Question

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.

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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`

Answer 1

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

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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`.