Laboratory for Atomic and Solid State Physics, Cornell University · 2019-01-14 · Brad Ramshaw...

Strain and symmetry breaking

Brad Ramshaw Laboratory for Atomic and Solid State Physics,

Cornell University

Symmetry breaking and order parameters

•  Disordered state preserves symmetry. •  Different symmetry-breaking ground states •  Need more information in the ordered state to describe

thermodynamics.

Probing order parameters with fields

•  Field has the same symmetry as the order parameter. •  Susceptibility diverges at Tc.

χ

Probing order parameters with fields

•  What about the AFM? •  “Wrong” field – need Pi,Pi – neutron.

Iron pnictide nematicity

Jiang et al., J. Phys. C. M. (2009), J.H. Chu et al., Science (2012)

+nema'c

Compression strain fields

•  Compression strains – change volume.

𝑥 

𝜖↓𝑥𝑥 

𝜖↓𝑥𝑥 = 𝑑(𝑥− 𝑥↓0 )/𝑑𝑥 

Compression waves

•  Compression strains supported in solids, liquids, and gasses.

Shear strain fields

𝝐=(█𝜖↓𝑥𝑥 &𝜖↓𝑥𝑦 &𝜖↓𝑥𝑧 @𝜖↓𝑦𝑥 &𝜖↓𝑦𝑦 &𝜖↓𝑦𝑧 @𝜖↓𝑧𝑥 &𝜖↓𝑧𝑥 &𝜖↓𝑧𝑧  )

𝑥 

𝑥 

𝑦 

𝜖↓𝑥𝑥 

𝜖↓𝑥𝑦 

𝜖↓𝑥𝑦 = 𝑑(𝑥− 𝑥↓0 )/𝑑𝑦 + 𝑑(𝑦− 𝑦↓0 )/𝑑𝑥 

Shear waves

•  Solids support shear and compression. •  Liquids support only compression.

Strain in lower symmetry systems

M. Rotter et al., PRL (2008)

•  1 compression •  1 shear



Resonant Ultrasound Spectroscopy


Drive at frequency ω Detect transmitted power

Temperature dependence of RUS on PuCoGa5

J. L. Sarrao et al., Nature (2002)

Temperature dependence of RUS on YBa2Cu3O7

A.Shekhter et al. Nature (2013)

T

Temperature dependence of RUS on URu2Si2

Complex spectra from complex strains

Symmetry to the rescue!

𝑖ℏ 𝜕/𝜕𝑡 Ψ(𝒓,𝑡)=− ℏ↑2 /2𝑚 𝛻↑2 Ψ(𝒓,𝑡)+V(r,t)Ψ(𝒓,𝑡)

?

SO(3)

D4h

Square (D4)

1 2

34

1 2

34

1 2

34

4 1

23

1 2

34

2 3

41

1 2

34

3 4

12

1 2

34

2 1

43

1 2

34

4 3

21

1 2

34

3 2

14

1 2

34

1 4

32

x1 x1 x2

I S4+ S2 S4

-

S4

R100 R010

R100

R110 R1-10

R110

Representations of D4

1 2

34

Ψ=(██2@3 @█2@1  )

•  Square is a “representation” of D4 that preserves all symmetries of the group.

•  What if we want to break symmetry? More complicated things on a square?


•  There are minimal, or “irreducible” representations.

1 2

34

1 2

34

Ψ=(██4@3 @█2@1  ) Ψ=(██2@1 @█4@3  )


•  There are minimal, or “irreducible” representations.

Ψ=+1 Ψ=−1

1 2

34

1 2

34


Ψ=+1

1 2

34

1 2

34

1 2

34

1 2

34

1 2

34x1 x2

I S2 S4 R100 R110

Ψ=−1

χ 1 -1 1 1 -1


Ψ=+1 Ψ=−1

x2-y2

B11 2

34

1 2

34

1 2

34

1 2

34

1 2

34x1 x2

I S2 S4 R100 R110

χ 1 -1 1 1 -1


xy

B2

Ψ=+1 Ψ=−1

1 2

34

1 2

34

1 2

34

1 2

34

1 2

34x1 x2

I S2 S4 R100 R110

χ 1 -1 1 -1 1


Ψ=+1

A1

x2+y2z2

1 2

34

1 2

34

1 2

34

1 2

34

1 2

34x1 x2

I S2 S4 R100 R110

χ 1 1 1 1 1


A2

xvy-yvx=Jz

Ψ=+1 Ψ=−1

z

1 2

34

1 2

34

1 2

34

1 2

34

1 2

34x1 x2

I S2 S4 R100 R110

χ 1 1 1 -1 -1


E

Ψ=+1 Ψ=+1

Ψ=−1 Ψ=−1

1 2

34

1 2

34

1 2

34

1 2

34

1 2

34x1 x2

I S2 S4 R100 R110

χ


E

Ψ=(█+1@0 )

