# The one-dimensional periodic Schrödinger equation

Jordan Bell
April 23, 2016

## 1 Translations and convolution

For $y\in\mathbb{R}$, let

 $\tau_{y}f(x)=f(x-y).$

To say that $f:\mathbb{R}\to\mathbb{C}$ is uniformly continuous means that $\left\|\tau_{h}f-f\right\|_{b}\to 0$ as $h\to 0$, where

 $\left\|g\right\|_{b}=\sup_{x\in\mathbb{R}}|g(x)|.$

Let $1\leq p<\infty$ and let $\mathscr{L}(L^{p}(\mathbb{R}))$ be the Banach algebra of bounded linear operators $L^{p}(\mathbb{R})\to L^{p}(\mathbb{R})$, with the strong operator topology: a net $T_{i}$ converges to $T$ in the strong operator topology if and only if for each $f\in L^{p}(\mathbb{R})$, $\left\|T_{i}f-Tf\right\|_{L^{p}}\to 0$.

###### Lemma 1.

$y\mapsto\tau_{y}$ is continuous $\mathbb{R}\to\mathscr{L}(L^{p}(\mathbb{R}))$, using the strong operator topology.

###### Proof.

For $y\in\mathbb{R}$ and $f\in L^{p}(\mathbb{R})$, $\left\|\tau_{y+h}f-\tau_{y}f\right\|_{L^{p}}=\left\|\tau_{h}f-f\right\|_{L^{p}}$. Take $\epsilon>0$ and let $\phi\in C_{c}(\mathbb{R})$ with $\left\|f-\phi\right\|_{L^{p}}<\infty$. Say $\mathrm{supp}\,\phi\subset[a,b]$. Let $K=[a-1,b+1]$. For $|h|\leq 1$, if $x\notin K$ then $x-h,x\not\in\mathrm{supp}\,\phi$, and hence

 $\displaystyle\left\|\tau_{h}\phi-\phi\right\|_{L^{p}}^{p}$ $\displaystyle=\int_{\mathbb{R}}|\phi(x-h)-\phi(x)|^{p}dx$ $\displaystyle=\int_{K}|\phi(x-h)-\phi(x)|^{p}dx$ $\displaystyle\leq(b-a+2)\left\|\tau_{h}\phi-\phi\right\|_{b}^{p}$ $\displaystyle=(b-a+2)\left\|\tau_{\phi}-\tau_{y}\phi\right\|_{b}^{p}.$

Because $\phi\in C_{c}(\mathbb{R})$, $\phi$ is uniformly continuous on $\mathbb{R}$, whence $\left\|\tau_{h}\phi-\phi\right\|_{L^{p}}\to 0$ as $h\to 0$, say $\left\|\tau_{h}\phi-\phi\right\|_{L^{p}}<\epsilon$ for $|h|\leq h_{\epsilon}$. Hence

 $\displaystyle\left\|\tau_{y+h}f-\tau_{y}f\right\|_{L^{p}}$ $\displaystyle=\left\|\tau_{h}f-f\right\|_{L^{p}}$ $\displaystyle\leq\left\|\tau_{h}f-\tau_{h}\phi\right\|_{L^{p}}+\left\|\tau_{h}% \phi-\phi\right\|_{L^{p}}+\left\|\phi-f\right\|_{L^{p}}$ $\displaystyle=2\left\|f-\phi\right\|_{L^{p}}+\left\|\tau_{h}-\phi\right\|_{L^{% p}}$ $\displaystyle<3\epsilon.$

Define $A:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ by

 $A(x_{1},x_{2})=x_{1}+x_{2}.$

If $\mu_{1},\mu_{2}$ are finite Borel measures on $\mathbb{R}$, let $\mu_{1}\otimes\mu_{2}$ be the product measure on $\mathbb{R}^{2}$, and let

 $\mu_{1}*\mu_{2}=A_{*}(\mu_{1}\otimes\mu_{2})$

be the pushforward of $\mu_{1}\otimes\mu_{2}$ by $A$, called the convolution of $\mu_{1}$ and $\mu_{2}$. If $f:\mathbb{R}\to[0,\infty]$ is measurable then applying the change of variables formula and then Tonelli’s theorem we obtain

 $\displaystyle\int fd(\mu_{1}*\mu_{2})$ $\displaystyle=\int f\circ Ad(\mu_{1}\otimes\mu_{2})$ $\displaystyle=\int\left(\int f\circ A(x_{1},x_{2})d\mu_{1}(x_{1})\right)d\mu_{% 2}(x_{2})$ $\displaystyle=\int\left(\int f(x_{1}+x_{2})d\mu_{1}(x_{1})\right)d\mu_{2}(x_{2% }).$

