Find #j(x)#. What type of wave this is describing?

#j(x)# is NOT a wave function! It does not describe a wave at all. In fact, we could choose any arbitrary wave, and it would not change the definition of #j(x)#.

The probability current, #j(x,t)#, is the flow of probability through a certain unit area per unit time. We could then predict that #j(x)# is a constant w.r.t. position, as the probability density does not vary if we only look at some initial state #j(x,0)#.

#(i)#

Well, if we want to show that #j(x)# is a constant w.r.t. #x#, we just need

#(dj(x))/(dx) = 0#.

So... we first find the form of #(dj(x))/(dx)#.

#(dj(x))/(dx) = d/(dx)[(iℏ)/(2m)(psi (dpsi^"*")/(dx) - psi^"*" (dpsi)/(dx))]#

#= (iℏ)/(2m)(psi (d^2psi^"*")/(dx^2) + cancel((dpsi)/(dx) (dpsi^"*")/(dx)) - psi^"*" (d^2psi)/(dx^2) - cancel((dpsi^"*")/(dx)(dpsi)/(dx)))#

The time-independent Schrodinger equation is

#-ℏ^2/(2m) (d^2psi)/(dx^2) + V(x)psi = Epsi#

so we have:

#(d^2psi)/(dx^2) = -(2m(E-V(x)))/(ℏ^2)psi#

#(d^2psi^"*")/(dx^2) = -(2m(E-V(x)))/(ℏ^2)psi^"*"#

Therefore:

#(dj(x))/(dx) = (iℏ)/(2m)(psi [-(2m(E-V(x)))/(ℏ^2)psi^"*"] - psi^"*" [-(2m(E-V(x)))/(ℏ^2)psi])#

#= -(iℏ)/(2m)((2m(E-V(x)))/(ℏ^2)psipsi^"*" - (2m(E-V(x)))/(ℏ^2)psi^"*"psi)#

Obviously, #psi# commutes with #psi^"*"#, so

#color(blue)((dj(x))/(dx)) = -(iℏ)/(2m) cdot (2m(E-V(x)))/(ℏ^2) cdot (psi^"*"psi - psi^"*"psi)#

#= color(blue)(0)#

Thus, #j(x)# is a constant w.r.t. #x#.

#(ii)#

We consider the plane wave #psi(x) = Ae^(ikx) + Be^(-ikx)#, with #k# being the usual wave number #k = p/ℏ#.

It has some probability of reflection or transmission (we do not know what, until we know the properties of #V# as a function of #x#), and its derivatives are:

#(dpsi)/(dx) = ikAe^(ikx) - ikBe^(-ikx)#

#(dpsi^"*")/(dx) = -ikA^"*"e^(-ikx) + ikB^"*"e^(ikx)#

Therefore, for this #psi(x)#:

#color(blue)(j_1(x)) = (iℏ)/(2m)[(Ae^(ikx) + Be^(-ikx))(-ikA^"*"e^(-ikx) + ikB^"*"e^(ikx)) - (A^"*"e^(-ikx) + B^"*"e^(ikx))(ikAe^(ikx) - ikBe^(-ikx))]#

#= (iℏ)/(2m)[-ikA A^"*" + cancel(ikAB^"*"e^(2ikx)) - cancel(ikBA^"*"e^(-2ikx)) + ikBB^"*" - ikA A^"*" + cancel(ikBA^"*"e^(-2ikx)) - cancel(ikAB^"*"e^(2ikx)) + ikBB^"*"]#

#= (iℏ)/(cancel(2)m)[cancel(2)ik(B B^"*" - A A^"*")]#

#= color(blue)((ℏk)/m(A A^"*" - B B^"*"))#

For #psi(x) = Fe^(ikx)#,

#(dpsi)/(dx) = ikFe^(ikx)#

#(dpsi^"*")/(dx) = -ikF^"*"e^(-ikx)#

Thus:

#color(blue)(j_2(x)) = (iℏ)/(2m)[Fcancel(e^(ikx))(-ikF^"*"cancel(e^(-ikx))) - F^"*"cancel(e^(-ikx))(ikFcancel(e^(ikx)))]#

#= (iℏ)/(2m)[-ikF F^"*" - ikF F^"*"]#

#= (iℏ)/(cancel(2)m)[-cancel(2)ikF F^"*"]#

#= color(blue)((ℏk)/(m)F F^"*")#

The terms #A A^"*"#, etc. indicate the probability of the event occurring corresponding to its #e^(pmikx)# term. Of course, this means that

• the first plane wave has some finite probability of moving forward (#e^(ikx)#) or moving backward (#e^(-ikx)#).
• the second plane wave obviously has a #100%# probability of moving forward (#e^(ikx#) (how else would it move?).

#(iii)#

This is a pretty roundabout way of simply saying to place a barrier of arbitrary height in front of the particle of width #2a# centered at #x = 0#:

#V(x) = {(0, |x| > a),(V_0, -a <= x <= a):}#

Here we define #k# on the left of the barrier and #k'# on the right of the barrier (#k ne k'#). That is, we define the wave functions for the left and right of the barrier as:

#psi_L(x) = Ae^(ikx) + Be^(-ikx)#

#psi_R(x) = Fe^(ik'x)#

where #psi_R(x)# necessarily has no #Ge^(-ik'x)# term so that the particle does not double back.

Therefore, what we got in #(ii)# is now specified:

#j_L(x) = (ℏk)/m(A A^"*" - B B^"*") = "const"#

#j_R(x) = (ℏk')/mFF^"*" = "const"#

From #(i)#, since #j(x)# is a constant w.r.t. #x# (which we see here) #j_L(x) = j_R(x) = "same const"#.

The reflection coefficient is defined as

#R = (B B^"*")/(A A^"*")#

And the transmission coefficient is defined as

#T = (k')/(k)(F F^"*")/(A A^"*")#

First, we can say that

#overbrace(cancel((ℏ)/m)k(A A^"*" - B B^"*"))^(j_L(x)) = overbrace(cancel((ℏ)/m)k'FF^"*")^(j_R(x))#

or that

#-(A A^"*" - B B^"*") = B B^"*" - A A^"*" = -(k')/k FF^"*"#

Therefore, we use our results to find:

#color(blue)(R + T) = (B B^"*")/(A A^"*") + (k')/(k)(F F^"*")/(A A^"*")#

#= (B B^"*" + A A^"*" - A A^"*")/(A A^"*") + (k')/(k)(F F^"*")/(A A^"*")#

#= 1 + (B B^"*" - A A^"*")/(A A^"*") + (k')/(k)(F F^"*")/(A A^"*")#

#= 1 - cancel((k')/k (FF^"*")/(A A^"*") + (k')/(k)(F F^"*")/(A A^"*"))#

#= color(blue)(1)#

This result should make sense.

Obviously, a particle can only reflect or transmit through a potential barrier. There is nothing else it can physically do, so this total probability should be #100%#.