Calculation of Generalized Pauli Constraints

Murat Altunbulak

Department of MathematicsDokuz Eylul University

April, 2016

Murat Altunbulak Oxford 2016

Notations

Quantum States

Quantum system A is described by a complex Hilbert space HA,

called state space.

Pure state = unit vector |ψ〉 ∈ HA or projector operator |ψ〉〈ψ|Mixed state = classical mixture of pure states

ρ =∑i

pi |ψi 〉〈ψi | ; pi ≥ 0 ;∑i

pi = 1

ρ is a non-negative (ρ ≥ 0) Hermitian operator with Trρ = 1, calledDensity matrix.

Notations

Quantum States

Quantum system A is described by a complex Hilbert space HA,called state space.

ρ =∑i

pi |ψi 〉〈ψi | ; pi ≥ 0 ;∑i

pi = 1

Notations

Quantum States

Pure state = unit vector |ψ〉 ∈ HA or projector operator |ψ〉〈ψ|

Mixed state = classical mixture of pure states

ρ =∑i

pi |ψi 〉〈ψi | ; pi ≥ 0 ;∑i

pi = 1

Notations

Quantum States

ρ =∑i

pi |ψi 〉〈ψi | ; pi ≥ 0 ;∑i

pi = 1

Notations

Quantum States

ρ =∑i

pi |ψi 〉〈ψi | ; pi ≥ 0 ;∑i

pi = 1

Notations

Superposition Principle

implies that the state space of composite system AB splits into tensorproduct of its components A and B

HAB = HA ⊗HB

Density Matrix of Composite Systems

Density matrix of composite system can be written as linear combination

ρAB =∑α

aαLαA ⊗ LαB

where LαA, LαB are linear operators on HA, HB , respectively.

Notations

Superposition Principle

implies that the state space of composite system AB splits into tensorproduct of its components A and B

HAB = HA ⊗HB

Density Matrix of Composite Systems

Density matrix of composite system can be written as linear combination

ρAB =∑α

aαLαA ⊗ LαB

where LαA, LαB are linear operators on HA, HB , respectively.

Notations

Reduced state

Its reduced matrices are defined by partial traces

ρA =∑α

aαTr(LαB)LαA := TrB(ρAB)

ρB =∑α

aαTr(LαA)LαB := TrA(ρAB)

they are called one-particle reduced density matrices.

Notations

Reduced state

ρA =∑α

ρB =∑α

Notations

Reduced state

ρA =∑α

ρB =∑α

Notations

Reduced state

ρA =∑α

ρB =∑α

Pauli Exclusion Principle

Pauli exclusion principle

Initial form (1925)

state that no two identical fermions may occupy the same quantumstate.

In the language of density matrices it can be stated as followsLet ρN : H⊗N , H⊗N = state space of N electrons.Define ρ = ρN1 + ρN2 + . . .+ ρNN sum of all reduced states ρNi : H. ρis called electron density matrix.In terms of electron density matrix ρ, Pauli principle amounts to〈ψ|ρ|ψ〉 ≤ 1⇐⇒ Specρ ≤ 1

Modern Version (1926)

Heisenberg-Dirac replaced the Pauli exclusion principle byskew-symmetry of multi-electron wave function

which implies that the state space of N electrons shrinks to∧NH ⊂ H⊗N .

This implies the original Pauli principle, because ψ ∧ ψ = 0.

Initial form (1925)

In the language of density matrices it can be stated as follows

Let ρN : H⊗N , H⊗N = state space of N electrons.Define ρ = ρN1 + ρN2 + . . .+ ρNN sum of all reduced states ρNi : H. ρis called electron density matrix.In terms of electron density matrix ρ, Pauli principle amounts to〈ψ|ρ|ψ〉 ≤ 1⇐⇒ Specρ ≤ 1

Initial form (1925)

In the language of density matrices it can be stated as followsLet ρN : H⊗N , H⊗N = state space of N electrons.

Define ρ = ρN1 + ρN2 + . . .+ ρNN sum of all reduced states ρNi : H. ρis called electron density matrix.In terms of electron density matrix ρ, Pauli principle amounts to〈ψ|ρ|ψ〉 ≤ 1⇐⇒ Specρ ≤ 1

Initial form (1925)

In the language of density matrices it can be stated as followsLet ρN : H⊗N , H⊗N = state space of N electrons.Define ρ = ρN1 + ρN2 + . . .+ ρNN sum of all reduced states ρNi : H. ρis called electron density matrix.

