leo tzou - purdue university

The Aharonov-Bohm Effect and the Geometry of Connections

Leo Tzou

Speaker is partially supported by the Academy of Finland

Speaker is partially supported by NSF Grant DMS-3861041

Double Slit Experiment

one shoots electrons through a double slit

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Double Slit Experiment

and obtains a symmetric distribution function

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The Aharonov-Bohm Experiment

One places a solenoid behind the slit

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The Aharonov-Bohm Experiment

without magnetic potential

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The Aharonov-Bohm Experiment

and scatter electrons

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The Aharonov-Bohm Experiment

and obtains again a symmetric distribution function

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The Aharonov-Bohm Experiment

•We turn on the magnetic potential A in such a way that the magnetic

field E = ∇×A is completely contained in the solenoid.

•Observe that electrons only pass through regions where E = 0

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The Aharonov-Bohm Experiment

• Surprisingly, the distribution is no longer symmetrical.

• Remember that electrons only passed through regions of vanishing

magnetic field. So what happened?

• The region where E = ∇×A vanishes is not simply connected.

• The potential A does not vanish up to gauge outside of the solenoid.

• What kind of A gives trivial interference patterns?9

The Magnetic Schrodinger Equation

Quantum mechanical effects involving the magnetism is modeled by

the magnetic Schrodinger equation:

LAu := (d+ iA)∗(d+ iA)︸ ︷︷ ︸=∆ if A=0

u = 0

• A is real valued 1-form represents the magnetic potential.

• The curl dA is the magnetic field.

• How does A effect the boundary behaviour of the solution?

• Classically only dA matters and not A.

• But when there is topology the Aharonov-Bohm experiment suggests

otherwise.

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This question was studied in the setting of Euclidean setting with cavities by Ballestero-Weder. In the

geometric setting the time dependent hyperpolic and boundary spectral data case was done by Kurylev

et al. Furthermore recovering the magnetic field from partial boundary measurement was done recently

by Emanuvilov-Yamamoto-Uhlmann on planar domains.

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Calderon Problem for (d+ iA)∗(d+ iA)

• Let (M, g) be a Riemannian manifold with boundary and f ∈ C∞(∂M)

• Assume well-posedness, there exists a unique uf solving

(d+ iA)∗(d+ iA)uf = 0

uf = f ∂M

• Define the Dirichlet-Neumann map by

ΛA : f 7→ in(d+ iA)uf

where n is the normal vector field along the boundary.

Does ΛA uniquely determine A?

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Calderon Problem for (d+ iA)∗(d+ iA)

• Let (M, g) be a Riemannian manifold with boundary and f ∈ C∞(∂M)

• Assume well-posedness, there exists a unique uf solving

(d+ iA)∗(d+ iA)uf = 0

uf = f ∂M

• Define the Dirichlet-Neumann map by

ΛA : f 7→ in(d+ iA)uf

where n is the normal vector field along the boundary.

Does ΛA uniquely determine A?

NO

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Gauge Invariance

• Let φ ∈ C∞(M) be a real function with φ |∂M= 0.

• Consider the operator LA+dφ = (d+ iA+ idφ)∗(d+ iA+ idφ)

• If LAu = 0 then LA+dφe−iφu = 0.

• So ΛA = ΛA+dφ.

Natural Conjecture (false in general):

If ΛA1= ΛA2

then A1 −A2 is exact.

This holds only on planar domains

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Simply Connected Domains

Lets suppose M =unit disk.

•Observe that LA = (d+iA)∗(d+iA) =(2nd order elliptic) +idA+︸ ︷︷ ︸magneticfield

|A|2.

• So by analytic techniques we can show that

ΛA1= ΛA2

⇒ dA1 = dA2

• By the fact that M is simply connected,

φ(z) =∫ z

z0

A1 −A2

is path independent and well defined so

A1 = A2 + dφ

Thus ΛA1= ΛA2

⇔ A1 −A2 = dφ

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This corresponds to the double slit experiment with no topology:

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Topological Obstructions

• On a surface M with genus similar analytic techniques will obtain

ΛA1= ΛA2

⇒ d(A1 −A2) = 0

• However, this does not imply A1 −A2 is exact. So,

A1 −A2 is exact ⇒ ΛA1= ΛA2

A1 −A2 is closed ⇐ ΛA1= ΛA2

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Topological Obstructions

• On a surface M with genus similar analytic techniques will obtain

ΛA1= ΛA2

⇒ d(A1 −A2) = 0

• However, this does not imply A1 −A2 is exact. So,

A1 −A2 is exact ⇒ ΛA1= ΛA2

lA1 −A2 is closed ⇐ ΛA1

= ΛA2

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Topological Obstructions

• On a surface M with genus similar analytic techniques will obtain

ΛA1= ΛA2

⇒ d(A1 −A2) = 0

• However, this does not imply A1 −A2 is exact. So,

A1 −A2 is exact ⇒ ΛA1= ΛA2

(cohomology of M)

A1 −A2 is closed ⇐ ΛA1= ΛA2

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Topological Obstructions

• On a surface M with genus similar analytic techniques will obtain

ΛA1= ΛA2

⇒ d(A1 −A2) = 0

• However, this does not imply A1 −A2 is exact. So,

A1 −A2 is exact ⇒ ΛA1= ΛA2

? ⇔ ?

