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Физика конденсированного тела

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Eschrig - Theory Of Superconductivity, A Primer

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Introdu tion of dimensionless quan

r=titi;

j j

;

jtjr

2 0

;

b = B.p2B ( ) = B

(66)

= j

p2B

(t) ;

a = A.p

( )

yields the dimensionless Ginsburg-Landau equations

i x

+ a

+ j j2

= 0;

(67)

b = is;

whi h ontain t

only

parameter .

4.4 The phase boundary

t T . T

in an homogeneous

eld B B

(T ) in

W onsider a homog

ous

z-dire tion. W

assume

pla

phase

ary in the y z-plane so that forextern!al 1 the material is

ill super ondu ting, and the magnetiboundel is expelled, but for x ! 1 the material is in the normal

state with the eld penetratisuperg.

ondu tor

b = 0

b = 1=

= 1

= 0

super ondu ting

phase

normal

boundary

We put

Figure 12: Geometry

of a plane phase boundary.

(x);

= b(x);

= b

= 0;

= a(x); a

= a

= 0; b(x) = 0(x):

Then, the super urrent i

ws in

phase of

on y. W

the y-dire tion, and hen e

real. Further, by xing another gauge onstant,dependswmay hoose

onsid r y = 0 and may then ohoose

a(0) = 0:

Then, Eqs. (67) redu e to

+ a2

3 = 0; a00

= a

= 1 (68)x),

1. We put

Let us rst onsider 1: F r large enough negative x we have a 0 and

and get from the rst equation (68)

2 x

1 1 + 3

= 2

; e

; x .

On the other ha

d, for large enough positive x we have b = 1=

1; hen e, again from

2; a = x

the rst equation (68),

22x2 ;

e x2

=2p2;

x2 1:

(x) where the

The se ond Eq. (68) yields a penetration depth

; where

denotes the value of

eld drops:

1 ep2 x

e 0x p12 1 1=p0 > p1 e xb2=2p2

1= =

Figure 13: The phase boundary of a type I super ondu tor.

In the opposite ase 1;

falls o for x

1; where b

2; and for x 1;

2; a x=

00 2x2 =2 :

p12 1

e bx2=12p2

1 =

< 1

4.5

Figure 14: The phase boundary of a type II super ondu tor.

The energy of

For B = B

T ); b = 1 in our units, the Free Energy of the normal phase is just equal to the Free

bwhiouhndary= 0;

= 1: If we integrate the Free Energy density

Energy of the super ondu ing phasein

variation (per unit area of they z-

we obtain

energy of the phase boundary per area:

(

(B B

)

s=n

0j2

the4

A2j

jtjj

(69)

+ 4m

2 j

Sin e

, w

= B

= B B

In our dimensionless

he external eld is B

havplane),used B

tot

ext

quantities this is (x is now measured in units of )

s=n

b p

1 2

(70)

and then (68) was inserted. We see that

First an inte ration per parts of

s=n

have both signs:

? 0

for

as performed,?

b m

s=n

in reases and

must

de rease if

= 0 at b = 1=

2, (b 1=

2) = (a

2) and

Sinhav eopposiuste signs whi h leads to the last ondition. If

= p2

would be a solution of (68), it would orrespond to

s=n

= 0:

We now show that this is indeed the ase for																													2	= 1=2: First we nd a rst integral of (68):
We now show that this is indeed the ase for																														= 1=2: First we nd a rst integral of (68):
				2	0			00									a2		1							+				3		;					3		0					=
				2	0			00							2					0		2						2		0		+ 2					3	0	0					=			0				3 0
										=					2		2			0	a	2		+		2		2		aa			0					0	a		00		2				0	+ 2			3 0
										=							2				a			+		2				aa				2a					a				2					+ 2
							02			=			2					2a2			a0					\|= 0 by{zt										he se }ond Eq. (67)																							(71)
																										\|= 0 by{zt
																										2					2		+				4		+ onst.														1
																			0						= 0 for a0												2		1														1
													sin e						0		=				= 0 for a0												=		p						=) onst. =								2	:
Now we use			(712)		1					1						2													p			p										2								1			2					2
and have from			(712)		p	2				1						2													p												1					2				1			0			p		2
		2	=	;	p				a0			=		p					) a00										=						2			0	= a							2		2	)				0		= a	p
		2	=	;			2		a0			=							) a00										=						2				= a										)						= a
						0	2			= 2			2		0	2 a02 1																2a0					+			p2					a0				+	2		;
whi h is indeed an id					ity.					= 2			2			2 a02 1																2a0					+								a0				+	2		;
whi h is indeed an id					ity.																						in the rst line of (70), it is lear that																																is positive
Sin e	02= 2 > 0 enters the integral for																						s=n				in the rst line of (70), it is lear that																															s=n	is positive
for 2 ! 0: Therefore, the nal result is																							s=n							1																												s=n
											s=n				? 0				for						7 p2												:		type							II										1			(72)
The			\type I" and \type II" for																						r ondu tors were oined																								y Abrikosov,							1	and it was the
existen e of ty			e II super ondu tors and a theoreti al predi tion by																																												Abrikosov, whi h paved the way
for te hninamesal ppli ations of super ondu tivity.