If $B$ is a Borel set in $\mathbb{R}$ then applying the above with $f=1_{B}$,

 $\displaystyle(\mu_{1}*\mu_{2})(B)$ $\displaystyle=\int 1_{B}d(\mu_{1}*\mu_{2})$ $\displaystyle=\int\left(\int 1_{B}(x_{1}+x_{2})d\mu_{1}(x_{1})\right)d\mu_{2}(% x_{2})$ $\displaystyle=\int\mu_{1}(B-x_{2})d\mu_{2}(x_{2}).$

## 2 Periodic functions

Let $\mathbb{T}=\mathbb{R}/\mathbb{Z}$, and let $\mathscr{S}(\mathbb{T})$ be the collection of $C^{\infty}$ functions $\phi:\mathbb{R}\to\mathbb{C}$ satisfying $\phi(x+1)=\phi(x)$ for all $x\in\mathbb{T}$. For $\phi,\psi\in\mathscr{S}(\mathbb{T})$, for $n\geq 1$ let

 $d_{n}(\phi,\psi)=\sup_{x\in[0,1]}|\phi^{(n)}(x)-\psi^{(n)}(x)|$

and

 $d(\phi,\psi)=\sum_{n=0}^{\infty}2^{-n}\frac{d_{n}(\phi,\psi)}{1+d_{n}(\phi,% \psi)}.$

With this metric, $\mathscr{S}(\mathbb{T})$ is a Fréchet space.

For $n\in\mathbb{Z}$, define

 $e_{n}(x)=e^{2\pi inx},\qquad x\in\mathbb{R}.$

For $f\in L^{1}(\mathbb{T})$, define $\widehat{f}:\mathbb{Z}\to\mathbb{C}$, for $n\in\mathbb{Z}$, by

 $\widehat{\phi}(n)=\int_{0}^{1}\phi(x)e_{-n}(x)dx=\int_{0}^{1}\phi(x)e^{-2\pi inx% }dx.$

Denote by $\mathscr{S}^{\prime}(\mathbb{T})$ the dual space of $\mathscr{S}(\mathbb{T})$, the collection of continuous linear maps $\mathscr{S}(\mathbb{T})\to\mathbb{C}$. For $L\in\mathscr{S}^{\prime}(\mathbb{T})$, define $\widehat{L}:\mathbb{Z}\to\mathbb{C}$ by

 $\widehat{L}(n)=Le_{-n}.$

For $x\in\mathbb{R}$, define $\delta_{x}:\mathscr{S}(\mathbb{T})\to\mathbb{C}$ by

 $\delta_{x}\phi=\phi(x).$

$\delta_{x}$ belongs to $\mathscr{S}^{\prime}(\mathbb{T})$, and

 $\widehat{\delta}_{x}(n)=\delta_{x}e_{-n}=e_{-n}(x)=e^{-2\pi inx}.$

For $f\in L^{1}(\mathbb{T})$, define $L_{f}\in\mathscr{S}^{\prime}(\mathbb{T})$ by

 $L_{f}\phi=\int_{0}^{1}f(x)\phi(x)dx,\qquad\phi\in\mathscr{S}(\mathbb{T}).$

For $n\in\mathbb{Z}$,

 $\widehat{L_{f}}(n)=L_{f}e_{-n}=\int_{0}^{1}f(x)e_{-n}(x)dx=\widehat{f}(n).$

## 3 The Poisson summation formula

If $f\in\mathscr{L}^{1}(\mathbb{R})$,

 $\displaystyle\int_{0}^{1}\sum_{n\in\mathbb{Z}}|f(x+n)|dx$ $\displaystyle=\sum_{n\in\mathbb{Z}}\int_{0}^{1}|f(x+n)|dx$ $\displaystyle=\sum_{n\in\mathbb{Z}}\int_{n}^{n+1}|f(x)|dx$ $\displaystyle=\int_{\mathbb{R}}|f(x)|dx.$

This implies that there is a Borel set $N_{f}$ in $\mathbb{R}$ with $\lambda(N_{f})=0$ such that for $x\in N_{f}^{c}$,

 $\sum_{n\in\mathbb{Z}}|f(x+n)|<\infty.$

We define $Pf(x)=\sum_{n\in\mathbb{Z}}f(x+n)$ for $x\in N_{f}^{c}$ and $Pf(x)=0$ for $x\in N_{f}$. Thus it makes sense to define $P:L^{1}(\mathbb{R})\to L^{1}(\mathbb{R})$ by

