Presentation - Uni of Bath

IntroductionESA of Schrodinger OperatorsL2-Hardy Inequalities and ESA

Essential Self-adjointness & the L2-HardyInequality

A.D Ward - NZ Institute of Advanced Study

Presentation at University of Bath - 12 March 2015

A.D. Ward - NZ Institute of Advanced Study Essential Self-adjointness & the L2-Hardy Inequality


DefinitionsMotivationSchrodinger Operators

Definitions

Definition: A Hilbert space H is a complex vector space that isendowed with an inner product

⟨·, ·⟩

: H×H → C and which is

complete in the induced norm || x || =⟨x , x

⟩ 12 .

Definition: An operator A is a linear mapping A : D(A)→ Hwhere D(A), the domain of A, is a subspace of H. If D(A) isdense in H, then the operator A is said to be densely defined.

The operator B is said to be an extension of A, A ⊆ B, ifD(A) ⊆ D(B) and Ax = Bx for all x ∈ D(A).




Definitions


⟨·, ·⟩



⟩ 12 .






Definitions


⟨·, ·⟩



⟩ 12 .






Definitions


⟨·, ·⟩



⟩ 12 .






Definition: If A is a closeable operator, we define its closure A by

D(A) =f ∈ H

∣∣ ∃ fn∞n=1 ⊆ D(A), fn → f , Afn∞n=1 converges

Af = limn→∞

Afn.

Definition: If A is densely defined we define it’s adjoint A∗ by

D(A∗) =g ∈ H

∣∣ ∃ g∗ ∈ H s.t⟨Ax , g

⟩=⟨x , g∗

⟩∀ x ∈ D(A)

A∗g = g∗ for all g ∈ D(A∗).




Definition: If A is a closeable operator, we define its closure A by

D(A) =f ∈ H

∣∣ ∃ fn∞n=1 ⊆ D(A), fn → f , Afn∞n=1 converges

Af = limn→∞

Afn.

Definition: If A is densely defined we define it’s adjoint A∗ by

D(A∗) =g ∈ H

∣∣ ∃ g∗ ∈ H s.t⟨Ax , g

⟩=⟨x , g∗

⟩∀ x ∈ D(A)

A∗g = g∗ for all g ∈ D(A∗).




Definition: A is symmetric if A ⊆ A∗ so that D(A) ⊆ D(A∗) andAx = A∗x for all x ∈ D(A), or equivalently, if

⟨Ax , y

⟩=⟨x ,Ay

⟩for all x , y ∈ D(A).

Definition: A is self-adjoint if A = A∗ so that D(A) = D(A∗) andAx = A∗x for all x ∈ D(A).

Definition: A is essentially self-adjoint (ESA) if it’s closure isself-adjoint, i.e. A = A

∗ ≡ A∗.

Demonstrating that A is ESA amounts to showing thatD(A∗) = D(A ).

Theorem (von Neumann - [1] Section 7.7)

If A is a densely defined symmetric operator, then

D(A∗) = D(A ) ⊕ ker(A∗ − i) ⊕ ker(A∗ + i)

s.t. A is ESA if and only if ker(A∗ ± i) = 0H.





⟨Ax , y

⟩=⟨x ,Ay




∗ ≡ A∗.




D(A∗) = D(A ) ⊕ ker(A∗ − i) ⊕ ker(A∗ + i)






⟨Ax , y

⟩=⟨x ,Ay




∗ ≡ A∗.




D(A∗) = D(A ) ⊕ ker(A∗ − i) ⊕ ker(A∗ + i)






⟨Ax , y

⟩=⟨x ,Ay




∗ ≡ A∗.




D(A∗) = D(A ) ⊕ ker(A∗ − i) ⊕ ker(A∗ + i)






⟨Ax , y

⟩=⟨x ,Ay




∗ ≡ A∗.




