70

CHAPTER 3. FERMIONS

which leads to

1

2

1

ln

=

+ 1

(3.199)

2

e

F0

2

2

) e =

e F0

(3.200)

e2

2

2

c0 = vF

1 +

e F0 vF

(3.201)

e2

compared to sound in the hydrodynamic regime:

0

1

¶p

1 ¶p

1

¶

2

3

c12 =

=

=

B

n

eFC

(3.202)

¶r

m

¶n

m

¶n

3

5

B

E

C

1

2

1 6

@

2

A

6

2

3

2

0

2

3

|{z}3

1 5 1 1

n

3

= m ¶n

5 n 2m

pg~

n

1

= m 3 5 m

pg~

(3.203)

@

A

¶

1

2

2

vF

pF

vF

=

=

)

c1 = p

(3.204)

3

m

3

3

This is ordinary hydrodynamic sound.

~

(3.205)

P = ~p1 +~p2 = 0

Here we can write the SCHRÖDINGER

equation as

~2

2

2

y(~r1

Ñ1

+ Ñ2

+UI

~r1 ~r2

;~r2) = Ey(~r1;~r2)

(3.206)

2m

~p(~r1

~r2)

y(~r1

;~r2) = åc~pe

i

~

(3.207)

3.5. BARDEEN-COOPER-SHIEFFER-THEORY

71

(e1 + e2)c~p + åUI(~p ~p 0)c~p 0

= Ec~p

(3.208)

~p 0

where

p2

e1 = e2 =

= ep

(3.209)

2m

since both particles are identical.

Rewriting the SCHRÖDINGER

equation once

more we have

(E 2ep)c~p = åUI(~p ~p 0)c~p 0

or

(3.210)

~p 0

c~p =

1

åUI(~p ~p 0)c~p 0

(3.211)

E

2

ep ~p 0

U

p

p

V0

¯

(3.212)

0 ep;ep0 w

I

(~

~ 0) =

(0

otherwise

In ordinary space this expression looks rather strange.

Using this model we have

1

¯

˜

c~p

= E

2

(3.213)

Θ(w

ep)V0åc~p 0

ep

~p 0

The tilde denotes that the sum obeys the constraint

ep0 w¯

(3.214)

To solve this expression, we sum over all coefficients

˜

˜

1 ˜

åc~p

= V0å

åc~p 0

(3.215)

~p

~p E 2ep ~p 0

˜

1

, 1 = V0å~p

(3.216)

E 2ep

V0 = jV0j E = 2

(3.217)

72

CHAPTER 3. FERMIONS

where is the binding energy per particle. Using this we have

1

˜

1

1 =

jV0jå

(3.218)

2

+ ep

~p

1

¯

1

w

=

jV0jZ0

de n(e)

(3.219)

2

+ e

1

w¯

mp

1

2m

=

jV0jZ0

de

pe

(3.220)

2

2p2~3

+ e

mp

Z0

p

2x2

=

1

2m

w¯

(3.221)

jV0j

dx

2

2p2~3

+ x2

mp

p

1

m

w¯

1

= jV0j

2

Z0

dx

(3.222)

2p2~3

+ x2

mp

p

m

¯

= jV0j

2

pw¯ p

arctan p

w

(3.223)

2p2~3

mp

p

2m

jV0j

npw¯

p

o

(3.224)

2p2~3

2

= jV0jn(w¯)(1

2 r

)

(3.225)

w¯

p

2

1

=

1

> 0

(3.226)

¯

V (

)

w

p

j

0jn eF

ep;ep0 eF

(3.227)

c~p =

1

~p

0

cp

(3.228)

~p 0 ~pF

0

E 2ep

å

UI ~p

We assume for the potential

I(~ ~ 0) = (0

otherwise

¯

U p p

V0

eF ep;ep0

eF + w

(3.229)

3.5. BARDEEN-COOPER-SHIEFFER-THEORY

73

Now we define

xp = ep eF 0

(3.230)

and note that the energy of the ground state is now

E = 2eF 2

(3.231)

where is the binding energy per particle.

Again we can rewrite the SCHRÖDINGER

equation in our case as

¯

c

=

Θ(w xp)

V

˜ c

(3.232)

~p

2( + x

) j

0jå ~p 0

p

~p 0

¯

¯

The tilde denotes xp0 w.

Using the same method with

w eF as before we get

1

˜

1

1

¯

1

w

1 =

jV0jå~p

=

jV0jZ0

dx n(eF + xp)

(3.233)

2

+ xp

2

+ xp

2 jV0jn(eF)ln

+ w

2 jV0jn(eF)ln w

(3.234)

1

¯

1

¯

In the last step we approximated again

¯

w

. solving this for the binding energy

= we

2

C

1956

(3.235)

jV0jn(eF)

OOPER

¯

vF p

,

p

(3.236)

vF

If we denote the size of the correlation as x we can approximate

x

~

~

vF

~ eF

~

(3.237)

p

pF

pF