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I'm trying to prove that $\forall\psi_0\in L^2(\mathbb{R})$,the integral $$\psi(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{ikx}\left[\hat{\psi_0}(k)e^{-i\omega(k)t}\right]dk$$ satisfies the Schrodinger equation in the weak sense, namely, $$\int_{\mathbb{R}^2}\psi(x,t)\left[\frac{\partial\chi}{\partial t}+\frac{i\hbar}{2m}\frac{\partial^2\chi}{\partial x^2}\right]dxdt=0$$ for any $\chi$ smooth and compactly supported.

My idea is that I can prove this for functions $\psi_0$ in the space $\phi(\mathbb{R})$ of smooth and rapidly decreasing functions (the Schwartz space) and extend it to all $L^2$ functions. Indeed, the proof for Schwartz functions is easy since they are decreasing fast enough so that $k^2\hat{\psi_0}$ is integrable, enabling us to apply the dominated convergence theorem and integration by part.

Now the Schwartz space $\phi(\mathbb{R})$ is dense in $L^2(\mathbb{R})$ and thus $\forall\psi_0\in L^2(\mathbb{R})$, $\exists(\psi_{0,n})\subseteq\phi(\mathbb{R})$ such that $\psi_{0,n}\rightarrow\psi_0$ in $L^2$ norm. However, I'm not sure in what sense $\psi_n(x,t)$, defined similar to the first integration, converges to $\psi(x,t)$, defined in the first equation. Can I say that $\psi_n(x,t)$ converges to $\psi(x,t)$ in the sense of $L^2(\mathbb{R}^2)$? Why?

By the way, it would be equally appreciated if one can give another proof, for example, a direct proof without utilizing the Schwartz space. Thanks for your time and patience in advance!

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First observe that in general the function $\psi(x,t)$, defined by you as a general weak solution of the free Schroedinger equation, doesn't belong to $L^2(\mathbb R^2)$. Indeed, we usually have $\int_{\mathbb R}\left|\psi(x,t)\right|^2dx=\left\|\psi_0\right\|^2_{L^2(\mathbb R)}$ $\forall t\in \mathbb R$, and then $\int_{\mathbb R^2}\left|\psi(x,t)\right|^2dx\,dt=\infty$ by the Fubini-Tonelli's theorem. We cannot hope that such functions could belong to $L^2(\mathbb R^2)$. At least you have to consider a finite time interval. Otherwise the function $\psi(x,t)$ could be seen as an element of the space $L^{\infty}(L^2(\mathbb R))$, i.e. the space of the function $f:t\in\mathbb R\rightarrow f(t)\in L^2(\mathbb R)$ with bounded sup-norm.

Nevertheless the function $\psi(x,t)$ is a weak solution of the free Schroedinger equations for every $\psi_0\in L^2(\mathbb R)$. Since $H_0=-\frac{\hbar^2}{2m}\Delta$ is a self-adjoint operator in $L^2(\mathbb R)$, by common results of the semi-group/group theory we have that $e^{-iH_0t/\hbar}$ is a unitary operators in $L^2(\mathbb R)$, and by varying $t$ in $\mathbb R$ they also form a strongly continuos group of operators. Then (see Engel - Nagel, One-Parameter Semigroups for Linear Evolution Equations, Lemma II.1.3 (iii)-(iv) for example) $\int_0^te^{-iH_0s/\hbar}\psi_0\, ds$ is in the domain of $H_0$ for every $t\in\mathbb R$ and $$ e^{-iH_0t/\hbar}\psi_0=-\frac{i}{\hbar} H_0\int_0^te^{-iH_0s/\hbar}\psi_0\, ds+\psi_0$$ for every $\psi_0\in L^2(\mathbb R)$.

Defined $\psi(t)=e^{-iH_0t/\hbar}\psi_0$, then $\psi(t)$ is a "mild solution" of the Schroedinger equation, i.e.

$$\psi(t)=-\frac{i}{\hbar}H_0\int_0^t \psi(s)ds+\psi_0$$

which correspond to the usual Schroedinger equation only when $\psi_0$ is in the domain of $H_0$ (in that case the operator $H_0$ can be moved inside the integral, and by time-differentiation we get such equation).

We now have to check that (i) $\psi(t)$ corresponds to the function $\psi(x,t)$ given by you and that (ii) a mild solution is a weak solution too.

Defined with $U_F$ the Fourier's transform operator, which is a unitary operator in $L^2(\mathbb R)$, by the spectral theorem we have $$ \psi(t)=U_F^{-1}\left[\left(U_F e^{-iH_0t/\hbar}U_F^{-1}\right)\left(U_F\psi_0\right)\right]=U_F^{-1}\left[e^{i\hbar k^2t/2m}\hat{\psi}_0(k)\right], \quad k\in\mathbb R$$ that corresponds to your expression of $\psi(x,t)$ with $\omega(k)=-\hbar k^2/2m$ when $\hat{\psi}_0\in L^1(\mathbb R)\cap L^2(\mathbb R)$ (as you probably know your expression is indeed "formal", since the integral $$\int_{-\infty}^{\infty} e^{ikx}\left[\hat{\psi}_0(k)e^{-i\omega(k)t}\right]dt$$ is defined only when the integrand function is absolutely integrable).