 $Pf(x)=\sum_{n\in\mathbb{Z}}f(x+n),$

in other words,

 $Pf=\sum_{n\in\mathbb{Z}}\tau_{-n}f.$

Then

 $\displaystyle\int_{0}^{1}Pf(x)e^{-2\pi imx}dx$ $\displaystyle=\int_{0}^{1}\left(\sum_{n\in\mathbb{Z}}f(x+n)\right)e^{-2\pi imx% }dx$ $\displaystyle=\sum_{n\in\mathbb{Z}}\int_{0}^{1}f(x+n)e^{-2\pi imx}dx$ $\displaystyle=\sum_{n\in\mathbb{Z}}\int_{n}^{n+1}f(x)e^{-2\pi imx}dx$ $\displaystyle=\int_{\mathbb{R}}f(x)e^{-2\pi imx}dx$ $\displaystyle=\widehat{f}(m).$

That is,

 $\widehat{Pf}(m)=\widehat{f}(m).$

Supposing that $Pf(x)=\sum_{n\in\mathbb{Z}}\widehat{Pf}(n)e^{2\pi inx}$,

 $Pf(x)=\sum_{n\in\mathbb{Z}}\widehat{f}(n)e^{2\pi inx}$

and supposing $Pf(x)=\sum_{n\in\mathbb{Z}}f(x+n)$,

 $\sum_{n\in\mathbb{Z}}f(x+n)=\sum_{n\in\mathbb{Z}}\widehat{f}(n)e^{2\pi inx},$

the Poisson summation formula.

For $N\geq 1$, let

 $L_{N}=\frac{1}{N}\sum_{j=0}^{N-1}\delta_{j/N}.$

For $n\in\mathbb{Z}$,

 $\widehat{L}_{N}(n)=\frac{1}{N}\sum_{j=0}^{N-1}\delta_{j/N}e_{-n}=\frac{1}{N}% \sum_{j=0}^{N-1}e_{-n}(j/N)=\frac{1}{N}\sum_{j=0}^{N-1}e^{-2\pi inj/N}.$

If $n\in N\mathbb{Z}$ then $\widehat{L}_{N}(n)=1$ and otherwise $\widehat{L}_{N}(n)=0$. That is,

 $L_{N}=\frac{1}{N}\sum_{j=0}^{N-1}\delta_{j/N}\sim\sum_{k\in\mathbb{Z}}\widehat% {L}_{N}(k)e_{k}=\sum_{k\in\mathbb{Z}}e_{Nk}.$

## 4 The heat kernel

For $x\in\mathbb{R}$ and $t>0$ define

 $H_{t}(x)=\int_{\mathbb{R}}e^{-4\pi^{2}t\xi^{2}}e^{2\pi i\xi x}d\xi.$

Using

 $\int_{\mathbb{R}}\exp\left(\frac{1}{2}iaw^{2}+iJw\right)dw=\sqrt{\frac{2\pi i}% {a}}\exp\left(-\frac{iJ^{2}}{2a}\right),$

for $\frac{1}{2}ia=-4\pi^{2}t$ we get $a=8i\pi^{2}t$ and $J=2\pi x$, and we calculate

 $\displaystyle H_{t}(x)$ $\displaystyle=\sqrt{\frac{2\pi i}{8\pi^{2}it}}\exp\left(-\frac{i}{16\pi^{2}it}% \cdot 4\pi^{2}x^{2}\right)$ $\displaystyle=\sqrt{\frac{1}{4\pi t}}\exp\left(-\frac{x^{2}}{4t}\right).$

By the Fourier inversion theorem,

 $\widehat{H_{t}}(\xi)=e^{-4\pi^{2}t\xi^{2}}.$

For $f\in L^{1}(\mathbb{R})$,

 $\widehat{\tau_{y}f}(\xi)=\int_{\mathbb{R}}f(x-y)e^{-2\pi i\xi x}dx=e^{-2\pi i% \xi y}\widehat{f}(\xi)=e_{-n}(y)\widehat{f}(\xi).$

## 5 The Schrödinger equation on ℝ

Let

 $\Gamma(t,x)=\sqrt{\frac{i}{t}}e^{-\pi ix^{2}/t},$

which satisfies

 $\partial_{x}\Gamma(t,x)=-\frac{2\pi ix}{t}\Gamma(t,x),\quad\partial_{x}^{2}% \Gamma(t,x)=-\frac{4\pi^{2}x^{2}}{t^{2}}\Gamma(t,x)-\frac{2\pi i}{t}\Gamma(t,x)$

and

 $\partial_{t}\Gamma(t,x)=-\frac{1}{2}t^{-1}\Gamma(t,x)+\pi ix^{2}t^{-2}\Gamma(t% ,x).$

