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Физика конденсированного тела

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Baranov.Introduction to Physics Of Ultra Cold Gases.pdf

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2.3. WEAKLY INTERACTING BOSE GAS

2.3Weakly interacting Bose gas

If we consider a unit volume at T = 0 the HAMILTONian reads

ˆ	†	1			† †
H = åepapap +			g	å	ap3 ap4 ap2 ap1	:	(2.64)

	p	2		~p1+~p2=~p3+~p4

Since ep e0 0 we cannot apply perturbation theory because we cannot guarantee g < ep e0. Explicit calculation show divergent terms already in second order. Our solution will also show the invalidity of perturbation theory (cf. (2.111)).

We note that almost all particles are in the ground state, i.e.

a†pap	n0		8p 6= 0	(2.65)
a0† a0	= n0 1			(2.66)
a0a0†	= 1 + n0 1:			(2.67)
	\|		}
		{zn0

This leads to the simpliﬁcation a0;a†0 pn0 1. Taking leading terms in n0, we can write (2.64) in the form

†

H =

åp

epapap +

gn0

~p6=0 n

apa

p + a

(2.68)

gn0

pap + 4apap :

The remaining momenta have to be equal in the last sum because of conservation

of total momentum.

Remembering that

n0 = n ånp

(2.69)

we can rewrite (2.68) with the considered accuracy as

1 2

†

~p6=0 n

†

H =

p6=0

2apap + a

pap + apa

+ å epapap +

gn å

(2.70)

˜ † ˜

(2.71)

= E0 +

å wpapap

p6=0

which is a HAMILTONoperator for quasi particles in a harmonic potential. To ﬁnd the relation between these quasi-particles and our original (or base) operators we assume according to BOGOLYUBOV

a˜ p = upap + vpa†	p	up = up	up = u p	(2.72)
a˜ †p = upa†p + vpa p		vp = vp	vp = v p	(2.73)

26	CHAPTER 2. BOSONS

and require them to obey the BOSE commutators:

ha˜ p;a˜ p0i = ha˜ †p;a˜ †p0i = 0

ha˜ p;a˜ †p0i = dpp0

(2.74)

Using this we get

hap;a†p0i+upvp0

hap;a p0i

ha˜ p;a˜ †p0i = upup0

}

{z0

}

d{zpp0

†

+ vpvp0

ha p;ap0i

+vpvp0

ha p;a p0i

(2.75)

{z0

{zdp

}

(2.76)

= dpp0(up vp) ) up vp = 1:

This equation is solved if we set

up = cosh(fp)

vp = sinh(fp):

(2.77)

The inverse relations are then given by

†

(2.78)

ap = upap

vpa p

†

˜ †

(2.79)

ap = upap

vpa p:

If we now deﬁne ep = ep +gn and insert (2.78) and (2.79) into the HAMILTONian

(2.70) we get

p6=0

†

H =

gn + å

apa

p + a

pap

(2.80)

epapap + gn

2 ˜ † ˜

†

˜ † ˜

†

p6=0 n h

p + a

gn +

ep upapap + vpa

upvp apa

pap

gnhu2pa˜ †pa˜ † 2p +† v†2pa˜ pa˜ p vpu† p a˜ †pa˜ p +†a˜ pa˜ †

+ upa˜ pa˜ p + vpa˜ pa˜ p upvp a˜ pa˜ p + a˜ pa˜ p io

(2.81)

p6=0

epvp

gnvpup

}

p p

å p p

p6=0

˜ † ˜

a a

u + v

2gnu v

˜ †

†

p6=0

+ å apa

p + a

pap

epupvp +

(2.82)

}

2.3. WEAKLY INTERACTING BOSE GAS					27
Thus fp has to be chosen such that
2	2	˜			(2.83)
gn(up	+ vp)	2epupvp = 0:			(2.83)
The simples way to ﬁnd the angle	f from this equation is to use the relation
4u2pv2p = (u2p + v2p)2 (u2p v2p)2:					(2.84)
		\|	{z	}

We get as the result

u2p

= 2 wp + 1

v2p = 2

wp 1

= ep + gn

= ep (gn)

with the HAMILTONian

†

H =

2 gn +

with

å (wp ep) + å wpapap

p6=0

˜ 2

wp = ep (gn) = (ep

gn)(ep + gn) = ep(ep + 2gn)

+ 2gn :

For p ! 0 we have w p.

