[proofplan] We prove the continuity equation by differentiating the density $\rho=|\psi|^2$ in time and using the Schrödinger equation, together with its complex conjugate, to replace the time derivatives. The real-valuedness of $V$ makes the potential terms cancel exactly. The remaining second-derivative expression is then identified as the negative spatial derivative of the current. For a stationary state, the density is time-independent and the current reduces to $j_\phi$, so the continuity equation forces $j_\phi$ to have zero derivative. [/proofplan]

[step:Differentiate the probability density and substitute the Schrödinger equation]Fix $(x,t)\in\mathbb R\times\mathbb R$. Since $\psi$ and $\partial_t\psi$ are continuous, the product rule gives \begin{align*} \partial_t\rho(x,t)=\partial_t\overline{\psi(x,t)}\,\psi(x,t)+\overline{\psi(x,t)}\,\partial_t\psi(x,t). \end{align*} Solving the Schrödinger equation for $\partial_t\psi(x,t)$ gives \begin{align*} \partial_t\psi(x,t)=\frac{i\hbar}{2m}\partial_x^2\psi(x,t)-\frac{i}{\hbar}V(x)\psi(x,t). \end{align*} Because $V(x)\in\mathbb R$, conjugating this identity gives \begin{align*} \partial_t\overline{\psi(x,t)}=-\frac{i\hbar}{2m}\partial_x^2\overline{\psi(x,t)}+\frac{i}{\hbar}V(x)\overline{\psi(x,t)}. \end{align*} Substituting these two identities into the formula for $\partial_t\rho(x,t)$, the two potential terms cancel: \begin{align*} \partial_t\rho(x,t)=\frac{i\hbar}{2m}\left(\overline{\psi(x,t)}\,\partial_x^2\psi(x,t)-\psi(x,t)\,\partial_x^2\overline{\psi(x,t)}\right). \end{align*}[/step]

[guided]Fix a point $(x,t)\in\mathbb R\times\mathbb R$. The density is the product \begin{align*} \rho(x,t)=\overline{\psi(x,t)}\,\psi(x,t). \end{align*} Since $\psi$ is continuously differentiable in $t$, its complex conjugate is also continuously differentiable in $t$, and the product rule applies: \begin{align*} \partial_t\rho(x,t)=\partial_t\overline{\psi(x,t)}\,\psi(x,t)+\overline{\psi(x,t)}\,\partial_t\psi(x,t). \end{align*} The Schrödinger equation gives the time derivative of $\psi$. Dividing \begin{align*} i\hbar \partial_t\psi(x,t)=-\frac{\hbar^2}{2m}\partial_x^2\psi(x,t)+V(x)\psi(x,t) \end{align*} by $i\hbar$ yields \begin{align*} \partial_t\psi(x,t)=\frac{i\hbar}{2m}\partial_x^2\psi(x,t)-\frac{i}{\hbar}V(x)\psi(x,t). \end{align*} We also need the time derivative of $\overline{\psi}$. Taking complex conjugates is legitimate pointwise, and the hypothesis $V:\mathbb R\to\mathbb R$ is used here: $\overline{V(x)}=V(x)$. Therefore \begin{align*} \partial_t\overline{\psi(x,t)}=-\frac{i\hbar}{2m}\partial_x^2\overline{\psi(x,t)}+\frac{i}{\hbar}V(x)\overline{\psi(x,t)}. \end{align*} Now substitute both expressions into the product-rule formula: \begin{align*} \partial_t\rho(x,t)=\left(-\frac{i\hbar}{2m}\partial_x^2\overline{\psi(x,t)}+\frac{i}{\hbar}V(x)\overline{\psi(x,t)}\right)\psi(x,t)+\overline{\psi(x,t)}\left(\frac{i\hbar}{2m}\partial_x^2\psi(x,t)-\frac{i}{\hbar}V(x)\psi(x,t)\right). \end{align*} The potential contribution is \begin{align*} \frac{i}{\hbar}V(x)\overline{\psi(x,t)}\psi(x,t)-\frac{i}{\hbar}V(x)\overline{\psi(x,t)}\psi(x,t)=0. \end{align*} This cancellation is exactly where the real-valuedness of the potential enters. What remains is \begin{align*} \partial_t\rho(x,t)=\frac{i\hbar}{2m}\left(\overline{\psi(x,t)}\,\partial_x^2\psi(x,t)-\psi(x,t)\,\partial_x^2\overline{\psi(x,t)}\right). \end{align*}[/guided]

