# 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)$\left[\hat{K} , {J}^{T}\right] = + s {J}^{T}$ 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)$\left[\hat{K} , {J}^{T}\right] = + s {J}^{T}$ 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 \left(x , t\right) = \frac{i h}{2 m} \cdot - 2 i \theta ' A \text{*} A$$= | A \left(x\right) {|}^{2} \frac{h}{m} \frac{d}{\mathrm{dx}} \left(\theta \left(x\right)\right)$but i know that this can be wrong as the derivative of ${e}^{i \theta}$cannot be $i \theta ' {e}^{i \theta}$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$ $B = C$##### 1 Answer Write your answer here... Start with a one sentence answer Then teach the underlying concepts Don't copy without citing sources preview ? #### Answer Write a one sentence answer... #### Answer: #### Explanation Explain in detail... #### Explanation: I want someone to double check my answer Describe your changes (optional) 200 2 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 $\psi = A \left(x\right) {e}^{i \theta \left(x\right)}$containing real functions, then ${\Psi}^{\text{*"= A(x) [e^(itheta(x)) e^(-iEt//ℏ)]^"*}}$, 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 ${\Psi}^{\text{*}}$... = -(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 $\psi \left(x\right) = B {e}^{\alpha x} + C {e}^{- \alpha x}$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 ${e}^{\alpha x}$and ${e}^{- \alpha x}$have the same energy. Therefore, to have no current, we consider: $\frac{B}{{B}^{\text{*") = C/(C^"*}}}$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 ($a + i \cdot \text{const}$), then it won't work as you cannot divide $\left(b + i \cdot \text{const")/(b - i cdot "const}\right)$to equal another $\left(c + i \cdot \text{const")/(c - i cdot "const}\right)$. 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. $R = \left(C {C}^{\text{*")/(BB^"*}}\right) = 1$Or if we went with the condition of $B \ne C$for $\frac{B}{{B}^{\text{*") = C/(C^"*}}}$but $B$and $C$are chosen real or imaginary, that still leads to $R = 1\$.

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