Spontaneous Symmetry Breaking in Superfluid Helium-4

0.13 MB·English

by  Junpei Harada

Spontaneous Symmetry Breaking in Superfluid Helium-4
Junpei Harada∗
Asia Pacific Center for Theoretical Physics, Pohang, 790-784, Korea
Physics division, History of Science, Kyoto University, Kyoto, 606-8501, Japan
(Dated: January 16, 2007)
Wederiveananalyticalexpressionforacriticaltemperatureofspontaneoussymmetrybreakingin
7
arepulsivehard-coreinteractingBosesystem. Weshowthatthecriticaltemperatureofspontaneous
0
symmetry breaking in a hard-core interacting Bose system is determined by the three physical
0
2 parameters: the density of Bose liquid at absolute zero (ρ0), the mass (m) and the hard sphere
diameter (σ) of a boson. The formula that we have derived is Tc = ρ0π~2σ/m2kB. We report Tc
n of liquid helium-4 is 2.194 K, which is significantly close to the λ-temperature of 2.1768 K. The
a deviation between thepredicted and experimental values of theλ-temperature is less than 1%.
J

5 PACSnumbers: 67.40.-w
2

] 1. Theyear2008isthe100thanniversaryof“LiquidHe- Although London’s work was revolutionary progress,
r
e lium Year”,in which Heike KamerlinghOnnes produced thereis adifference ofapproximately0.95Kbetweenhis
h liquid helium. We are in a truly memorial moment in theoretical prediction and the experimental value. This
t
o the history of low temperature physics. Liquid helium deviation is approximately 44% and it is not negligible.
. has been studied for the past 100 years, and there is Itiswidelybelievedthatthisdiscrepancyarisesfromthe
t
a no question that liquid helium is of central importance neglect of interactions between helium atoms in his the-
m
in this field. (I would like to turn readers’ attention to oretical calculation. This conjecture is quite reasonable
- Ref.[1]. ItpresentsthefirstcompleteEnglishtranslation because liquid helium is not an ideal Bose gas but a re-
d
oftheinauguralspeechofHeikeKamerlinghOnnesatthe pulsive interacting Bose system. However, in general, it
n
o University of Leiden in 1882. Although his speech is not is difficult to consider interactions between atoms in an-
c related to helium, it is quite interesting and gives us a alytical calculations of the BEC. Although this problem
[ lesson. I believe that it attracts the interest of readers.) has been discussed [4], it remains very important.

2 In spite of greatefforts,an important problemstill re- In this paper, we study the “Why 2.2 K?” problem
v mains to be solved. The problem is as follows. Liquid
via an alternative approach. The key concept in our ap-
5 helium-4 undergoes a phase transition, known as the λ-
proach is spontaneous symmetry breaking. We derive an
6
transition, to a superfluid phase at the λ-temperature.
3 analytical formula for a critical temperature of sponta-
The experimental value of the λ-temperature T is ap-
1 λ neous symmetry breaking in a repulsive hard-core inter-
0 proximatelyequalto2.2K.The problemisquite simple. actingBosesystem. Weshowthatacriticaltemperature
7 WhyisT 2.2K?Whatphysicalparametersdetermine
λ ≃ is determined by the threephysicalparameters: the den-
0 the value of the λ-temperature? This is the “Why 2.2
sityofBoseliquidatabsolutezero(ρ ),themass(m)and
/ 0
at K?” problem, which is the subject of the present paper. the hard sphere diameter (σ) of a boson. Our formula
Itisalong-standingdreaminlowtemperaturephysicsto
m predicts that a critical temperature of liquid helium-4 is
derive a formula for the liquid helium-4 λ-temperature.
2.194 K, which is significantly close to the experimental
-
d In 1938, Fritz London proposed that the λ-transition value of the λ-temperature (T = 2.1768 K [5]).
λ
n of liquid helium-4 probably has to be regarded as the
o Bose-Einstein condensation (BEC) [2]. He calculated a 2. Spontaneoussymmetrybreakingisasymmetrybreak-
c critical temperature of the BEC in an ideal Bose gas: ing by the ground state of a system. The symmetry can
:
v bediscreteorcontinuous,andbelocalorglobal. Sponta-
i 2π~2 n 2/3 neous symmetry breaking is a phenomenon that occurs
X
T = , (1)
BE in many systems. In condensed matter physics, ferro-
mk ζ(3/2)
ar B (cid:18) (cid:19) magnetism is a primary example. For superconductiv-
where m is a boson mass, n is a number density of Bose ity, the importance of spontaneous symmetry breaking
gas and ζ(3/2) = 2.612375 . Thus, the Bose-Einstein was firstly emphasized by Yoichiro Nambu [6]. In parti-
···
temperature T is determined by the two physical pa- cle physics, the electroweaksymmetry breaking is estab-
BE
rameters, m and n. London reported T of helium is lishedintheStandardModel. Incosmology,anumberof
BE
3.13 K [3], which is the same order of magnitude as the spontaneous symmetry breaking play an important role
λ-temperatureof2.18K.Hisproposalisquiteimportant, inthe historyofthe earlyuniverse. Furthermore,the re-
because it was the first suggestion that the λ-transition centlyproposedghostcondensation[7]intheaccelerating
was intrinsically related to the Bose-Einstein condensa- universe can be regarded as a kind of spontaneous sym-
tion. Recall that it was not clear in the 1930s whether metry breaking. Thus, spontaneous symmetry breaking
the BEC was an actually physical phenomenon [3]. is a very important concept in modern physics.
2
Spontaneous symmetry breaking is crucial in liquid V
helium-4. In short, the λ-transition is the spontaneous
symmetry breaking. This is the basis of our approach.
We derive a formula for a critical temperature of spon-

