1

VI. TÓPICOS AVANZADOSPuntos Críticos cuánticos. Transiciones de fase cuánticas. Fases de Griffiths.

Frustración: Spin Ice. Fases de Shastry Sutherlnad.. Dyslochinsky-Moriya.

2

Magn. Ord.

Paramagn.

which bed is better to hibernate?

‘ A QCP is defined when a 2nd order transition is driven to T = 0 by a non-thermal parameter ‘ T. Vojta, Annal. Phys. 2000

< 2K : Q vs Therm fluct.

free energy

Quantum critical point like a rainbow end-point

3

keep away from my buried treasure !

0 .0 0 .2 0 .4 0 .6 0 .80

1

2

3

4

5

6CePd

1-x R h

x

TC [

K]

x [conc.]

QCP

Sereni et al, Phys.Rev.B 2007

Magn. Ord.

Paramagn.

4

0.01

0.1

1

10

100

1000

Universe temp.

Adiabatic demagn.

50mK dilution He3/He4 frig.

450mK-Bariloche

Liquid He

Liquid H2

Liquid O2 & N

2

Room (bio) temp.

TC of Fe

Log

(T /

K)

Sereni J Low Temp Phys 2007

Thermal energy scale ~kBT

a) pre-critical region: change of TN,C(x) curvature at ~ 2K

Fingerprints for the proximity of a QCP in exemplary magnetic phase diagrams

0.0 0.2 0.4 0.6 0.8 1.00

2

4

6

8

10

0.6 0.7 0.8 0.9 1.0

0.1

1

0.6 0.7 0.8 0.9 1.0

0.0

0.5

1.0

1.5

2.0

TN

,C [

K ]

x / xcr

CeIn3 - xSn

x

CePd1 - xRh

x

CeCu2 (Ge

1-xSix )2

(a)

(c)

pre-critical

T* C

[K]

Cmax

/ Tχ'

ac

xcr

CePd1-xRh

x

x [Rh conc.]

(b)

TN [K

]

x / xcr

CeIn3-x Sn

x

CeCu2 (Ge

1-xSix )2

“The southernmost

cosmological singularity”

Prof. Dieter Wollhardt”

Bche. Feb.25,2011

5

0.0

0.4

0.8

1.2

1 10

10.0

0.5

1.0

1.5

2.0

0.60

0.500.55 0.47

0.37 0.300.41

0.25

x=0.15

CeIn3-x

Snx

T [K]

T [K]

x=0.50

0.65 0.600.75

0.70

CePd1-x

Rhx

Cm

/ T

[J/

mol

K2 ]

Sereni J Low Temp Phys 2007

0.1 10

1

2

3

Cm

/ T

[J/

mol

K2 ]

0.85

1.0

0.9

0.80.7

0.5 z=0.35

CeAu1-z

Cu5+z

T [K]

b) maximum value of (specific heat) Cmax / T => const.c) Cm / T ~ - log(T/T0)

0.35 0.5 1 2 4 80.0

0.5

1.0

1.5

2.0 x= 0 x=0.05 x=0.1 x=0.2 x=0.3 x=0.4 x=0.5

CeCu2(Si

1-xGe

x)2

T (K)

CP

/ T

(J/

mol

K2 )

6

Entropy

SS = ∫ Cm / T dT from Cm (T) = Q/∆T

A) 3rd. Law of Thermodynamics 3rd. Law of Thermodynamics ::

i) SS(T → 0) => SS0 , usually (but not necessarily) SS0 = 0 ii) no singularity at T = 0, then ∂SS / ∂T = Cm / T → const.

B) Cosequences Cosequences ::

i) Once in the Ground State → no further modifications are possible (singlet: SS0 = 0)ii) In metastable phases (e.g. quenched alloys) → SS0 ≠ 0 , but for t→ ∞: S0 → 0

~ µeVspectrometry

iii) Quantum fluctuationsQuantum fluctuations or frustration: different scenario since ≠ f f (T)

How to determine whether S = or ≠ 0?If SS0 ≠ 0 but unknown, S(T → ∞) is the alternative reference

since all states will be equally occupied

7

AlternativeAlternative:: If SS0 is unknown, S(T → ∞) is the alternative reference

00.0

0.5

1.0

0.2

Paramagn.

Magn. Ord.

∆ S

/ R

Ln2

T / ∆

∆ = splitting of Ce-magnetic levelsin cubic FCC Ce(In,Sn)3

Simplest case:a 2 level system

“Wie hoch ist der plateau ?”

“Wie tief ist die Elbe”

8

Experimental results from Ce(In1-xSnx)3

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.70

2

4

6

8

10

?

QCP

xcr

Anti Ferro.

xcCeIn

3-xSn

x

TN [K

]

x [Sn conc.]

