Présentation PowerPoint - EPFL...anisotropy energy orbital magnetic moment The substrate and the...

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Magnetic recording technology 1 1 1 0 0 B~50nm W~500nm The grain (particle) can be described as a single macrospin μ = Σ i μ i All spins in the grain are ferromagnetically aligned Exchange length 10 nm 1 bit = 300 grains

Transcript of Présentation PowerPoint - EPFL...anisotropy energy orbital magnetic moment The substrate and the...

Page 1: Présentation PowerPoint - EPFL...anisotropy energy orbital magnetic moment The substrate and the atomic coordination are the key to the MAE and magnetic moment Co/K MAE = 0 P. Gambardella

Magnetic recording technology

1 1 10 0

B~50nm

W~500nm

The grain (particle) can be described as a single macrospin

μ = Σi μi

All spins in the grain are ferromagnetically aligned

Exchange length ≈ 10 nm

1 bit = 300 grains

Page 2: Présentation PowerPoint - EPFL...anisotropy energy orbital magnetic moment The substrate and the atomic coordination are the key to the MAE and magnetic moment Co/K MAE = 0 P. Gambardella

Limiting factor for magnetic recording

B~50nm

W~500nm

a) Bit coupling:

Due to the dipolar field Hd, a bit switch can induce the switch of a closed bit (impossible to approach indefinitely two opposite magnetic poles)

Demagnetizing field effects:

Hd ∝μ = Σi μiMaterials with low magnetic moment

Co μ = 1.7 μBCo67 Cr13 Pt20 μ = 0.6 μB

Record for in-plane longitudinal recording ≈ 100 Gbits/in2

b) Transition Width: • Demagnetizing field increase the transition (bit) width. • Grains have non-uniform boundaries that increase the magnetic noise in the read out process.

The future -> perpendicular recording

Page 3: Présentation PowerPoint - EPFL...anisotropy energy orbital magnetic moment The substrate and the atomic coordination are the key to the MAE and magnetic moment Co/K MAE = 0 P. Gambardella

Signal-to-Noise Ratio:SNR ∝ number of grains in each bit (~1000).

To keep the SNR constant the number of

grains is kept fixed

Smaller bits imply smaller

grain size

Reduced stability of magnetization vs thermal fluctuations (Superparamagnetic limit)

Αvg. time (relaxation time) taking to jump from one direction to the other: s10 10

0−≈ττ = τ0 exp(K/kT)

In grain with a size of a few tens of nanometers K ∝ grain volume

Page 4: Présentation PowerPoint - EPFL...anisotropy energy orbital magnetic moment The substrate and the atomic coordination are the key to the MAE and magnetic moment Co/K MAE = 0 P. Gambardella

Magnetic anisotropy energy (MAE) K

Exchange interaction is isotropic in space

However, real materials are not fully isotropic in space ->There exist preferred spatial directions (easy axes) along

which it is easier to align the magnetization

No preferred orientation for the total moment M = Σi Si

Bulk systems: K depends on the crystallographic structure

Mag

netiz

atio

n (e

mu/

cm3 )

Magnetic field (Oe)

5 31 4.1 10 /

45 /K J m

eV atomμ= ×=

easy axis: (0001)Co hcp

4 31 4.8 10 /

2.4 /K J m

eV atomμ= ×=

easy axis: (100)Fe bcc

Magnetic field (Oe)

Mag

netiz

atio

n (e

mu/

cm3 )

Eeasy ∝ - ∫ B ·dM is minimum

Page 5: Présentation PowerPoint - EPFL...anisotropy energy orbital magnetic moment The substrate and the atomic coordination are the key to the MAE and magnetic moment Co/K MAE = 0 P. Gambardella

Microscopic origin of the MAE

Two principal sources:1) Shape anisotropy -> it is due to the dipolar interaction

of the magnetic moment μ of each atom with the magnetic field generated by all the other atoms at the atom locationKshape ∝ Σj Σi grad(μi · rij/rij

