Griffiths Quantum Mechanics 2nd Edition: Problem 1.14

This is a fairly straightforward problem, using concepts that we have already seen multiple times. The probability $P_{ab}(t)$ of finding a particle at some position $x$ with $a \ < \ x \ < \ b$ at some time $t$ is simply equal to the integration over the probability density between $a$ and $b$:

$$P_{ab}(t) \ = \ \displaystyle\int_{a}^{b} \ |\Psi(x, \ t)|^2 \ \text{dx}$$
Now that we have this, we need to show that:

$$\frac{\text{d}P}{\text{d}t} \ = \ J(a, \ t) \ - \ J(b, \ t) \ \Rightarrow \ J(x, \ t) \ = \ \frac{i\hbar}{2m}\Big( \Psi\frac{\partial \Psi^{*}}{\partial x} \ - \ \Psi^{*}\frac{\partial \Psi}{\partial x} \Big)$$
This problem simply comes down to expanding and simplifying the time derivative of the integral that we just found as being equal to $P_{ab}(t)$. Let us start by expanding:

$$\frac{\text{d}P}{\text{d}t} \ = \ \frac{\text{d}}{\text{d}t} \ \displaystyle\int_{a}^{b} \ |\Psi(x, \ t)|^2 \ \text{dx} \ = \ \displaystyle\int_{a}^{b} \ \frac{\partial}{\partial t} \ \big( \Psi^{*}\Psi \big) \text{dx} \ = \ \displaystyle\int_{a}^{b} \ \frac{\partial \Psi^{*}}{\partial t}\Psi \ + \ \frac{\partial \Psi}{\partial t}\Psi^{*} \ \text{dx}$$
Now, remembering the definition of the Schrödinger equation, we can rewrite this integral:

$$\Rightarrow \ \frac{\partial \Psi}{\partial t} \ = \ \frac{i\hbar}{2m} \ \frac{\partial^2 \Psi}{\partial t} \ - \ \frac{i}{\hbar}V\Psi$$ $$\Rightarrow \ \displaystyle\int_{a}^{b} \ \Psi\Big( -\frac{i\hbar}{2m} \ \frac{\partial^2 \Psi^{*}}{\partial t} \ + \ \frac{i}{\hbar}V\Psi^{*} \Big) \ + \ \Psi^{*}\Big( \frac{i\hbar}{2m} \ \frac{\partial^2 \Psi}{\partial t} \ - \ \frac{i}{\hbar}V\Psi \Big) \ \text{dx} \ \Rightarrow \ \frac{i\hbar}{2m}\displaystyle\int_{a}^{b} \ \Psi^{*} \frac{\partial^2 \Psi}{\partial t} \ - \ \Psi\frac{\partial^2 \Psi^{*}}{\partial t} \ \text{dx}$$ $$\Rightarrow \ \frac{i\hbar}{2m}\displaystyle\int_{a}^{b} \ \Psi^{*}\frac{\partial^2 \Psi}{\partial t} \ - \ \Psi\frac{\partial^2 \Psi^{*}}{\partial t} \ \text{dx} \ = \ -\frac{i\hbar}{2m}\displaystyle\int_{b}^{a} \ \Psi^{*}\frac{\partial^2 \Psi}{\partial t} \ - \ \Psi\frac{\partial^2 \Psi^{*}}{\partial t} \ \text{dx} \ = \ \frac{i\hbar}{2m}\displaystyle\int_{b}^{a} \ \frac{\partial}{\partial x} \Big(\Psi\frac{\partial \Psi^{*}}{\partial t} \ - \ \Psi^{*}\frac{\partial \Psi}{\partial t} \Big) \ \text{dx}$$ $$\Rightarrow \ \frac{i\hbar}{2m}\displaystyle\int_{b}^{a} \ \frac{\partial}{\partial x} \Big(\Psi\frac{\partial \Psi^{*}}{\partial t} \ - \ \Psi^{*}\frac{\partial \Psi}{\partial t} \Big) \ \text{dx} \ = \ \frac{i\hbar}{2m} \Big( \Psi\frac{\partial \Psi^{*}}{\partial t} \ - \ \Psi^{*}\frac{\partial \Psi}{\partial t} \Big) \biggr\rvert_{b}^{a} \ = \ J(a, \ t) \ - \ J(b, \ t)$$
Now that we have proven the posed problem in part $(a)$ to be true, we can apply it to the wavefunction in Problem 1.9. This wavefunction is defined as follows:

$$\Psi(x, \ t) \ = \ Ae^{-a[(mx^2/\hbar) \ + \ it]}$$
We can now calculate the value of the probability current, $J(x, \ t)$. So for the posed wavefunction, we have:

$$J(x, \ t) \ = \ \frac{i\hbar}{2m}\Big( Ae^{-a[(mx^2/\hbar) \ + \ it]}\frac{\partial}{\partial x} \Big( Ae^{-a[(mx^2/\hbar) \ - \ it]} \Big) \ - \ Ae^{-a[(mx^2/\hbar) \ - \ it]} \frac{\partial}{\partial x} \Big( Ae^{-a[(mx^2/\hbar) \ + \ it]} \Big) \Big)$$ $$\Rightarrow \ \frac{i\hbar}{2m}\Big( Ae^{-a[(mx^2/\hbar) \ + \ it]} e^{ait}\frac{\partial}{\partial x} \Big( Ae^{-a[(mx^2/\hbar)]} \Big) \ - \ Ae^{-a[(mx^2/\hbar) \ - \ it]} e^{-ait}\frac{\partial}{\partial x} \Big( Ae^{-a[(mx^2/\hbar)]} \Big) \Big)$$ $$\Rightarrow \ \frac{i\hbar}{2m}\Big( A^2e^{-a[(mx^2/\hbar) \ + \ it]} e^{ait} \Big( e^{-a[(mx^2/\hbar)]} \Big ( -\frac{2amx}{\hbar} \Big) \Big) \ - \ A^2e^{-a[(mx^2/\hbar) \ - \ it]} e^{-ait} \Big( e^{-a[(mx^2/\hbar)]} \Big( -\frac{2amx}{\hbar} \Big) \Big) \Big)$$ $$\Rightarrow \ \frac{i\hbar}{2m} A^2\Big( e^{-a[(mx^2/\hbar)]} \Big( e^{-a[(mx^2/\hbar)]} \Big ( -\frac{2amx}{\hbar} \Big) \Big) \ - \ e^{-a[(mx^2/\hbar)]} \Big( e^{-a[(mx^2/\hbar)]} \Big( -\frac{2amx}{\hbar} \Big) \Big) \Big) \ = \ 0$$
And so we have found that the probability current of this particular wavefunction is equal to 0.