taneoussymmetrybreakinginthe frameworkofeffective 0 √µ/λ
field theory described by the order parameters. In the Reϕ
case of liquid helium-4, the order parameter is a macro-
scopic wavefunction ϕ, which is a one-component com-
plex scalar field with both amplitude and phase. We
begin with the theory described by this scalar field ϕ. FIG. 1: The potential V in thecase that µ>0.
3. The Lagrangiandensity (= ) of the system is
L K−V
givenby the nonrelativisticGoldstonemodel ofthe form
4. The ground state of a system is given by solving the
λ 2π~2σ condition ∂ /∂ϕ = 0. First, we consider the case that
= µϕ∗ϕ+ (ϕ∗ϕ)2, λ= . (2) V
V − 2 m a chemical potential µ is negative (µ 0). In this case,
≤
there is only a trivial solution ϕ = 0. This solution is
The kinetic part = i~ϕ∗∂ ϕ + ~2ϕ∗ 2ϕ/2m is not
important here, anKd thereforetwe concent∇rate on the po- the stable ground state, because ∂2V/∂ϕ∗∂ϕ|ϕ=0 ≥ 0.
HencetheglobalU(1)symmetryisunbrokeninthiscase.
tential . InEq.(2),µisthechemicalpotential,λisthe
V Second, we consider the case that µ is positive (µ > 0).
coupling constant for helium interactions and ϕ∗ is the
Inthis case,incontrasttothe casethatµ 0,there is a
complexconjugateofϕ. Theinteractionbetweenhelium ≤
nontrivialsolutioninadditiontoatrivialsolutionϕ=0:
atoms is repulsive at short distances (hard-core interac-
tions). Therefore, the coupling constant λ = 2π~2σ/m µ
ϕ = . (5)
is positive, where m and σ are the mass and the hard | | λ
r
sphere diameter of a helium atom, respectively.
Fig.1showsthepotential inthecasethatµ>0. Inthe
We first consider the physical dimensions of parame- V
present case, the trivial solution ϕ = 0 is unstable and
ters of the potential (2). In units of ~ = 1 and k = 1
B it is not the ground state of the system. The nontrivial
(~ is the reduced Planck’s constant and k is the Boltz-
B solution ϕ = µ/λ is the stable ground state, because
mannconstant),foranyphysicalquantityQ,itsphysical | |
the curvature of the potential at two solutions satisfies
dimension[Q]iswrittenastheproductofthedimensions p
of length and temperature: ∂2 ∂2
V <0, V >0. (6)
∂ϕ∗∂ϕ ϕ=0 ∂ϕ∗∂ϕ |ϕ|=√µ/λ
[Q]=[Length]α[Temperature]β, (3) (cid:12) (cid:12)
Therefore,i(cid:12)fachemicalpotent(cid:12)ialµispositive (µ>0),
(cid:12) (cid:12)
whereαandβ arenumericalconstants. The physicaldi- theglobalU(1)symmetryisspontaneouslybroken. From
mensionofthe Lagrangiandensity is [ ]=[L]−3[T], and above arguments, we reach a following view. A physical
L
that of the scalar field ϕ is [ϕ] = [L]−3/2. (Here [L] and systemundergoesaphasetransitionatacriticaltemper-
[T] represent [Length] and [Temperature], respectively.) ature T , if a chemical potential satisfies the conditions:
c
Hence it is straightforwardto derive the physicaldimen- µ 0 (T T ), µ > 0 (T < T ). Spontaneous symme-
c c
≤ ≥
sions of parameters in Eq. (2): try breaking occurs only if a chemical potential becomes
positive. Therefore,incontrasttothe Bose-Einsteincon-
[µ]=[T], [λ]=[L]3[T], [m]=[L]−2[T]−1, [σ]=[L].(4) densation, spontaneous symmetry breaking never occurs
in an ideal Bose gas, in which a chemical potential is
Note here that in the present case the coupling constant
always negative (µ 0). This is a crucial difference be-
λisadimensionfulparameter,incontrasttoarelativistic ≤
tweenspontaneoussymmetry breakingandthe standard
case in which λ is dimensionless. These physical dimen-
