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Probability current inside the barrier of a finitie square potential well

  1. Feb 26, 2012 #1
    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
     
  2. jcsd
  3. Feb 27, 2012 #2
    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
     
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