[PPT] - Integrable states, exact overlaps, and the Boundary Yang-Baxter PowerPoint Presentation

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Integrable states, exact overlaps, and the Boundary Yang-Baxter relation

Balázs Pozsgay

„Premium Postdoctoral Program” Hungarian Academy of Sciences and BME Statistical Field Theory Research group, Budapest University of Technology and Economics

RAQIS, Annecy, 13. September 2018

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Work with Lorenzo Piroli (SISSA) and Eric Vernier (Oxford)

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Outline

Non-equilibrium dynamics: |Ψ0 → |Ψ(t) = e−iHt|Ψ0 → O(t) =? Overlaps: Ψ0|{λ}N Integrable initial states |Ψ0 Relation to boundary integrability Exact solutions for quantum quenches (root densities ρ(λ))

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Outline

Non-equilibrium dynamics: |Ψ0 → |Ψ(t) = e−iHt|Ψ0 → O(t) =? Overlaps: Ψ0|{λ}N Integrable initial states |Ψ0 Relation to boundary integrability Exact solutions for quantum quenches (root densities ρ(λ))

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Outline

Non-equilibrium dynamics: |Ψ0 → |Ψ(t) = e−iHt|Ψ0 → O(t) =? Overlaps: Ψ0|{λ}N Integrable initial states |Ψ0 Relation to boundary integrability Exact solutions for quantum quenches (root densities ρ(λ))

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Outline

Non-equilibrium dynamics: |Ψ0 → |Ψ(t) = e−iHt|Ψ0 → O(t) =? Overlaps: Ψ0|{λ}N Integrable initial states |Ψ0 Relation to boundary integrability Exact solutions for quantum quenches (root densities ρ(λ))

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Outline

Non-equilibrium dynamics: |Ψ0 → |Ψ(t) = e−iHt|Ψ0 → O(t) =? Overlaps: Ψ0|{λ}N Integrable initial states |Ψ0 Relation to boundary integrability Exact solutions for quantum quenches (root densities ρ(λ))

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Integrable initial states

Overlaps display the pair structure, are non-zero only if {λ}N = {λ+, −λ+}N/2

r

{λ+, −λ+}(N−1)/2 ∪ {0} They take a factorized form. In simple cases |Ψ0|{λ}N|2 {λ}N|{λ}N =

N/2

j=1

v(λ+

j ) × det G +

det G − More generally some linear combination of these, with different v-functions

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Integrable initial states

Overlaps display the pair structure, are non-zero only if {λ}N = {λ+, −λ+}N/2

r

{λ+, −λ+}(N−1)/2 ∪ {0} They take a factorized form. In simple cases |Ψ0|{λ}N|2 {λ}N|{λ}N =

N/2

j=1

v(λ+

j ) × det G +

det G − More generally some linear combination of these, with different v-functions

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Integrable initial states

In spin-1/2 XXZ: all two site states |Ψ = ⊗L/2

j=1|ψ,

|ψ ∈ C2 ⊗ C2 In higher spin and higher rank models: A subset of two-site states Matrix Product States (from AdS/CFT) ω(i), i = 1, . . . , N |MPS =

N

i1,...,iL=1

tr0

ω(i1)ω(i2) . . . ω(iL)

|i1, i2, . . . , iL

M de Leeuw, C. Kristjansen, K. Zarembo, I. Buhl-Mortensen, S. Mori, G. Linardopoulos, 2015-2018

Overlaps calculated in SO(6)-symmetric spin chain

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Integrable initial states

In spin-1/2 XXZ: all two site states |Ψ = ⊗L/2

j=1|ψ,

|ψ ∈ C2 ⊗ C2 In higher spin and higher rank models: A subset of two-site states Matrix Product States (from AdS/CFT) ω(i), i = 1, . . . , N |MPS =

N

i1,...,iL=1

tr0

ω(i1)ω(i2) . . . ω(iL)

|i1, i2, . . . , iL

M de Leeuw, C. Kristjansen, K. Zarembo, I. Buhl-Mortensen, S. Mori, G. Linardopoulos, 2015-2018