3) · μjEx: A cylinder with h >> d prefers to be magnetized along the cylinder axis, while a cylinder with d >> h (i.e. a disk) prefers to be magnetized on the surface plane

2) Magnetocrystalline anisotropy (MC) -> it is due to the interaction of the magnetic moments with the crystal field which is in general non-isotropica) surface anisotropy -> at surface or at the interface the symmetry breaking induces an asymmetry in exchange interactionb) spin-orbit interaction -> the electron spin is coupled with the orbital moment which in turn is coupled with the crystal field

d-elements:KMC ∝ mL// - mL⊥

The MAE of a system is the summation over all the contributions

Page 6: Présentation PowerPoint - EPFL...anisotropy energy orbital magnetic moment The substrate and the atomic coordination are the key to the MAE and magnetic moment Co/K MAE = 0 P. Gambardella

Uni-axial monodomain system

20E( , , ) cos ( )Kθ θ ϕ = − ⋅ − ⋅μ B easy μ

K

If K << kT the magnetization vectorisotropically fluctuates in the space.

θ

E K << kT

K >>kT

θ

EIf K >> kT the magnetization vector can not switch the versus

B = 0

τ = 1 yearτ = 1 second

K = 40 kTK = 23 kT

Αvg. time (relaxation time) taking to jump from one minimum to the other: s10 10

0−≈ττ = τ0 exp(K/kT)

μ ·B

20E( , , ) cos ( )Kθ θ ϕ = − ⋅easy μ

The MAE determines the thermal stability of the magnetization direction

Page 7: Présentation PowerPoint - EPFL...anisotropy energy orbital magnetic moment The substrate and the atomic coordination are the key to the MAE and magnetic moment Co/K MAE = 0 P. Gambardella

Applied magnetic field

Ω−

Ω−θμ=θ

μ=

∫∫

d)k/(xpe

d)k/(xpecoscos

TE

TE

VN

VNM z

20E( , , ) cos ( )Kθ θ ϕ = − ⋅ − ⋅μ B easy μ

Mz

H

B // Easy axis and K > kTB // Easy axis and K < kTB // hard axis and K < kT

K ∝ ∫0Mmax Beasy ·dM – ∫0

Mmax Bhard ·dM

Magnetization loops

Η = 0Η < K/μΗ ≥ K/μ

θ

E K

T = 0

Loop shape -> direction of the easy axis

Page 8: Présentation PowerPoint - EPFL...anisotropy energy orbital magnetic moment The substrate and the atomic coordination are the key to the MAE and magnetic moment Co/K MAE = 0 P. Gambardella

eq2

1( ) ( )1 ( )

T Tχ χωτ

=+

2 Beq

B B

exp( / k ) 1( )2k Erfi( / k )

K TT MKK T K T

χπ

⎡ ⎤= −⎢ ⎥

⎢ ⎥⎣ ⎦O. Fruchart et al., J. Magn. Magn. Mater. 239, 224 (2002)S. Rusponi et al., Nat. Mater. 2, 546 (2003)

Zero field coolingMeasure of the zero field susceptibility χ = dM/dH

Experimentally: measure of the Mz variation induced by a small

AC field (frequency ν0) as a function of the temperature

Mz

H

Mz

H

Mz

H

Blocking: K/kT >>30 Activation Superparamagnetic: K/kT <<30

Blocking temperature:temperature at which the

peak occurs ->It is a measure of the MAE

When the condition 1/ ν0 = τ0 exp(K/kT) is

satisfied M starts to flip -> peak in the χ vs T curve

Page 9: Présentation PowerPoint - EPFL...anisotropy energy orbital magnetic moment The substrate and the atomic coordination are the key to the MAE and magnetic moment Co/K MAE = 0 P. Gambardella