Ψ=(█−1@0 )

Ψ=(█0@+1 )

Ψ=(█0@−1 )

(█1&0@0&1 )

(█0&1@1&0 )

(█−1&0@0&−1 )

(█1&0@0&−1 )

(█0&−1@1&0 )

1 2

34

1 2

34

1 2

34

1 2

34

1 2

34x1 x2

I S2 S4 R100 R110

χ


E

Ψ=(█+1@0 )

(xz,yz)(x,y)(Jx,Jy)Ψ=(█−1

@0 )

Ψ=(█0@+1 )

Ψ=(█0@−1 )

tr(█1&0@0&1 )=2

tr(█0&1@1&0 )=0

tr(█−1&0@0&−1 )=-2

tr(█1&0@0&−1 )=0

tr(█0&−1@1&0 )=0

1 2

34

1 2

34

1 2

34

1 2

34

1 2

34x1 x2

I S2 S4 R100 R110

χ

Reducible representation

?

= +

E B1

Character table of D4

1 2

34

1 2

34

1 2

34

1 2

34

1 2

34x1 x2

I S2 S4 R100 R110

A1 1 1 1 1 11 1 1 -1 -1A21 -1 1 1 -1B11 -1 1 -1 1B22 0 -2 0 0ETableis5x5,notacoincidence!

Representations in 3D – a simple example

𝐸 =(█𝐸↓𝑥 @𝐸↓𝑦 @𝐸↓𝑧  )

ℓ=1

𝐸=(█𝐸↓𝑥 @𝐸↓𝑦  )⊕ 𝐸↓𝑧 𝐸↓𝑢 

𝐴↓2𝑢 

SO(3)

D4h

Representations in 3D – a simple example

𝐵 =(█𝐵↓𝑥 @𝐵↓𝑦 @𝐵↓𝑧  )

𝐵=(█𝐵↓𝑥 @𝐵↓𝑦  )⊕ 𝐵↓𝑧 𝐸↓𝑔 

𝐴↓2𝑔 

D4h

Strain in 3D

𝝐=(█𝜖↓𝑥𝑥 &𝜖↓𝑥𝑦 &𝜖↓𝑥𝑧 @𝜖↓𝑦𝑥 &𝜖↓𝑦𝑦 &𝜖↓𝑦𝑧 @𝜖↓𝑧𝑥 &𝜖↓𝑧𝑥 &𝜖↓𝑧𝑧  )

𝝐=(𝜖↓𝑥𝑥 + 𝜖↓𝑦𝑦 )⊕ 𝜖↓𝑧𝑧 ⊕(𝜖↓𝑥𝑥 − 𝜖↓𝑦𝑦 )⊕ 𝜖↓𝑥𝑦 ⊕{𝜖↓𝑥𝑧 , 𝜖↓𝑦𝑧 }𝐴↓1𝑔 

𝐵↓1𝑔 

SO(3)𝝐= ( 𝜖↓𝑥𝑥 + 𝜖↓𝑦𝑦 + 𝜖↓𝑧𝑧 )⊕ {█𝜖↓𝑥𝑦 @𝜖↓𝑥𝑧 @█𝜖↓𝑦𝑧 @𝜖↓𝑥𝑥 − 𝜖↓𝑦𝑦 @𝜖↓𝑥𝑥 − 𝜖↓𝑧𝑧   }

ℓ=0

𝐴↓1𝑔 

𝐵↓2𝑔 

𝐸↓𝑔 

ℓ=2

D4h

Basis of strains

Compressional

Shear

(c11+c12)/2 c33

(c11-c12)/2 c66 c44

(𝜖↓𝑥𝑥 + 𝜖↓𝑦𝑦 )

{𝜖↓𝑥𝑧 , 𝜖↓𝑦𝑧 }

𝜖↓𝑧𝑧 

(𝜖↓𝑥𝑥 − 𝜖↓𝑦𝑦 )

𝜖↓𝑥𝑦 

Strains of different symmetry

Compressional

+

+

+

=

62more

Shear

Symmetry-resolved temperature dependence in PuCoGa5

B. J. Ramshaw et al., PNAS (2015)

A1g

B2g B1g Eg

Symmetry-resolved temperature dependence in URu2Si2

Universal features at Tc

URu2Si2 PuCoGa5

How do order parameters and strains couple?