1A. A. Abrikosov, Sov. Phys.{JETP 5, 1174 (1957). 24

5 INTERMEDIATE STATE, MIXED STATE

, B < B ; where w

2 w

in a suÆ iently weak

the ideal diamagnetism, the Meissn

e t, and

ux quantization.

ChapterIn

ter 3

wonsifoundderedthat the di er n e between the thermodynami potentials in the normal

foundthe super ondu ting homogensuper onduphasestor

per volume without

elds may be

expressed

as ( f. (35))

ous

magneti 2

thermo

G (p; T ) G (p; T )

F (V; T ) F (V; T )

(T )

(73)

riti al eld B (T ): (W

negle t

here ag in the e e ts of pressure or of orre-

additional

energy eld rB

=2 0 for the

sup

phase

( f. (56a)) and

f makineti energy

spondi

g volume

hanges on B :)

in the

where the super urrents ow.

> B ;

density

~e=dynami4e t)j( = + 2ieA=~)

(Meissnerthe F Energy of

sup r ondu ti

g state

larger than that of the no

l state inan

mag eti

is appl ed to some v

part of

tor, it

y be expelled

eating an

ternal

eld

= B throughsuper ond

rrents, on the ost of

orrespond

lume

thermodynami stabl

states. The external eld B at whi h the phase transition appolumernsar

and orresponding domain patt

long range

tera tions lik in ferro le tri s andrsurfaonduerromagnets,ting

homogeneous situati

. However, B itself may

tain

part re ted by urrents in another v

of the

and phase boundary energibe omesust also be onsidered.

There are

therefor

dependssupertheonduometrytor,

and on the phase boundary energy.

the

super ondu termediateting state b omes instable there.

On ould think of a normal-state on athev island

5.1

The

of a type I

depends

sh pe

Apply a hom geneous external eld B

to a super o super. ondu= B tor+ B

of the

ndu tor. There isstateertain point, at whi h B = B

> B

(Fig. 15). If B

> B ;

forming (Fig. 16).

ext

max

ext

max

Bext

Bext + Bm

SC max

BmagnetimaxFIG. 15: Totaleld around(externalSCtypeplus indIsupered)-

FIG. 16.

ondu tor.

This, however, nnot be

either: the point of B

= B has now moved

to the super-

means that in the shaded norm

area B < B ; this area ust be ome super ondu tingagain (Fig. 16).

Forming of a onvex island woustabled ause the same problem

(Fig. 17).

max

phases, whi h

ondu or to a point of the ph se boundary between the normal and

Bmax

What

FIG. 17.

FIG. 18.

ally

forms

3 B

Bext

ternating

super ondu ting

ompli at d lamellous or

lamentous stru ture of al-

the

eld

penetrates

and normal phases through

long rod

whiTherv oftruea ype I super on

sphere

(Fig. 18).

magnetizationthe geome

transition

du tor

di ren

thedepenshi lding super urreated

rst order

ries

Fig.

19.

be ause the eld

3.C,

tryb

oes.

Se tion

of the stray eld +

B =6 0

he phase tran ition

der.

the mov

+ energyof phase boundariesg

hinderedobtaineddefe ts, hen e

was

to be

rst for

there Generallyis h steresis

around

Figure 19:

men

of p

boundaries is

B .

5.2

Mixed state of

type II super ondu tor

; at the lower

If the phase boundary

egative, germs of

phase may form well below B

riti al eld B

; and germs of super ondu ting phasenormaly form well above B

; at the upper riti al

eld B

energy.

; in both asesenergyb gaining phase

;

small germs

1 andboundaryonst. The dimensionless Ginsburg-Landau equations

At B / B

(74)

(67) may be linearized:

i x

the external B- eld in z-dire tion (Fig. 20),

= b

= 0;

= b;

= by;

= a

= 0;

xapply

and assume germ laments along the eld lines:

Then, (74) is" ast into

(x; y):

With

(x; y

i x

= eipx

(y); p=(b ) = y0 this equation simpli es

to [ (1= 2)(

2)+b(y y0)2

= ; or, after multiplying

with =2 and b=dy

(y y

Figure 20:

de1ning

b2u2

(75)

This is the S hr•odinger equation for the ground state of a

2 du

2 B :

os illator with

! = b = =) =

Hen e,

(T ) = p2 B

harmoni

(76)

(T ):

If a g rm lose to

surfa e of a super2ondu tor at

= onst. is onsidered, then the u-

must be ut at some nite value. There, the boundary

ondition

(58) yields d =du = 0 (sin e oordinate= 0).

Theref

e, instead of the

boundary problem, the

symmetri ground state in a double os illator

with

a mirror plane may be onsidered (Fig. 21).