Overlaps calculated in SO(6)-symmetric spin chain

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Integrable initial states

In spin-1/2 XXZ: all two site states |Ψ = ⊗L/2

j=1|ψ,

|ψ ∈ C2 ⊗ C2 In higher spin and higher rank models: A subset of two-site states Matrix Product States (from AdS/CFT) ω(i), i = 1, . . . , N |MPS =

N

i1,...,iL=1

tr0

ω(i1)ω(i2) . . . ω(iL)

|i1, i2, . . . , iL

M de Leeuw, C. Kristjansen, K. Zarembo, I. Buhl-Mortensen, S. Mori, G. Linardopoulos, 2015-2018

Overlaps calculated in SO(6)-symmetric spin chain

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Integrable initial states

S. Ghoshal and A. B. Zamolodchikov, Int. J. Mod. Phys., A9 3841, hep-th/9306002

Ψ0| |Ψ0

ξn . . . ξ2 ξ1

1 2 3 4 . . . . . . L-1 L

Z = Ψ0|

n

j=1

T(ξj)|Ψ0

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Integrable initial states

S. Ghoshal and A. B. Zamolodchikov, Int. J. Mod. Phys., A9 3841, hep-th/9306002

Ψ0| |Ψ0

ξn . . . ξ2 ξ1

1 2 3 4 . . . . . . L-1 L

Z = Ψ0|

n

j=1

T(ξj)|Ψ0

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|Ψ0 Ψ0|

ξ1 ξ2 . . . ξn

=

K + K − K + K − K + K − K + K − K + K − K + K − K + K −

ξ1 ξ2 . . . ξn

For two site product states in XXZ: boundary transfer matrix τ(u) = Tr0{K+(u)T(u)K−(u)T −1(−u)}

BP, J. Stat. Mech. (2013) P10028

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Integrable MPS

|MPS =

N

i1,...,iL=1

tr0

ω(i1)ω(i2) . . . ω(iL)

|i1, i2, . . . , iL i1 ω i2 ω i3 ω i4 ω Let’s baxterize them! Looking for matrices ψab(u) with a, b = 1 . . . N a b ψ(u) u −u Initial condition: ψab(0) = ω(a)ω(b) ψ(0) = ω ω

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Integrable MPS

|MPS =

N

i1,...,iL=1

tr0

ω(i1)ω(i2) . . . ω(iL)

|i1, i2, . . . , iL i1 ω i2 ω i3 ω i4 ω Let’s baxterize them! Looking for matrices ψab(u) with a, b = 1 . . . N a b ψ(u) u −u Initial condition: ψab(0) = ω(a)ω(b) ψ(0) = ω ω

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Boundary Yang-Baxter relation: Relation to (operator valued) K-matrices K(u) ∈ End(V) K(u)b

a = ψab(u)

Twisted BYB relation in End(V ⊗ V) K2(v)Rt1

21(−u−v)K1(u)R12(u−v) = R21(u−v)K1(u)Rt1 12(−u−v)K2(v)

Soliton non-preserving boundary conditions Extended Twisted Yangian Integrability properties follow

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Boundary Yang-Baxter relation: Relation to (operator valued) K-matrices K(u) ∈ End(V) K(u)b

a = ψab(u)

Twisted BYB relation in End(V ⊗ V) K2(v)Rt1

21(−u−v)K1(u)R12(u−v) = R21(u−v)K1(u)Rt1 12(−u−v)K2(v)

Soliton non-preserving boundary conditions Extended Twisted Yangian Integrability properties follow

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Boundary Yang-Baxter relation: Relation to (operator valued) K-matrices K(u) ∈ End(V) K(u)b

a = ψab(u)

Twisted BYB relation in End(V ⊗ V) K2(v)Rt1

21(−u−v)K1(u)R12(u−v) = R21(u−v)K1(u)Rt1 12(−u−v)K2(v)

Soliton non-preserving boundary conditions Extended Twisted Yangian Integrability properties follow

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Boundary Yang-Baxter relation: Relation to (operator valued) K-matrices K(u) ∈ End(V) K(u)b

a = ψab(u)

Twisted BYB relation in End(V ⊗ V) K2(v)Rt1

21(−u−v)K1(u)R12(u−v) = R21(u−v)K1(u)Rt1 12(−u−v)K2(v)