D(A∗) = D(A ) ⊕ ker(A∗ − i) ⊕ ker(A∗ + i)





Technicalities - God is in the Details

Take a densely defined, symmetric linear operator A. We have...A ⊆ A∗ ⇒ A ⊆ A

∗ ≡ A∗ ⇒ D(A) ⊆ D(A∗).

Bounded Operators: Symmetry ⇔ ESAnessA defined on the whole space ⇒ D(A) = H ⊇ D(A∗) ⇒D(A) = D(A∗) ⇒ A is ESA.

Theorem ( Hellinger-Toeplitz - [2] Theorem III.12 )

An everywhere defined, linear, symmetric operator is closed if andonly if it is bounded.

Unounded Operators: Symmetry ; ESAnessA cannot be defined on whole space. Possible that D(A) ( D(A∗).Symmetric without being ESA.





Take a densely defined, symmetric linear operator A. We have...

A ⊆ A∗ ⇒ A ⊆ A∗ ≡ A∗ ⇒ D(A) ⊆ D(A∗).









Take a densely defined, symmetric linear operator A. We have...A ⊆ A∗

⇒ A ⊆ A∗ ≡ A∗ ⇒ D(A) ⊆ D(A∗).










∗ ≡ A∗

⇒ D(A) ⊆ D(A∗).










∗ ≡ A∗ ⇒ D(A) ⊆ D(A∗).










∗ ≡ A∗ ⇒ D(A) ⊆ D(A∗).

Bounded Operators: Symmetry ⇔ ESAnessA defined on the whole space

⇒ D(A) = H ⊇ D(A∗) ⇒D(A) = D(A∗) ⇒ A is ESA.









∗ ≡ A∗ ⇒ D(A) ⊆ D(A∗).

Bounded Operators: Symmetry ⇔ ESAnessA defined on the whole space ⇒ D(A) = H ⊇ D(A∗)

⇒D(A) = D(A∗) ⇒ A is ESA.









∗ ≡ A∗ ⇒ D(A) ⊆ D(A∗).

Bounded Operators: Symmetry ⇔ ESAnessA defined on the whole space ⇒ D(A) = H ⊇ D(A∗) ⇒D(A) = D(A∗)

⇒ A is ESA.









∗ ≡ A∗ ⇒ D(A) ⊆ D(A∗).










∗ ≡ A∗ ⇒ D(A) ⊆ D(A∗).










∗ ≡ A∗ ⇒ D(A) ⊆ D(A∗).




Unounded Operators: Symmetry ; ESAness

A cannot be defined on whole space. Possible that D(A) ( D(A∗).Symmetric without being ESA.






∗ ≡ A∗ ⇒ D(A) ⊆ D(A∗).




Unounded Operators: Symmetry ; ESAnessA cannot be defined on whole space. Possible that D(A) ( D(A∗).

Symmetric without being ESA.






∗ ≡ A∗ ⇒ D(A) ⊆ D(A∗).








So the obvious question is...

Why would anyone be interested in the ESAness of linearoperators?




Axioms of Quantum Mechanics

QM is the study of (essentially) self-adjoint operators!

AXIOM I For each quantum system there is a correspondingHilbert space L2(Ω).

AXIOM II States of the system are unit vectors in L2(Ω).

AXIOM III Observable ↔ self-adjoint operator. Measurementsof observables take values in the spectrum ofcorresponding SA operator.

AXIOM IV Dynamics of system governed by strongly continuousone parameter groups of unitary operators U(t).These are generated by SA operators (Stone).








































As Reed & Simon [2] put it...

In typical applications, physical reasoning gives a formalexpression for an operator. We use the word formalbecause domains are not specified. It is usually easy tofind a domain on which this operator is dense &symmetric. The first problem is to prove essentialself-adjointness [so that the operator defines a uniqueself-adjoint extension], or if the operator is not essentiallyself-adjoint to investigate the various self-adjointextensions.

Reed & Simon, Mathematical Physics, page 303.




