spin waves and magnons unit 20.pptx

8/17/2019 Spin waves and magnons unit 20.pptx

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Spin waves and magnons

Consider an almost perfectly ordered ferromagnet at low temperatures T


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0: Bh g H µ µ =

Derivation of spin waves in the classical limit&or simplicity let's consider classical Heisenberg spin chain

J

Classical spin vectors of length S S =

nS 1nS +1nS −

J

(round state : all spins parallel with energy 20 E NJS NhS = − −)eviations from ground state are spin wave e!citations which can be pictured as


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Tor#ue changes angular momentum

Deriving the spin wave dispersion relation

pin is an angular momentumClassical mechanics d L T

dt =

Here: n And S

T S H

dt

= = ×

( )1 1 A n n H J S S − += +*!change field" e!change interaction with neighbors can effectively be consideredas a magnetic field acting on spin at position n

( )1 1n n nS J S S − += × + nS 1nS +1nS −

J J

( )1 1n n n n nd S

J S S S S dt − +

= × + ×

1 1 1 1 1 1

xn

x y z x y z y

x y z x y z nn n n n n n

x y z x y z z n n n n n n

n

dS dt e e e e e e

dS J S S S S S S

dt S S S S S S dS

dt

− − − + + +

÷ ÷ ÷= + ÷ ÷

÷ ÷


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Ta+es care of the fact that spins are at discrete lattice positions ! n,n a

-et's write down the !.componentthe rest follows from cyclic permutation/be careful with the signs though01

1 1 1 1 1 1

xn

x y z x y z y

x y z x y z nn n n n n n

x y z x y z z n n n n n nn

dS dt e e e e e e

dS J S S S S S S

dt S S S S S S dS

dt

− − − + + +

÷

÷ ÷= + ÷ ÷ ÷

÷

( ) ( )1 1 1 1 1 1 1 1 x

y z z y y z z y y z z z y ynn n n n n n n n n n n n n n

dS J S S S S S S S S J S S S S S S

dt − − + + − + − + = − + − = + − +

%e consider e!citations with small amplitude ,

, z x yn nS S S S ≈


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%ith

( )

( )

xi nka t n

yi nka t n

dS i uSe

dt dS

i vSedt

ω

ω

ω

ω

−

−

= −

= −into

1 12 x

y y ynn n n

dS JS S S S

dt − + = − −

1 12 y

x x xnn n n

dS JS S S S

dt − + = − − −

( )

( )

i nka t xn

i nka t yn

S uSe

S vSe

ω

ω

−

−

==

and

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2

2

i nka t i nka t i nka t i nka t ika ika

i nka t i nka t i nka t i nka t ika ika

i uSe JS vSe vSe e vSe e

i vSe JS uSe uSe e uSe e

ω ω ω ω

ω ω ω ω

ω

ω

− − − −− +

− − − −− +

− = − − − = − − −

( )( )

2 1 cos

2 1 cos

i u vJS ka

i v uJS ka

ω

ω

− = −− = − −

( )( )

2 1 cos0

2 1 cos

i JS ka u

JS ka i v

ω ω

− = ÷ ÷− −

Non-trivial solution meaning other than u=v=0 for:

( )( )

2 1 cos0

2 1 cos

i JS ka

JS ka i

ω ω

− =− − ( )2 1 cos JS kaω = −

Magnon dispersion relation


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Thermodynamics of magnons

Calculation of the internal energy:12k k k

E nω = + ÷ ∑h 1

2k k k U E nω = = + ÷ ∑h 0 1k

k

k

E eβ ω

ω = +−∑ h

h

in complete analogy to the photonsand phonons

3

3... ...(2 )k

V

d k π =∑ ∫

%e consider the limit T.23:

4nly low energy magnons near +,3 e!cited ( ) 2 22 1 cos JS ka JSa k ω = − ≈%ith

2 2

2 22

0 3 4(2 ) 1 JSa k V JSa k

U E k dk eβ

π π

= +−∫ h

h

%ith

2

B B

JSa D x k k k T k T = =

h

and hence B

Ddx dk k T =


7/12

( )

2

2

24

0 3

5 / 23 / 2 40 2

4(2 ) 1

2 1

B B x

B x

V k T D k T U E x dx

D De

V dx E D k T x

e

π π

π −

= + ÷ −

= +−

∫

∫

Just a number which becomes with integration to infinity

2

4

0

3 3(5 / 2) 1.3419

8 81 xdx

xe

π π ς

∞

= ≈ ×−∫

3/ 2

BV

V

U k T C

T D∂ = µ ÷ ÷∂

*!ponent different than for phonons due to difference in dispersion

1

1( ) s

k

sk

ς ∞

==∑


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56 $hysi+ ! " 738 /9 ;31:

0 ( ) k k

U E k nω = +∑hThe internal energy

can alternatively be e!pressed as2 2

01

( )2

x yk k

k

U E k S S S

ω = + +∑hwhere nik r n k

k

S S e=∑2 21

2 x yk k k S S nS

+ =

Intuitive hand.waving interpretation:= of e!citations in a mode = < > + < >= < >& nω ↔ < >h

2 x n< > ↔ < >

>agnetization and its deviation from full alignment in z.direction is determined as

( ) z B

nn

g M T S V

µ

= ∑ ( ) ( )( )2 22 x y B

n nn

g

S S S V

µ

= − +∑

Magneti"ation and the cele#rated T $%& 'loch law:


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-et's closer inspect ( ) ( )( )2 22 x yn nS S S − + and remember ( ) ( )2 2 2 x yn nS S S +


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Now let(s calculate M)T* with magnon dispersion at T-+0

( ) B

k k

g M T NS nV

µ = − ÷ ∑

3

3... ...(2 )k

V

d k π =∑ ∫ 2 2

JSa k ω ≈with and

2 2

23( ) 4(2 ) 1

B JSa k

g V dk M T NS k

V eβ µ

π π

= − ÷− ∫ h

Again with2

B B

JSa D x k k

k T k T = =

hand hence

B

Ddx dk

k T =

2

3/ 22

2( ) 2 1 B B

x

g V k T dx M T NS xV D e

µ π

= − ÷ ÷ ÷ − ∫

3/ 2

2

(3/2)( ) ( 0) 1

2 4 BV k T M T M T

NS Dπ ς

π

= = − ÷ ÷ ÷ B

g NS V

µ Felix Bloch(1905 - 1983)Nobel Prize in 1952 for


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Modern research e,ample:'loch(s T $%&-law widely applica#le also in e,otic systems


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Spin waves and phase transitions: oldstone e,citations

. sta#ility analysis against long wavelength fluctuations giveshints for the possi#le e,istence of a long range ordered phase

( , )n m

n m

H J S S = − ∑% Heisenberg Hamiltonian e!ample for continuous rotational symmetrywhich can be spontaneously bro+en depending on the dimension" dd,9

d,7-et's have a loo+ to spin wave approach for

( 0) ( ) M M T M T ∆ = = −in various spatial dimensions d

( ) B k k

g M T NS n

V µ = − ÷

∑ ... ...(2 ) d k d

d L d k π

=∑ ∫ &rom and1

2

d dk k M

k

−=

∆ µ∫ 2 2 2 2

01 JSa k

k e JSa k β β

→− ≈h h

%hen a continuous symmetry is bro+en there must e!ista (oldstone mode /boson1 with ω→3 for + →3

In low dimensions d,9 and d,7 integral diverges at thelower bound +,3

/nphysical result indicates a#sence of orderedlow temperature phase in d=! and d=&

spin waves and magnons unit 20.pptx

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