the orbital symmetry. It is because on its orbital path the electrons experience two opposite directions of H_{SO} due the different angles between electron movement direction and the electrical field of the nuclear on different parts of the orbital path. For example, in the case of a spherical orbital the two contributions to H_{SO} are equal. As a result, the electron of the spherical orbital does not experience any spinorbit interaction. The case is the different when the orbital is deformed from the spherical shape. In the case when the orbital is elliptical or/and the orbital center is shifted from the nuclear position, the H_{SO} may be substantial and it is proportional to the degree of the orbital deformation. Additional condition for the H_{SO} to be a nonzero is that an external magnetic field H_{ext} should be applied to the orbital. The H_{SO} is linearly proportional to the H_{ext} and the direction of H_{SO} is along the direction of the H_{ext}. The H_{SO} is a nonzero only when the H_{ext} is applied along the direction of the orbital deformation. Figure 1 describes the spinorbit interaction for different mutual directions of orbital deformation and H_{ext}. When the orbital is spherical (Fig.1(a)), 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 along the orbital deformation, (Fig.1(b)), the H_{SO} is non zero and in direction of the H_{ext}. When the orbital is deformed and the H_{ext} is applied perpendicularly the orbital deformation (Fig.1(c)), the H_{SO} becomes zero again, because along the H_{ext}. the orbital is symmetrical. In the cases of Figs. 1(a,c), the magnetic energy E_{mag} of the electron equals to
where S is the electron spin and m_{B }is the Bohr magneton. In the cases of Figs. 1(b), the absolute value of E_{mag} is larger. The E_{mag} equals to
The electron magnetic energy is larger in the case Fig.1(b) and smaller in cases Fig.1 (a) and Fig.1 (c). Therefore, the deformation of the orbital changes substantially the electron magnetic energy and therefore the magnetic properties of a magnetic film. This property is used in disclosed invention to detect the proximity of the tested molecule near surface of a nanomagnet. (feature 2) The origin of the PMA
Case 1: Magnetization is inplane. There is no PMAFigure 2 shows the crosssection of a nanomagnet is shown as an array of its electronic orbitals. The arrow shows the equilibrium magnetization direction of the nanomagnet. Figure 2(a) shows the case when the orbitals at interface are spherical and not deformed. In this case, the equilibrium magnetization of the magnetic film is inplane. It is because the total magnetic energy of all the bulk and interface electrons are smaller when magnetization direction is inplane magnetization than perpendiculartoplane. The electron experience two types of magnetic fields. The first field is the intrinsic magnetic field H_{M}, which is directed along the magnetization. The second field is the demagnetization field H_{demag}, which is directed perpendicular to interface and along the directed perpendicular to the interface component of the magnetization. Therefore, the magnetic energy in the case of magnetization along and perpendicular to interface is calculated as
Since, the magnetization for the case of Fig.2(a) is aligned inplane. Case 2: PMA. Magnetization is perpendiculartoplane.Figure 2(b) shows the case when the orbitals at interface are deformed perpendicularly to the interface. In this case, the equilibrium magnetization of the magnetic film is perpendicularly to the plane. It is because the magnetic energy of the interface electrons becomes smaller in the case of the perpendicular magnetization. The magnetic energy of the interface electrons in the case of magnetization along and perpendicular to interface is calculated as
Even for the case of a small deformation. As a result, and the magnetization may be aligned perpendicularto plane. It depends on the film thickness. The orbitals in the bulk of the film are not deformed and their magnetic energy is described by Eqs.(3). The total magnetic energy E_{mag,total} is the sum of the magnetic energies of the bulk and interface electrons and can be calculated as [2]
where t is the film thickness, is E_{mag,interface} is the magnetic energy of the interface electrons and E_{mag,bulk} is the magnetic energy of the bulk electrons per film thickness Figure 3 shows the magnetic energy of a FeB as function of its thickness for two cases of different E_{mag,interface}. As was explained above, the value of E_{mag,interface} depends on the orbital deformation. The magnetization of a thinner film is perpendicularto plane. The magnetization of a thicker film is in plane. The thickness, at which the magnetization changes the direction, depends on the value of E_{mag,interface} , therefore on the degree of the orbital deformation at the interface. For example at the thickness of 1 nm, the magnetization direction can be switched between inplane and perpendicularto –plane by the modulation of the orbital deformation. [2] M.T. Johnson,et.al., Reports Prog. Phys. 59, 1409 (1996)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}. 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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