Schrodinger Operators

Classically, the energy of the system is the sum of its kinetic andpotential energy, E = T + V. For a one particle system

E = x2(t) + V(x(t)

). (1)

In QM the energy levels of a system are given by spectrum of theSchrodinger operator

H = −∆ + V(x). (2)

Define H on C∞0 (Ω). Since C∞0 (Ω) is dense in L2(Ω), H is denselydefined. H is clearly linear. Integration by parts twice gives⟨(−∆ + V ) u, ρ

⟩=⟨u, (−∆ + V ) ρ

⟩, so H is also symmetric.

Question - Is H essentially self-adjoint?






E = x2(t) + V(x(t)

). (1)


H = −∆ + V(x). (2)


⟩=⟨u, (−∆ + V ) ρ








E = x2(t) + V(x(t)

). (1)


H = −∆ + V(x). (2)


⟩=⟨u, (−∆ + V ) ρ








E = x2(t) + V(x(t)

). (1)


H = −∆ + V(x). (2)

Define H on C∞0 (Ω). Since C∞0 (Ω) is dense in L2(Ω), H is denselydefined.

H is clearly linear. Integration by parts twice gives⟨(−∆ + V ) u, ρ

⟩=⟨u, (−∆ + V ) ρ








E = x2(t) + V(x(t)

). (1)


H = −∆ + V(x). (2)

Define H on C∞0 (Ω). Since C∞0 (Ω) is dense in L2(Ω), H is denselydefined. H is clearly linear.

Integration by parts twice gives⟨(−∆ + V ) u, ρ

⟩=⟨u, (−∆ + V ) ρ








E = x2(t) + V(x(t)

). (1)


H = −∆ + V(x). (2)


⟩=⟨u, (−∆ + V ) ρ








E = x2(t) + V(x(t)

). (1)


H = −∆ + V(x). (2)


⟩=⟨u, (−∆ + V ) ρ





Schrodinger Operators on Rm

Physical InterpretationSchrodinger Operators on Ω ( Rm

Main Result for Ω = Rm

Theorem ( Berezin & Schubin [3] )

Let H = −∆ + V be a Schrodinger operator define on C∞0 (Rm)where V (x) is a real valued, measurable, locally bounded functionon Rm. Then, for any b > 0,

i) If V (x) ≥ − |x |2 − b, then H is ESA.

ii) If V (x) ≤ − |x |2+ε − b, then H is not ESA

There is a very nice physical interpretation that can be attached tothis result...





Main Result for Ω = Rm

Theorem ( Berezin & Schubin [3] )

Let H = −∆ + V be a Schrodinger operator define on C∞0 (Rm)where V (x) is a real valued, measurable, locally bounded functionon Rm. Then, for any b > 0,

i) If V (x) ≥ − |x |2 − b, then H is ESA.

ii) If V (x) ≤ − |x |2+ε − b, then H is not ESA

There is a very nice physical interpretation that can be attached tothis result...





Physical Interpretation

Classical energy of the system is given by E =(dxdt

)2+ V (x).

dt

dx=(E − V (x)

)− 12







)2+ V (x).

dt

dx=(E − V (x)

)− 12







)2+ V (x).∫ ∞

0

dt

dxdx =

∫ ∞0

(E − V (x)

)− 12 dx






Classical energy of the system is given by E =(dx(t)

dt

)2+ V (x).

t(∞) − t(0) =

∫ ∞0

(E − V (x)

)− 12 dx







)2+ V (x).

t(∞) − t(0) =

∫ ∞0

(E − V (x)

)− 12 dx

Represents time taken by a particle under the influence of V (x) toreach infinity.

If V (x) ≥ − |x |2 − b

t(∞) − t(0) ≥∫ ∞

0

(E + x2 + b

)− 12 dx = ∞

Time taken to reach infinity is infinite.







)2+ V (x).

t(∞) − t(0) =

∫ ∞0

(E − V (x)

)− 12 dx

Represents time taken by a particle under the influence of V (x) toreach infinity.If V (x) ≥ − |x |2 − b

t(∞) − t(0) ≥∫ ∞

0

(E + x2 + b

)− 12 dx = ∞

Time taken to reach infinity is infinite.







