3.1. FREE FERMIONS

51

Can we understand this solution physically ? We have T n(e) states available for excitation with energy of the order of T , which is exactly what the calculation gives in 0th order.

The specific heat is

			dE			p2
	c		=		=			n(eF)T	(3.35)
		V		dT		3

unlike the BOSE gas where cV T 3.

3.1.2Pressure of degenerated Fermi gas

For classical gases

p = nT

(3.36)

holds meaning at T = 0 pressure vanishes. For quantum gases (BOSE as well as

FERMI)

	p =	2		E	(3.37)
		3 V

holds. Refer to appendix (A) for the derivation of (3.37). Here E = E(T ) for the quantum system considered. For FERMIons we get using (3.4) and (3.6)

	2 3								2 pF2							1 n 6p~3n							2
																							3
p =			n			eF		=			n			=									(3.38)
		3		5					5			2m				5 m				g
	1						6p~		3			2
=												3	5		6= 0								(3.39)
													n 3
	5m						g

This pressure is sometimes called FERMI-Pressure. It stabilizes the nucleons against strong interaction as well as neutron stars against gravitational forces.

If we look at high temperatures T eF we should get the classical behavior plus some quantum corrections. To get the classical behavior we have to require

jmj	and m negative	(3.40)
e T 1

To calculate the classical behavior we note that independently of the statistics of the particle we have

e+jmj

e		1	for large T	(3.41)
	T

52

CHAPTER 3. FERMIONS

and for FERMIons

nf(e) =

1

=

1

(3.42)

exp

e

m

exp

e+ m

+ 1

j + 1

m eT

m +e

Tj

e

jT j

1 e

j

Tj

(3.43)

T

Bearing in mind, that we work with a fixed

n we get

n = g Z

d3 p

g Z

d3 p

m

e

1 e

m

e

(3.44)

nF

e

T

T

T

T

(2p~)3

(2p~)3

=

g

mp

Z

de p

e

m

e

1 e

m

e

(3.45)

2m

e

T

T

T

T

2p~3

gmp

Z

2m

3 m

m

=

dx pxe x

1 e

x

(3.46)

T 2 e

T

T

2p2~3

gmp

3 m p

m

1

2m

1 e

=

T 2 e

T

p

T

2p

(3.47)

2p2~3

2

2

Here we used the definition of the

Γ-Function

∞

Γ(z) = Z

dt tz 1e t

(3.48)

0

with its properties

1

Γ(1 + z) = zΓ(z)

Γ

= pp

(3.49)

2

Therefore our assumption (3.40) must hold true for the last term in (3.47) to be small. Neglecting the last term we solve for the chemical potential in the lowest order (BOLTZMANN case)

mB

n 2 2p2~3 1

n 2p

3

2

3

e T

=

p

mp

=

~

(3.50)

g

T 2

g

mT

p

2m

eF

3

3

2

1

(3.51)

T

Similarly we can calculate the first correction to the classical equation for the pressure:

2

∞

p =

Z0

d3 p n(e)nf(e)e

(3.52)

3

2 mp

g Z0

∞

2m

3 m

e

m e

=

de e 2 e

1

e

(3.53)

T

T

T

T

3

2p2~3

2

mp

∞

m

2m

m 5

3

=

g

e

T 2 Z0

dx x 2 e x

1 e xe

(3.54)

T

T

3

2p2~3

3.1. FREE FERMIONS

53

=

3 g

T

2 e T Γ

2

1 e T

4p2

(3.55)

2p2~3

2 mp2m

5

m

5

m

1

1 e T

= g

2p2

2~3

T

2 e T

4p2

(3.56)

2p

mp m

5

m

p

m

1

1

4p2 e T

= T g

+

(3.57)

2p2~3 T

2 e T

2p2 e T

2p

mp2m

3

m p

1

m

1

m

mp

p

1

2m

mB

= T n + T g

p

3

e2

(3.58)

T 2

4p

T

2p2~3

2

2

= T n + T

n

n

2

2p2~3

1

(3.59)

3

4p2 g pp mp2m T 2

= nT 1 +

p

23

n

3

~

(3.60)

mT

2g

For T % ∞ the correction vanishes of course. Note, that this increase in pressure is due to only the FERMI statistics as no interaction was considered. The correction

		eF		3
is of the order of				2	. In the case of BOSons a similar term appears but it is
			T
subtracted from	the classical value.

3.1.3Excitations of Fermions at T=0

Figure 3.2: Small excitations at the FERMI surface (left)

If we look for excitations at T = 0 the system is in its ground state before the excitation. Afterwards one particle is above the FERMI surface leaving a hole below the FERMI surface. Condense matter physicists call this a "particle hole pair". Since we currently consider no interaction the particle and hole are uncorrelated and can be treated separately. Even if interaction is considered the correlation between particle and hole can be usually neglected since their number is small (for T eF) and, therefore they are "far away".

p

54	CHAPTER 3. FERMIONS

particle excitations We add a particle into a state ~p with j~pj > pF. This gives the excitation energy

ep = E0(N) +

p2

E0(N + 1)

(3.61)

2m

p2

p2

p2

=

(

E0(N + 1) E0(N)) =

F

> 0

(3.62)

2m

2m

2m

|

=m{zeF

}

=

hole excitations We remove a particle from a state ~p with j~pj < pF. This gives

the excitation energy

p2

p2

p2

ep = E0(N)

E0(N + 1) =

F

> 0

(3.63)

2m

2m

2m

The energy gain is therefore in both cases

e =

p2

p2

> 0

(3.64)

2m

2m

F

Obviously you can get the same result

for a particle

hole pair if you simply calcu-

late the energy difference before (p2 < pF) and after (p1 > pF) the excitation

p2

p2

p1

p2

+

p2

p2

= e1 + e2

(3.65)

de = 2m

2m =

2m 2m

2m 2m

1

2

1

F

F

2

We can also discuss this in operator language. If we define

˜ †

†

+ vpa p

(3.66)

with

ap

= upap

up = (0

vp = (1

p < pF

p < pF

(3.67)

1

p > pF

0

p > pF

where u adds a particle and v adds a hole we can rewrite the HAMILTONian (1.84 with g = 0) as

Hˆ 0 = å	2		m a†pap = E00 + åp		2	m	a˜ †pa˜ p	(3.68)
	2pm				2pm

E00 is the energy of the ground state and the energy of excitations is always positive (which was not satisfied before the transformation). Of course the new operators obey

˜	(3.69)
apjgi = 0

where jgi is the ground state (filled F ERMI sphere).