In this note, we consider the Hamiltonian 1/2m (p-e/c A)(p-e/c A) W = E W ((1)) in the situation where there is no magnetic field even though A(x) still exists. Thus, if there is no force, one might expect to have a free particle which satisfies -1/2m d/dx d/dx W = E W and yields a free particle spatial density. Thus, one might guess the solution to ((1)) should be a free particle wavefunction exp(ikx) with perhaps a phase added. Such a phase does not affect spatial density W*W. One may solve ((1)) using the assumption that d/dx A(x) = 0 (used in the literature (1)) to find: W= Wfree(x) exp(ie/hbar c Integral x1=0 to x A(x1)dx1). Thus, one treats this phase as if one had exp(i K x) where K=A(x). A second derivative d/dx brings down i A(x) again. In previous notes, we suggested that a potential is stochastic and may be written as: V(x)= Sum over k V(k) exp(ikx). If one expands the terms in ((1)) and uses d/dx A =0, one obtains a momentum dependent potential. This potential should be physical, we argue, because it applies to cases where there is a magnetic field as well as to the case here where there is not. Thus, the potential should give hits of the form V(k) exp(ikx) as well as couple to d/dx W. In other words, we argue these hits are real whether or not average force is zero. The effect of such hits is to create a momentum distribution at each x: P(p/x) = a(p) exp(ipx) / W(x). There is forward and backward motion, but the average kinetic energy(x) + V(x)= E at each point where: kinetic energy(x) = [Sum over p p*p/2m a(p) exp(ipx)]/W(x). Given that average kinetic energy should be p*p/2m for a free particle (if the potential yields no hits), one may consider a phase exp(i K(x) x) multiplied by Wfree(x) i.e. exp(ipx) . This seems to imply a shifted velocity i.e. a local collective velocity of K(x) where K(x) happens to equal A(x) in this case. Matters, however, seem to be more complicated as we consider a microscopic coupling of exp(ikx) terms from both Wfree(x) and exp(i Bohm Aharonov) phase just as there may be a coupling between W=WoW1 where Wo(x) is the ground state and W1(x) represents the effects of an interacting substance (e.g. a photon or phonon).
In this note, we consider the Hamiltonian 1/2m (p-e/c A)(p-e/c A) W = E W ((1)) in the situation where there is no magnetic field even though A(x) still exists. Thus, if there is no force, one might expect to have a free particle which satisfies -1/2m d/dx d/dx W = E W and yields a free particle spatial density. Thus, one might guess the solution to ((1)) should be a free particle wavefunction exp(ikx) with perhaps a phase added. Such a phase does not affect spatial density WW. One may solve ((1)) using the assumption that d/dx A(x) = 0 (used in the literature (1)) to find: W= Wfree(x) exp(ie/hbar c Integral x1=0 to x A(x1)dx1). Thus, one treats this phase as if one had exp(i K x) where K=A(x). A second derivative d/dx brings down i A(x) again. In previous notes, we suggested that a potential is stochastic and may be written as: V(x)= Sum over k V(k) exp(ikx). If one expands the terms in ((1)) and uses d/dx A =0, one obtains a momentum dependent potential. This potential should be physical, we argue, because it applies to cases where there is a magnetic field as well as to the case here where there is not. Thus, the potential should give hits of the form V(k) exp(ikx) as well as couple to d/dx W. In other words, we argue these hits are real whether or not average force is zero. The effect of such hits is to create a momentum distribution at each x: P(p/x) = a(p) exp(ipx) / W(x). There is forward and backward motion, but the average kinetic energy(x) + V(x)= E at each point where: kinetic energy(x) = [Sum over p pp/2m a(p) exp(ipx)]/W(x). Given that average kinetic energy should be p*p/2m for a free particle (if the potential yields no hits), one may consider a phase exp(i K(x) x) multiplied by Wfree(x) i.e. exp(ipx) . This seems to imply a shifted velocity i.e. a local collective velocity of K(x) where K(x) happens to equal A(x) in this case. Matters, however, seem to be more complicated as we consider a microscopic coupling of exp(ikx) terms from both Wfree(x) and exp(i Bohm Aharonov) phase just as there may be a coupling between W=WoW1 where Wo(x) is the ground state and W1(x) represents the effects of an interacting substance (e.g. a photon or phonon).
