Probability current inside the barrier of a finitie square potential well

[StephenD420]

if ψ=C*e^(kx) + D*e^(-kx)
show that the probability current density is
Jx=(i*k*hbar/m)[c*conj(D) - conj(C)*D]

since Jx= (i*hbar/2m)*[ψ * derivative of conj(ψ) - conj(ψ)*derivative of ψ]
ψ=C*e^(kx) + D*e^(-kx)
conj(ψ)= conj(C)*e^(-kx) + conj(D)*e^(kx)
ψ ' = C*k*e^(kx) - D*K*e^(-kx)
derivative of conj(ψ) = -conj(C)*k*e^(-kx) + conj(D) *k*e^(kx)

plugging in and simplifying I get

Jx = (i*hbar/2m)*[-C*conj(C)*k - k*conj(C)*D*e^(-2kx) + C*conj(D)*k*e^(2kx) + D*conj(D)*k -C*conj(C)*k -C*conj(D)*k*e^(2kx) + conj(C)*D*k*e^(-2kx) +D*conj(D)*k]

which simplifies to
Jx = (i*hbar/2m)*[-2*c*conj(C)*k + 2*D*conj(D)*k]
Jx = (i*hbar/m)*[D*conj(D) - C*conj(C)]

how do I get from here to
Jx=(i*k*hbar/m)[c*conj(D) - conj(C)*D]

Thanks so much for any help you guys can provide. I am really stuck as to what to do next.
Thanks.
Stephen

 

[StephenD420]

bump...
Please help me with a nudge to finish this problem up.

Thank you for any help you guys can give me. I really appreciate it.
Stephen