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For simplicity, let's look at the case of one particle in one dimension. We usually think of the wave function as a function \begin{align} \Psi\colon\mathbb R\times[0,\infty[&\to\mathbb C\\ (x,t)&\mapsto\Psi(x,t) \end{align} and Schrödinger's equation has the form \begin{equation}\tag{1} \mathrm{i}\hbar\frac{\partial}{\partial t}\Psi=\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x{}^2}+V\right)\Psi\quad \end{equation}

On the other hand, currying allows us to think of the wave function as a function of time: \begin{align} \psi\colon[0,\infty[&\to L^2(\mathbb{R},\mathbb{C}) \end{align} In other words, $\psi(t)$ is the function defined by $(\psi(t))(x)=\Psi(x,t)$ - the particle's wave function at the time $t$. That being said, the following equation makes sense: \begin{equation}\tag{2} \mathrm{i}\hbar\frac{\mathrm{d}}{\mathrm{d}t}\psi=\left(-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x{}^2}+V\right)\psi \end{equation} Please notice that I replaced the partial derivatives by total derivatives.

Q: Are both equations - (1) and (2) - valid? Are they equivalent?

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  • $\begingroup$ Your two equations / interpretations seem to be completely equivalent mathematically, why would you think they're not? I assume the first interpretation is augmented with the square integrability condition. $\endgroup$ Nov 7, 2020 at 18:06
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    $\begingroup$ The existence of the standard $t$-derivative is a $x$-pointwise condition, whereas the derivative in the sense of $L^2$ space is $x$-global (integral) condition.They are quite independent from each other. The former does not have global implications whereas the latter sees the functions up to zero $x$- measure sets. However a link exists arising from some results in the $L^p$ convergence theory. What matters in QM is the second notion due to its connection with the Stone theorem. $\endgroup$ Nov 7, 2020 at 18:20

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The two approaches are different and inequivalent. The physically correct one is the latter since the Schroedinger equation is nothing but the application of the Stone theorem to the time evolution and there the relevant topology to compute derivatives is the Hilbert space one. In other words, the S. equation arises by taking a suitable type of $t$ derivative on both sides of the identity $$\psi_t = e^{-itH/\hbar} \psi_0\:.$$ This type of derivative is the one described below in (1) which is used in the said theorem eventually leading to the wanted equation $$i\hbar \dot{\psi}_t = H \psi_t\:.$$ The other notion of $t$-derivative, relying on standard calculus only, is however interesting from the viewpoint of PDE theory.

The correct notion of $t$-derivative in QM is therefore the one, I indicate by $\dot{\psi}_t$, in the $L^2$ space performed with respect to the norm of that space:

$$\left|\left|\dot{\psi}_t - \frac{\psi_{t+h}-\psi_t}{h}\right|\right|_{L^2(\mathbb{R})} \to 0 \quad \mbox{for}\quad h\to 0\:.\tag{1}$$ In integral terms $$\int_{\mathbb{R}}\left|\dot{\psi}_t(x) - \frac{\psi_{t+h}(x)-\psi_t(x)}{h} \right|^2 dx \to 0 \quad \mbox{for}\quad h\to 0\:.$$

The existence of the standard $t$-derivative $\partial_t \psi_t(x)$ is a $x$-pointwise condition, whereas the derivative $\dot{\psi}_t$ in the sense of $L^2$ space is $x$-global (integral) condition.They are quite independent from each other. The former does not have global implications whereas the latter sees the functions up to zero $x$-measure sets.

However if $t \mapsto \psi_t$ admits $L^2(\mathbb{R})$ $t$-derivative $\dot{\psi}_t$ and, simultaneously, $(t,x) \mapsto \psi_t(x)$ also admits standard partial $t$ derivative $\partial_t\psi_t(x)$, then the two derivatives coincide for almost all $x\in \mathbb{R}$ at the given $t$.

PROOF Suppose that $\partial_t\psi_t(x)$ exists for all $x\in \mathbb{R}$ and the said $t$. Suppose that also $\frac{d}{dt}_{L^2}\psi_t =: \dot{\psi}_t \in L^2(\mathbb{R})$ exists as in (1). Hence, if $h\to 0$, $$\frac{\psi_{t+h}-\psi_t}{h} \to \dot{\psi}_t \quad \mbox{in $L^2(\mathbb{R})$ sense as in (1)}\:.$$ Due to a known theorem of $L^p$ convergence, there is a sequence $h_n \to 0$ such that, as $n\to +\infty$, $$\frac{\psi_{t+h_n}(x)-\psi_t(x)}{h_n} \to \dot{\psi}_t(x) \quad \mbox{for almost all $x \in \mathbb{R}$}\:.$$ However the existence of $\partial_t\psi_t(x)$ requires $$\frac{\psi_{t+h_n}(x)-\psi_t(x)}{h_n} \to \partial_t\psi_t(x) \quad \mbox{for all $x \in \mathbb{R}$}\:.$$ In summary $$\partial_t\psi_t(x) = \dot{\psi}_t(x) \quad \mbox{for almost all $x \in \mathbb{R}$.}$$

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Nov 8, 2020 at 13:19
  • $\begingroup$ In other words: If $\dot{\psi}_t\in L^2$ and $\partial_t\Psi(\,\cdot\,,t)$ exist, then $\partial_t\Psi(\,\cdot\,,t)$ is square integrable (that is, $\partial_t\Psi(\,\cdot\,,t)\in\mathcal L^2$) and its equivalence class is $\dot{\psi}_t$. Is that correct? $\endgroup$
    – Filippo
    Nov 11, 2020 at 12:19
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    $\begingroup$ Yes, it is correct. $\endgroup$ Nov 11, 2020 at 12:21
  • $\begingroup$ Okay, this means that we have proven $\left[\mathrm{i}\hbar\frac{\partial}{\partial t}\Psi(\,\cdot\,,t)\right]=\mathrm{i}\hbar\dot{\psi}_t$. My guess would be that we can also prove $\left[\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\Psi+V\Psi\right)(\,\cdot\,,t)\right]=H\psi_t$. If this is true, the second Schrödinger equation follows from the first one. $\endgroup$
    – Filippo
    Nov 11, 2020 at 13:06
  • $\begingroup$ Yes, it is true also that. In this case you can handle the S. equation as a standard PDE. $\endgroup$ Nov 11, 2020 at 13:14

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