A1 −A2 is closed ⇐ ΛA1= ΛA2

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A Satisfactory Answer (Guillarmou - LT, GAFA 2011)

ΛA1= ΛA2

⇔ (A1 −A2) ∈ H1(M,∂M ; N)

This means that

• d(A1 −A2) = 0 (ie. (A1 −A2) ∈ H1(M,∂M ; R))

•∫γ(A1 −A2) ∈ 2πN for all closed loops γ.

Corollary

ΛA = Λ0 IFF dA = 0 and∫γ A ∈ 2πN for all loops γ.

What motivated us to this condition?

The answer is in the geometry of connection.

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Point of View of Parallel Transport

Let E = C×M be the trivial complex line bundle over M .

• ∇A := d+ iA is a connection acting on this line bundle.

• Let γ be a closed loop and z0 ∈ γ

• Fix v ∈ Ez0

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Point of View of Parallel Transport

Let E = C×M be the trivial complex line bundle over M .

• ∇A := d+ iA is a connection acting on this line bundle.

• Let γ be a closed loop and z0 ∈ γ

• Parallel transport v along γ by ∇A

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Point of View of Parallel Transport

Let E = C×M be the trivial complex line bundle over M .

• ∇A := d+ iA is a connection acting on this line bundle.

• Let γ be a closed loop and z0 ∈ γ

• Parallel transport v along γ by ∇A

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Point of View of Parallel Transport

Let E = C×M be the trivial complex line bundle over M .

• ∇A := d+ iA is a connection acting on this line bundle.

• Let γ be a closed loop and z0 ∈ γ

• Parallel transport v along γ by ∇A

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Point of View of Parallel Transport

Let E = C×M be the trivial complex line bundle over M .

• ∇A := d+ iA is a connection acting on this line bundle.

• Let γ be a closed loop and z0 ∈ γ

• Parallel transport v along γ by ∇A

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Point of View of Parallel Transport

Let E = C×M be the trivial complex line bundle over M .

• ∇A := d+ iA is a connection acting on this line bundle.

• Let γ be a closed loop and z0 ∈ γ

• Parallel transport v along γ by ∇A

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Point of View of Parallel Transport

Let E = C×M be the trivial complex line bundle over M .

• ∇A := d+ iA is a connection acting on this line bundle.

• Let γ be a closed loop and z0 ∈ γ

• Parallel transport v along γ by ∇A

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Point of View of Parallel Transport

Let E = C×M be the trivial complex line bundle over M .

• ∇A := d+ iA is a connection acting on this line bundle.

• Let γ be a closed loop and z0 ∈ γ

• Parallel transport v along γ by ∇A

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Point of View of Parallel Transport

Let E = C×M be the trivial complex line bundle over M .

• ∇A := d+ iA is a connection acting on this line bundle.

• Let γ be a closed loop and z0 ∈ γ

• to obtain v′ ∈ Ez0

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Point of View of Parallel Transport

Let E = C×M be the trivial complex line bundle over M .

• ∇A := d+ iA is a connection acting on this line bundle.

• Let γ be a closed loop and z0 ∈ γ

• to obtain v′ ∈ Ez0

• Solving the ODE for parallel transport yields v′ = (ei∫γ A)v

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• Therefore that holonomy of d+ iA is equal to that of ∇0 = d iff∫γA ∈ 2πN

for all closed loops γ

• From the geometric point of view, it is not the exactness of A but

rather the isomorphism of the connections ∇A = d+ iA and ∇0 = d that

matters.

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Proof of Result

ΛA = Λ0 IFF dA = 0 and∫γ A ∈ 2πN for all loops γ.

• Analytic methods show that ΛA = Λ0 ⇒ dA = 0 and A = 0∂M

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Proof of Result

ΛA = Λ0 IFF dA = 0 and∫γ A ∈ 2πN for all loops γ.

• Analytic methods show that ΛA = Λ0 ⇒ dA = 0 and A = 0∂M

Consider a closed loop γ on M :

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Proof of Result

ΛA = Λ0 IFF dA = 0 and∫γ A ∈ 2πN for all loops γ.

