Quantum Average Momentum Compared with Classical Statistical Pressure

   Classical statistical pressure is calculated by averaging 2pv where 2p is the momentum change related to an impulse hit Integral F dt (i.e. the particle bounces back) and v is related to flux which in turn is linked to time or the number of particles with momentum p in a certain volume which strike a surface in a given time interval. The time units in the flux term v cancel dt time units in the impulse.

     We have argued in previous notes that within a quantum wavelength hbar/p (from exp(ipx)) there is no internal time (although there is an external one linked with x=p/m t.). The two dimensions of exp(ipx) account for the direction of motion and so time is not needed internally. [Given that exp(ipx) contains p which in turn is linked to v, there is an external time in the picture, but not an internal one, we argue.] The quantum bound state problem is statistical, but there is only one particle and no time. Thus instead of considering an average of pv, one may consider an average of the impulse i.e. average momentum given by  -2idW/dx / W where W is the wavefunction.(A bound state wavefunction is real so the i factor indicates that the momentum is not associated with the entire system translating i.e. it is a bound state.)

 Impulse, however, is Integral F dt in classical physics, so an average 2p treated as an average impulse contains some kind of average time term. We argue that this seems to be hbar/E in a quantum system. In order for the quantum system to mimic a Maxwell-Boltzmann gas, we argue that the force -dV/dx at x should match hbar/E * average impulse at x. There is no spatial density in W(x) (W*(x)W(x) is spatial density) so one does not need to consider a little box, as in the classical statistical mechanical case. Interestingly, there is a quantum case in which this situation arises, namely the oscillator ground state.