In terms of electron density matrix ρ, Pauli principle amounts to〈ψ|ρ|ψ〉 ≤ 1⇐⇒ Specρ ≤ 1

Initial form (1925)

Statement of The Problem

Statement of the problem

In the latter case, electron density matrix becomes ρ = NρN1 ,Trρ = N.

Problem

What are the constraints on electron density matrix ρ beyond the originalPauli principle Specρ ≤ 1.

Pure N-representability

The above problem became known as (pure) N-representability problemafter A.J. Coleman (1963).

Mixed N-representability

More generally we have mixed N-representability problem:“What are the constraints on the spectra of a mixed state and itsreduced matrix?”

Problem

What was known before 2008?

Initial constraints

Ordering inequalities λ1 ≥ λ2 ≥ . . . ≥ λr ≥ 0

Normalization condition Trρ =∑

i λi = N

Two-particle (∧2Hr ) and Two-hole (∧r−2Hr ) systems

Constraints on electron density matrix ρ is given by even degeneracyof its eigenvalues, i.e. λ1 = λ2 ≥ λ3 = λ4 ≥ . . .Similar results hold for two-hole system.

Initial constraints

i λi = N

Initial constraints

i λi = N

Constraints on electron density matrix ρ is given by even degeneracyof its eigenvalues, i.e. λ1 = λ2 ≥ λ3 = λ4 ≥ . . .

Similar results hold for two-hole system.

Initial constraints

i λi = N

Borland-Dennis system

For the system ∧3H6 of three electrons of rank 6, theN-representability conditions are given by the following (in)equalities:

λ1 + λ6 = λ2 + λ5 = λ3 + λ4 = 1, λ4 ≤ λ5 + λ6,

Sufficiency proved by Borland-Dennis (1972)

Necessity proved by Ruskai (2007)

λ1 + λ6 = λ2 + λ5 = λ3 + λ4 = 1, λ4 ≤ λ5 + λ6,

Formal Solution of Mixed N-representability

Formal Solution

Solution of Mixed N-representability

Main Theorem. Let ρN : ∧NHr (TrρN = 1) be mixed state and ρ : Hr

(Trρ = N) be its particle density matrix.

Then all constraints onSpecρ = λ : λ1 ≥ λ2 ≥ . . . ≥ λr and SpecρN = µ : µ1 ≥ µ2 ≥ . . . ≥ µR

(R =(rN

)= dim∧NHr ) are given by the following linear inequalities

(Generalized Pauli Constraints)∑i

aiλv(i) ≤∑k

(∧Na)kµw(k) (a, v ,w)

Here, ∧Na consists of all sums ai1 + ai2 + · · ·+ aiN ,1 ≤ i1 < i2 < · · · < iN ≤ r arranged in decreasing order, v ∈ Sr andw ∈ SR subject to topological condition cwv (a) 6= 0 explained soon.

Formal Solution

(Trρ = N) be its particle density matrix. Then all constraints onSpecρ = λ : λ1 ≥ λ2 ≥ . . . ≥ λr and SpecρN = µ : µ1 ≥ µ2 ≥ . . . ≥ µR

(R =(rN

(Generalized Pauli Constraints)

aiλv(i) ≤∑k

Formal Solution

(R =(rN

aiλv(i) ≤∑k

Formal Solution

(R =(rN

aiλv(i) ≤∑k

Remark

Solution of Pure N-representability

Solution of pure N-representability problem can be deduced from theabove theorem by specialization µi = 0 for i 6= 1, because pure statescorresponds projector operators |ψ〉〈ψ|.

In this case all costraints together with ordering ineqaulities andnormalization condition describe a convex polytope, called MomentPolytope.

Remark

Solution of Pure N-representability

Solution of pure N-representability problem can be deduced from theabove theorem by specialization µi = 0 for i 6= 1, because pure statescorresponds projector operators |ψ〉〈ψ|.In this case all costraints together with ordering ineqaulities andnormalization condition describe a convex polytope, called MomentPolytope.