ℱ=∑𝑘↑▒1/2 𝑐↓𝑘 𝜖↓𝑘↑2  𝑐↓𝑘 = 𝜕↑2 ℱ/𝜕𝜖↓𝑘↑2  

ℱ=∑𝑘↑▒1/2 𝑐↓𝑘 𝜖↓𝑘↑2  

•  What are we measuring? Elastic moduli – curvature of free energy with respect to strain.

•  How does free energy transform?

+𝛽𝜖↑𝜈 𝜂↑𝜌 + 1/2 𝛼(𝑇− 𝑇↓𝑣 )𝜂↑2 + 1/4 𝛾𝜂↑4 

𝐴↓1𝑔 

𝐴↓1𝑔 

𝐴↓1𝑔 

𝐴↓1𝑔 

𝐴↓1𝑔 

𝐴↓1𝑔 


ℱ=∑𝑘↑▒1/2 𝑐↓𝑘 𝜖↓𝑘↑2  𝑐↓𝑘 = 𝜕↑2 ℱ/𝜕𝜖↓𝑘↑2  

ℱ=∑𝑘↑▒1/2 𝑐↓𝑘 𝜖↓𝑘↑2  +𝛽𝜖↑𝜈 𝜂↑𝜌 

+ 1/2 𝛼(𝑇− 𝑇↓𝑣 )𝜂↑2 + 1/4 𝛾𝜂↑4 


ℱ=∑𝑘↑▒1/2 𝑐↓𝑘 𝜖↓𝑘↑2  +𝛽𝜖↑𝜈 𝜂↑𝜌 

+ 1/2 𝛼(𝑇− 𝑇↓𝑣 )𝜂↑2 + 1/4 𝛾𝜂↑4 

•  Order parameters also break other symmetries – time reversal, gauge, translation – that prevent linear coupling.

•  Lowest-order term is often linear in A1g strain, quadratic in OP.

𝛽𝜖↓𝐴↓1𝑔  𝜂↑2  𝛽𝜖↓𝐵↓1𝑔  𝜂↑2 


ℱ=∑𝑘↑▒1/2 𝑐↓𝑘 𝜖↓𝑘↑2  +𝛽𝜖↓𝐴↓1𝑔  𝜂↑2 

+ 1/2 𝛼(𝑇− 𝑇↓𝑣 )𝜂↑2 + 1/4 𝛾𝜂↑4 

𝜕ℱ/𝜕𝜂 =𝛼(𝑇− 𝑇↓𝑣 )𝜂+𝛾𝜂↑3 +2𝛽𝜖↓𝐴↓1𝑔  𝜂=0

𝜂 =0

𝜂 =√𝛼(𝑇↓𝑣 −𝑇)−2𝛽𝜖↓𝐴↓1𝑔  /𝛾  

𝑇> 𝑇↓𝑣 

𝑇< 𝑇↓𝑣 

𝑐↓𝐴↓1𝑔  (𝑇)= 𝜕↑2 ℱ/𝜕𝜖↓𝐴↓1𝑔 ↑2  = 𝑐↓𝐴↓1𝑔  𝑇> 𝑇↓𝑣 

= 𝑐↓𝐴↓1𝑔  −2𝛽↑2 /𝛾 𝑇< 𝑇↓𝑣 

Universal features at Tc

URu2Si2 PuCoGa5

𝑐↓𝐴↓1𝑔  (𝑇)= 𝜕↑2 ℱ/𝜕𝜖↓𝐴↓1𝑔 ↑2  = 𝑐↓𝐴↓1𝑔  𝑇> 𝑇↓𝑣 

= 𝑐↓𝐴↓1𝑔  −2𝛽↑2 /𝛾 𝑇< 𝑇↓𝑣 

Jumps show up in other thermodynamic parameters

K.A. Modic et al., Nat.Comm. (2018)

