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Probability current inside the barrier of a finitie square potential well 
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#1
Feb2612, 03:45 AM

P: 100

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
Feb2712, 12:53 AM

P: 100

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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