AFI

-4 0 4 80.0

0.2

0.4

0.6

0.8

1.0

0.5 Klimit

∆T = T-TN [K]

MO

0.80

0.4

CeIn3-x

Snx

Cm

/ T

[J /

mol

K2 ]

x 0.450.470.500.600.70Within linear (pre-critical) region

Cm / T curves coincide if plottedvs. ∆T =T-TN (T shifted by TN)

2 4 6 8

2 4 6 80.0

0.2

0.4

0.6

0.8

1.0

CeIn3-xSn

x

Sn conc [x]

x=0.25

0.30

0.47

0.60

0.50

0.37

0.55

0.450.41

T [K]

Cm

/ T

[J/m

olK

2 ]

linear TN (x) dependence: 0.4 < x < 0.65

Thermod. T = 0 shifts towards 0 as TN => 0

Magnetic Specific heat contribution as Cm / T

Pedrazzini, PhD Bariloche 2004

0.4

9

0.0

0.2

0.4

0.6

0.8

1.0-4 0 4 8 12

MO0.80

0.4

CeIn3-x

Snx

∆T = T-TN [K]

Cm

/ T

[J /

mol

K2 ]

x 0.450.470.500.550.600.70

0 4 8 12 16-0.2

0.0

0.2

0.4

0.6

0.8

1.0

Sm [

R L

n2 ]

x=0.41

T [K]

0 4 8 12 16 200.0

0.2

0.4

0.6

0.8

1.0

SM

O

S

m [

R L

n2 ]

∆T + 4 K SS(T) of x = 0.4 as SSm + SSMOMO

Full RLn2 entropy computed from ∆T = - 4K

Computing the Entropy

60% RLn2 – paramag .20% RLn2 - MO20%RLn2 – missing entropy

SS(T) of x = 0.7 as SSm + SSMOMO

60% RLn2 – paramag .. 0% RLn2 - MO40%RLn2 – missing entropy

10

0 1 2 3 4 5 6-50

-40

-30

-20

-10

0

0 1 2 3 4

-4

-2

0

2

0.80

0.70CeIn

3-xSn

x

0.65

0.55

(VT -

V 6K

) /

V

T [K]

0.80

( Vx -

V0.

8 ) /

V

[10

-6 ]

T [K]

TN

0.55

0.650.70

0.0 0.2 0.4 0.6 0.80.0

0.2

0.4

0.6

0.8

1.0

5 10 15 200.0

0.2

0.4

0.6

0.8

1.0

S0 = RLn2 - S

m

Sm

(20K)

Sm /

RLn

2

x [Sn Conc.]

0.38

x = 0.70

x = 0.15

CeIn3-x

Snx

Sm /

RLn

2

T [K]

Sms(x,T) + S0 (x) = RLn 2

Independent observation on thermal expansion:Independent observation on thermal expansion:Since Smeas(T) = ST→0 + S(x,T)from Maxwell relations- ∂S/∂P = ∂V/ ∂T = β (therm. expan.)

V(T) = ∫ β dT thenVmeas = V0 (x) + V(x,T)

V(T) normalizaed at 8K

Beyond x = xcr, Kondo effect takes over

Corresponding Vol. anomalyComputing the Entropy ........

β(T) after Kuechler PHD Darmstad 2003

11

0.1

1

100.0 0.2 0.4 0.6 0.8

0.0 0.2 0.4 0.6 0.8

Therm. vs Q.fluctuations

Pre-critical region

xcr

TI

TN

V 0(x

)

x [Sn conc.]T

N [K

]

S 0 (x)

x [Sn conc.]

Conclusions

S0(x) increases as x → xcr up to 0.38 RLn2

S=RLn2 RLn3/2new degree of freedom introduced by quantum fluctuations between MO & Paramag. statesas x → xcr & T → 0

There is a pre-critical region with Quantum Critical features at T ≤ 2K

Corresponding V0(x) anomalous dependence

100% - 60% => 40%

Alternative scenario:

12

EntropySS = ∫ Cm / T dT from Cm (T) = Q/∆T

A) 3rd. Law of Thermodynamics 3rd. Law of Thermodynamics ::

i) SS(T → 0) => SS0 , usually (but not necessarily) SS0 = 0 ii) no singularity at T = 0, then ∂SS / ∂T = Cm / T → const.

B) Cosequences Cosequences ::

i) Once in the Ground State → no further modifications are possible (singlet: SS0 = 0)ii) In metastable phases (e.g. quenched alloys) → SS0 ≠ 0 , but for t→ ∞: S0 → 0

thermal fluct.

~ µeVspectrometry

for t→ ∞

iii) Quantum fluctuations: different scenario since tuneling ≠ f f (T)

How to evaluate whether S = or ≠ 0?If SS0 ≠ 0 but unknown, S(T → ∞) is the alternative reference

since all states will be equally occupied

however

13

Divergencias (teória) vs.comportamientos observados

Régimen cuántico