Soliton non-preserving boundary conditions Extended Twisted Yangian Integrability properties follow

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Z = Ψ0|

n

j=1

T(ξj)|Ψ0 = Tr (TQTM(0))L/2 . ξ1 ξ2 ξ3 ξ4 TBA side: Quench Action method e−E(λ)/T ∼ v(λ+) Data: ρ(h)

x (λ)

ρx(λ) = eεx(λ) = Yx(λ) Y -system relations follow from factorized overlaps QTM side: Fusion of double row transfer matrices Data: Tx(λ), Yx(λ)

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Z = Ψ0|

n

j=1

T(ξj)|Ψ0 = Tr (TQTM(0))L/2 . ξ1 ξ2 ξ3 ξ4 TBA side: Quench Action method e−E(λ)/T ∼ v(λ+) Data: ρ(h)

x (λ)

ρx(λ) = eεx(λ) = Yx(λ) Y -system relations follow from factorized overlaps QTM side: Fusion of double row transfer matrices Data: Tx(λ), Yx(λ)

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Z = Ψ0|

n

j=1

T(ξj)|Ψ0 = Tr (TQTM(0))L/2 . ξ1 ξ2 ξ3 ξ4 TBA side: Quench Action method e−E(λ)/T ∼ v(λ+) Data: ρ(h)

x (λ)

ρx(λ) = eεx(λ) = Yx(λ) Y -system relations follow from factorized overlaps QTM side: Fusion of double row transfer matrices Data: Tx(λ), Yx(λ)

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Z = Ψ0|

n

j=1

T(ξj)|Ψ0 = Tr (TQTM(0))L/2 . ξ1 ξ2 ξ3 ξ4 TBA side: Quench Action method e−E(λ)/T ∼ v(λ+) Data: ρ(h)

x (λ)

ρx(λ) = eεx(λ) = Yx(λ) Y -system relations follow from factorized overlaps QTM side: Fusion of double row transfer matrices Data: Tx(λ), Yx(λ)

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Z = Ψ0|

n

j=1

T(ξj)|Ψ0 = Tr (TQTM(0))L/2 . ξ1 ξ2 ξ3 ξ4 TBA side: Quench Action method e−E(λ)/T ∼ v(λ+) Data: ρ(h)

x (λ)

ρx(λ) = eεx(λ) = Yx(λ) Y -system relations follow from factorized overlaps QTM side: Fusion of double row transfer matrices Data: Tx(λ), Yx(λ)

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Z = Ψ0|

n

j=1

T(ξj)|Ψ0 = Tr (TQTM(0))L/2 . ξ1 ξ2 ξ3 ξ4 TBA side: Quench Action method e−E(λ)/T ∼ v(λ+) Data: ρ(h)

x (λ)

ρx(λ) = eεx(λ) = Yx(λ) Y -system relations follow from factorized overlaps QTM side: Fusion of double row transfer matrices Data: Tx(λ), Yx(λ)

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Results

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XXZ spin-1/2 chain, generic two-site states: |Ψ0 = ⊗L/2

j=1|ψ,

ψjk = (σyK(−η/2))j

k,

j, k = 1, 2 K11(u, α, β, θ) =2(sinh(α) cosh(β) cosh(u) + cosh(α) sinh(β) sinh(u)) K12(u, α, β, θ) =eθ sinh(2u) K21(u, α, β, θ) =e−θ sinh(2u) K22(u, α, β, θ) =2(sinh(α) cosh(β) cosh(u) − cosh(α) sinh(β) sinh(u)).

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XXZ spin-1/2 chain, generic two-site states: |Ψ0 = ⊗L/2

j=1|ψ,

ψjk = (σyK(−η/2))j

k,

j, k = 1, 2 K11(u, α, β, θ) =2(sinh(α) cosh(β) cosh(u) + cosh(α) sinh(β) sinh(u)) K12(u, α, β, θ) =eθ sinh(2u) K21(u, α, β, θ) =e−θ sinh(2u) K22(u, α, β, θ) =2(sinh(α) cosh(β) cosh(u) − cosh(α) sinh(β) sinh(u)).