(2.85)

(2.86)

(2.87)

(2.88)

(2.89)

Figure 2.2: wp as function of p
More precisely, if we deﬁne 3
pc2 = mgn	v2 =	gn	(2.90)
pc2 = mgn	v2 =	m	(2.90)
		m

3In this context we assume n n0

28						CHAPTER 2. BOSONS
we get	( p2				pc:
wp =	( p2		+ gn	p	pc:	(2.91)
	vp			p	pc

		2m

Reconsidering the requirement for superﬂuidity (2.30) we get

	c =	p	wp	=		6=		:
	c =	p	p	=		6=		:
v		min			v		0		(2.92)

If p pc i.e.

wp =

for p pc

we get

8(gn0)2

wp e˜ p =

2 m

˜ 2

(gn0)2

= gn

(gn0)

gn0

wp + ep

= (gn0)

p pc

2ep

E0 =

åp p2

gn0

p2 )

2 gn0

(gn0)

2 åp (wp + e˜ p

(gn )2

()

(gn

n02

g g2 åp

åp

wp + e˜ p

|{z }

			independent of n
	1			(gn	)2	1		m
=		n02	Γ	0		åp			:
=	2	n02	Γ	2		åp	wp + e˜ p	p2	:

Here Γ is the quantum mechanical scattering amplitude deﬁned by

Γ = g g å m Γ and

p p2

Γ = 4p~2 a: m

The solution is solved iteratively:

Γ = g g2 å m + :::

p p2

(2.93)

(2.94)

(2.95)

(2.96)

(2.97)

(2.98)

(2.99)

(2.100)

2.3. WEAKLY INTERACTING BOSE GAS

The sum (2.97) is convergent with the dominant contribution coming from p pc. Looking at each term of E0 we get

d p3

påpc

d3 p

(2.101)

wp + e˜ p

(2p~)3

wp + ep

gn0

gn0~3

mpc

(2.102)

d3 p

påpc

m c

(2.103)

p/pc

Using this the second term of (2.97) for E0 can be estimated (up to some numerical

constant C resp. C, c.f. (2.106)) as

	2p~2	2		1	(gn0)	2 mpc			(2.104)
E0		an0	C
	m			2			~3
	2	2
=	2p~ an0		1 C˜ qa3n0					:	(2.105)
	m

Here a3n0 is our old small parameter (cf. (1.1) with a r0). The exact calculation leads to

E0 =	m~	2	an2	"1 +	15	r		#:	(2.106)
							p
	2p				128		a3n

Note, that the equation contains n, not n0.

Lastly the number of particles outside the condensate (i.e. p 6= 0) in the ground state considering a˜ pj0i = 0 is

˜ 2

a†

e˜ p wp

(2.107)

h å

å wp(e˜ p + wp)

p6=0

(gn)2

(2.108)

wp(e˜ p + wp)

vpgn

p6=0

p/pc

1 pc3

1 pc2

with

(2.109)

n na

(2p~)3

g =

4p~2a

(2.110)

This result ( a 2 is not possible by perturbation because in perturbation only integer powers of the interaction constant are possible).

30				CHAPTER 2. BOSONS
By exact calculations we get
håa†papi =	3 nr		= n0 n	with	(2.111)
håa†papi =	3 nr	p	= n0 n	with	(2.111)
	8 na3
a r0 n 31			:		(2.112)

Using the deﬁnition of g (1.83) we calculate the chemical potential in leading order using (2.105) or (2.106)

m =	¶E0		¶E0		= gn0		ng > 0	(2.113)
	¶n		¶n0
because (cf. (2.69))
n0 = n		1		3 r			! n:	(2.114)
						p
				8		na3

This is valid for the ground state at T = 0 and no excitations present.