In a previous note (1), we argued that a potential V(x) may be written as Sum over k V(k) exp(ikx). Thus, -dV/dx is only an average force which may in fact be zero, yet -d/dx V(k) exp(ikx) is not zero leading to possible forward and backward motion/acceleration (due to each V(k) exp(ikx) which act at a different times). This gives rise to a momentum distribution P(p/x) =a(p) exp(ikx)/W(x) (W(x)=wavefunction) at each x. In this note, we wish to link these ideas to the Bohm Aharonov effect by considering two recent (2018, 2020) examples from the literature. In the first (2), the magnetic potential A(r) is written in a Fourier series in the same manner as V(k)exp(ikx). Thus, although grad A = B (magnetic field) = 0, grad A(k) exp(ikx) should not be zero and so there may be hits from “virtual photons”. (2) applies perturbation theory to find a correction to energy E and then considers exp(i Integral over t delta E dt) as in adiabatic theory. The second approach (3) applies the usual Hamiltonian containing the magnetic potential A, but imagines a little box at various xi positions such that A(xi) is a constant at each xi. In such a case, one may show directly that W(x)= W(x=0) exp( ie/hbar Integral xo to x1 A(x) dx) which yields the Ehrenberg–Siday–Aharonov–Bohm effect. In this approach, one essentially has a different constant potential at each point and hence a different energy. In a previous note, we argued that even though average kinetic energy may remain the same, a change in constant potential using Sum over k V(k) exp(ikx) gives rise to different physical effects which try to argue are related to the Bohm Aharonov effect. We also try to consider a more general case.
The Bohm Aharonov phase shift for a free quantum particle with momentum p moving through a magnetic potential A(x) with no magnetic force is well known as exp(i Integral dx1 (a,x) e/cA(x1)dx1. In the literature this has been obtained in two very different ways. The first way (standard) (1) is to use the classical equation for a charged particle in an electromagnetic field with electrostatic potential i.e. (p-e/cA)(p-e/cA)/2m = E with p=-id/dx representing an overall momentum, not just that of the particle. In such a case, there is coupling at the momentum level, but E is the energy of a free particle with momentum p1=p-e/cA (classically) i.e. p1p1/2m. In other words, momentum considerations are treated differently from energy ones. The time dependent wavefunction is exp(-iEt) where E shows only a noninteracting free particle. The second approach (2) is the opposite. It ignores the Schrodinger equation and the momentum picture, but retains the idea that for a nonacclerating free particle p/m = velocity=dx/dt. It then considers an interaction which changes E namely delta E = p e/cA(x)/m. This leads to Integral pA(x)/m dt = Integral pA/m dx/v which yields the same Bohm Aharonov phase shift. In a previous note (3) we considered a free particle classical action as A=-Et+px. We argued that Et and px should be kept separate leading to exp(-iEt) and exp(ipx) (not combined) if one creates an eigenvalue equation. We argued that scaling exists in both time (by E) and space (by p). At the same time, the particle moves classically with a speed v=p/m, but has an associated exp(ipx) momentum wave (and independently an energy wave exp(-iEt)). Even if one uses v=dx/dt as in (2) these waves are different i.e. Et= Integral E dt = Integral .5pv dx/v = .5 px while px is the phase from the momentum consideration. Thus one might argue that it is unusual that a change in E i.e. delta E = p/m A(x) from (2) applied to exp(-iEt) yields the same Bohm Aharonov phase as the Schrodinger equation with p, A coupling, but E left untouched In this note, we consider the classical action A= -Et+px and try to explain this feature directly from this expression.
If a Lagrangian contains a potential dependent on dx/dt, it is possible this does not contribute to the classical equations of motion. Nevertheless, the fact that the potential is present suggests there is a physical impact. A different Lagrangian yielding the same equations of motion, does not necessarily represent the same physical situation i.e. the equations of motion do not uniquely determine a physical situation. In particular, using the Lagrange formalism yields momentum=p = dL / d (dx/dt) which may contain the usual m dx/dt term plus terms from velocity related potentials. We argued in previous notes, that for quantum mechanics there is a momentum distribution at each x point which yields an average pp/2m at x. Moving from one x to another yields a change in pp indicating an average acceleration. The potential V(x), it was argued, may be written as Sum over k Vk exp(ikx). Thus, it is stochastic, yielding momentum hits of k which only yield a force -dV/dx on average. Given that average acceleration is stochastic, average momentum should also be stochastic. For the case of a potential A(x)dx/dt, the Lagrangian momentum dL/d (dx/dt)= mdx/dt +A(x). dx/dt is an average of stochastic value i.e. Sum over p a(p) exp(ipx), but A(x) is also stochastic i.e. A(x)= Sum over k Ak exp(ikx). One may combine the two terms to create a term of the form bk exp(i j x). We argue that this contains the combined stochasticity of p and A(x). Special considerations show that if W(x)=Wo(x)W1(x), then one may break out W1(x) as a phase related directly to A(x) i.e. W1(x)= exp{ i (Integral dx1 (0,x) A(x1) } which is experimentally observable.