)2+ V (x).

t(∞) − t(0) =

∫ ∞0

(E − V (x)

)− 12 dx

Represents time taken by a particle under the influence of V (x) toreach infinity.If V (x) ≤ −|x |2+ε − b

t(∞) − t(0) ≤∫ ∞

0

(E + x2+ε + b

)− 12 dx < ∞

Time taken to reach infinity is finite.





Theorem ( Eastham, Evans & Mcleod [4] )

Let H = −∆ + V be a Schrodinger operator defined on C∞0 (Rm).If there is a sequence of ‘sufficiently thick’ annuli that occur‘sufficiently regularly’, and if on these annuli the potential is‘sufficiently large’ then H is ESA. But, if one inserts a tube of lowpotential that extends to infinity then H is not ESA irrespective ofthe potential elsewhere.

A Schrodinger operator is ESA if a particle under theinfluence of the associated potential is unable to come into

contact with the boundary of the domain!





Theorem ( Eastham, Evans & Mcleod [4] )

Let H = −∆ + V be a Schrodinger operator defined on C∞0 (Rm).If there is a sequence of ‘sufficiently thick’ annuli that occur‘sufficiently regularly’, and if on these annuli the potential is‘sufficiently large’ then H is ESA. But, if one inserts a tube of lowpotential that extends to infinity then H is not ESA irrespective ofthe potential elsewhere.

A Schrodinger operator is ESA if a particle under theinfluence of the associated potential is unable to come into

contact with the boundary of the domain!





Schrodinger Operators on Ω ( Rm

Now we consider the ESAness of Schrodinger operators ondomains with non-empty boundary. This situation is a lot morecomplicated.

Consider classical problem of confining a particle to the interval(−1, 1). If the potential V (x)→∞ as x → ±1 then the particlewill not be able to reach the boundary.

However, in QM must take into account two opposing effects...

Quantum Tunneling - potential must inflate at a sufficientrate as x → ∂Ω.

Uncertainty Principle - cannot say with certainty that aparticle is located at the boundary.







Consider classical problem of confining a particle to the interval(−1, 1).

If the potential V (x)→∞ as x → ±1 then the particlewill not be able to reach the boundary.














































DefinitionsESA of Schrodinger Operators on Ω ( Rm

Quantum Tunneling & The Uncertainty Principle

Immortality is a silly word, but a mathematician has thebest chance of whatever it may mean - G.H.Hardy

A good mathematical joke is better than a dozenmediocre papers - J.E. Littlewood





Definitions

Definition: Ω admits an L2-Hardy inequality if ∃ C > 0 s.t∫Ω

|ω(x) |2

d(x)2dx ≤ C

∫Ω| ∇ω(x) |2 dx (3)

for all ω(x) ∈W 12,0(Ω).

Definition: Define the variational constant

µ2(Ω) = infω ∈W 1

2,0(Ω)

∫Ω | ∇ω(x) |2 dx∫

Ω|ω(x) |2d(x)2 dx

(4)

µ2(Ω) > 0 if and only if Ω admits an L2-Hardy inequality.





Definitions


|ω(x) |2

d(x)2dx ≤ C

∫Ω| ∇ω(x) |2 dx (3)




2,0(Ω)

∫Ω | ∇ω(x) |2 dx∫

Ω|ω(x) |2d(x)2 dx

(4)






Definitions


|ω(x) |2

d(x)2dx ≤ C

∫Ω| ∇ω(x) |2 dx (3)




2,0(Ω)

∫Ω | ∇ω(x) |2 dx∫

Ω|ω(x) |2d(x)2 dx

(4)






Main Theorem - Extending Nenciu & Nenciu [5]

Theorem

Let Ω be a domain with non-empty boundary. Let H = −∆ + Vbe a Schrodinger operator defined on the domain D(H) = C∞0 (Ω)and where V ∈ L∞,loc(Ω) is a real potential of the form

V (x) ≥ 1−µ2(Ω)d(x)2 .