Bose-Einstein condensation.
sionshelpustounderstandthephysicsofliquidhelium-4.
5. Before deriving a formula for a critical temperature,
We next consider the symmetry of Lagrangian. The
we consider the temperature dependence of a chemical
potential of Eq. (2) is invariant under the U(1) phase
V potential µ. Although the temperature dependence of µ
transformation of the field ϕ: ϕ eiθϕ, ϕ∗ e−iθϕ∗.
→ → is notnecessaryfor derivinga formula,it is usefulto un-
Furthermore, if this is a global transformation in which
derstandaphasetransition. Forthis reason,weconsider
the transformation parameter θ does not depend on the
the temperature dependence of µ in the following. The
space and time coordinates, the kinetic part is in-
K quantity ϕ2 is equivalent to the number density n of
variant under the U(1) phase transformation. Hence, | |
superfluid. Therefore, the following relation is satisfied
the system described by the Lagrangian density has
L below the λ-temperature
aglobalU(1) O(2)symmetry. ThisglobalU(1)symme-
≃
try is spontaneously broken at low temperature as you µ ρ
ϕ = =√n= s, (7)
see below. | | λ m
r r
3
whereρ isthesuperfluiddensityandmisabosonmass.
From thsis relation, the superfluid density is given by 1 - (T/Tλ)5
1 1 - (T/Tλ)6
mµ 1 - (T/Tλ)7
ρs = . (8) Ref.[5]
λ 0.8
1
This expression indicates that the temperature depen- ρ
dence of µ is obtained from that of ρs. ρ/s0.6 0.9
Here we take a minimal model:
0.4
0.8
6
ρ T
s
=1 , (9)
ρ − T 0.2
(cid:18) λ(cid:19) 0.71 1.1 1.2 1.3 1.4 1.5 1.6 1.7
where ρ(=ρ +ρ ) is the total density of liquid helium,
n s 0
and ρ , ρ are the normal fluid and superfluid densities, 0 0.5 1 1.5 2 T 2.5
n s Temperature [K] λ
respectively. It should be emphasized that this model
doesnotaffectaformulaforacriticaltemperature. Fig.2 FIG.2: Superfluiddensityratioforliquidhelium-4asafunc-
shows the superfluid density ratio as a function of tem- tion oftemperature. Theinset showsthetemperatureregion
perature. Two functions 1 (T/Tλ)5 and 1 (T/Tλ)7 1.0 K ≤ T ≤ 1.7 K. Key: (dotted line) 1−(T/Tλ)5; (solid
are plotted, for comparison.−Fig. 2 shows th−at Eq. (9) line) 1−(T/Tλ)6; (dashed line) 1−(T/Tλ)7; (+) the data
from Ref. [5].
and experimental values are consistent except near the
λ-transition, at which the superfluid density ratio is of
the form (1 T/T )2/3. The form of Eq. (9) is the same V
λ
−
asthatoftheBose-EinsteincondensationinanidealBose
gas, 1 (T/T )3/2. Although the exponent “6” has not
c
− T>T
yet been derived, we do not consider this problem here. λ
From Eqs. (8) and (9), we obtain the expression
T=T
0 λ
T 6 λρ Reϕ
µ=µ0 1 , µ0 = 0, (10) T<Tλ
−(cid:18)Tλ(cid:19) ! m
T=0
whereµ isachemicalpotentialatabsolutezeroandρ is
0 0
the total density at absolute zero. We have used the ap-
FIG.3: Theschematictemperaturedependenceofthepoten-
proximationρ ρ inEq.(10),becausethetotaldensity
≃ 0 tial V for a second-order phasetransition.
ρ is approximately a constant below the λ-temperature.
Therefore, Eq.(10) is a good approximationat low tem-
perature (T < 1.7 K). Fig. 3 shows the schematic tem- where n0 = ρ0/m is the number density of Bose liquid
peraturedependence ofthe potential . ForT >Tλ,the atabsolutezero. Thisquantityµ0n0/2representsthere-
V
potentialhasonlyoneminimum atϕ=0 andthe curva- quiredenergy density to recoverthe globalU(1) symme-