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Y-system: Yj(u + iη/2)Yj(u − iη/2) = (1 + Yj−1(u))(1 + Yj+1(u)) Solution for Y1: 1 + Y1(λ) = N(λ + iη/2)N(λ − iη/2) χ(λ) where N(λ) = Tr

K+(λ + η/2)K−(λ − η/2)
and

χ = 16 v s

ηv c η

v s

η/2v c η/2

v s

αv s α∗v c βv c β∗

with v s

κ(λ) = sin(λ + iκ) sin(λ − iκ)

v c

κ(λ) = cos(λ + iκ) cos(λ − iκ)

L. Piroli, BP, E. Vernier, arXiv:1611.06126

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Y-system: Yj(u + iη/2)Yj(u − iη/2) = (1 + Yj−1(u))(1 + Yj+1(u)) Solution for Y1: 1 + Y1(λ) = N(λ + iη/2)N(λ − iη/2) χ(λ) where N(λ) = Tr

K+(λ + η/2)K−(λ − η/2)
and

χ = 16 v s

ηv c η

v s

η/2v c η/2

v s

αv s α∗v c βv c β∗

with v s

κ(λ) = sin(λ + iκ) sin(λ − iκ)

v c

κ(λ) = cos(λ + iκ) cos(λ − iκ)

L. Piroli, BP, E. Vernier, arXiv:1611.06126

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For the overlaps we get |Ψ0|{±λ+}N/2|2 = |eθ(L−2N)| sinhL(η) |N|L/2

N/2

j=1

u(λ+

j ) ×

detN/2 G +

jk

detN/2 G −

jk

, with u(λ) = v s

αv s α∗v c βv c β∗

v s

η/2v c η/2v s 0v c

, and v s

κ(λ) = sin(λ + iκ) sin(λ − iκ)

v c

κ(λ) = cos(λ + iκ) cos(λ − iκ)

BP, arxiv:1801.03838

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SU(3)-symmetric chain, R(u) = P + u, H =

j Pj,j+1

Nested Bethe vectors: |{λ}N, {µ}M Integrable two-site states: Symmetric or anti-symmetric Main example: ψjk = δjk Overlaps: |Ψδ|{λ}N, {µ}M|2 Ψδ|Ψδ = 3−L/2

N/2

j=1

u(λ+

j ) M/2

j=1

u(µ+

j ) × det G+

det G− u(λ) = λ2 λ2 + 1/4

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SU(3)-symmetric chain, R(u) = P + u, H =

j Pj,j+1

Nested Bethe vectors: |{λ}N, {µ}M Integrable two-site states: Symmetric or anti-symmetric Main example: ψjk = δjk Overlaps: |Ψδ|{λ}N, {µ}M|2 Ψδ|Ψδ = 3−L/2

N/2

j=1

u(λ+

j ) M/2

j=1

u(µ+

j ) × det G+

det G− u(λ) = λ2 λ2 + 1/4

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SU(3)-symmetric chain, R(u) = P + u, H =

j Pj,j+1

Nested Bethe vectors: |{λ}N, {µ}M Integrable two-site states: Symmetric or anti-symmetric Main example: ψjk = δjk Overlaps: |Ψδ|{λ}N, {µ}M|2 Ψδ|Ψδ = 3−L/2

N/2

j=1

u(λ+

j ) M/2

j=1

u(µ+

j ) × det G+

det G− u(λ) = λ2 λ2 + 1/4

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Solution of the quench problem: ρ(λ)

j

(u), ρ(µ)

j

(u), Y (λ)

j

(u), Y (µ)

j

(u) Y -system: Y (λ)

j

(u + iη/2)Y (λ)

j

(u − iη/2) = (1 + Y (λ)

j−1(u))(1 + Y (λ) j+1(u))

1 + 1/Y (µ)

j

(u) Y (µ)

j

(u + iη/2)Y (µ)

j

(u − iη/2) = (1 + Y (µ)

j−1(u))(1 + Y (µ) j+1(u))

1 + 1/Y (λ)

j

(u) 1 + Y (λ)

1

(u) = 1 + Y (µ)

1

(u) = 3u2 + 1/4 u2 ρ(λ)