[step:Define the current and identify its derivative with the remaining term]Define the probability current $j_\psi:\mathbb R\times\mathbb R\to\mathbb R$ by \begin{align*} j_\psi(x,t)=\frac{\hbar}{m}\operatorname{Im}\!\left(\overline{\psi(x,t)}\,\partial_x\psi(x,t)\right). \end{align*} Define the complex-valued function $A:\mathbb R\times\mathbb R\to\mathbb C$ by \begin{align*} A(x,t)=\overline{\psi(x,t)}\,\partial_x^2\psi(x,t). \end{align*} Then $\psi(x,t)\,\partial_x^2\overline{\psi(x,t)}=\overline{A(x,t)}$, so \begin{align*} \partial_t\rho(x,t)=\frac{i\hbar}{2m}\left(A(x,t)-\overline{A(x,t)}\right)=-\frac{\hbar}{m}\operatorname{Im}A(x,t). \end{align*} Because $\psi$ is $C^2$ in $x$, the product rule applies to $\overline{\psi}\,\partial_x\psi$. Differentiating the current in $x$ gives \begin{align*} \partial_x j_\psi(x,t)=\frac{\hbar}{m}\operatorname{Im}\!\left(\partial_x\overline{\psi(x,t)}\,\partial_x\psi(x,t)+\overline{\psi(x,t)}\,\partial_x^2\psi(x,t)\right). \end{align*} The first term satisfies \begin{align*} \partial_x\overline{\psi(x,t)}\,\partial_x\psi(x,t)=|\partial_x\psi(x,t)|^2\in\mathbb R, \end{align*} so it has zero imaginary part. Hence \begin{align*} \partial_x j_\psi(x,t)=\frac{\hbar}{m}\operatorname{Im}A(x,t). \end{align*} Combining the two displayed identities gives \begin{align*} \partial_t\rho(x,t)+\partial_x j_\psi(x,t)=0. \end{align*} Since $(x,t)$ was arbitrary, the continuity equation holds on $\mathbb R\times\mathbb R$.[/step]

[guided]The previous step reduced the time derivative of the density to the imaginary part of one second-derivative expression. We now define the current as the quantity whose spatial derivative produces the same expression with the opposite sign. Define $j_\psi:\mathbb R\times\mathbb R\to\mathbb R$ by \begin{align*} j_\psi(x,t)=\frac{\hbar}{m}\operatorname{Im}\!\left(\overline{\psi(x,t)}\,\partial_x\psi(x,t)\right). \end{align*} This is a real-valued function because the imaginary-part operator has values in $\mathbb R$. Define $A:\mathbb R\times\mathbb R\to\mathbb C$ by \begin{align*} A(x,t)=\overline{\psi(x,t)}\,\partial_x^2\psi(x,t). \end{align*} Since conjugation commutes with the $x$-derivatives under the assumed $C^2$ regularity in $x$, we have \begin{align*} \psi(x,t)\,\partial_x^2\overline{\psi(x,t)}=\overline{A(x,t)}. \end{align*} Thus the identity obtained in the first step becomes \begin{align*} \partial_t\rho(x,t)=\frac{i\hbar}{2m}\left(A(x,t)-\overline{A(x,t)}\right). \end{align*} For any complex number $z$, $z-\overline z=2i\operatorname{Im}z$. Applying this with $z=A(x,t)$ gives \begin{align*} \partial_t\rho(x,t)=-\frac{\hbar}{m}\operatorname{Im}A(x,t). \end{align*} Now compute the spatial derivative of the current. The product rule is valid because $\psi$ is $C^2$ in $x$, so $\overline{\psi}$ is $C^2$ in $x$ and $\partial_x\psi$ is $C^1$ in $x$. Therefore \begin{align*} \partial_x j_\psi(x,t)=\frac{\hbar}{m}\operatorname{Im}\!\left(\partial_x\overline{\psi(x,t)}\,\partial_x\psi(x,t)+\overline{\psi(x,t)}\,\partial_x^2\psi(x,t)\right). \end{align*} The first product is real because \begin{align*} \partial_x\overline{\psi(x,t)}\,\partial_x\psi(x,t)=\overline{\partial_x\psi(x,t)}\,\partial_x\psi(x,t)=|\partial_x\psi(x,t)|^2\in\mathbb R. \end{align*} Its imaginary part is therefore zero, and the remaining term is exactly $A(x,t)$. Hence \begin{align*} \partial_x j_\psi(x,t)=\frac{\hbar}{m}\operatorname{Im}A(x,t). \end{align*} The two formulas have opposite signs, so \begin{align*} \partial_t\rho(x,t)+\partial_x j_\psi(x,t)=0. \end{align*} Since the point $(x,t)$ was arbitrary, this pointwise identity holds on all of $\mathbb R\times\mathbb R$.[/guided]