Nature of Topological Condition cvw(a) 6= 0

Consider flag variety

Fa(Hr ) = {X : Hr → Hr | a = Spec(X )}

and morphism

ϕa : Fa(Hr ) → F∧Na(∧NHr )

X 7→ X (N)

Here, X (N) : ψ1 ∧ ψ2 ∧ . . . ∧ ψN 7→∑

i ψ1 ∧ ψ2 ∧ . . . ∧ Xψi ∧ . . . ∧ ψN .

and morphism

X 7→ X (N)

Here, X (N) : ψ1 ∧ ψ2 ∧ . . . ∧ ψN 7→∑

i ψ1 ∧ ψ2 ∧ . . . ∧ Xψi ∧ . . . ∧ ψN .

and morphism

X 7→ X (N)

Here, X (N) : ψ1 ∧ ψ2 ∧ . . . ∧ ψN 7→∑

i ψ1 ∧ ψ2 ∧ . . . ∧ Xψi ∧ . . . ∧ ψN .

The coefficients cvw (a) are defined via induced morphism of cohomology

ϕ∗a : H∗(F∧Na(∧NHr )) → H∗(Fa(Hr )),

σw 7→∑v

cvw (a)σv

written in the basis of Schubert cocycles σw .

ϕ∗a : H∗(F∧Na(∧NHr )) → H∗(Fa(Hr )),

σw 7→∑v

cvw (a)σv

ϕ∗a : H∗(F∧Na(∧NHr )) → H∗(Fa(Hr )),

σw 7→∑v

cvw (a)σv

Connection with Representation Theory

Plethysm

Consider the m-th symmetric power of ∧NHr , called Plethysm

It splits into irreducible components Hλ, parameterized by Young

diagrams λ = of size N.m, with multiplicity mλ

Sm(∧NHr ) =∑λ

mλHλ.

Problem

“Which irreducible representations Hλ of U(Hr ) can appear in thedecomposition of Sm(∧NHr )?”

Surprisingly solution of this problem coincides with that of pureN-representability problem.

Plethysm

Sm(∧NHr ) =∑λ

mλHλ.

Problem

Plethysm

Sm(∧NHr ) =∑λ

mλHλ.

Problem

Plethysm

Sm(∧NHr ) =∑λ

mλHλ.

Problem

Connection

treat the diagrams λ as spectra.λ : λ1 ≥ λ2 ≥ . . . ≥ λr .

normalize them to a fixed size λ̃ = λ/m s.t.

Trλ̃ = N.

Representation theoretical solution

Theorem. Every λ̃ obtained from irreducible componentHλ ⊂ Sm(∧NHr ) is a spectrum of one point reduced matrix ρ of a purestate ψ ∈ ∧NHr . Moreover every one point reduced spectrum is a convexcombination of such spectra λ̃ with bounded m ≤ M.

Connection

treat the diagrams λ as spectra.λ : λ1 ≥ λ2 ≥ . . . ≥ λr .normalize them to a fixed size λ̃ = λ/m s.t.

Trλ̃ = N.

Connection

Trλ̃ = N.

Connection

Trλ̃ = N.

Theorem. Every λ̃ obtained from irreducible componentHλ ⊂ Sm(∧NHr ) is a spectrum of one point reduced matrix ρ of a purestate ψ ∈ ∧NHr .

Moreover every one point reduced spectrum is a convexcombination of such spectra λ̃ with bounded m ≤ M.

Connection

Trλ̃ = N.

Practical Algorithm

For a fixed M the convex hull of the spectra λ̃ from previous Theoremgives an inner approximation to the moment polytope,

while any set ofinequalities of Main Theorem amounts to its outer approximation. Thissuggests the following approach to the pure N-representability problem,which combines both theorems.

The Algorithm

Find all irreducible components Hλ ⊂ Sm(∧NHr ) for m ≤ M.

Calculate the convex hull of the corresponding spectra λ̃ which givesan inner approximation P in

M ⊂ P for the moment polytope P.

Identify the facets of P inM that are given by the inequalities of the

Main Theorem. They cut out an outer approximation PoutM ⊃ P.

Increase M and continue until P inM = Pout

Practical Algorithm

For a fixed M the convex hull of the spectra λ̃ from previous Theoremgives an inner approximation to the moment polytope,while any set ofinequalities of Main Theorem amounts to its outer approximation.

Thissuggests the following approach to the pure N-representability problem,which combines both theorems.

The Algorithm

Practical Algorithm

For a fixed M the convex hull of the spectra λ̃ from previous Theoremgives an inner approximation to the moment polytope,while any set ofinequalities of Main Theorem amounts to its outer approximation. Thissuggests the following approach to the pure N-representability problem,which combines both theorems.

The Algorithm

Practical Algorithm

The Algorithm

Practical Algorithm

The Algorithm

Practical Algorithm

The Algorithm

Practical Algorithm

The Algorithm

How it works?