Unique couplings for two-component order parameters

ℱ=∑𝑘↑▒1/2 𝑐↓𝑘 𝜖↓𝑘↑2  +𝛽𝜖↑𝜈 𝜂↑𝜌 

+ 1/2 𝛼(𝑇− 𝑇↓𝑣 )𝜂↑2 + 1/4 𝛾𝜂↑4 

𝜂↓𝐸 = {𝜂↓𝑥 , 𝜂↓𝑦 } ∴ 𝜂↓𝐸↑2 =( 𝜂↓𝑥↑2 + 𝜂↓𝑦↑2 )⊕( 𝜂↓𝑥↑2 − 𝜂↓𝑦↑2 )

⊕ 𝜂↓𝑥 𝜂↓𝑦 E

A↓1𝑔 

B↓1𝑔 

B↓2𝑔 

Unique couplings for two-component order parameters

ℱ=∑𝑘↑▒1/2 𝑐↓𝑘 𝜖↓𝑘↑2  + 𝛽↓1 𝜖↓𝐵↓1𝑔  𝜂↓𝐸↑2 + 𝛽↓2 𝜖↓𝐵↓2𝑔  𝜂↓𝐸↑2 + 1/2 𝛼(𝑇− 𝑇↓𝑣 )𝜂↑2 + 1/4 𝛾𝜂↑4 +

•  Same mathematics now predicts jumps in cA1g, cB1g, and cB2g.

𝛽𝜖↓𝐴↓1𝑔  𝜂↓𝐸↑2 

URu2Si2

Linear coupling

ℱ=∑𝑘↑▒1/2 𝑐↓𝑘 𝜖↓𝑘↑2  𝑐↓𝑘 = 𝜕↑2 ℱ/𝜕𝜖↓𝑘↑2  

ℱ=∑𝑘↑▒1/2 𝑐↓𝑘 𝜖↓𝑘↑2  +𝛽𝜖↑𝜈 𝜂↑𝜌 

+ 1/2 𝛼(𝑇− 𝑇↓𝑣 )𝜂↑2 + 1/4 𝛾𝜂↑4 

Linear coupling

ℱ=∑𝑘↑▒1/2 𝑐↓𝑘 𝜖↓𝑘↑2  +𝛽𝜖↓𝑘 𝜂↓𝑘 

+ 1/2 𝛼(𝑇− 𝑇↓𝑣 )𝜂↑2 + 1/4 𝛾𝜂↑4 

𝜕ℱ/𝜕𝜂 =𝛼(𝑇− 𝑇↓𝑣 )𝜂+𝛽𝜖↓𝑘 =0+𝛽𝜖↓𝑘 =0

𝜂 =−𝛽𝜖↓𝑘 /𝛼(𝑇− 𝑇↓𝑣 ) 

𝑐↓𝑘 (𝑇)= 𝜕↑2 ℱ/𝜕𝜖↓𝑘↑2  = 𝑐↓𝑘 − 𝛽↑2 /𝛼(𝑇− 𝑇↓𝑣 ) 

Soft A1g moduli in PuCoGa5

B. J. Ramshaw et al., PNAS (2015)

𝑐↓𝑘 (𝑇)= 𝑐↓𝑘 − 𝛽↑2 /𝛼(𝑇− 𝑇↓𝑣 ) 

𝑇↓𝑣 ≈9𝐾Whatdoeslinearcouplingsay?•  Q=0•  Non-magne'c•  C4preserving•  etc…

Valence transition in YbInCu4

B. Kindler et al., PRB (1994), B. J. Ramshaw et al., PNAS (2015)

Mixed plutonium valence in PuCoGa5

J. L. Smith et al., JLCM (1983), J. M. Lawrence et al., Rep. Prog. Phys. (1981), C. H. Booth et al., JESRP (2014)

Valence fluctuations as a mechanism for high-Tc

H. Q. Yuan et al., Science (2003), K. Miyake, J. Phys. C. M. (2008), E.D. Bauer et al., J.Physics C. M. (2012)

Summary

Singlet order parameter

Physics

URu2Si2

PuCoGa5

•  A1g order parameter •  Preserves time reversal •  Q=0 •  Valence fluctuation driven

superconductivty

Laboratory for Atomic and Solid State Physics, Cornell University · 2019-01-14 · Brad Ramshaw...

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