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58

CHAPTER 3. FERMIONS

3.3Weakly interacting Fermi gas

3.3.1 Ground state

Again we work in second quantization and assume there to be g different types of FERMIons, i.e.

ˆ

(~r)

j = 1;:::;g with

(3.101)

yj

nyˆ j(~r);yˆ j0(~r 0)o = nyˆ †j (~r);yˆ †j0(~r 0)o

= 0

(3.102)

no

yˆ

j

(~r);yˆ

†

(~r 0) = d

j j0

d(~r

~r 0)

(3.103)

 

 

j0

 

 

 

 

This means that for each type of FERMIon each state is either occupied once or is not occupied at all, which is of course the PAULI principle.

The HAMILTONian now reads

Hˆ =

åj

Z

d3r

 

ˆ

†

(~r)

~2

 

2

ˆ

(~r)

 

 

 

 

 

 

 

 

 

 

 

 

 

yj

 

2m

Ñ yj

 

 

 

 

 

 

 

 

 

 

 

 

+

1

 

åj j0

Z

 

d3rd3r0

yˆ †j (~r

)yˆ j (~r

)UI(~r ~r 0)yˆ †j0(~r 0

)yˆ j0(~r 0

)

(3.104)

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

åj

Z

d3r

 

 

†

 

~

2 |

2

nˆ{zj (~r)

}

|

nˆ j{z0(~r 0)

}

 

=

 

ˆ

(~r)

 

 

ˆ

(~r)

 

 

 

 

 

 

 

 

 

 

 

 

 

yj

 

2m

Ñ yj

 

 

 

 

 

 

 

 

 

 

 

 

+

1

åj j0

Z

 

d3rd3r0 yˆ †j0(~r 0)yˆ †j (~r)UI(~r ~r 0)yˆ j(~r)yˆ j0(~r 0)

 

(3.105)

 

2

 

 

We use the following assumption:

UI = V0d(~r ~r 0)

(3.106)

This means we consider only short range interactions. This is reasonable because we consider gases with r0 pF 1 and pF density 13 ((1.1) and (1.3)). We wrote pF here instead of p because only momenta p pF are relevant for any physical process.

3.3. WEAKLY INTERACTING FERMI GAS

 

 

 

 

 

 

 

59

Using this we get

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Hˆ =

åj

Z

d3r ˆ

†

(~r)

~2

 

2

ˆ

(~r)

 

 

 

 

 

 

 

 

 

 

yj

 

 

2m

Ñ yj

 

 

 

 

 

 

 

 

 

 

 

 

2

 

j= j0

Z

 

 

 

0

 

 

 

 

 

 

 

 

0

 

 

 

+

1

 

å

 

 

 

3

 

ˆ †

 

 

 

ˆ †

ˆ

 

ˆ

 

 

(~r)

(3.107)

 

 

V0

 

 

d

r yj

 

(~r)yj

(~r)yj

(~r)yj

 

 

 

Z

 

 

6

 

 

~2

 

 

 

 

 

 

 

 

 

 

 

 

 

=

åj

d3r ˆ

†

(~r)

 

2

ˆ

(~r)

 

 

 

 

 

 

 

 

 

 

yj

 

 

2m

Ñ yj

 

 

 

 

 

 

 

 

 

 

 

 

 

j< j0

Z

 

 

 

ˆ †

0

 

 

ˆ

 

ˆ

 

ˆ

0

 

 

 

 

 

+V0 å

 

 

3

 

 

 

 

†

 

 

(~r)

(3.108)

 

 

 

d

r yj (~r)yj (~r)yj

(~r)yj

 

where we have to exclude the case of j = j0 due to the PAULI principle. If we have a spatial homogeneous system we can write our wave function as

ˆ

(~r) = åa~p; je

i

~p ~r

 

~

yj

 

 

 

~p

 

 

In this case we also have momentum conservation, i.e.

~p1 +~p2 = ~p3 +~p4

This will be denoted by a tick at the appropriate summations.

ˆ

p2

†

 

 

0

† †

H = å

 

a~p; ja~p; j

+V0

å

 

å ap4 j0ap3 jap1 jap2 j0

2m

 

j;~p

 

 

 

p1 p2 p3 p4 j< j0

(3.109)

(3.110)

(3.111)

In this expression we obviously see that if we have only one type of FERMIons the interaction term vanishes. This is because we consider only s-wave scattering (i.e. lowest order). If we consider for example the two particle wave function which has to fulfill the P AULI principle

j(1;2) = ji j(~r1 ~r2) = jji(~r2 ~r1)

(3.112)

If we look at the state with a given relative angular momentum l of these two particles we have

ji j(~r1

~r2) = ( 1)l ji j(~r2 ~r1) leading to

(3.113)

ji j(~r1

~r2) = ( 1)l jji(~r1 ~r2)

(3.114)

60

 

 

 

 

 

 

 

 

 

 

CHAPTER 3. FERMIONS

Thus we have two cases to consider

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

j ij

 

i j

ij i

 

 

 

 

 

 

 

2

 

 

 

l = 0;2;:::

ji j

=

jji

= p

 

 

( i

 

 

j

 

 

j

 

i

)

(3.115)

 

 

 

 

1

 

j ij

 

i

 

 

j

ij i

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

l = 1;3;:::

ji j

= jji = p

 

( i

j

 

+

j

i

)

 

(3.116)

This means we have to have g 2 for even relative angular momentum (3.115) and g 1 for odd relative angular momentum (3.116). Since we consider only l = 0 we therefore need at least two types of FERMIons to have a non vanishing interaction.