Figure 2.3: Energy spectra in BOSE condensates. If in the free case e m would become zero n¯ (2.4) cannot be ﬁxed. If a repulsive interaction is present, the average number of particles can be ﬁxed because Eint an2

2.4Mean ﬁeld approximation

Again we note, that

apa†p ! np + 1 1 and	(2.115)
a†pap ! np 1:	(2.116)

2.4. MEAN FIELD APPROXIMATION

If we are interested only in quantities proportional to n, we can neglect [ap;a†p] = 1. Our theory then becomes a classical ﬁeld description:

ˆ		iΦ(r)
	p		with	(2.118)
y† y = n(r):
y(~r) ! y(r) =	n(r)e			(2.117)

This works only if the classical ﬁeld is slowly varying or by using F OURIERtransformation the momenta are small (slow motion). The reason behind this is, that n and Φ behave similar to p and x in ordinary space. The wave function can always be multiplied with a phase without changing the physics:

yn (~r) ! eif yn (~r)				single particle	(2.119)
Ψ(~r1;:::;~rN ) ! eiNf Ψ(~r1;:::;~rN )					(2.120)
ˆ	¶		ˆ	= f which leads to
This implies the operators N = i	¶f and f
fˆ ;Nˆ = i		)		4N4f 1:	(2.121)
But in a sufﬁciently large volume		V one may have N 1 (but still			N	1).
					N¯

Therefor, it follows from (2.121), that in this case f 1. As a result, for such a

¯	¯	are well deﬁned quantities. By dividing our system into blocks
volume N and f
with volume	V	¯	¯
		, we can deﬁne N and f for each block and these quantities vary

slowly from block to block.

If we consider the thermodynamic limit, i.e. N ! ∞ while VN remains ﬁxed we get

hN 1jy(~r)jNi = y(~r)

and i~¶t y

= [H;y]

lim N

H ˆ

) ~¶t y =

:::

y y

)j i

lim N

E N

:::

(

))j i

:::

(

)

(

))h

jyj i

lim

E N

}

=: m

(2.122)

(2.123)

(2.124)

(2.125)

= my:

(2.126)

This differential equation can be solved by

y(~r;t) = e i	mt	y(~r):	(2.127)
	~

ˆ ˆ 0 = ˆ If we replace H by H H

m we absorb this trivial phase in our HAMILTONian.

32						CHAPTER 2. BOSONS
This leads to4 the GROSS-PITAJEWSKI equation (GP):
i~	¶	y =	~2	4y + (gjyj	2	m)y	(2.128)
	¶t		2m

Here jyj2 = n n0. Therefore we will not distinguish between n and n0.

If we are looking for a stationary, homogeneous solution, then all derivatives (in GP) become zero and if y 6= 0 we ﬁnd

m = gn:

(2.129)

Setting the possible phase to 0, we can describe small ﬂuctuations around the ground state by a wave function

y = p		+ dy(~r;t):	(2.130)
	n0

Substituting this wave function into the GP (2.128), taking only terms up to linear order in dy and using (2.129) we get

dy =

dy + gn

(dy + dy ):

(2.131)

~¶t

To solve this, we make the ansatz

dy(~r;t) = A exp(i(

~p ~r

wt)) + B exp( i(

~p ~r

wt))

(2.132)

and insert it into (2.131):

~wA = (ep + gn0)A + gn0B

(2.133)

~wB = (ep + gn0)B + gn0A

(2.134)

Solving this for w we get wp =

˜ 2

(gn0)

= ep + gn0

again.

and ep

Calculating the expectation

value

qand remembering (2.129) we get a functional

for y:

d3r

Efyg = Z

2 gjyj4 mjyj2

(2.135)

2~m jÑyj2 +

= Z

jÑyj2 +

g jyj

gn0

(2.136)

Since the functional does not depend on a space independent F, i.e. Efyg = EfeiΦyg we can choose a phase. For the previous calculations F = 0. Once the phase is chosen, the symmetry is broken, because

y0 6= eiΦy0

(2.137)

If F is a slowly varying function, then the functional will not change much.