The above algorithm works perfectly only for some small systems.

Borland-Dennis System (N = 3, r = 6)

We only need to calculate symmetric powers of ∧3H6 up to 4th degree.

That is, M = 4.Symmetric powers contains the following components

m λ1 [1, 1, 1, 0, 0, 0]2 [2, 2, 2, 0, 0, 0], [2, 1, 1, 1, 1, 0]3 [2, 2, 2, 1, 1, 1], [3, 3, 3, 0, 0, 0], [3, 2, 2, 1, 1, 0]4 [2, 2, 2, 2, 2, 2], [3, 3, 3, 1, 1, 1], [4, 4, 4, 0, 0, 0],

[3, 3, 2, 2, 1, 1], [4, 3, 3, 1, 1, 0], [4, 2, 2, 2, 2, 0]

How it works?

We only need to calculate symmetric powers of ∧3H6 up to 4th degree.That is, M = 4.

Symmetric powers contains the following components

m λ1 [1, 1, 1, 0, 0, 0]2 [2, 2, 2, 0, 0, 0], [2, 1, 1, 1, 1, 0]3 [2, 2, 2, 1, 1, 1], [3, 3, 3, 0, 0, 0], [3, 2, 2, 1, 1, 0]4 [2, 2, 2, 2, 2, 2], [3, 3, 3, 1, 1, 1], [4, 4, 4, 0, 0, 0],

[3, 3, 2, 2, 1, 1], [4, 3, 3, 1, 1, 0], [4, 2, 2, 2, 2, 0]

How it works?

We only need to calculate symmetric powers of ∧3H6 up to 4th degree.That is, M = 4.Symmetric powers contains the following components

m λ1 [1, 1, 1, 0, 0, 0]2 [2, 2, 2, 0, 0, 0], [2, 1, 1, 1, 1, 0]3 [2, 2, 2, 1, 1, 1], [3, 3, 3, 0, 0, 0], [3, 2, 2, 1, 1, 0]4 [2, 2, 2, 2, 2, 2], [3, 3, 3, 1, 1, 1], [4, 4, 4, 0, 0, 0],

[3, 3, 2, 2, 1, 1], [4, 3, 3, 1, 1, 0], [4, 2, 2, 2, 2, 0]

How it works?

We only need to calculate symmetric powers of ∧3H6 up to 4th degree.That is, M = 4.Symmetric powers contains the following components

m λ1 [1, 1, 1, 0, 0, 0]2 [2, 2, 2, 0, 0, 0], [2, 1, 1, 1, 1, 0]3 [2, 2, 2, 1, 1, 1], [3, 3, 3, 0, 0, 0], [3, 2, 2, 1, 1, 0]4 [2, 2, 2, 2, 2, 2], [3, 3, 3, 1, 1, 1], [4, 4, 4, 0, 0, 0],

[3, 3, 2, 2, 1, 1], [4, 3, 3, 1, 1, 0], [4, 2, 2, 2, 2, 0]

How it works?

After the normalization we obtain the spectra λ̃

m λ̃1 [1, 1, 1, 0, 0, 0]2 [1, 1, 1, 0, 0, 0], [1, 12 ,

12 , 0]

3 [ 23 ,23 ,

13 ], [1, 1, 1, 0, 0, 0], [1, 23 ,

13 , 0]

4 [ 12 ,12 ,

12 ], [ 34 ,

14 ], [1, 1, 1, 0, 0, 0],

[ 34 ,34 ,

14 ], [1, 34 ,

14 , 0], [1, 12 ,

12 , 0]

Taking convex hull of these spectra we obtain the full moment polytopewhich is described by 3 equalities and 1 inequality.

N = 3, r = 7

M = 8.

N = 4, r = 8

M = 10.

How it works?

m λ̃1 [1, 1, 1, 0, 0, 0]2 [1, 1, 1, 0, 0, 0], [1, 12 ,

12 , 0]

3 [ 23 ,23 ,

13 ], [1, 1, 1, 0, 0, 0], [1, 23 ,

13 , 0]

4 [ 12 ,12 ,

12 ], [ 34 ,

14 ], [1, 1, 1, 0, 0, 0],

[ 34 ,34 ,

14 ], [1, 34 ,

14 , 0], [1, 12 ,

12 , 0]

N = 3, r = 7

M = 8.

N = 4, r = 8

M = 10.