tureatϕ=0ispositive. WhenT =T ,thecurvatureat try of the system. Therefore, the corresponding thermal
λ
the minimum is zero, ∂2 /∂ϕ∗∂ϕϕ=0 = 0. For T <Tλ, energy density kBTcn0 is equivalent to µ0n0/2. Conse-
V |
the curvature at ϕ = 0 is negative and the scalar field quently, we obtain the following relation
rolls down to the ϕ = 0 minimum. Thus, there is no µ
| | 6 k T = 0. (12)
barrier of potential between the minimum at ϕ = 0 and B c 2
ϕ =0, indicating a second-order phase transition.
6|.|W6 e now derive a formula for a criticaltemperature of Fig. 4 shows the present situation. From above argu-
ments, we have the following relations:
spontaneous symmetry breaking. When T =0, the field
ϕ is located at the nonzerominimum ϕ =0. Hence the µ0 ρ0 µ0 2π~2σ
global U(1) symmetry is broken in th|is|6phase. As the kBTc = 2 , m = λ , λ= m . (13)
temperature increases above a critical temperature, the
Therefore,acriticaltemperatureofspontaneoussymme-
potentialhas only one minimum atϕ=0 and the global
try breaking in a repulsive interacting Bose system is:
U(1)symmetryisrecovered. Fromtheseobservations,we
reach a following conclusion; the depth of the potential ρ0π~2σ
T = , (14)
well at absolute zero determines a critical temperature. c m2kB
When T =0, the depth of the potential well is given by where ρ is the density of Bose liquid at absolute zero,
0
µ n m and σ are the mass and the hard sphere diame-
0 0
(ϕ = µ0/λ)= , (11) ter of a boson, respectively. Substituting the values of
−V | | 2
p
4
V/n 2.22
0
T=0
2.21 Eq. (14)
2.2
K]
√µ/λ ure [2.19 2.194 +1%
0 0 Reϕ mperat2.18 2.1768 -0.8%
kBTC λ-te2.17 -1%
2.16
-µ/2
0 2.15 2.639
2.592 2.618 2.644
FIG. 4: The absolute zero potential per numberdensity. 2.124.58 2.6 2.62 2.64 2.66
the hard sphere diameter of a helium atom [Å]
liquid helium-4 into Eq. (14), ρ = 0.1451 g/cm3 [5], FIG. 5: The λ-temperature as a function of the hard sphere
0
m=6.6465 10−24 g, and σ =2.639˚A [8, 9], we obtain diameter σ of a helium atom.
×
the value T = 2.194 K. This prediction is significantly
c
close to the experimental value of the λ-temperature
cable to the present study, because it is not sufficiently
2.1768K[5]. Thedeviation∆betweenthepredictedand
a good approximation. For these reasons, we plot the
experimentalvaluesoftheλ-temperatureislessthan1%:
λ-temperature as a function of σ. Fig. 5 shows that the
T(theory) T(exp) deviationbetweenthepredictedandexperimentalvalues
∆ 100 c − λ 0.8%<1%. (15) is less than 1% for the region 2.592 ˚A σ 2.644 ˚A.
≡ × Tλ(exp) ≃ 7. In conc±lusion, we have derived an≤ana≤lytical for-
mulaforacriticaltemperatureofspontaneoussymmetry
Thus, our prediction is much closer to the experimen-
breakinginarepulsiveinteractingBosesystem. Thefor-
tal value than London’s prediction. It is because in the
mulathatwehavederived,Eq.(14),isasimpleanalytical
present approach the repulsive interactions between he-
expressiontopredictT 2.2K.Wehopethatthiswork
lium atoms are included in the derivation of a formula. λ ≃
contributes to progress of low temperature physics.
While the Bose-Einstein temperature T in an ideal
BE
Bose gas is determined by the two physical parameters,
ourformulaforacriticaltemperature,Eq.(14),is deter-
mined by the three physical parameters: the density of
superfluid (ρ ), the mass (m) andthe hard sphere diam- ∗
0 E-mail:[email protected]
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