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Aharanov-Bohm Effect and Free Particle Probability Francesco R. Ruggeri Hanwell, N.B. Oct. 23, 2025 In previous notes, we argued that one may obtain the quantum free particle wavefunction exp(-iEt + i p dot r) by considering the notion of a free particle probability in Newtonian mechanics, in particular in two body elastic scattering. We argued that given initial (e1,e2) energies and (p1,p2) momentum vectors, any outcome pair (ei,ej) and (pi, pj) should be equally likely and so there should exist a probability for P(p) such that P(p1)P(p2) = P(p1+p2). If one considers a particular p along the x axis, one would anticipate exp(iE) and exp(ip) as the probabilities. This, however, leads to a problem as changing the direction of the x-axis changes exp(ip) to exp(-ip), in other words the value of the probability is changed which should not be. We argued that this may be remedied by using exp(ipx) which may be generalized to exp(-iEt+i p dot r) to make the probability Lorentz invariant. We note that a free particle has a probability which is a unit modulus complex number because it has no real value weight. Here we consider the following. If one writes exp(ip1x)exp(ip2x), it is as if a resonance occurs with exp(i (p1+p2) x). The point we make is that in a two body collision, p2 is considered as a force source for p1, yet given that p2 contains momentum, one may consider a probability linked to momentum (and space as well as argued above). We argue that to be consistent with this notion, one should be able to predict the Aharanov-Bohm effect a priori in the following manner. Usually B= grad x A (B=magnetic field, A=vector magnetic potential) is considered in terms of the Lorentz force F = q v x B. In other words, the B is only considered in Newtonian mechanics if it creates a force. We have argued above, however, that exp(ipx) is about momentum and Lorentz invariance. P2 ultimately delivers a force on p1 in p1+p2 = p3+p4, but p2 also represents momentum and so when considering momentum in space and probability which is Lorentz invariant, one may write exp(ip1x)exp(ip2x). We suggest that A (magnetic vector potential) represents momentum and transforms like one under a Lorentz transformation. Thus, if exp(i (p1+p2) x) is a resonance of p1 and p2 (ignoring any force effects of p2 on p1), then one should have a resonance of the total momentum, i.e. p(particle) and A(x). If one considers exp(i (p1+p2) x) as the total momentum, then one should have a probability exp(i ptotal x) = exp(i [p(particle) x + Integral (0,x) A(x’)dx’]). For the region in which A=0, the phase exp(i integral (0, x1) A(x’)dx’) must remain as probability is a continuous function. In dealing with the Aharaonov-Bohm situation one may use Lagrangian formalism which leads to a momentum of p= mv+qA for L=.5mv dot v +e v dot A + ephi ,i.e. the Lagrangian momentum is not the particle nonrelativistic momentum mv, but a total momentum as advocated above. The Hamiltonian is: H= mvv/2 + e phi. If one wishes to use -id/dx, this must represent the total momentum, as there is a resonance between mv and e/c A we argue. This leads to H= (p-e/c A) dot (p-e/c A) + e phi. This means the wavefunction for the system cannot be exp(i p (particle) x) even though p (particle is not changed by any force, but one must have exp(i[ p(particle)x+ Integral (0,x) A(x’)dx]). Thus, it is the resonance total momentum which appears in the wavefunction which shows that the particle is linked with a phase shift after it leaves the region of A (i.e moves into one in which there is no A). The wavefunction is not simply exp(i p (particle) x) then as might be expected because no force occurred.
We have argued in (1) that even though E and p are the variables of interest in Newtonian mechanics, exp(ipx) is a probability in space and that resonances may occur which are linked to probability issues and may give a wavefunction a phase as in 1-D reflection-refraction at an n1-n2 junction (x=0) for n1<n2. Here it is momentum considerations in the resonance and not flux ones which lead to the phase. The probability phase which occurs in the resonance region then does not disappear when the particle is free from A, just as the -1 phase that a reflected photon needs at x=0 does not disappear. Free Particle Probability We have argued in previous notes that one may define a free particle particle probability for a Newtonian elastic two body scattering event. A free particle has no real value weight (unlike a particle in an ideal gas which has, i.e. p(ei)). The probability is associated with energy and momentum. In particular, one may postulate that for any initial (e1,e2) energies and (p1,p2) momentum vectors, that any pair (ei,ej) and (pi,pj) (vectors) have the same product weight. This leads to the notion of a probability of: exp(iE) and exp(i |p|) ((1)) If p lies along the x-axis and one changes the direction of this axis, then p→ -p and one has exp(ip) change to exp(-ip). This cannot be, i.e. one cannot have different probabilities. This is equivalent to stating that one must time reversal symmetry and this may be achieved by using: exp(i p x) ((2)) To make the probability Lorentz invariant, one may ultimately use: exp(-iEt+i p dot r) ((3)) A probability linked to p and E (Newtonian variables) is now also linked to space. A key point we wish to make is that for a two body collision one has: exp(ip1x) exp(ip2x) if p1 and p2 lie along the x-axis initially ((4)) We argue that a kind of resonance probability is created, i.e. exp(i (p1+p2) x). This resonance contains the full momentum present, even though p2 is a source of the force which will act on p2 and vice versa. In other words, one may think of a force in terms of its momentum value and use this in this free particle probability. Extension to the Case of a Magnetic Field If electric and magnetic fields are present, one has the well-known Lagrangian:
