If #psi=A(x)e^(itheta(x))#where #A(x) and theta(x)# are real functions prove that the probability current #J(x,t) = |A(x)|^2(ℏ/m)d/(dx)(theta(x))#.?

Calculate the current in a region where the wavefunction is given by #Be^(αx)+Ce^(−αx)#, where α is a real constant and B, C are complex numbers, i.e. B, C are not position-dependent. Is it correct to say that since #e^(+-αx)# are real functions, the current inside the barrier must be zero? Find a condition on B and C such that the current vanishes.


e)#[hatK,J^(T)] =+sJ^(T)#
iii. Suppose ...

Calculate the current in a region where the wavefunction is given by #Be^(αx)+Ce^(−αx)#, where α is a real constant and B, C are complex numbers, i.e. B, C are not position-dependent. Is it correct to say that since #e^(+-αx)# are real functions, the current inside the barrier must be zero? Find a condition on B and C such that the current vanishes.


e)#[hatK,J^(T)] =+sJ^(T)#
iii. Suppose #Kϕ_k = kϕ_k#.

Show that #(Jϕ_ k)# is an eigenfunction of K with eigenvalue
#(k − s)#.

T indicates the hermitian adjoint of #J# Attempt at first question
#psi$d/(dx)psi = Ae^(itheta)itheta'Ae^(itheta)=itheta'A"*"A#

Ive used $ to indicate complex conjugate at some places because the #"*# would ruin the formatting

#psid/(dx)psi$ = -itheta'A"*"A#

#j(x,t) =(ih)/(2m) * -2itheta'A"*"A#

#=|A(x)|^2h/md/(dx)(theta(x))#
but i know that this can be wrong as the derivative of #e^(itheta)# cannot be #itheta'e^(itheta)# but this is the only way i found out

Second question

#psid/(dx)Be^(alphax)+Ce^(-alphax)=(alphaBB$ - alphaC C $)#
#psid/(dx)Be^(-alphax)+Ce^(alphax)= -alphaBB$+alphaC C$#
then #j(x,t)=(ih)/(2m)(-alphaBB$+alphaC C$ -alphaBB$-alphaC C$)#

#-ibarhaBB$ + ibarhaC C$#

if we want the current to vanish

#-ibarhaBB$ = -ibarhaC C$#

#BB$ =C C$#

#B=C#

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Mar 12, 2018

I'm only going to answer the first two questions because that's all you tried...


For a given #Psi#, the probability current in one dimension is defined as:

#-(iℏ)/(2m) (Psi^"*" (del Psi)/(delx) - Psi (delPsi^"*")/(delx))#

If some #psi = A(x)e^(itheta(x))# containing real functions, then #Psi ^"*"= A(x) [e^(itheta(x)) e^(-iEt//ℏ)]^"*"#, so that:

#color(blue)(j(x,t)) = -(iℏ)/(2m) (A(x)e^(-itheta(x)) cancel(e^(iEt//ℏ)) cdot [i(d theta(x))/(dx)A(x)e^(itheta(x)) + (dA(x))/(dx)e^(itheta(x))]cancel(e^(-iEt//ℏ)) - A(x)e^(itheta(x)) cancel(e^(-iEt//ℏ)) cdot [-i(d theta(x))/(dx)A(x)e^(-itheta(x)) + (dA(x))/(dx)e^(-itheta(x))]cancel(e^(iEt//ℏ)))#

Distribute in the #Psi# and #Psi^"*"#...

#= -(iℏ)/(2m) ([i(d theta(x))/(dx)|A(x)|^2cancel(e^(-itheta(x))e^(itheta(x))) + (dA(x))/(dx)A(x)cancel(e^(-itheta(x))e^(itheta(x)))] - [-i(d theta(x))/(dx)|A(x)|^2 cancel(e^(itheta(x))e^(-itheta(x))) + (dA(x))/(dx)A(x)cancel(e^(itheta(x))e^(-itheta(x)))])#

Cancel some terms...

#= -(iℏ)/(2m) (i(d theta(x))/(dx)|A(x)|^2 + cancel((dA(x))/(dx)A(x)) + i(d theta(x))/(dx)|A(x)|^2 - cancel((dA(x))/(dx)A(x)))#

#= -(iℏ)/(2m) (2i(d theta(x))/(dx)|A(x)|^2)#

#= color(blue)((ℏ)/(m) |A(x)|^2(d theta(x))/(dx))#

From your wording, you're doing a tunnelling problem where

#psi(x) = Be^(alphax) + Ce^(-alphax)#

#Psi(x,t) = [Be^(alphax) + Ce^(-alphax)]e^(-iEt//ℏ)#

where #alpha = sqrt((2m(V_0-E))/ℏ^2)#, #V_0 > E#.

As a result, #B# is related to the amplitude of backwards propagation and #C# is related to the amplitude of forwards propagation.

The probability current result from before has to be modified, because #B# and #C# are complex, and #alpha# is real... We've done this before inside a potential well, and you should get:

#color(blue)(j(x,t)) = -(iℏ)/(2m) [(B^"*"e^(alphax) + C^"*"e^(-alphax))e^(iEt//ℏ)(alphaBe^(alphax) - alphaCe^(-alphax))e^(-iEt//ℏ) - (Be^(alphax) + Ce^(-alphax))e^(-iEt//ℏ)(alphaB^"*"e^(alphax) - alphaC^"*"e^(-alphax))e^(iEt//ℏ)]#

#= -(iℏ)/(2m) [cancel(alphaBB^"*"e^(2alphax)) - alphaB^"*"C + alphaBC^"*" - cancel(alphaC C^"*"e^(-2alphax)) - cancel(alphaBB^"*"e^(2alphax)) + alphaBC^"*" - alphaB^"*"C + cancel(alphaC C^"*"e^(-2alphax))]#

#= -(iℏ)/(2m) [ - alphaB^"*"C + alphaBC^"*" + alphaBC^"*" - alphaB^"*"C]#

#= -(iℏ)/(2m) [2 alpha (BC^"*" - B^"*"C)]#

#= color(blue)(-(ialphaℏ)/(m) [BC^"*" - B^"*"C])#

The time-dependent part (#e^(-iEt//ℏ)#) cancels out to be #1# because #e^(alphax)# and #e^(-alphax)# have the same energy. Therefore, to have no current, we consider:

#B/(B^"*") = C/(C^"*")#

And one choice for this is if #B# and #C# are both real or both imaginary (but not complex), if #B ne C#.

If #B# and #C# are individually complex (#a + i cdot "const"#), then it won't work as you cannot divide #(b + i cdot "const")/(b - i cdot "const")# to equal another #(c + i cdot "const")/(c - i cdot "const")#. Instead, an option that works is that #B = C#.

If we have #B = C#, that clearly leads to a percent reflection of #100%#, i.e.

#R = (C C^"*")/(BB^"*") = 1#

Or if we went with the condition of #B ne C# for #B/(B^"*") = C/(C^"*")# but #B# and #C# are chosen real or imaginary, that still leads to #R = 1#.

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