(a) Angular dependence of the XMCD spectra (T = 10 K, B = 5 T) for a Ni2–Pt2 multilayer sample at normal (θ = 0°) and grazing x-ray incidence (θ = 75°). (b) Angular dependence of the orbital moment μL determined from the angular dependence of the XMCD spectra. The solid line is a fit according to the equation μL(θ) = μL⊥+ (μL // - μL ⊥ ) sin2θ -> KMC ∝ μL // - μL ⊥

X-ray absorption spectra for a Ni2–Pt2multilayer:

(a) x-ray absorption coefficients for R (µ+(E)) and L (µ-(E)) circularly

polarized x-rays at the Ni L2,3 edges in the soft x-ray range,

(b) (c) XMCD spectra at the Ni and Pt L2,3 edges

Ni/Pt multilayers

Out-of-plane magnetocrystalline anisotropy

Page 10: Présentation PowerPoint - EPFL...anisotropy energy orbital magnetic moment The substrate and the atomic coordination are the key to the MAE and magnetic moment Co/K MAE = 0 P. Gambardella

STM image 85 x 85 Å2

out-of-planein-plane

0z sat

0

cos exp [ , , / k ] d dM M

exp [ , , / k ] d d

E( ) T

E( ) T

θ θ θ ϕ θ ϕ

θ θ ϕ θ ϕ

−=

−∫ ∫

∫ ∫

Numerical fit of the magnetization:

Magnetization loops by XMCD

Co/Pt(111)

mtot = mS + mLmS = -3 nh μB (A3 - 2A2) + mTmL = -2 nh μB (A3 + A2)mT ≈ 0.34 μB << mS ≈ 2.1 μB

mtot ∝ A3 (A2 ≈ 0)

20E( , , ) cos ( )Kθ θ ϕ = − ⋅ − ⋅μ B easy μ

K ∝ ∫0Mmax Beasy ·dM – ∫0

Mmax Bhard ·dM

MAE

Page 11: Présentation PowerPoint - EPFL...anisotropy energy orbital magnetic moment The substrate and the atomic coordination are the key to the MAE and magnetic moment Co/K MAE = 0 P. Gambardella

Co particles on Pt(111) with average size n

magnetic anisotropy

energy

orbital magnetic moment

The substrate and the atomic coordination are the key to the MAE and magnetic moment

Co/K

MAE = 0

P. Gambardella et al., Science 300, 1130 (2003)

(B) Hard-sphere representation of the Co particles considered in the theoretical calculations. The labels indicate the values of L for nonequivalent Co sites. S, L, ΔL, and K in the table are averaged over all Co sites.

40 Å

n = 140 Å20 Å

n = 3 n = 8

Page 12: Présentation PowerPoint - EPFL...anisotropy energy orbital magnetic moment The substrate and the atomic coordination are the key to the MAE and magnetic moment Co/K MAE = 0 P. Gambardella

SMOKE: Surface Magneto Optical Kerr Effect

Polar Longitudinal Transverse

Monochromatic linearly polarized light (generated by a laser) issent on the sample

Depending on the magnetic state of the sample the polarization of the reflected light is slightly canted (some mrad) in respect to the

incident light

I = I0(1+ a θ)θ ∝ MI0 and I are the light intensity sent on the sample and measured by the detector, respectively

Heuristic argument: 1) The media electrons are accelerated by the incident electrical field

and deviated by the Lorentz force due to the sample magnetization2) The electrons motion contributes to the outgoing wave.3) Therefore the outgoing wave has an addition component which

rotates its polarization plane

Page 13: Présentation PowerPoint - EPFL...anisotropy energy orbital magnetic moment The substrate and the atomic coordination are the key to the MAE and magnetic moment Co/K MAE = 0 P. Gambardella

Plane electromagnetic wave propagating in the positive z direction in a birefringent medium. The indices of refraction for RCP and LCP light are different (n+ and n-, respectively). If the wave has a wavelength of λ0in vacuum, then the wavelengths of its circular components in the medium will be λ0/n±. If at the point z = 0 the wave is linearly polarized along the x axis, then in the medium the wave is given by

Right circular polarization

Left circular polarization

The light propagation is described by the Maxwell equations:

