Transcript Paramagnetism

Magnetic model systems
Goal: Using magnetic model systems to study interaction phenomena
Let’s start with a reminder and a word of caution:
If, without further thought, we use our standard approach
Hamiltonian
Eigenenergies
canonical partition function Z
Helmholtz free energy, F
Contradiction
for magnetic
systems
We know from thermodynamics that Gibbs free energy: G=G(T,H)
dG  SdT  0VMdH
 G 
S  

 T  H
and
M 
1  G 
0V  H T
Next we show:
G (T , H )  k BT ln Z rather than F, which is very tempting to assume, in fact confused
frequently in the literature but nevertheless wrong !!!
Note: F=F(T,M)
Helmholtz free energy is a function of T and M
Let’s consider for simplicity N independent Ising spins i=1 in a magnetic field H
The magnetic energy of the N particle system reads
N
E  m0 0 H   i
where the microstate  depends on (1, 2, …, N)
and m0 is the magnitude of the magnetic moment of a spin
i 1
Z  e
 e
  E
N
 i
i 1


d ln Z 1

dH
Z
 m0 0 H
N
 m   e


0
0
i 1
i
 m0 0 H
N
 i
 N

i 1
 0    m0 i     0 N m  0 VM
  i 1

with
1
M 
0V
 G 


 H T
From this we expect
1  G 
1 d ln Z
 

0  H T 0 dH
G(T , H )  kBT ln Z
d ln Z
 G 


k
T
B


dH
 H T
Paramagnetism
Note, we will use the result of the interaction free paramagentism to find an approximate solution for the 3d interacting case
Rather than restricting to the Ising case where the spins can only point up or down
relative to the field direction
mJ=+J=+5/2
+3/2
+1/2
we consider the general case of N independent atoms
with total angular momentum quantum number J
-1/2
-3/2
mJ=-J=-5/2
Energy of microstate 
here example for J=5/2
N
E   g J B 0 H  mJ ,i where now m0  g J B
Z  e


J

mJ ,1  J
e
i 1
  E

 g J  B 0 H

 mJ ,1 , mJ ,2 ..., mJ , N 
 g J B 0 HmJ ,1
J

mJ ,2  J
e
e
 g J  B 0 HmJ ,2
N
 mJ ,i
i 1
J
...

mJ , N  J
e
 g J  B 0 HmJ , N

 g J  B 0 HmJ 
  e

 mJ  J

J
N
To streamline the notation we define:
Z   Z1 
N
J
B 0
N


mJ
J
 J 1
J N
 ...       J 1    ...   2 J 
        
 mJ  J

J

1
Z   J
1

2 J 1




1
1


  2  2

 1
 J  
 2
  e g   H



N
 1
 J  
 2

 

 1
J
N
J 1




N
1   2 J 1
1
N
N
N





  sinh   g J  B 0 H  2 J  1 / 2  


1
1
 
 


sinh   g J  B 0 H / 2 

  2  2
 




1
sinh x   e x  e  x 
2
 1
 J  
 2
 1
 J  
 2
 sinh   g J  B 0 H  2 J  1 / 2  
 sinh  x  2 J  1 / 2  
G  k BT N ln 
  k BT N ln 


sinh   g J  B 0 H / 2 
sinh  x / 2 




Since the thermodynamics is obtained from derivatives of G with
respect to H and T let’s explore
 sinh  x  2 J  1 / 2  
sinh  x / 2 
d LnZ1 d
d

ln 


dx
dx 
sinh  x / 2 
 sinh  x  2 J  1 / 2  dx
 sinh  x  2 J  1 / 2  


sinh
x
/
2
  

 2 J  1 / 2 cosh  x  2 J  1 / 2  sinh  x / 2   1 / 2 cosh  x / 2  sinh  x  2 J  1 / 2 

sinh  x  2 J  1 / 2  sinh  x / 2 
2 J  1


coth
2
 x  2 J  1 / 2   2 coth  x / 2 
1
: J BJ ( x )
Where the Brillouin function BJ is therefore reads:
1
BJ ( x ) 
2J
  2 J  1 coth  x  2 J  1 / 2   coth  x / 2  
Discussion of the Brillouin function
BJ ( x ) 
1
1
2J
  2 J  1 coth  x  2 J  1 / 2   coth  x / 2  
J=1/2
B1/2 ( x )   2 coth  x   coth  x / 2  
with z : e
x /2
2z  2z   z  z 
z z
zz
2 2


