Classical Elastic Scattering Probabilities and the Necessity of a Wavelength

   In previous notes, we argued that one may introduce a complex probability into Newtonian mechanics to describe elastic scattering. In particular, even in exercises in Newtonian mechanics, given an elastic two-body collision, energy and momentum are conserved and so one accepts that any E, p sets that satisfy this conservation have equal probability. In other words, deterministic mechanics is not deterministic enough in this case to predict the exact outcome of a collision of two (e1,e1), (p1,p2) vectors. This justifies the introduction of energy and momentum probabilities.

   In previous notes, we suggested exp(ip) as the momentum probability. A complex number with unit modulus is used because each free particle has the same weight. Here we point out a problem with this definition which was not  mentioned in previous notes. Given the periodicity of exp(ip), exp(ip) = exp(i (p+2*n*3.14) which means that p has the same probability as p+ 2*n*31.4. One, however, wishes to have a unique probability. In previous notes we went on to argue that one ultimately requires exp(-iEt+ip dot r) as this is Lorentz invariant and exp(ipx) yields the same value for the x-axis pointing in one direction or the other.

   If one considers exp(ipx), however, the same problem mentioned above occurs, because any px1+ 2*n*3.14 yields the same value for exp(ipx) and so any pnew= p + 2*n*3.14/x1 has the same probability as p when what one desires is a unique probability. We note that for changing x1 values, the p(new) values change. In fact, one may obtain a uniqueness for exp(ipx) if one considers a range of x values, i.e. the wavelength=hbar/|p|. This leads to the notion of orthogonality, i.e. Integral dx exp(-ip1x) exp(ip2x) = 0 if p1 not= p2. By sampling x-space, i.e by taking the notion of wavelength seriously, one may have exp(ipx) represent a unqiue p value over a range of x because in such a case one has a unique number, namely the wavelength hbar/|p|. Thus, it is the wavelength and the positive and negative values of sin(px) and cos(px) which in fact define the probability associated with p. P represents a momentum impulse hit, but in terms of probability, it is actually the wavelength and the adding and subtracting of probability in space which defines p, we argue. This probability in space with addition and removal of probability then becomes important in a calculation of a probabilistic interaction with a 2-slit apparatus in which the slits are separated by about a wavelength.