• Analytic methods show that ΛA = Λ0 ⇒ dA = 0 and A = 0∂M

Consider a closed loop γ on M :

Since dA = 0 we can choose any representative of the homology class.

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Proof of Result

ΛA = Λ0 IFF dA = 0 and∫γ A ∈ 2πN for all loops γ.

• Analytic methods show that ΛA = Λ0 ⇒ dA = 0 and A = 0∂M

Consider a closed loop γ on M :

so we deform the curve as such so that γ = Γ1 + Γ2 with Γ2 ⊂ ∂M

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Proof of Result

ΛA = Λ0 IFF dA = 0 and∫γ A ∈ 2πN for all loops γ.

• Analytic methods show that ΛA = Λ0 ⇒ dA = 0 and A = 0∂M

Consider a closed loop γ on M :

We need that∫γ A =

∫Γ1A+

∫Γ2

A︸ ︷︷ ︸=0

∈ 2πN

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Proof of Result

ΛA = Λ0 IFF dA = 0 and∫γ A ∈ 2πN for all loops γ.

• Analytic methods show that ΛA = Λ0 ⇒ dA = 0 and A = 0∂M

Consider z0, z1 ∈ ∂M ∩ Γ1 :

We need that∫γ A =

∫Γ1A+

∫Γ2

A︸ ︷︷ ︸=0

∈ 2πN

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Proof of Result

ΛA = Λ0 IFF dA = 0 and∫γ A ∈ 2πN for all loops γ.

• Analytic methods show that ΛA = Λ0 ⇒ dA = 0 and A = 0∂M

Consider z0, z1 ∈ ∂M ∩ Γ1Let f ∈ C∞(∂M) such that f(z1) = f(z0) = 1 :

We need that∫γ A =

∫Γ1A+

∫Γ2

A︸ ︷︷ ︸=0

∈ 2πN

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Proof of Result

• Let L0w = 0 and LAv = 0 such that

v = w = f ∂M

• Since ΛA = Λ0,

∂νw = ∂νv

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Proof of Result

• Let ∆w = 0 and LAv = 0 such that

v = w = f ∂M

• Since ΛA = Λ0,

∂νw = ∂νv

Consider an open neighbourhood U of Γ1

We will show that in U one can transform v to w via parallel transport.41

Proof of Result

• Let ∆w = 0 and LAv = 0 such that

v = w = f ∂M

• Since ΛA = Λ0,

∂νw = ∂νv

Consider an open neighbourhood U of Γ1

We will show that in U one can transform v to w via parallel transport.42

L0w = LAv = 0, w = v = f on ∂M , ∂νw = ∂νv

• For z ∈ Γ1 define

φ(z) =∫ z

z0

A

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L0w = LAv = 0, w = v = f on ∂M , ∂νw = ∂νv

• Since U is simply connected and dA = 0, we can extend φ to U

with

dφ = A

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L0w = LAv = 0, w = v = f on ∂M , ∂νw = ∂νv

• Since U is simply connected and dA = 0, we can extend φ to U

with

dφ = A

Therefore, if L0w = 0 then e−iφw solves LA(e−iφw) = 0 in U .

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L0w = LAv = 0, w = v = f on ∂M , ∂νw = ∂νv

• Since U is simply connected and dA = 0, we can extend φ to U

with

dφ = A

Therefore, if L0w = 0 then e−iφw solves LA(e−iφw) = 0 in U .

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I claim that e−iφw = v in U51

Since A = 0 on ∂M52

Since A = 0 on ∂M53

Since A = 0 on ∂M54

Since A = 0 on ∂M55

Both v and e−iφw solve LA(v) = LA(e−iφw) = 0 in U .56

Unique continuation gives e−iφw = v in U .57

In particular, e−iφ(z1)w(z1) = v(z1).58

In particular, e−iφ(z1)f(z1) = f(z1).59

In particular, e−iφ(z1)1 = 1.60

Therefore, φ(z1) =∫ z1z0A ∈ 2πN

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Generalization to Higher Rank Bundles

Let E be a complex bundle of rank n over a Riemann surface M with

boundary. Let ∇ be a connection acting on sections of E and V an

endomorphism of E. Define the elliptic operator

L := ∇∗∇+ V

and consider the boundary value problem

Luf = 0 M,u = f ∂M

Let Λ∇,V : f 7→ ∇νuf |∂M be the Dirichlet-Neumann operator.

Thm (Albin-Guillarmou-LT)

If Λ∇1,V1= Λ∇2,V2

then there exists a unitary endomorphism F : E → E

such that

F−1∇1F = ∇2, F−1V1F = V2

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