Let us to prove the second assertion. Given $\chi(x,t)$ a smooth real function with compact support, then for every $t\in \mathbb R\,$ it results that $\chi(t):=\chi(\cdot,t)$ and $\dot{\chi}(t):=\partial \chi(\cdot,t)/\partial t$ are both in $L^2(\mathbb R)$ and in the domain of $H_0$ too. Denoted by $\left<\cdot,\cdot\right>$ the inner product in $L^2(\mathbb R)$, i.e. $$\left<\alpha,\beta\right>=\int_{\mathbb R}\alpha(x)^*\beta(x)dx,\quad \alpha, \beta\in L^2(\mathbb R)$$ from the above definition of mild solution we get

$$\int_{\mathbb R^2}\frac{\partial \chi(x,t)}{\partial t}\psi(x,t)dxdt =\int_{\mathbb R}\left<\dot{\chi}(t),\psi(t)\right>dt=\int_{\mathbb R}\left<\dot{\chi(t)},-\frac{i}{\hbar}H_0\int_0^t e^{-iH_0s/\hbar}\psi_0\,ds\right>dt=\int_{\mathbb R}\left(-\frac{i}{\hbar}\right)\left<H_0\dot{\chi}(t),\int_0^t e^{-iH_0s/\hbar}\psi_0\,ds\right>dt=\int_{\mathbb R}\frac{i}{\hbar}\left<H_0\chi(t), e^{-iH_0t/\hbar}\psi_0\right>dt=\int_{\mathbb R}\frac{i}{\hbar}\left<H_0\chi(t), \psi(t)\right>dt=-\int_{\mathbb R^2}\frac{i\hbar}{2m}\frac{\partial^2\chi(x,t)}{\partial x^2}\psi(x,t)dx\,dt$$ where in the forth equality we have commuted $\partial/\partial t$ with $H_0$ being $\chi$ smooth enough to change the order of derivation and then we have applyied an integration by parts.

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  • $\begingroup$ @taritgoswami It looks a better answer, anyway, although I myself have to recap my memory very hard for relavant details. But I like this answer! $\endgroup$ – Zheng Liu Oct 12 '18 at 2:37
  • $\begingroup$ @ZhengLiu Ooh, fortunately you are using MATH.SE , probably your last seen have motivated to answer it then :-) Users usually not wait for answers for a question asked years ago. $\endgroup$ – tarit goswami Oct 12 '18 at 4:41
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Hint: use the fact that $$ \|\psi(\cdot,t)-\psi_n(\cdot,t)\|_{L^2(\mathbb{R})} =\|\hat{\phi}_0-\hat{\phi}_{0,n}\|_{L^2(\mathbb{R})}. $$

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  • $\begingroup$ Hi. Thanks for your answer. Just want to clarify, the function $\psi(\cdot, t)$ seems to be a function in $L^2(\mathbb{R})$, not in $L^2(\mathbb{R}^2)$. Actually this is my question. The defining integral of the weak solution above seems to be an inner product on $L^2(\mathbb{R}^2)$, not on $L^2(\mathbb{R})$. I know that $||\psi(\cdot, t)-\psi(\cdot, t)||_{L^2(\mathbb{R})}=||\hat{\psi}_0-\hat{\psi_{0,n}}||_{L^2(\mathbb{R})}$, but why is it true that I could replace $\mathbb{R}$ by $\mathbb{R}^2$? Thanks. $\endgroup$ – Zheng Liu Jan 12 '16 at 6:26
  • $\begingroup$ Sorry it was a typo, and I meant $L^2(\mathbb{R})$. If you really want a space $\psi_n$ converges to $\psi$, you can choose $L^\infty(L^2(\mathbb{R}),\mathbb{R})$, such that $\|\psi\|_{L^\infty(L^2(\mathbb{R}))}=\sup_{t \in\mathbb{R}} \|\psi(\cdot,t)\|_{L^2(\mathbb{R})}$. In general, you can not replace the space to be $L^2(\mathbb{R}^2)$. $\endgroup$ – yhhuang Jan 12 '16 at 20:00
  • $\begingroup$ Oh. It seems there is another weak topology to deal with then. I wonder how it helps to prove that $\psi$ is the weak solution to Schrodinger equation then. Could you offer me more detail to the proof? Thanks a lot. $\endgroup$ – Zheng Liu Jan 14 '16 at 3:11

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