0

0

HEt

EHt

μ μ

ε ε

∂∇ × = −

∂∂

∇ × =∂

At the optical frequency the

SMOKE: theoretical basis

permeability tensor μ can be assumed to be equal to the unit tensor, while the dielectric tensor ε is given by

The linear polarization direction turns of δ = π(n+ - n-) z/λ0

M.J. Freiser, IEEE Trans. Magn. 4, 152 (1968); Z.Q. Qiu et al., Rev. Sci. Instrum. 71, 1243 (2000); M. Mansuripur, J. Appl. Phys. 67, 6466 (1990)

Page 14: Présentation PowerPoint - EPFL...anisotropy energy orbital magnetic moment The substrate and the atomic coordination are the key to the MAE and magnetic moment Co/K MAE = 0 P. Gambardella

In situ STM -> morphology

SMOKE -> magnetism

Pressure during MBE growth < 8 x 10-11 mbar Base pressure 3 x 10-11 mbar

UHV system for MBE growth

Page 15: Présentation PowerPoint - EPFL...anisotropy energy orbital magnetic moment The substrate and the atomic coordination are the key to the MAE and magnetic moment Co/K MAE = 0 P. Gambardella

250 Å

0.4 ML Co T = 130 K Tann = 300 K

1 atomic layer high islandsSize about 1000 atoms

S. Rusponi et al., Nature Mat. 2, 546 (2003).

Compared to pure Co islands: 1) same total MAE2) reduced magnetic moment

Pt core and Co shell

Co islands on Pt(111): edge atoms determine the MAE

Page 16: Présentation PowerPoint - EPFL...anisotropy energy orbital magnetic moment The substrate and the atomic coordination are the key to the MAE and magnetic moment Co/K MAE = 0 P. Gambardella

Ultra-high density Co clusters superlattices on Au(788)

A A

A

B B

B

N. Weiss et al., Phys. Rev. Lett. 95, 157204 (2005)

0.75 ML Co

1.1 ML Co

Two island populations -> two blocking temperature

Page 17: Présentation PowerPoint - EPFL...anisotropy energy orbital magnetic moment The substrate and the atomic coordination are the key to the MAE and magnetic moment Co/K MAE = 0 P. Gambardella

Magneto-optical data storage

To store a bit, the magnetization has to be oriented in a certain direction in a well defined area on the disc. The process used for this is so-called thermo-magnetic writing. As it is not possible to apply a field only locally to a portion on the disc as small as the desired bit size, one combines the magnetizing field with application of a heating step which can be applied locally by using a semiconductor laser focused to a spot of about 0.5 μm. The laser radiation (>10 mW) raises the temperature locally reducing the coercive field below the magnitude of the applied field. Therefore, the region heated by the laser pulse changes its magnetization, whereas the surrounding remains unchanged. The same laser, operated at lower power (about 1mW), is used to readback the stored information. The Kerr rotation is about 0.1°

A. Partovi et al., Appl. Phys. Lett. 75, 1515 (1999); H. F. Hamann et al. 84, 810 (2004)

Page 18: Présentation PowerPoint - EPFL...anisotropy energy orbital magnetic moment The substrate and the atomic coordination are the key to the MAE and magnetic moment Co/K MAE = 0 P. Gambardella

1D magnetism: Co/Pt(997)

THEORY

heisen,

1H2 ij i j i

i j iJ μ= − ⋅ − ⋅∑ ∑s s s H

Heisenberg model -> no FM state in 1D and 2D

z z zising i j i

,

1H J s s s2 i j i

Hμ= − −∑ ∑Ising model -> no FM state in 1D

• Finite system (N localized moments):

ground state: E=E0= -J(N-1)

change in free energy: ΔF= Δ(U-TS) = 2 J - kT ln(N-1)

ΔF<0 no ferromagnetism for N-1 > exp(2J/kT)

lowest excited state: E=E0 + 2 J( N-1 such states)