Returning to the non interacting HAMILTONian we get the ground state energy

 

p2

p2

3

 

 

3

 

 

Eg = å

 

ha†p jap ji = å

 

np j = ån j

 

eF

= n

 

eF

(3.117)

2m

2m

5

5

j;~p

 

j~p

 

j

 

 

 

 

Here we used (3.6) and considered the ordinary case for which E does not depend on j. Treating the interaction as a perturbation we get

E1 = V0åpi 0 jå0 jha†p4 j0a†p3 jap1 jap2 j0i

<

= V0 jå0 j på1 p2ha†p2 j0ap2 j0iha†p1 jap1 ji

<

(3.118)

(3.119)

= V

å å

n

 

n

 

= V

g(g 1)

n

n

 

= V

1

 

g 1

n2

(3.120)

p2 j

p1 j0

 

2

 

 

0

 

 

0 2 1

 

0 2 g

 

 

j0< j p1 p2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

In (3.118) the operators create two particles and two holes. The resulting state will usually be orthogonal to the ground state, unless p4 = p2 and p3 = p1. We further assume that n1 = n2 = = ng which is true of course only if no external fields are present.

In second order we get

 

ˆ

2

 

E2 = å

jhejHintjgij

 

(3.121)

Eg Ee

 

e

 

 

Here jgi is the ground state and jei is any state possible with two holes (p1 j and p2 j0) and two particles (p3 j and p4 j0).

E

g

E

e

=

 

p42 + p32 p12 p22

(3.122)

2m

 

 

 

 

As required for any energy difference in respect to the ground state, (3.122) is negative. Now (3.121) becomes (cf. appendix (B))

3.3. WEAKLY INTERACTING FERMI GAS

61

E2

= V0

å å0

p2

+p2

p2

p2

 

 

 

(3.123)

 

 

2

 

 

 

np1 jnp2 j0

1

np3 j

 

1

 

np4 j0

 

 

 

j< j0

p

i

 

4

3 1

2

 

 

 

 

 

 

 

 

2m

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

In respect to p1; p2 < pF we have no problem to integrate (sum up). The conservation of momentum fixes p4. p3 however can take any value larger than pF. The third sum (or integral)

∞

1

 

ZpF

 

d3 p3 p32

(3.124)

does not converge. To eliminate this divergency, we can use the same technique as in section (2.3) page 28 and use BORN approximation1:

4p~2a

 

 

 

å

 

 

1

 

 

 

 

 

 

4p~2a

 

 

 

 

 

 

 

 

 

= V0

V0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(3.125)

 

m

 

 

p42+p32

 

p12

 

p22

 

 

 

m

 

 

 

 

 

 

 

 

 

 

 

 

p

p

4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

 

 

 

2m

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4p~2a

 

 

 

4p~2a

 

2

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

)

 

V0

 

 

 

+

 

 

 

 

 

 

 

 

3 4

 

 

 

 

 

(3.126)

 

 

m

 

 

m

 

 

 

 

2m

 

p22

 

 

 

 

 

 

 

 

 

 

å 0

 

p42+p32

 

p12

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

p p

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Inserting this expression for V0 into (3.120) and (3.123) we can get rid of the divergent sums

E1 + E2 =

4p~2a

å å np1 jnp2 j0

 

 

 

 

 

 

 

(3.127)

m

 

 

 

 

 

 

 

 

 

 

 

 

 

j< j0 p1 p2

å å0

h

 

 

p2

 

 

i

 

 

 

 

 

 

 

p2+p2

p2

 

 

4p~2a

2

 

 

np1 jnp2 j0

1 np3 j

 

1 np4; j0

1

 

 

 

 

 

 

 

j< j0

 

i

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2m

2

 

 

 

 

 

 

 

m

 

p

 

 

 

4 3

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

[:::] = 1 np3 j np4 j0 + np3 jnp4 j0 1 = fnp3 jnp4 j0g np3J np4 j0

The addend of (3.127) with f:::g is symmetric if p1 p2 is interchanged with p3 p4. On the other hand the denominator is antisymmetric under interchange. Since the sum runs over all pi it has to vanish thus preventing the ultraviolet divergence since now three out of 4 momenta are inside the sphere and the fourth momentum is fixed.

E1 + E2 =

4p~2a

å å np1 jnp2 j0

 

 

 

 

 

 

 

 

 

 

m

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+

 

 

 

j< j0

p1 p2

å å0

 

 

+p2

 

 

 

 

(3.128)

 

 

 

 

 

~

 

 

 

p2

p2

p2

4

 

 

 

4p

 

2a

 

2

 

 

np1 jnp2 j0

np3 j + np ; j0

 

 

 

 

 

 

 

j< j0

 

 

 

 

 

 

 

 

m

 

 

i

 

 

 

2m

2

 

 

 

 

 

 

 

 

 

 

 

 

p

 

 

4

 

3

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1The summation runs over two momenta as we have two particles in the intermediate state

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