1 (u) =

16(80u4 + 168u2 + 53) (4u2 + 1)(8u2 + 3)(4u2 + 9)2 ρ(µ)

1 (u) = (4u2 + 1)(5u4 + 18u2 + 8)

(u2 + 1)2(u2 + 4)2(8u2 + 3)

L. Piroli, BP, E. Vernier, arXiv:18xx.xxxxx

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Solution of the quench problem: ρ(λ)

j

(u), ρ(µ)

j

(u), Y (λ)

j

(u), Y (µ)

j

(u) Y -system: Y (λ)

j

(u + iη/2)Y (λ)

j

(u − iη/2) = (1 + Y (λ)

j−1(u))(1 + Y (λ) j+1(u))

1 + 1/Y (µ)

j

(u) Y (µ)

j

(u + iη/2)Y (µ)

j

(u − iη/2) = (1 + Y (µ)

j−1(u))(1 + Y (µ) j+1(u))

1 + 1/Y (λ)

j

(u) 1 + Y (λ)

1

(u) = 1 + Y (µ)

1

(u) = 3u2 + 1/4 u2 ρ(λ)

1 (u) =

16(80u4 + 168u2 + 53) (4u2 + 1)(8u2 + 3)(4u2 + 9)2 ρ(µ)

1 (u) = (4u2 + 1)(5u4 + 18u2 + 8)

(u2 + 1)2(u2 + 4)2(8u2 + 3)

L. Piroli, BP, E. Vernier, arXiv:18xx.xxxxx

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Solution of the quench problem: ρ(λ)

j

(u), ρ(µ)

j

(u), Y (λ)

j

(u), Y (µ)

j

(u) Y -system: Y (λ)

j

(u + iη/2)Y (λ)

j

(u − iη/2) = (1 + Y (λ)

j−1(u))(1 + Y (λ) j+1(u))

1 + 1/Y (µ)

j

(u) Y (µ)

j

(u + iη/2)Y (µ)

j

(u − iη/2) = (1 + Y (µ)

j−1(u))(1 + Y (µ) j+1(u))

1 + 1/Y (λ)

j

(u) 1 + Y (λ)

1

(u) = 1 + Y (µ)

1

(u) = 3u2 + 1/4 u2 ρ(λ)

1 (u) =

16(80u4 + 168u2 + 53) (4u2 + 1)(8u2 + 3)(4u2 + 9)2 ρ(µ)

1 (u) = (4u2 + 1)(5u4 + 18u2 + 8)

(u2 + 1)2(u2 + 4)2(8u2 + 3)

L. Piroli, BP, E. Vernier, arXiv:18xx.xxxxx

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χ(k)

=

N

i1,...,iL=1

tr0 [Si1Si2 . . . SiL] |i1, i2, . . . , iL Here Sj are the SU(2)-generators in the k-dimensional representation

M. de Leeuw, C. Kristjansen, G. Linardopoulos, Phys.Lett. B781 (2018) 238

Qλ(x) =

N

j=1

(x − λj) Qµ(x) =

M

j=1

(x − µj) |χ(k)|{λ}N, {µ}M|2 = (Tk−1)2 Qλ(0)Qλ(i/2) Qµ(0)Qµ(i/2) × det G+ det G− Tn = Qλ(i(n + 1)/2)

n/2

a=−n/2

(ia)L Qµ(ia) Qλ(i(a + 1/2))Qλ(i(a − 1/2))

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χ(k)

=

N

i1,...,iL=1

tr0 [Si1Si2 . . . SiL] |i1, i2, . . . , iL Here Sj are the SU(2)-generators in the k-dimensional representation

M. de Leeuw, C. Kristjansen, G. Linardopoulos, Phys.Lett. B781 (2018) 238

Qλ(x) =

N

j=1

(x − λj) Qµ(x) =

M

j=1

(x − µj) |χ(k)|{λ}N, {µ}M|2 = (Tk−1)2 Qλ(0)Qλ(i/2) Qµ(0)Qµ(i/2) × det G+ det G− Tn = Qλ(i(n + 1)/2)

n/2

a=−n/2

(ia)L Qµ(ia) Qλ(i(a + 1/2))Qλ(i(a − 1/2))