Then H is essentially self-adjoint.






Theorem


V (x) ≥ 1

d(x)2

[1− µ2(Ω)− 1

ln(1/d)− 1

ln(1/d) ln ln(1/d)−

. . .− 1

ln(1/d) ln ln(1/d) . . . ln ln . . . ln︸︷︷︸M times

(1/d)

]







Theorem


V (x) ≥ 1−µ2(Ω)d(x)2 .


The idea of the proof is to show that ker(H∗ ± i) = 0H usingthe following crucial estimate.






Theorem


V (x) ≥ 1−µ2(Ω)d(x)2 .


The idea of the proof is to show that ker(H∗ ± i) = 0H usingthe following crucial estimate.





Main Estimate Choose E ∈ R

⟨(H + E ) u , u

⟩−∫

Ω

1

d(x)2| u(x) |2 dx

= −∫

Ω∆u(x) u(x) dx +

∫ΩV (x) | u(x) |2 dx + E || u ||2L2

−∫

Ω

1

d(x)2| u(x) |2 dx

≥∫

Ω| ∇ u(x) |2 dx + E || u ||2L2

− µ2(Ω)

∫Ω

| u(x) |2

d(x)2dx

≥∫

Ω| ∇ u(x) |2 dx + E || u ||2L2

−∫

Ω| ∇ u(x) |2 dx

= E || u ||2L2





Main Estimate Choose E ∈ R⟨(H + E ) u , u

⟩−∫

Ω

1

d(x)2| u(x) |2 dx

= −∫

Ω∆u(x) u(x) dx +

∫ΩV (x) | u(x) |2 dx + E || u ||2L2

−∫

Ω

1

d(x)2| u(x) |2 dx

≥∫

Ω| ∇ u(x) |2 dx + E || u ||2L2

− µ2(Ω)

∫Ω

| u(x) |2

d(x)2dx

≥∫

Ω| ∇ u(x) |2 dx + E || u ||2L2

−∫

Ω| ∇ u(x) |2 dx

= E || u ||2L2






⟩−∫

Ω

1

d(x)2| u(x) |2 dx

= −∫

Ω∆u(x) u(x) dx +

∫ΩV (x) | u(x) |2 dx + E || u ||2L2

−∫

Ω

1

d(x)2| u(x) |2 dx

≥∫

Ω| ∇ u(x) |2 dx + E || u ||2L2

− µ2(Ω)

∫Ω

| u(x) |2

d(x)2dx

≥∫

Ω| ∇ u(x) |2 dx + E || u ||2L2

−∫

Ω| ∇ u(x) |2 dx

= E || u ||2L2






⟩−∫

Ω

1

d(x)2| u(x) |2 dx

= −∫

Ω∆u(x) u(x) dx +

∫ΩV (x) | u(x) |2 dx + E || u ||2L2

−∫

Ω

1

d(x)2| u(x) |2 dx

≥∫

Ω| ∇ u(x) |2 dx

+ E || u ||2L2− µ2(Ω)

∫Ω

| u(x) |2

d(x)2dx

≥∫

Ω| ∇ u(x) |2 dx + E || u ||2L2

−∫

Ω| ∇ u(x) |2 dx

= E || u ||2L2






⟩−∫

Ω

1

d(x)2| u(x) |2 dx

= −∫

Ω∆u(x) u(x) dx +

∫ΩV (x) | u(x) |2 dx + E || u ||2L2

−∫

Ω

1

d(x)2| u(x) |2 dx

≥∫

Ω| ∇ u(x) |2 dx + E || u ||2L2

− µ2(Ω)

