Implications of Average Momentum On Quantum Bound State Behaviour

In a previous note (1) we argued that the translational generator d/dx being associated with momentum (px) already appears in Fermat’s principle for both reflection and refraction. We suggested that d/dx is linked to an invariant spatial variable which in turn is associated with momentum conservation in the same direction. This may be written as:  d/dx (px)x = (px). Alternatively (px)x = -i ln(exp(i px x) ) with exp(i px x) being a directional probability distribution (i.e. values for p and -p differ) and ln(exp(i px x)) being of the form of information in information theory.

   One may then ask: How does one link average momentum in the case of a bound state with a potential V(x)? In general, such a bound system is solved using the time-independent Schrodinger equation i.e. conservation of energy using d/dx twice and a p(rms) not a p average obtain from  -id/dx ln(W) = p(x)  where W(x) is the wavefunction and pave= W dW/dx. After all, the form exp(i (px) x) may be obtained from considerations of d/dx and p, not energy conservation (even though E=pc for light). 

   We argue that exp(ipx) is a directional probability. If both p and -p are present then:  p exp(ipx) -p exp(-ipx) is not zero because the spatial probability distinguishes between the two. Only by multiplying exp(ipx) and exp(-ipx) does one remove motion. If p and -p are present, however, as in an OR situation, i.e. exp(ipx) + exp(-ipx), then it is as if the particle is bound or localized and P(p,x)+P(-p,x) = C cos(px).

  We consider the two components of exp(ipx) in: Integral dx cos(px) sin(px). Why does this equal 0 from a momentum perspective? sin(px) = 1/p d/dx cos(px). Thus the integral is:

Integral dx  { P(p)+P(-p)} momentum(P(p)+P(-p)) } or the average momentum at x. Overall the particle is bound and so there can be no momentum globally so the integral must be zero and hence cos(px) and sin(px) must be orthogonal over a period even though they represent the same p value. 

   This same idea holds for a linear combination of  {P(p,x)+P(-p,x)} sums for different p’s. (We consider overall positive or negative parity.) Thus: Integral dx W dW/dx = 0 holds for even-odd reasons, but also for the physical overall zero momentum. Even with a potential present: -i WW d/dx ln(W) = Pave(x). In other words, bound state quantum mechanics is linked to W dW/dx integrating to 0 even though there are positive and negative pockets at various x values. This follows from the orthogonality of cos(px) and sin(px) over a wavelength from a free particle which conserves momentum. This result holds for any V(x) even though V(x) fixes the coefficients of the linear combination  of  Sum over p  a(p) {P(p,x)+P(-p,x)} using energy considerations.  

  Zero average momentum follows from momentum conservation found in exp(i (px) x). In fact we have argued that the unusual probability wave form exp(i (px) x) exists to ensure constant momentum (momentum conservation) for a free particle.

    We argue that considering average momentum equal to 0 overall leads to ideas of zero point energy, orthogonal energy states and a crest/trough pattern without considering the time-independent Schrodinger equation which is based on energy conservation. Thus we argue that ideas related to d/dx and momentum carry over from Fermat’s principle into a bound quantum state and its average momentum which is often not considered.