In previous notes, we argued one may formulate bound state quantum mechanics by assuming a stochastic potential which delivers “hits” of momentum k V(k) exp(ikx), while having an average form of V(x). A main constraint seems to be that the different k V(k) exp(ikx) hits may occur at any x point and that V(x) be of the form Sum over k b(kx) so that -d/dx b(kx) = -k b(kx). In other words, conservation of energy exists at the level of each hit as well as for the overall average. To force all of these constraints, a solution of b(kx)=V(k) exp(ikx) emerges which introduces periodicity and complex numbers. It should be noted that V(x)= Sum V(k) exp(ikx) so each term is actually a “piece of potential energy”. Furthermore, we argue that V(x) is decomposed for physical reasons i.e. V(x) is a low resolution result which incorporates all of the “pieces”. In a quantum mechanical situation, a “piece” V(k)exp(ikx) may act at t1 and V(k2)exp(i k2x) at t2. We argue that exp(ikx) should have physical meaning and leads to sin(kx) terms for the “k” and “-k” pieces. Thus, there is forward and backward motion which “cancel” i.e. have nothing to do with V(x). For a particle, however, in this region, there may be forward and backward motion, both of which have kinetic energy which should not be part of the average kinetic energy at x. The various physical (resolved) pieces of the potential lead to momentum distribution at each x point with P(p/x)= a(p) exp(ipx)/ W(x) (W(x)=wavefunction). In this case, one has a probability, not “pieces” of KE(x). At first, one may be alarmed by complex probabilities, but they are necessary to allow for cancellation of kinetic energy pieces due to the positive/negative values combined V(k)exp(ikx) pieces. Thus, even though KE(x) (average) matches classical results, the actual motion is more complicated. The potential seems to be the driver of the ensemble picture and should be linked to the microscopic physical nature of the potential. In this note, we wish to examine features and consequences of such a stochastic potential, including the idea of tunneling and infinite potential walls.
In a previous note (1), we argued that W(x+dx)= W(x) exp(i dx ) where = - i d/dx lnW mimicking the behaviour of exp(ipx) for a free quantum particle. In this note, we consider that for the conditional probability P(p/x)= a(p) exp(ipx)/W(x), a shift from x to x+dx leads to: P(p/ x+dx) = P(p/x) exp(i dx (p- ) ) = P(p/x)(1+ idx (p- )). We note that in shifting from x to x+dx the overall sum of probability should not change, so P(p / x+dx)= P(p/x) + P(p/x) g(p,x). If one assumes that g(p,x) is a function of p alone, the simplest solution is p- . If one considers Sum over p pp/2m P(p/x+dx) = Sum over p pp/2m P(p/x) = delta KE ave then this should be equal to Force(x) dx. We also consider classical hydrodynamical equations and compare these with modified equations applying to the bound state quantum situation. The two sets of equations are very similar with density in the classical case being replaced with W(x)exp(-iEt), the wavefunction, and E energy or V(x) (potential) terms replaced with 2E or 2V (i.e. also 2Force). In a previous note, we had considered hydrodynamical equations with the restriction 2m=1 which is too strong. Here any m (mass) may be used.
In Part 1, we suggested that exp(ipx) represents a spatial probability for the rendering of momentum hits. A single particle has a center-of-mass which undergoes x=vt, but momentum hits are not rendered from this point, but are rather delivered in x according to the complex probability exp(ipx) which is linked with the resolution delta x = hbar/p. This accounts for the fact that a particle can probabilistically interact with each slit of a 2-slit apparatus if they are about hbar/apart. If one is only interested in energy, pp/2m, the classical result holds for both exp(-ipx) and exp(ipx). In a classical bound state, one considers only energy p(x)p(x)/2m + V(x)=E because each p(x) exists at x in dx->0. One cannot distinguish any probabilistic hit distribution. exp(ipx) and exp(-ipx) differ even though they are linked with the same kinetic energy pp/2m (nonrelativistic case). We argued in Part 1, that the Newtonian conservation of energy equation actually includes a force profile linked with V(x). The catch is that for V(x) one has a force at each x, dx->0. Such is not the case for exp(ipx). We argued in Part 1 that one must consider not only pp/2m, but the associated spatial probability distribution for hits (i.e. energy together with force). Here, we note that this idea gives rise to an equilibrium calculation. This is somewhat startling because classically, and even quantum mechanically for higher energy levels, a particle moves to the right part of the time and then to the left for other times. As a result, a calculation which respects time would not average over exp(ipx) and exp(-ipx). Only an equilibrium calculation in time would do so. We suggest that one must consider the Newtonian conservation of energy equation as applying to both p(x) and -p(x) and so from an energy point of view one may consider it to be an average. The catch is that each term p(x) and -p(x) yields the same kinetic energy profile in the Newtonian case and the same V(x). For a quantum particle the kinetic energy with probabilistic force distribution is not the same for p and -p. As a result, to use V(x) which applies to forward and backwards motion, one is forced to consider the average and this leads to an equilibrium scenario. We show that the average kinetic with probabilistic force hit profile in space is not the same for +p and -p. Measurements, however, take time and may be averages of much data and so the notion of an average or equilibrium seems to make sense. We note, however, that an average is very different from the Newtonian forwards-backwards motion picture. Only in the high energy limit does the energy pp/2m with probabilistic force profile become the same for both p and -p. Thus, the main point of this note is to stress the notion of a single particle bound state as an average and to point out the consequences of this.