2
1
z z
zz
 z  z 1  z  z 1 
2
B1/2
e x  e  x e x /2  e  x /2
2 x
 x /2  x /2
x
e e
e e
2
1
2
2z  2z  z  2  z

 z  z 1  z  z 1 
2
zz  e



z  z  e
1
1
2
2
2
1 2
2
zz 

z  z 2


1
1
 z  z  z  z   z  z 1  z  z 1 
2
2
 e  x /2 sinh  x / 2 

 tanh( x / 2)
x /2
 x /2
e
cosh  x / 2 
x /2
1 2
X<<1 with
2
e y  e y
1 y
coth y  y
 
 y y 0
e e
y 3
1
BJ ( x ) 
2J
Expand each exponential up to 3rd order, factor out the essential singular term
1/y and expand the remaining factor in a Taylor series up to linear order.
  2 J  1 coth  x  2 J  1 / 2   coth  x / 2  
1
2J
3
Note, this expansion is not completely trivial.
X<<1


x  2 J  1  2 x 
2

  2 J  1 
   
6
 x  2 J  1
 x 6

2
2



x
2
J

1
x
2
J

1
x




1 2
2 x
1

  
 






2J  x
6
x 6  2J 
6

1  4 xJ 2  4 xJ  x  J  1




2J 
6
3

e y  e y
X>>1 coth y  y
1
 y y 1
e e
1
BJ ( x ) 
 2 J  1 coth  x  2 J  1 / 2   coth  x / 2  X<<1
2J


1
BJ ( x )
J=10
J=7/2
J=3/2
J=1/2
J
4
BJ ( x ) 
1
2J
e y  e y
1
coth y  y

e  e  y y 0 y
  2 J  1 coth  x  2 J  1 / 2   coth  x / 2  
1
coth  Jx  
coth  x / 2 
2J
negligible except
for x0
J
1
coth  Jx  
: L( x )
Jx
J
Langevin function
X=g B0H/kBT
Average magnetic moment <m> per particle
1
0V
k BT d ln Z1
 G 

N
and



H

V
dH

T
0
k T d ln Z1 k BT d ln Z dx
m  B N

0 N
dH
0 dx dH
 g  H 
m  gB J BJ  B 0 
 k BT 
m   gB  0 H
2
m  MV / N
k BT
d

J BJ ( x )
0
dH
 gB 0 H 


k
T
B


In the limit of H0 we recall BJ ( x ) 
J  J  1
3k BT
J  J  1
dM N d m
N
2

    gB  0
dH V dH
V
3k BT
x  J  1
3
4

M 
2
Curie law
0
0
2
4
T
Entropy per particle s=S/N
d
d ln Z1 dx
1  G 

k
T
ln
Z

k
ln
Z

k
T


B
1
B
1
B


dx dT
N  T  H dT
d  gB 0 H 
 k B ln Z1  k BTJ BJ ( x )


dT  k BT 
 G 
S  

 T  H
sS/N 
 g  H  g  H
s  k B ln Z1  k BTJ BJ  B 0  B 02
 k B T  k BT
In the limit of H0
0  sinh   g   H  2 J  1 / 2  
J B 0
s  k B ln Z1 ( H  0)  k B lim ln 

H 0

sinh  y 
1  y  2 J  1  1  y  2 J  1

1 y 1 y
 2J  1
multiplicity
ln2
0.7
0.6
0.5
H increases
0.4
s  k B ln  2 J  1
with


Example for J=1/2
s
sinh  y  2 J  1 
sinh   g J  B 0 H / 2 
0.3
0.2
0.1
0.0
0
2
4
6
T
8
10