How can the momentum be non-zero if the probability current density is zero?

[neelakash]

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|?

 

[neelakash]

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?

 

[vanesch]

If the two terms are present, then your particle is in a superposition of having momentum +hbar k and momentum - hbar k.

This is the momentum equivalent of being in a superposition of positions (like the wavefunction indicates you).