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χ(k)

=

N

i1,...,iL=1

tr0 [Si1Si2 . . . SiL] |i1, i2, . . . , iL Here Sj are the SU(2)-generators in the k-dimensional representation

M. de Leeuw, C. Kristjansen, G. Linardopoulos, Phys.Lett. B781 (2018) 238

Qλ(x) =

N

j=1

(x − λj) Qµ(x) =

M

j=1

(x − µj) |χ(k)|{λ}N, {µ}M|2 = (Tk−1)2 Qλ(0)Qλ(i/2) Qµ(0)Qµ(i/2) × det G+ det G− Tn = Qλ(i(n + 1)/2)

n/2

a=−n/2

(ia)L Qµ(ia) Qλ(i(a + 1/2))Qλ(i(a − 1/2))

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Baxterization: ψab(u) = (1 + u)SaSb − uSbSa − u2δab Fusion, etc. to be done!

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Solutions

For SO(N) we have R(u) = (u + c)P + u(u + c)I − uK, c = N−2

2

Completely SO(N) symmetric solution: ψab(u) = (−2u + c)γaγb + 2u(1 + u)δab Solution with SO(D) ⊗ SO(N − D) Initial condition: ω(j) =

γj,

for j = 1, . . . , D for j = D + 1, . . . , N Solution: ψab(u) = (−2u + c)γaγb + u(2u + D − 2c)δab ψaI(u) = ψIa = 0 ψIJ(u) = u(D − 2u)δIJ Relevant to AdS/CFT

M. de Leeuw, C. Kristjansen, G. Linardopoulos, J.Phys. A50 (2017) 254001

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Solutions

For SO(N) we have R(u) = (u + c)P + u(u + c)I − uK, c = N−2

2

Completely SO(N) symmetric solution: ψab(u) = (−2u + c)γaγb + 2u(1 + u)δab Solution with SO(D) ⊗ SO(N − D) Initial condition: ω(j) =

γj,

for j = 1, . . . , D for j = D + 1, . . . , N Solution: ψab(u) = (−2u + c)γaγb + u(2u + D − 2c)δab ψaI(u) = ψIa = 0 ψIJ(u) = u(D − 2u)δIJ Relevant to AdS/CFT

M. de Leeuw, C. Kristjansen, G. Linardopoulos, J.Phys. A50 (2017) 254001

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Solutions

For SO(N) we have R(u) = (u + c)P + u(u + c)I − uK, c = N−2

2

Completely SO(N) symmetric solution: ψab(u) = (−2u + c)γaγb + 2u(1 + u)δab Solution with SO(D) ⊗ SO(N − D) Initial condition: ω(j) =

γj,

for j = 1, . . . , D for j = D + 1, . . . , N Solution: ψab(u) = (−2u + c)γaγb + u(2u + D − 2c)δab ψaI(u) = ψIa = 0 ψIJ(u) = u(D − 2u)δIJ Relevant to AdS/CFT

M. de Leeuw, C. Kristjansen, G. Linardopoulos, J.Phys. A50 (2017) 254001

SLIDE 47

Solutions

For SO(N) we have R(u) = (u + c)P + u(u + c)I − uK, c = N−2

2

Completely SO(N) symmetric solution: ψab(u) = (−2u + c)γaγb + 2u(1 + u)δab Solution with SO(D) ⊗ SO(N − D) Initial condition: ω(j) =

γj,

for j = 1, . . . , D for j = D + 1, . . . , N Solution: ψab(u) = (−2u + c)γaγb + u(2u + D − 2c)δab ψaI(u) = ψIa = 0 ψIJ(u) = u(D − 2u)δIJ Relevant to AdS/CFT

M. de Leeuw, C. Kristjansen, G. Linardopoulos, J.Phys. A50 (2017) 254001

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Solutions

For SO(N) we have R(u) = (u + c)P + u(u + c)I − uK, c = N−2

2

Completely SO(N) symmetric solution: ψab(u) = (−2u + c)γaγb + 2u(1 + u)δab Solution with SO(D) ⊗ SO(N − D) Initial condition: ω(j) =

