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-386104 1
Double Slit Experiment one shoots electrons through a double slit 2
Double Slit Experiment and obtains a symmetric distribution function 3
The Aharonov-Bohm Experiment One places a solenoid behind the slit 4
The Aharonov-Bohm Experiment without magnetic potential 5
The Aharonov-Bohm Experiment and scatter electrons 6
The Aharonov-Bohm Experiment and obtains again a symmetric distribution function 7
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 8
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 Schr¨ odinger Equation Quantum mechanical effects involving the magnetism is modeled by the magnetic Schr¨ odinger equation: L A u := ( d + iA ) ∗ ( d + iA ) u = 0 � �� � =∆ A =0 if • 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. 10
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. 11
on Problem for ( d + iA ) ∗ ( d + iA ) Calder´ • Let ( M, g ) be a Riemannian manifold with boundary and f ∈ C ∞ ( ∂M ) • Assume well-posedness, there exists a unique u f solving ( d + iA ) ∗ ( d + iA ) u f = 0 u f = f ∂M • Define the Dirichlet-Neumann map by Λ A : f �→ i ˆ n ( d + iA ) u f where ˆ n is the normal vector field along the boundary. Does Λ A uniquely determine A ? 12
on Problem for ( d + iA ) ∗ ( d + iA ) Calder´ • Let ( M, g ) be a Riemannian manifold with boundary and f ∈ C ∞ ( ∂M ) • Assume well-posedness, there exists a unique u f solving ( d + iA ) ∗ ( d + iA ) u f = 0 u f = f ∂M • Define the Dirichlet-Neumann map by Λ A : f �→ i ˆ n ( d + iA ) u f where ˆ n is the normal vector field along the boundary. Does Λ A uniquely determine A ? NO 13
Gauge Invariance • Let φ ∈ C ∞ ( M ) be a real function with φ | ∂M = 0. • Consider the operator L A + dφ = ( d + iA + idφ ) ∗ ( d + iA + idφ ) • If L A u = 0 then L A + dφ e − iφ u = 0. • So Λ A = Λ A + dφ . Natural Conjecture (false in general): If Λ A 1 = Λ A 2 then A 1 − A 2 is exact. This holds only on planar domains 14
Simply Connected Domains Lets suppose M =unit disk. • Observe that L A = ( d + iA ) ∗ ( d + iA ) =(2nd order elliptic) | A | 2 . + idA + � �� � magneticfield • So by analytic techniques we can show that Λ A 1 = Λ A 2 ⇒ dA 1 = dA 2 • By the fact that M is simply connected, � z φ ( z ) = A 1 − A 2 z 0 is path independent and well defined so A 1 = A 2 + dφ Thus Λ A 1 = Λ A 2 ⇔ A 1 − A 2 = dφ 15
This corresponds to the double slit experiment with no topology: 16
Topological Obstructions • On a surface M with genus similar analytic techniques will obtain Λ A 1 = Λ A 2 ⇒ d ( A 1 − A 2 ) = 0 • However, this does not imply A 1 − A 2 is exact. So, A 1 − A 2 is exact ⇒ Λ A 1 = Λ A 2 A 1 − A 2 is closed ⇐ Λ A 1 = Λ A 2 17
Topological Obstructions • On a surface M with genus similar analytic techniques will obtain Λ A 1 = Λ A 2 ⇒ d ( A 1 − A 2 ) = 0 • However, this does not imply A 1 − A 2 is exact. So, A 1 − A 2 is exact ⇒ Λ A 1 = Λ A 2 � A 1 − A 2 is closed ⇐ Λ A 1 = Λ A 2 18
Topological Obstructions • On a surface M with genus similar analytic techniques will obtain Λ A 1 = Λ A 2 ⇒ d ( A 1 − A 2 ) = 0 • However, this does not imply A 1 − A 2 is exact. So, A 1 − A 2 is exact ⇒ Λ A 1 = Λ A 2 (cohomology of M ) A 1 − A 2 is closed ⇐ Λ A 1 = Λ A 2 19
Topological Obstructions • On a surface M with genus similar analytic techniques will obtain Λ A 1 = Λ A 2 ⇒ d ( A 1 − A 2 ) = 0 • However, this does not imply A 1 − A 2 is exact. So, A 1 − A 2 is exact ⇒ Λ A 1 = Λ A 2 ? ⇔ ? A 1 − A 2 is closed ⇐ Λ A 1 = Λ A 2 20
A Satisfactory Answer (Guillarmou - LT, GAFA 2011) Λ A 1 = Λ A 2 ⇔ ( A 1 − A 2 ) ∈ H 1 ( M, ∂M ; N ) This means that • d ( A 1 − A 2 ) = 0 (ie. ( A 1 − A 2 ) ∈ H 1 ( M, ∂M ; R )) � • γ ( A 1 − A 2 ) ∈ 2 π N for all closed loops γ . Corollary � Λ A = Λ 0 IFF dA = 0 and γ A ∈ 2 π N for all loops γ . What motivated us to this condition? The answer is in the geometry of connection. 21
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 z 0 ∈ γ • Fix v ∈ E z 0 22
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 z 0 ∈ γ • Parallel transport v along γ by ∇ A 23
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 z 0 ∈ γ • Parallel transport v along γ by ∇ A 24
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 z 0 ∈ γ • Parallel transport v along γ by ∇ A 25
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 z 0 ∈ γ • Parallel transport v along γ by ∇ A 26
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 z 0 ∈ γ • Parallel transport v along γ by ∇ A 27
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 z 0 ∈ γ • Parallel transport v along γ by ∇ A 28
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 z 0 ∈ γ • Parallel transport v along γ by ∇ A 29
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 z 0 ∈ γ • to obtain v ′ ∈ E z 0 30
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 z 0 ∈ γ • to obtain v ′ ∈ E z 0 • Solving the ODE for parallel transport yields v ′ = ( e i � γ A ) v 31
• Therefore that holonomy of d + iA is equal to that of ∇ 0 = d iff � γ A ∈ 2 π N 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. 32
Proof of Result � Λ A = Λ 0 IFF dA = 0 and γ A ∈ 2 π N for all loops γ . • Analytic methods show that Λ A = Λ 0 ⇒ dA = 0 and A = 0 ∂M 33
Proof of Result � Λ A = Λ 0 IFF dA = 0 and γ A ∈ 2 π N for all loops γ . • Analytic methods show that Λ A = Λ 0 ⇒ dA = 0 and A = 0 ∂M Consider a closed loop γ on M : 34
Proof of Result � Λ A = Λ 0 IFF dA = 0 and γ A ∈ 2 π N for all loops γ . • Analytic methods show that Λ A = Λ 0 ⇒ 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. 35
Proof of Result � Λ A = Λ 0 IFF dA = 0 and γ A ∈ 2 π N for all loops γ . • Analytic methods show that Λ A = Λ 0 ⇒ dA = 0 and A = 0 ∂M Consider a closed loop γ on M : so we deform the curve as such so that γ = Γ 1 + Γ 2 with Γ 2 ⊂ ∂M 36
Proof of Result � Λ A = Λ 0 IFF dA = 0 and γ A ∈ 2 π N for all loops γ . • Analytic methods show that Λ A = Λ 0 ⇒ dA = 0 and A = 0 ∂M Consider a closed loop γ on M : � � � We need that γ A = Γ 1 A + ∈ 2 π N A Γ 2 � �� � =0 37
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