∫Ω

| u(x) |2

d(x)2dx

≥∫

Ω| ∇ u(x) |2 dx + E || u ||2L2

−∫

Ω| ∇ u(x) |2 dx

= E || u ||2L2






⟩−∫

Ω

1

d(x)2| u(x) |2 dx

= −∫

Ω∆u(x) u(x) dx +

∫ΩV (x) | u(x) |2 dx + E || u ||2L2

−∫

Ω

1

d(x)2| u(x) |2 dx

≥∫

Ω| ∇ u(x) |2 dx + E || u ||2L2

− µ2(Ω)

∫Ω

| u(x) |2

d(x)2dx

≥∫

Ω| ∇ u(x) |2 dx + E || u ||2L2

−∫

Ω| ∇ u(x) |2 dx

= E || u ||2L2






⟩−∫

Ω

1

d(x)2| u(x) |2 dx

= −∫

Ω∆u(x) u(x) dx +

∫ΩV (x) | u(x) |2 dx + E || u ||2L2

−∫

Ω

1

d(x)2| u(x) |2 dx

≥∫

Ω| ∇ u(x) |2 dx + E || u ||2L2

− µ2(Ω)

∫Ω

| u(x) |2

d(x)2dx

≥∫

Ω| ∇ u(x) |2 dx + E || u ||2L2

−∫

Ω| ∇ u(x) |2 dx

= E || u ||2L2






⟩−∫

Ω

1

d(x)2| u(x) |2 dx

= −∫

Ω∆u(x) u(x) dx +

∫ΩV (x) | u(x) |2 dx + E || u ||2L2

−∫

Ω

1

d(x)2| u(x) |2 dx

≥∫

Ω| ∇ u(x) |2 dx + E || u ||2L2

− µ2(Ω)

∫Ω

| u(x) |2

d(x)2dx

≥∫

Ω| ∇ u(x) |2 dx + E || u ||2L2

−∫

Ω| ∇ u(x) |2 dx

= E || u ||2L2






H is ESA if V (x) ≥[

1− µ2(Ω)]d(x , ∂Ω)−2.

If µ2(Ω) > 0 (i.e. Ω admits L2-Hardy inequality) this relaxesconditions for ESAness. Why?

Recall physical interpretation of ESAness - particle cannot comeinto contact with boundary! µ2(Ω) puts limits on the certaintywith which we can say a particle is located at the boundary.

1 = ||ω || 2L2(Ω) =

∫Ω

d(x) ω(x)ω(x)

d(x)dx







1− µ2(Ω)]d(x , ∂Ω)−2.

If µ2(Ω) > 0 (i.e. Ω admits L2-Hardy inequality) this relaxesconditions for ESAness.

Why?


1 = ||ω || 2L2(Ω) =

∫Ω

d(x) ω(x)ω(x)

d(x)dx







1− µ2(Ω)]d(x , ∂Ω)−2.



1 = ||ω || 2L2(Ω) =

∫Ω

d(x) ω(x)ω(x)

d(x)dx







1− µ2(Ω)]d(x , ∂Ω)−2.


Recall physical interpretation of ESAness - particle cannot comeinto contact with boundary!

µ2(Ω) puts limits on the certaintywith which we can say a particle is located at the boundary.

1 = ||ω || 2L2(Ω) =

∫Ω

d(x) ω(x)ω(x)

d(x)dx







1− µ2(Ω)]d(x , ∂Ω)−2.



1 = ||ω || 2L2(Ω) =

∫Ω

d(x) ω(x)ω(x)

d(x)dx







1− µ2(Ω)]d(x , ∂Ω)−2.



1 = ||ω || 2L2(Ω) =

∫Ω

d(x) ω(x)ω(x)

d(x)dx







1− µ2(Ω)]d(x , ∂Ω)−2.



1 ≤( ∫

Ωd(x)2 |ω(x) |2 dx

) 12

·( ∫

Ω

|ω(x) |2

d(x)2dx

) 12







1− µ2(Ω)]d(x , ∂Ω)−2.



