# 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))#.?

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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#

Calculate the current in a region where the wavefunction is given by

e)

iii. Suppose

Show that

#(Jϕ_ k)# is an eigenfunction of K with eigenvalue

#(k − s)# .

T indicates the hermitian adjoint of

Ive used $ to indicate complex conjugate at some places because the

but i know that this can be wrong as the derivative of

Second question

then

if we want the current to vanish

##### 1 Answer

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

For a given **probability current in one dimension** is *defined* as:

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

If some

#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

#= -(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,

The probability current result from before has to be modified, because

#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 (

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

And one choice for this is if

If

If we have

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

Or if we went with the condition of