How it works?

m λ̃1 [1, 1, 1, 0, 0, 0]2 [1, 1, 1, 0, 0, 0], [1, 12 ,

12 , 0]

3 [ 23 ,23 ,

13 ], [1, 1, 1, 0, 0, 0], [1, 23 ,

13 , 0]

4 [ 12 ,12 ,

12 ], [ 34 ,

14 ], [1, 1, 1, 0, 0, 0],

[ 34 ,34 ,

14 ], [1, 34 ,

14 , 0], [1, 12 ,

12 , 0]

N = 3, r = 7

M = 8.

N = 4, r = 8

M = 10.

How it works?

m λ̃1 [1, 1, 1, 0, 0, 0]2 [1, 1, 1, 0, 0, 0], [1, 12 ,

12 , 0]

3 [ 23 ,23 ,

13 ], [1, 1, 1, 0, 0, 0], [1, 23 ,

13 , 0]

4 [ 12 ,12 ,

12 ], [ 34 ,

14 ], [1, 1, 1, 0, 0, 0],

[ 34 ,34 ,

14 ], [1, 34 ,

14 , 0], [1, 12 ,

12 , 0]

N = 3, r = 7

M = 8.

N = 4, r = 8

M = 10.Murat Altunbulak Oxford 2016

How it works?

N = 3, r = 8

Using the above algorithm we manage to calculate symmetric powers upto M = 24. Resulting polytope was not the full moment polytope.That is, P in

24 6= Pout24 .

There was one problematic facet.To get rid of this, we resort to use a numerical minimization of the linearform obtained from the problematic facet over all particle densitymatrices.It turns out that the form attains its minimum at a single point.Adding this new point gives a polytope P whose all facets are covered bythe Main Theorem.Thus the polytope is the genuine moment polytope for ∧3H8 which isgiven by 31 independent inequalities.

How it works?

N = 3, r = 8

Using the above algorithm we manage to calculate symmetric powers upto M = 24. Resulting polytope was not the full moment polytope.

That is, P in24 6= Pout

24 .There was one problematic facet.To get rid of this, we resort to use a numerical minimization of the linearform obtained from the problematic facet over all particle densitymatrices.It turns out that the form attains its minimum at a single point.Adding this new point gives a polytope P whose all facets are covered bythe Main Theorem.Thus the polytope is the genuine moment polytope for ∧3H8 which isgiven by 31 independent inequalities.

How it works?

N = 3, r = 8

24 6= Pout24 .

How it works?

N = 3, r = 8

24 6= Pout24 .

There was one problematic facet.

To get rid of this, we resort to use a numerical minimization of the linearform obtained from the problematic facet over all particle densitymatrices.It turns out that the form attains its minimum at a single point.Adding this new point gives a polytope P whose all facets are covered bythe Main Theorem.Thus the polytope is the genuine moment polytope for ∧3H8 which isgiven by 31 independent inequalities.

How it works?

N = 3, r = 8

24 6= Pout24 .

There was one problematic facet.To get rid of this, we resort to use a numerical minimization of the linearform obtained from the problematic facet over all particle densitymatrices.

It turns out that the form attains its minimum at a single point.Adding this new point gives a polytope P whose all facets are covered bythe Main Theorem.Thus the polytope is the genuine moment polytope for ∧3H8 which isgiven by 31 independent inequalities.

How it works?

N = 3, r = 8

24 6= Pout24 .

There was one problematic facet.To get rid of this, we resort to use a numerical minimization of the linearform obtained from the problematic facet over all particle densitymatrices.It turns out that the form attains its minimum at a single point.

Adding this new point gives a polytope P whose all facets are covered bythe Main Theorem.Thus the polytope is the genuine moment polytope for ∧3H8 which isgiven by 31 independent inequalities.

How it works?

N = 3, r = 8

24 6= Pout24 .

There was one problematic facet.To get rid of this, we resort to use a numerical minimization of the linearform obtained from the problematic facet over all particle densitymatrices.It turns out that the form attains its minimum at a single point.Adding this new point gives a polytope P whose all facets are covered bythe Main Theorem.

Thus the polytope is the genuine moment polytope for ∧3H8 which isgiven by 31 independent inequalities.

How it works?

N = 3, r = 8

24 6= Pout24 .

Number of Constraints for systems of rank ≤ 10

r N Number of constraints7 3 48 3 318 4 159 3 529 4 60

10 3 9310 4 12510 5 161

Thank You for your attention!!!

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