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I am working on part (b) of exercise 1 in Chapter 4 of Brian C. Hall's book. It states:

$\textbf{Exercise 1}$: A $\textbf{locally integrable}$ function $\psi(x,t)$ satisfies the free Schr$\ddot{\textrm{o}}$dinger equation in the weak (or distributional) sense if for each smooth compactly supported function $\chi$, we have $$ \int_{\mathbb{R}^2}\ \psi(x,t)\left[\frac{\partial\chi}{\partial t}+\frac{i\hbar}{2m}\frac{\partial^2\chi}{\partial x^2}\right]dxdt=0. $$ $\textbf{(b)}$ For any $\psi_0\in L^2(\mathbb{R})$, define $\psi(x,t)$ by Definition 4.4. Show that $\psi$ satisfies the free Schr$\ddot{\textrm{o}}$dinger equation in the weak sense.

$\textbf{Definition 4.4}$: For any $\psi_0\in L^2(\mathbb{R})$, define, for each $t\in\mathbb{R}$, $\psi(x,t)$ to be the unique element of $L^2(\mathbb{R})$ that has a Fourier transform (with respect to $x$) given by: $$ \hat{\psi}(k,t)=\hat{\psi}_0(k)\exp\left[-i\frac{\hbar k^2t}{2m}\right]. $$

Hint: First show that the function $\psi_A$ given by $$ \psi_A(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-A}^A\hat{\psi}_0(k)e^{i(kx-\omega(k)t)}\ dk $$ satisfies the free Schr$\ddot{\textrm{o}}$dinger equation, for each $A$ (where $\omega(k)=\hbar k^2/2m$).

$\textbf{My Question}$

I solved part (a) which is easily used to obtain the hint. However, I absolutely cannot for the life of me figure out how to demonstrate $\psi$, as defined above for arbitrary $\psi_0\in L^2(\mathbb{R})$, is $\textbf{locally integrable}$ in $\mathbb{R}^2$.

My first thought for solving part (b) was to use the dominated convergence theorem to bring the limit as $A\rightarrow\infty$ into the integral over $t$, but I can't obtain any useful bounding function - the issue is I don't seem to have any control over how $\psi$ or its local $x$-norm varies with respect to $t$.

Then I tried looking at Jensen's inequality for $\psi_A$ but that leaves me with a factor of $\sqrt{A}$ when trying to use $||\hat{\psi}_0||_2$ to bound each $\psi_A$, so no luck.

I tried using Fourier transform identities but again nothing helps, I just seem to circle back to the $\psi_A$ function.

Am I just supposed to assume $\psi$ is locally integrable? This seems wrong since Hall states:

$\psi(x,t)$, as defined by Definition 4.4, $\textbf{always}$ satisfies the Schr$\ddot{\textrm{o}}$dinger equation in the weak (distributional) sense.

Can someone help? I'm losing my mind over this problem, I feel as if either I'm overlooking something obvious or I am not supposed to actually prove local integrability of $\psi$. Once I can demonstrate/utilize local integrability then I can effortlessly solve the problem using the hint, but it doesn't seem so easy.

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If $f\in L^2(\Bbb{R})$ then for all interval $[a,b]\subseteq\Bbb{R}$ by Cauchy-Schwarz $$ \int_a^b |f(x)|dx\le \left(\int_a^b |f(x)|^2 dx\right)^{\frac{1}{2}}\left(\int_a^b 1 dx\right)^{\frac{1}{2}}\le \|f\|_2\sqrt{b-a}\tag{1} $$ Now, since $$ \hat{\psi}(k,t)=\hat{\psi}_0(k)\exp\left[-i\frac{\hbar k^2t}{2m}\right]\tag{2} $$ and $\psi(x,t)$ is the inverse Fourier transform of this, by the Plancherel theorem $$ \|\psi(-,t)\|_2=\|\hat\psi(-,t)\|_2=\|\hat\psi(-,0)\|_2\tag{3} $$ where the right equality in $(3)$ is because the $\hat\psi$ have $L^2$ norm that is independent of $t$ (that multiplier with $t$ in $(2)$ has absolute value $1$.)

So, combining with $(1)$ $$ \int_c^d\int_a^b|\psi(x,t)|dx dt\le \|\hat\psi(-,0)\|_2 (\sqrt{b-a}) (d-c)<+\infty $$

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  • $\begingroup$ Great, thank you. I was failing to utilize the Plancherel theorem, it seems so obvious now but I didn't think of it for some reason. $\endgroup$
    – Derek H.
    Commented Mar 20 at 18:17

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