L= .5m v dot v + e v dot A(x) + e phi(x) ((5)) A(x) is the magnetic vector potential and B= grad x A(x) ((6)) B creates a force on a charged particle through the Lorenz equation: F = e (El+ vxB) ((7)) where El is the electric field and v, the velocity of the particle. Newtonian mechanics is described by forces creating changes, but we argued above that even though p2 represents a force which acts on p1, it also represents momentum and so if one considers the resonance momentum, one has exp(i (p1+p2) x ) ((7)) Even if B does not create a force on a charged particle, the A field is present and represents momentum because it (e/cA) transforms with e phi as a Lorentz 4-vector. (charge * phi) is energy and so e/c A must be a kind of momentum. Thus, according to the recipe of ((7)) one should write a priori: exp(i ( p(particle) x + integral (0,x) A(x’)dx’ ) ((8)) Now A may only exist in a certain region, say (0, x2). In such a case, the probability of the resonance becomes exp(i p(particle)x + integral (0,x1) A(x’) dx’ ) ((9)) outside of (0,x1). This must hold because the exp(i function) form is a probability and one must have continuity of probability in space. One cannot have the phase created during the resonance simply disappear. After the resonance is finished one has ((9)) and the free particle is no longer the same as a particle which did not form the resonance. Such a particle would only have: exp(i p(particle) x) ((10)) The resonance has an effect on the particle beyond simply the effects of force. ((9)) is called the Aharanov-Bohm effect, but we argue that a similar effect occurs in one-dimensional reflection-refraction at an n1-n2 junction for n1<n2. One Dimensional Reflection-Refraction We argued above that a particle forms a resonance with a vector magnetic field and that this is described by a total momentum, i.e. p (particle) + e/c A. This leads to a physical effect, namely a
phase, when the particle leaves the resonance region, i.e. enters a region in which A=0. Thus, a physical effect results even though A is not linked with an B field which creates a force on the particle. A similar phase results in the resonance formed in 1-D reflection-refraction, even though this interaction is described in a very different manner. If one has an incident photon, it may reflect or refract at an n1-n2 junction at x=0. We suggest that there is a dx region in which one does not know whether one has an incident, reflected or refracted photon. Thus, probabilities for all three must be considered. We write this in terms of continuity of exp(ipx)s and d/dx of exp(ipx)s i.e. A exp(ipx) + B exp(-ipx) = C exp(ip2 x) at x=0 ((11a)) Ap exp(ipx) - pB exp(-ipx) = Cp2 exp(ip2 x) at x=0 ((11b)) If n1<n2, then for n1=1, p2=n2 p and c2=c/n2. Here AA, BB, CC are the fluxes of the incident, reflected and refracted photon. One has the equation: 1= P(reflected)+P(refracted) —> AA/c = BB/c + CC/c2 ((12)) For AA/c=1, CC/c2<1 c2=c/n2 so CC/c2= CC n2 /c. Thus, C<A. For this to hold, B<0 and so a reflected photon in the dx region must acquire a phase of: -1 = exp(i 3.14) ((13)) This phase must manifest itself outside of the resonance region, i.e the reflected photon receives a phase which has nothing to do with the fact that it has -p. In the case of probabilities of the form exp(ipx), there is more physics than that linked with Newtonian mechanics. One cannot simply say that because no force occurs, nothing happens. Lagrangian-Hamiltonian Formalism We argued above that one should use a total resonance momentum to describe the case of a charged particle moving in A(x) even if there is no B field causing a force. This notion of momentum already appears in classical Lagrangian formalism: L = .5m v dot v + e v dot A + e phi ((13)) p = dL/dv = mv + e A ((14)) p is not simply the momentum of the particle. The Hamiltonian is: H= pv - L = .5 m v dot v + e phi ((15)) Writing this in terms of p total one has: H = (p - eA) (p-eA) /2m ((16))
In quantum mechanics, one uses p= -id/dx and so one requires a wavefunction of ((8)) to satisfy this equation. This is consistent with the Aharonov-Bohm effect in the resonance region. As argued, the phase term linked to A cannot disappear when A=0 because of continuity of probability. Conclusion In conclusion, we argue that the notion of introducing a free particle probability into Newtonian mechanics for two-body elastic scattering in which any (e1,e2) (p1,p2) pair has equal product probability to produce (ei,ej) (pi,pj) as long as conservation occurs ultimately leads to: exp(-iEt+i p dot r). We note that in a 2-body collision with p1 and p2 along the x-axis one has the resonance exp(i (p1+p2) x). We also note that p2 is the source of the force acting on p1, but it represents momentum as well and so appears in the exp() resonance probability. To be consistent with this scenario, we argue that a vector magnetic potential represents momentum because it is part of a 4-vector (e/cA , ephi), where ephi is energy. B= grad x A and F= e v xB, but even if there is no force, there is still a momentum linked with A and just like exp(i (p1+p2) x), one should be able to write the resonance probability: exp(i p(particle) x + Integral (0,x) A(x’)dx’). Outside of the resonance region (where A=0), one should still have the phase exp(i Integral(0,x1) A(x’)dx’). Thus, the resonance has introduced a physical effect even though no force has occurred. We argue that this is not the only place where one has such a resonance effect. We note that in 1-D reflection-refraction at an n1-n2 index of refraction junction, one has the reflected photon pick up a phase -1=exp(i 3.14) if n1<n2. This has nothing to do with -p, the reflected momentum, but is an extra effect linked to the resonance and described above. Thus, it is not only Newtonian force considerations which lead to physical effects when one considers the free particle probability introduced in the first section. References 1. Ruggeri, Francesco R. Complex Probabilities in Free Particle Quantum Mechanics Part 3 (preprint, zenodo, 2025)