4not shown here

2.4. MEAN FIELD APPROXIMATION

Theorem 1 (Goldstone) If a global continuous symmetry is spontaneously broken (the ground state is not invariant under symmetry operations) then there exists

p!0

a soft mode, i.e. a wp with wp ! 0.

We have such a situation. If p ! 0 then ep ! 0 and therefore wp ! 0. This leads (2.131) to B = A and a purely imaginary change in the phase:


dy = Aei() A e i()									(2.138)
= iep				F(~r;t) and (2.130) becomes					(2.139)
		n0
y(~r;t) = p			+ iep			F(~r;t) p		eieΦ	(2.140)
	n0				n0		n0

If we substitute this solution into the deﬁnition of the probability ﬂow (=superﬂuid ﬂow) we get

~j = n~vs =			i~	(y (Ñy) (Ñy )y)		(2.141)
~j = n~vs =			2m	(y (Ñy) (Ñy )y)		(2.141)
~
~		ÑF		with y(~r;t) = pn(~r;t)eiΦ(~r;t):
= n						(2.142)
= n	m					(2.142)
Here ~vs is the velocity of the superﬂuid ﬂow. It is a potential ﬂow:
				~	~
				~vs =	ÑF	(2.143)
				~vs =	ÑF	(2.143)
				m

therefore the rotation Ñ ~vs = 0. If we compare this to solid body rotation with

~	(2.144)
~vSB = W ~r we get
~	(2.145)
Ñ ~vSB = 2W 6= 0:

Therefore the ﬂuid must stay at rest even if the vessel is rotated. On the other hand

and thus rotation must occur To solve this we keep Ñ ~vs

Calculating

Erot = E M~ ~W	(2.146)
(M becomes nonzero) if W is large enough.
= 0 everywhere except for a line where
yjline = 0:	(2.147)

I ~vs dl~	~		I	~		~
	= const = 2pG =			ÑFdl~ =		dF =		2p k:	(2.148)
		m			m		m

34	CHAPTER 2. BOSONS

Figure 2.4: Top view on rotating superﬂuid liquid

Here Γ is the vorticity or circulation and k 2 Z the circulation quantum number.

~
With ~vs ~ef and dl =~ef r df we get for the critical velocity
vs =	~	k	:	(2.149)

m r

This expression becomes inﬁnite for r ! 0 (while being nice for r ! ∞) therefore superﬂuidity has to break down at some distance r x with

vs = vc = m

= m pmgn0

= r

(2.150)

For k = 1 we get

(2.151)

mvc

mgn0

By using (2.151), (1.83) and (1.1) we get for a number of particles in a volume x3 the macroscopic value

nx3		1	1			1:	(2.152)
		2			2
				1
	~		pan 3

and therefore, the length x is much larger than the average interparticle distance. If L is the length of one vortex, we now get for the energy of the vortex

E(k)

d2r

Z d2r rvs

= pnm

R dr

k2 ln

= pn

and for the angular momentum

M(k)	= Z	~		k2p Zx	R
		d2r rvsr = mn			dr r pn~kR2:
L			m

(2.153)

(2.154)

(2.155)

2.4. MEAN FIELD APPROXIMATION

Figure 2.5: Radial part f (r) of the wave function in superﬂuid B OSE gas

Conclusions:

If M is ﬁxed then we either have k repeated vortices or one vortex with circulation k5. Since kE(1) < E(k) it is energetically favorable to have vortices with k = 1 only.

The critical velocity, when at least one vortex exists, is with

	~ ~		!				(2.156)
Erot = E (M ΩC) = 0						:
ΩC	= M(1) = mR2 ln			x			(2.157)
		E(1)	~		R

ΩC is very small, usually Ω ΩC and many vortices exist. They repel each other and a lattice is created where phonons can be observed.

Figure 2.6: Many vortices in superﬂuid B OSE gas

5or a proper combination of both

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