[step:Apply the continuity equation to a stationary state]Assume now that $E\in\mathbb R$, that $\phi:\mathbb R\to\mathbb C$ is $C^2$, and that $\phi$ satisfies the stationary Schrödinger equation \begin{align*} -\frac{\hbar^2}{2m}\phi''(x)+V(x)\phi(x)=E\phi(x) \end{align*} for every $x\in\mathbb R$. Define $\psi:\mathbb R\times\mathbb R\to\mathbb C$ by \begin{align*} \psi(x,t)=e^{-iEt/\hbar}\phi(x). \end{align*} This $\psi$ satisfies the hypotheses of the time-dependent part and solves the time-dependent Schrödinger equation, because \begin{align*} i\hbar\partial_t\psi(x,t)=E e^{-iEt/\hbar}\phi(x) \end{align*} and \begin{align*} -\frac{\hbar^2}{2m}\partial_x^2\psi(x,t)+V(x)\psi(x,t)=e^{-iEt/\hbar}\left(-\frac{\hbar^2}{2m}\phi''(x)+V(x)\phi(x)\right)=E e^{-iEt/\hbar}\phi(x). \end{align*} Moreover \begin{align*} \rho(x,t)=|e^{-iEt/\hbar}\phi(x)|^2=|\phi(x)|^2, \end{align*} so $\partial_t\rho(x,t)=0$. Define $j_\phi:\mathbb R\to\mathbb R$ by \begin{align*} j_\phi(x)=\frac{\hbar}{m}\operatorname{Im}\!\left(\overline{\phi(x)}\,\phi'(x)\right). \end{align*} The current associated to $\psi$ is \begin{align*} j_\psi(x,t)=\frac{\hbar}{m}\operatorname{Im}\!\left(\overline{e^{-iEt/\hbar}\phi(x)}\,e^{-iEt/\hbar}\phi'(x)\right)=\frac{\hbar}{m}\operatorname{Im}\!\left(\overline{\phi(x)}\,\phi'(x)\right)=j_\phi(x). \end{align*} Applying the continuity equation already proved gives \begin{align*} 0=\partial_t\rho(x,t)+\partial_x j_\psi(x,t)=\partial_x j_\phi(x). \end{align*} Thus $j_\phi'(x)=0$ for every $x\in\mathbb R$. Since $\mathbb R$ is connected and $j_\phi$ is differentiable with zero derivative everywhere, $j_\phi$ is independent of $x$.[/step]

[guided]We now specialize the identity just proved to the separated form of a stationary state. Assume $E\in\mathbb R$, let $\phi:\mathbb R\to\mathbb C$ be $C^2$, and assume that $\phi$ solves the stationary equation \begin{align*} -\frac{\hbar^2}{2m}\phi''(x)+V(x)\phi(x)=E\phi(x) \end{align*} for every $x\in\mathbb R$. Define $\psi:\mathbb R\times\mathbb R\to\mathbb C$ by \begin{align*} \psi(x,t)=e^{-iEt/\hbar}\phi(x). \end{align*} Because the exponential factor is smooth in $t$ and independent of $x$, while $\phi$ is $C^2$ in $x$, this $\psi$ has the regularity required in the first part. We verify the time-dependent equation directly. Differentiating in $t$ gives \begin{align*} i\hbar\partial_t\psi(x,t)=E e^{-iEt/\hbar}\phi(x). \end{align*} Differentiating twice in $x$ gives $\partial_x^2\psi(x,t)=e^{-iEt/\hbar}\phi''(x)$. Therefore \begin{align*} -\frac{\hbar^2}{2m}\partial_x^2\psi(x,t)+V(x)\psi(x,t)=e^{-iEt/\hbar}\left(-\frac{\hbar^2}{2m}\phi''(x)+V(x)\phi(x)\right). \end{align*} Using the stationary equation, the right-hand side becomes \begin{align*} E e^{-iEt/\hbar}\phi(x), \end{align*} which equals $i\hbar\partial_t\psi(x,t)$. Thus the continuity equation applies to this $\psi$. The density is independent of time because the phase has modulus one: \begin{align*} \rho(x,t)=|e^{-iEt/\hbar}\phi(x)|^2=|\phi(x)|^2. \end{align*} Hence $\partial_t\rho(x,t)=0$. Define $j_\phi:\mathbb R\to\mathbb R$ by \begin{align*} j_\phi(x)=\frac{\hbar}{m}\operatorname{Im}\!\left(\overline{\phi(x)}\,\phi'(x)\right). \end{align*} The current for $\psi$ reduces to this function because the conjugate phase and the phase cancel: \begin{align*} j_\psi(x,t)=\frac{\hbar}{m}\operatorname{Im}\!\left(\overline{e^{-iEt/\hbar}\phi(x)}\,e^{-iEt/\hbar}\phi'(x)\right)=j_\phi(x). \end{align*} Substituting $\partial_t\rho(x,t)=0$ and $j_\psi(x,t)=j_\phi(x)$ into the continuity equation gives \begin{align*} 0=\partial_t\rho(x,t)+\partial_x j_\psi(x,t)=\partial_x j_\phi(x). \end{align*} Therefore $j_\phi'(x)=0$ for every $x\in\mathbb R$. Since $\mathbb R$ is connected and $j_\phi$ is differentiable with zero derivative everywhere, $j_\phi$ is constant on $\mathbb R$, so it is independent of $x$.[/guided]