I am reading One-dimensional examples from Bransden and Joachain.For the free particle solutions:Ψ=A exp [i(kx-ωt)] +B exp [–i(kx+ωt)] they say that for |A|=|B|,the probability current density=0.This is OK.Then they say we can associate the standing wave with a free particle along the x axis with a momentum whose magnitude is p=ћk but the direction is unknown... My problem is I cannot understand what they say regarding momentum.If j=0,how can momentum be non-zero? In fact, if A=0 or, B=0 I can see there is a momentum of precise value p=ћk.There j is non-zero and j=vP where P is the probability... Also, in the very next example of a potential step they show j=0 everywhere and concludes that no net momentum in the state... I tried to solve the problem by thinking that in the latter case, j=0 for the entire state.So, by conservation of momentum, p=0 everywhere... But in the former case, j is not zero everywhere.So, p must conserve its non-zero value...!!!Or, may be that They meant momentum direction is unknown as there is no net momentum in free particle if |A|=|B|?
Let me approach mathematically: Ψ=A exp [i(kx-ωt)] +B exp [–i(kx+ωt)] For A=B, Ψ=C cos kx exp(-iωt) Operating this with momentum operator,we are getting -(ћk/i) C sin(kx) exp(-iωt) Sin and cos differ by a phase factor only... So,momentum value matches...and the direction is "unknown" for the "i"... what about the potential step?