(0) = 0

min E

= 0:59

FIG. 21: The ground state of a double os illator.

A. A. Abrikosov, unpublished 1955; Sov. Phys.{JETP 5, 1174 (1957).

Hen e, at

B 2

= 1:7B 2

= 2:4 B

(77)

y may set in in a

0:59

In a

layer of thi kness .

germs with

small

I super ondu tor, B

< B .

However, only below B

- alu

ould

where the

surfa2

phase

stable in

ype I super-

ondu tor. Here,form

< B <superB germsondu tian only form with

non-zero minimal

-value, whi h is

superinhibitedonduypb etivitpositiv

energy. In this

region

theisuper ooled normal phase is arbitrarilyme astable. For

> 0:59=

2; B

> B ; and surfa e super ondu tivity maalreadyexist absolutelyabov B :

5.3

The

surfa e

ype II

;

line

is onsidered. In the h mogeneous

st te

To determine B

opposite

ith

ux1

eld is tuned

until the

rmal state g rm is forming. Ag in

= 1; the external

ay suppose that

germ is forming along a eld line in z-

. Sin superthe phaseondu ting

energy is

egative, theremagnetiust besituatendention

form many fa e boundaries. However, the

ot form arbit arily small sin e w

w from (17)rst direthetionux onne ted withboundarynormal

germ inansuper ondu tor is quantized and anno

be smaller than

2 ~

super ondu tor

(78)

= 2

We onsider a ux line of total ux

along the z-dire tion (Fig. 22):

j j

= j jei ; B = ezB( ); 2 = x2 + y2

FIG. 22: An isolated ux line.

e ould try to solv

the G nsburg-L n

au equations for that ase. However, there is no general

analyti solution, and the equat

are valid

lose to T

only, where

j is small. I

stead w

assume

j j

. Then, from Ampere's

1; that is, ; and onsider only the region ;

law and

(6),

0js =

2 wh2ere

j2A

= onst

2 0e2j j2

;

hen e,

r A

A + 2

B =

integrate this

along a ir le

the ux line with radius and use Stokes'

theorem,

A) :

dsA = equationd (

(79)

n B +arounds

The phase must in rease by 2 on a ir le around one uxoid.

Now, we take : Then, the rst term in (79) may be negle ted. We nd

2 2 dB

B( ) = 2 2

;

(81)

The integration onstan

was hosen su h that (81) vanishes for & ; where a more a urate

By applying Stokes' theorem also to the se ond integral of (79), w

hav

for all

analysis of (79) is

to get the orre t asymptoti s.

2 B

= 0:

ne essaryB +

d n

Sin e the right hand side does not hange if we v

the area of integration,

B 2

B = 0

must hold. In ylindri oordinates,

ary

;

hen e,

B + 2

B = 0

B00

(82)

2 B = 0:

This equation is of the Bessel type. Its for large de aying solution is

(83)

B( ) =

! p

e =;

2 2

where K is M D nald's fun tion (Hankel's fun tion with ima inary argument), and the oeÆ ient

;

j =energy2

kineti= j

j ;

has be n hosen to meet (80) for :

of eld

and

energy of the super ur-

The energy per length of the ux line

rent:

4e onsists

Bj2

d2r

B =

2 0

d2rB

B + 2

!ds

B :

2 0

rst integrand was shown to vanish for ; and the last ontour integral vanishes for ! 1.

TheW negle t the ontribution from . ; and nd with

)

(84)

2 B dB

)

(0)

(80

With (81), the nal result with logarithmi a ura y (ln(=) 1) is

ln :

(85)

This result also proves that the total en

g one xoid ea h:

is minimum for ux lines ontaini

For a ux

line on

uxoids the

ld be

while for

ux lines it would only be

n ( > 0).

the super urren

around the vertex line. Its intera tion

In this analysis, B was the eld

energy per length withtaininghomogeneous exte nal eld in the same dire tion is

BBext

0Bext

(86)

reatedenergy

(85) and (86) are equal at the low

= B

riti al eld B

ext

(87)

4 2

ln = B p

;

The phase diagr m of

II super ondu tor

ig. 23. (There migh be a

phase with

shown in

modulated order

ameter under ertain onditions, theoreti ally

predi ted independently

spatiallyb Fulde and Ferell and bypeLarkin and Ovshinnikov; this

FFLO phase has not yet beenotherlea ly observed

experimentally.)

surfa e

B 2

B 3

FFLO?

mixed

phase

s B 1 B

Figure 23: The phase diagram of a type II super ondu tor.

nergy is minim

A more d ailed numeri al al ulation shows that in

isotro

material

for a regular tria

gular latti e of the ux

in the

plane

perpendi ular to them.

From (87), at

B = B

the

sity of ux lines is ln

4 2

;

that is,

the

latti e

onstanthea

is obtained froum

= ln :

a1 =linesp8

& :

(88)

3=2 = 4

3 ln

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