# Two different versions of Schrödinger's equation - are they equivalent?

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 $$\begin{equation}\tag{1} \mathrm{i}\hbar\frac{\partial}{\partial t}\Psi=\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V\right)\Psi\quad \end{equation}$$

But we could also think of the wave function as a function of time: \begin{align} \psi\colon[0,\infty[&\to L^2\\ t&\mapsto\Psi(\,\cdot\,,t)=:\psi_t \end{align} That is, $$\psi_t\colon\mathbb R\to\mathbb C,\ x\mapsto\Psi(x,t)$$ is the particle's wave function at the time $$t$$. One might conclude that Schrödinger's equation is $$\begin{equation}\tag{2} \mathrm{i}\hbar\dot{\psi}=\left(-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2}+V\right)\psi=H\psi\quad \end{equation}$$ Please notice that I switched from partial derivatives to total derivatives.

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

Explanation of the notation:

$$\dot{\psi}$$ is the derivative of $$\psi$$ and $$H\psi$$ is the map $$[0,\infty[\to L^2,\ t\mapsto H(\psi_t)$$.

• 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. – Prof. Legolasov Nov 7 '20 at 18:06
• 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. – Valter Moretti Nov 7 '20 at 18:20

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 $$\partial_t\psi_t(x) = \dot{\psi}_t(x) \quad \mbox{for almost all x \in \mathbb{R}.}$$

• Comments are not for extended discussion; this conversation has been moved to chat. – ACuriousMind Nov 8 '20 at 13:19
• 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? – Filippo Nov 11 '20 at 12:19
• Yes, it is correct. – Valter Moretti Nov 11 '20 at 12:21
• 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. – Filippo Nov 11 '20 at 13:06
• Yes, it is true also that. In this case you can handle the S. equation as a standard PDE. – Valter Moretti Nov 11 '20 at 13:14