= Solution
For the <Schrödinger equation> with a real <scalar potential> and no <vector potential>, the <probability current> is
$$
\mathbf j=\frac{\hbar}{2mi}(\Psi^*\nabla\Psi-\Psi\nabla\Psi^*)=\frac{\hbar}{m}\operatorname{Im}(\Psi^*\nabla\Psi).
$$
Together with the <probability density> $\rho=|\Psi|^2$, it satisfies the <continuity equation> $\partial_t\rho+\nabla\cdot\mathbf j=0$. Substituting $\partial_x\Psi=ik\Psi$ gives
$$
\boxed{j=\frac{\hbar k}{m}|A|^2,\qquad \rho(x)=|A|^2,\qquad dP=|A|^2\,dx.}
$$
A complex <wavefunction> requires both its real and <imaginary parts> in the sketch. Choosing the irrelevant constant <phase> so that $A$ is real, they are $A\cos kx$ and $A\sin kx$, while the <probability density> is horizontal. It is the \b[<spatial phase gradient of a Schrodinger wavefunction>] $\partial_x\arg\Psi=k$ that determines the sign of the <probability current>. In particular, $e^{ikx}$ and $e^{-ikx}$ have the same <real part> but opposite <probability currents>, so the slope of the <real part> cannot determine the direction.
A nonzero <plane wave> on the whole real line is not square integrable. The displayed density is understood with box <normalization>, generalized <normalization>, or as a local description of a broad <wave packet>; it does not define a normalized <probability distribution> on the whole line.
\Image[/past-exam-of-the-quantum-physics-course-of-the-university-of-oxford/2016-plane-wave-current-phase.png]
{title=Complex plane wave and its constant probability density}
{description=The plotted example has positive wavevector. Reversing the wavevector reverses the imaginary part and the probability current while preserving the real part and the probability density.}
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