1 ≤ 1

µ2(Ω)12

( ∫Ωd(x)2 |ω(x) |2 dx

) 12

·( ∫

Ω| ∇ω(x) |2 dx

) 12







1− µ2(Ω)]d(x , ∂Ω)−2.



µ2(Ω) ≤( ∫

Ωd(x)2 |ω(x) |2 dx

)·( ∫

Ω| ∇ω(x) |2 dx

)

Cannot confine a particle to an arbitrarily small nbhood of theboundary without increasing it’s total momentum indefinitely!







1− µ2(Ω)]d(x , ∂Ω)−2.



µ2(Ω) ≤( ∫

Ωd(x)2 |ω(x) |2 dx

)·( ∫

Ω| ∇ω(x) |2 dx

)

Cannot confine a particle to an arbitrarily small nbhood of theboundary without increasing it’s total momentum indefinitely!





Optimality of Conditions for ESAness

If V (x) ≥[

1− µ2(Ω)]d(x)−2 then H is ESA (5)

but D(H∗) = D(H ) ⊕ ker(H∗ − i) ⊕ ker(H∗ + i)

⇒ ker(H∗ ± i) = 0H

What is ker(H∗ ± i) ?

⇒ No weak, square integrable solutions of the equation

−∆Ψ(x) + V (x) Ψ(x) ± i Ψ(x) = 0 (6)

Q: Can we replace 1− µ2(Ω) in (5) with a smaller constant?A: No - Produces square integrable solutions to (6) which would

mean that H is not ESA!






If V (x) ≥[



⇒ ker(H∗ ± i) = 0H



−∆Ψ(x) + V (x) Ψ(x) ± i Ψ(x) = 0 (6)








If V (x) ≥[



⇒ ker(H∗ ± i) = 0H



−∆Ψ(x) + V (x) Ψ(x) ± i Ψ(x) = 0 (6)








If V (x) ≥[



⇒ ker(H∗ ± i) = 0H



−∆Ψ(x) + V (x) Ψ(x) ± i Ψ(x) = 0 (6)








If V (x) ≥[



⇒ ker(H∗ ± i) = 0H



−∆Ψ(x) + V (x) Ψ(x) ± i Ψ(x) = 0

(6)








If V (x) ≥[



⇒ ker(H∗ ± i) = 0H



−∆Ψ(x) + V (x) Ψ(x) ± i Ψ(x) = 0 (6)

Q: Can we replace 1− µ2(Ω) in (5) with a smaller constant?

A: No - Produces square integrable solutions to (6) which wouldmean that H is not ESA!






If V (x) ≥[



⇒ ker(H∗ ± i) = 0H



−∆Ψ(x) + V (x) Ψ(x) ± i Ψ(x) = 0 (6)







Question Related to Metamaterials

If V (x) ≥[

1− µ2(Ω)]d(x)−2 then H is ESA

What is the value of µ2(Ω)?

Depends on dimension of boundary -What if boundary is fractal?

Quantum principle dependant on scalessmaller than wavelength of an electron!

Metamaterials: Fractal-like structure

Coherent reflections?

Uncertainty principle?






If V (x) ≥[













If V (x) ≥[












L.A. Ljusternik & V.J. Sobolev - Elements of Functional Analysis, Hindustan Publishing Corporation, (1961).

M.C. Reed & B. Simon - Methods of Modern Mathematical Physics: Vol. 1, Academic press, (1972).

F. A. Berezin & M. A. Schubin - The Schrodinger equation, Kluwer, Dordrecht, (1991).

M.S.P. Eastham, W.D. Evans & J.B. Mcleod - The Essential Self-adjointness of Schrodinger Type

Operators, Archive for Rational Mechanics and Analysis, 60, 2, (1976).

G. Nenciu & I. Nenciu - On Confining Potentials and Essential Self-Adjointness for Schrodinger Operators

on Bounded Domains in Rn , Annales Henri Poincare, Volume 10, 377 - 394, (2009).


Documents

Presentation - Uni of Bath