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Aharanov-Bohm Effect and Free Particle Probability Part 2 Francesco R. Ruggeri Hanwell, N.B. Oct. 24, 2025 In Part 1, we argued that a charged particle moving through a region with a magnetic vector potential acquires a phase even if it is not acted upon by a magnetic field. Classically, one might expect that only a force changes the particle physically, i.e. changes its momentum and that there is nothing else to change. In fact, the Hamiltonian for a particle in the presence of an electric potential phi and a magnetic vector potential is: H= pp/2m + e phi (nonrelativistically). One could write such a simple equation without any knowledge of a magnetic vector potential and so one may ask: Why did we state in Part 1 that there exists a resonance between the particle and A, the magnetic vector potential? In particular, why is the Lagrangian: L = .5mvv + A dot v - e phi and not L= .5mvv - ephi which would yield the same Hamiltonian? We note that the first Lagrangian yields: p = mv + A (one dimension). In other words, there are two pieces to momentum. We show that this is compatible with special relativity, but it still does not explain why a free particle should form a resonance with an A field. We argue that there seems to be a physical interaction occurring because: Electric field = - grad phi - dA/dt partial. If one only has an A(x) and a nonrelativistic particle moving with v, from the point of view of the particle’s rest frame, the A field is moving and one should have Afunction( x-vt) and so there is a d/dt partial A function term which contributes to an electric field acting on the particle. Formally, one should transform: (eA, 0) using a Lorentz transformation, obtaining phi’ = gvA and A’ = gA. A(x) may be written as A(x’-vt’). Thus: Electric field = - grad’ phi’ - d/dt’ partial A’ = 0 which must be the case because the particle does not accelerate. Nevertheless it interacts with both a phi’ and an A’ field and so physically there may very well be a resonance occurring even though the net force is zero. If quantum mechanics is focused on exp(i Integral(0,x) p(x)dx) where p (total) is the total momentum of the resonance, then p(particle) x + integral (0,x) A(x) dx represents the total resonance momentum which one may write because the particle and the A field are physically interacting (just as the particle at rest would interact with the A’ and phi’ moving fields). The fact that the classical Lagrangian yields ptotal = mov + A, yet the Hamiltonian is movv/2 + e phi seems to mean that the particle is in a resonance with A and that both momenta are present even though the particle’s momentum is not changing. Given that the particle is in a resonance state, it seems it would make no sense to use exp(i p(particle) x) as if the particle were moving through free space with no resonance, we argue. Vector Magnetic Potential as a Momentum It is well-known that a magnetic vector potential is part of a Lorentz 4-vector: (e/c A, phi) ((1)) If a particle interacts with phi, one has an energy e phi (e=charge of the particle), and so if phi is thought of as energy in ((1)), e/c A must be thought of as momentum. This, however, is a formal statement. Furthermore, one may argue that even if A is a momentum, this does not mean that
a particle moving with constant speed through a region with A(x) present such that there is no B (magnetic field) and no Electric field should form a resonance with A(x). If this is in fact the case, one would expect the wavefunction of the particle to be: exp(i p(particle) x) ((2)) From a Newtonian point of view, it is as if there is no A(x) present. This is in fact exemplified by: H= p (particle) p(particle) /2m (or H=p (particle) p(particle)/2m + e phi if an electrostatic potential is present) ((3)) If this is the case, one would postulate the following Lagrangian: L = .5mvv - e phi(x) ((4)) This yields the Hamiltonian ((3)), but does not properly describe a particle which is present in both a magnetic and electric field. In such a case, one has: Electric field = -grad partial phi - d/dt partial A(x,t) ((5)) This electric field acts on the particle as a force when multiplied by e, but ((4)) suggests that: dp/dt = - grad phi ((6)) Thus, from the point of view of the electric field alone, ((4)) must be wrong. An equation for dp/dt must contain A, even if there is no B field acting on the particle. We note that: B= magnetic field = grad x A ((7)) If A appears in a Lagrangian for a particle, i.e as: L = .5mvv + v dot A - e phi ((8)) then the particle must interact with both an A field and a phi one. Thus, we suggest that there is in fact a resonance which occurs between the particle and the A field. Second Way to Explain a Resonance between a Particle and A One may imagine a particle with constant speed moving through an A field which is a function of x alone. One may, however, enter the rest frame of the particle and so transform A as: | g gv | | e/c A| | gv g| |0 | ((9)) Here g=1/sqrt(1-vv/cc)
Then: phi’ = gvA(x) = gvA(x’-vt’) and A’= gA(x’-vt’) ((10)) Creating Electric field = - d/dx’ phi’ - d/dt’ partial A =0 ((11)) This means that the particle does not accelerate, but it is as if it interacts with both a phi’ and A’. We argue that this is equivalent to it interacting with A when it moves (just as the Lagrangian indicates). The catch is that its overall acceleration is not changed. Classically, one might argue that as long as there is no net force, nothing happens. Quantum mechanics, however, describes a particle with constant momentum by exp(i p(particle) x). One cannot, however, write exp(i Integral p(x) dx) if the particle is in a V(x) which can change p i.e. exp(i Integral p(x) dx) does not describe the wavefunction in a bound state of V(x). The resonance of the constant p(particle) particle with A, however, does not involve any physical force. Thus, one cannot write: Sum over p a(p) exp(ipx) ((12)) it seems, because various p values should be measurable. In the resonance, there is only p(particle) which does not change and A(x). We suggest: exp(i p(particle) x + integral (0,x) A(x1)dx1) ((13)) As a result a phase is picked up if the particle moves from 0 to x2 at which point A=0. Given that ((13)) represents a free probability, it must be continuous and so one has at x2: exp(i p(particle) x + integral(0,x2) A(x1)dx1) ((14)) ((14)) must hold for all points beyond x2. It seems a physical interaction, i.e. resonance may change the phase of exp(ipx) even if no force acts on the particle. The point we wish to make is that there do seem to be physical interactions which occur even though the force is overall 0. Conclusion In conclusion, Newtonian mechanics focuses on changes in momentum and or energy due to the action of a force. One, however, has many cases of net forces of 0 which are not the same as no force present at all. In Part 1, we suggested that a charged particle which passes through a region with no electrostatic potential and a vector magnetic potential A whose B field renders no force on the particle, still interacts with the particle to form a resonance. Here, we try to verify that such a resonance occurs. We first note that even though the Hamiltonian is H= p(partilce)p(particle)/2mo (+ e phi if an electrostatic potential is present), the Lagrangian is not L= .5movv - e phi, but L = .5movv + v dot A - ephi. This, we argue, suggests that the particle is interacting with A, regardless of whether the B field acts on the particle. In particular, the second Lagrangian is required to be consistent with Electric field = -grad partial phi - d/dt partial A. We provide a second argument. We consider a time independent A(x) and a moving particle (v=constant). In the rest frame of the particle, one may Lorentz transform A(x) to yield: phi’ = gvA(x-vt) and A’=gA(x-vt). This then leads to Electric field = 0, which must be the case, but there
are two interactions occurring, one with a phi’ and the other with A’. Thus, we suggest that A(x) must also interact with the particle. In quantum mechanics, one uses exp(i p x) to describe a free particle with constant p. Here one has a resonance of p(particle) with A (which mathematically behaves as a momentum under a Lorentz transformation). We suggest that an interaction between the particle and A is physically real and so one may postulate writing: exp(i p(particle)x + i integral(0,x) A(x1)dx1). If A=0 beyond x2, then the exp(i function) must be continuous at this point, as it is a probability, and so there is an acquired phase.