γj,

for j = 1, . . . , D for j = D + 1, . . . , N Solution: ψab(u) = (−2u + c)γaγb + u(2u + D − 2c)δab ψaI(u) = ψIa = 0 ψIJ(u) = u(D − 2u)δIJ Relevant to AdS/CFT

M. de Leeuw, C. Kristjansen, G. Linardopoulos, J.Phys. A50 (2017) 254001

SLIDE 49

Solutions

For SO(N) we have R(u) = (u + c)P + u(u + c)I − uK, c = N−2

2

Completely SO(N) symmetric solution: ψab(u) = (−2u + c)γaγb + 2u(1 + u)δab Solution with SO(D) ⊗ SO(N − D) Initial condition: ω(j) =

γj,

for j = 1, . . . , D for j = D + 1, . . . , N Solution: ψab(u) = (−2u + c)γaγb + u(2u + D − 2c)δab ψaI(u) = ψIa = 0 ψIJ(u) = u(D − 2u)δIJ Relevant to AdS/CFT

M. de Leeuw, C. Kristjansen, G. Linardopoulos, J.Phys. A50 (2017) 254001

SLIDE 50

Thank you for the attention!!!

SLIDE 51

Overlaps

|Ψ0|{λ}N|2 {λ}N|{λ}N =

N/2

j=1

v(λ+

j ) × det G +

det G − Interpretation of the Gaudin-like determinants: Non-trivial density of states due to the pair structure In QFT: M. Kormos, BP, arXiv:1002.2783 In the XXZ chain in the TDL: BP, arXiv:1801.03838 1 = Ψ0|Ψ0 =

|Ψ0|{λ+, −λ+}N/2|2

BEC overlap in Lieb-Liniger model: Proof from coordinate Bethe Ansatz

SLIDE 52

Overlaps

|Ψ0|{λ}N|2 {λ}N|{λ}N =

N/2

j=1

v(λ+

j ) × det G +

det G − Interpretation of the Gaudin-like determinants: Non-trivial density of states due to the pair structure In QFT: M. Kormos, BP, arXiv:1002.2783 In the XXZ chain in the TDL: BP, arXiv:1801.03838 1 = Ψ0|Ψ0 =

|Ψ0|{λ+, −λ+}N/2|2

BEC overlap in Lieb-Liniger model: Proof from coordinate Bethe Ansatz

SLIDE 53

Overlaps

|Ψ0|{λ}N|2 {λ}N|{λ}N =

N/2

j=1

v(λ+

j ) × det G +

det G − Interpretation of the Gaudin-like determinants: Non-trivial density of states due to the pair structure In QFT: M. Kormos, BP, arXiv:1002.2783 In the XXZ chain in the TDL: BP, arXiv:1801.03838 1 = Ψ0|Ψ0 =

|Ψ0|{λ+, −λ+}N/2|2

BEC overlap in Lieb-Liniger model: Proof from coordinate Bethe Ansatz

SLIDE 54

Decomposition: ψ(u) = u −u v K(u)

SLIDE 55

. ξ1 ξ2 ξ3 ξ4 ψ(u1) ψ(u2) ψ(u3) ψ(−u1) ψ(−u2) ψ(−u3) −u1 u1 −u2 u2 −u3 u3 y x R(x − y)

SLIDE 56

. ξ1 ξ2 ξ3 ξ4 ψ(u1) ψ(u2) ψ(u3) ψ(−u1) ψ(−u2) ψ(−u3) −u1 u1 −u2 u2 −u3 u3 y x R(x − y) T (u) =

N

a1,a2,b1,b2=1

ψ(1)

a2a1(u) ⊗

Ta2b2(−u)Ta1b1(u)
⊗ ψ(2)

b2b1(−u)

SLIDE 57

. ξ1 ξ2 ξ3 ξ4 ψ(u1) ψ(u2) ψ(u3) ψ(−u1) ψ(−u2) ψ(−u3) −u1 u1 −u2 u2 −u3 u3 y x R(x − y) T (u) =

N

a1,a2,b1,b2=1

ψ(1)

a2a1(u) ⊗

Ta2b2(−u)Ta1b1(u)
⊗ ψ(2)

b2b1(−u)