Mtot ≠ 0

Mtot = 0

Infinite chain -> no magnetism (Mtot = 0) at any temperature ≠ 0

Page 19: Présentation PowerPoint - EPFL...anisotropy energy orbital magnetic moment The substrate and the atomic coordination are the key to the MAE and magnetic moment Co/K MAE = 0 P. Gambardella

Interatomic exchange energy for Co 2J ≈ 15 meV

ferromagnetism is allowed only for N < 50 atoms at T = 50 K

( )2z z z easyising i j i

,

1 1H J s s s K s2 2 i

i j i iHμ= − − −∑ ∑ ∑

Ising model + anisotropy-> FM state in 1D

Spin blocks oriented along the easy axis by the external field have to overcome the MAE barrier to reverse the orientation

Finite chain

The MAE stabilizes the direction of the spin blocks -> Mtot ≠ 0 when the external field is put back to 0

K

Page 20: Présentation PowerPoint - EPFL...anisotropy energy orbital magnetic moment The substrate and the atomic coordination are the key to the MAE and magnetic moment Co/K MAE = 0 P. Gambardella

0.1 ML of Co deposited at 260 K: array of parallel cobalt monatomic chains

10 nm 2 nm

Pt(997)

P. Gambardella et al., Nature 416, 301 (2002).

FM coupled atoms:N = 15

short-range FM

K = 2.1 meV/atom

Page 21: Présentation PowerPoint - EPFL...anisotropy energy orbital magnetic moment The substrate and the atomic coordination are the key to the MAE and magnetic moment Co/K MAE = 0 P. Gambardella

Co wire magnetization M measured at a field Br in the plane perpendicular to the wire axis (left column) and parallel to the wire axis and the (111) direction right column). The data points represent the XMCD signal at Co L3 edge (779 eV). The solid lines evidence a ⏐cos(x-x0)⏐ behavior with x = Φ, Θ, respectively, measured with respect to the (111) normal direction, as expected for uniaxialanisotropy. (a) 1-wires, T =10 K, Br = 0.25 T; (b) 2-wires, T =10 K, Br = 1 T; (c) 4-wires, T =10 K, Br = 2.5 T. No data were recorded in the plane parallel to the wires; (d) 1.3 ML, T = 45 K, Br =1.5 T.

P. Gambardella et al., Phys. Rev. Lett. 93,

077203 (2004).

Easy axis oscillates with the chain thickness

90°

Θ0°

Φ

-90°

Page 22: Présentation PowerPoint - EPFL...anisotropy energy orbital magnetic moment The substrate and the atomic coordination are the key to the MAE and magnetic moment Co/K MAE = 0 P. Gambardella

Magnetization loop of

(a) 1-wires, Φ1 = + 43° (solid squares), Φ2 = - 57° (open circles);

(b) 2-wires, Φ1 = - 67°, Φ2 = + 23°;

(c) 3-wires, Φ1 = - 7°, Φ2 = +63°;

(d) 1.3 ML, Φ1 = - 7°, Φ2 = +63°.

The data points represent the XMCD at the Co L3 edge (779 eV) normalized by the L3 absorption edge jump; solid lines are fits to the data.

MAE oscillates with the chain thickness

P. Gambardella et al., Phys. Rev. Lett. 93, 077203 (2004).

K = 2.0 meV/atomK = 0.33 meV/atom

K = 0.45 meV/atom

K = 0.15 meV/atom

Page 23: Présentation PowerPoint - EPFL...anisotropy energy orbital magnetic moment The substrate and the atomic coordination are the key to the MAE and magnetic moment Co/K MAE = 0 P. Gambardella

0.68 μBmL 0.37 μB 0.33 μB 0.31 μB

Orbital moment decreases increasing chain thickness

A. B. Shick et al., Phys. Rev. B 69, 212410 (2004).

LSDA: Local Spin Density ApproximationLSDA+U: LSDA + electron correlations

chain in respect to bulk hcp:- Narrowing of the d-band- increasing of the spin splitting