This satisfies

 $\displaystyle\partial_{t}\Gamma(t,x)$ $\displaystyle=\frac{1}{2}\left(-\frac{1}{t}+\frac{2\pi ix^{2}}{t^{2}}\right)% \Gamma(t,x)$ $\displaystyle=\frac{1}{4\pi i}\left(-\frac{2\pi i}{t}-\frac{4\pi^{2}x^{2}}{t^{% 2}}\right)\Gamma(t,x)$ $\displaystyle=\frac{1}{4\pi i}\partial_{x}^{2}\Gamma(t,x).$

For $f:\mathbb{R}\to\mathbb{C}$, let

 $\psi(f)(t,x)=f*\Gamma(t,\cdot)(x)=\int_{\mathbb{R}}f(y)\Gamma(t,x-y)dy.$

This satisfies

 $\displaystyle\partial_{t}\psi(f)(t,x)$ $\displaystyle=\int_{\mathbb{R}}f(y)\cdot\partial_{t}\Gamma(t,x-y)dy$ $\displaystyle=\int_{\mathbb{R}}f(y)\cdot\frac{1}{4\pi i}\partial_{x}^{2}\Gamma% (t,x-y)dy$ $\displaystyle=\frac{1}{4\pi i}\partial_{x}^{2}\psi(f)(t,x).$

We also calculate

 $\displaystyle\psi(f)(t,x)$ $\displaystyle=\int_{\mathbb{R}}f(y)\cdot\Gamma(t,x-y)dy$ $\displaystyle=\int_{\mathbb{R}}f(y)\cdot\sqrt{\frac{i}{t}}e^{-\pi i(x-y)^{2}/t% }dy$ $\displaystyle=\int_{\mathbb{R}}f(y)\cdot\sqrt{\frac{i}{t}}\exp\left(-\frac{\pi ix% ^{2}}{t}+\frac{2\pi ixy}{t}-\frac{\pi iy^{2}}{t}\right)dy$ $\displaystyle=\Gamma(t,x)\cdot\int_{\mathbb{R}}f(y)\exp\left(-\frac{\pi i}{t}(% y^{2}-2xy)\right)dy.$

Let

 $\widehat{f}(y)=\int_{\mathbb{R}}f(x)e^{-2\pi ixy}dx.$

Using

 $\int_{\mathbb{R}}\exp\left(\frac{1}{2}iaw^{2}+iJw\right)dw=\sqrt{\frac{2\pi i}% {a}}\exp\left(-\frac{iJ^{2}}{2a}\right),$

we get, with $a=2\pi t$ and $J=2\pi u$,

 $\begin{split}&\displaystyle\Gamma(t,x)\cdot\psi(\widehat{f})(-1/t,-x/t)\\ \displaystyle=&\displaystyle\Gamma(t,x)\cdot\int_{\mathbb{R}}\widehat{f}(y)% \Gamma\left(-\frac{1}{t},-\frac{x}{t}-y\right)dy\\ \displaystyle=&\displaystyle\Gamma(t,x)\cdot\int_{\mathbb{R}}\widehat{f}\left(% -\frac{x}{t}-y\right)\Gamma\left(-\frac{1}{t},y\right)dy\\ \displaystyle=&\displaystyle\sqrt{\frac{i}{t}}e^{-\pi ix^{2}/t}\cdot\int_{% \mathbb{R}}\left(\int_{\mathbb{R}}f(u)e^{-2\pi iu\left(-\frac{x}{t}-y\right)}% du\right)\cdot\sqrt{-it}e^{\pi ity^{2}}dy\\ \displaystyle=&\displaystyle e^{-\pi ix^{2}/t}\int_{\mathbb{R}}f(u)e^{2\pi iux% /t}\left(\int_{\mathbb{R}}e^{2\pi iuy+\pi ity^{2}}dy\right)du\\ \displaystyle=&\displaystyle e^{-\pi ix^{2}/t}\int_{\mathbb{R}}f(u)e^{2\pi iux% /t}\cdot\sqrt{\frac{2\pi i}{2\pi t}}\exp\left(-\frac{i}{4\pi t}(2\pi u)^{2}% \right)du\\ \displaystyle=&\displaystyle e^{-\pi ix^{2}/t}\sqrt{\frac{i}{t}}\int_{\mathbb{% R}}f(u)e^{2\pi iux/t}\exp\left(-\frac{\pi iu^{2}}{t}\right)du\\ \displaystyle=&\displaystyle\sqrt{\frac{i}{t}}\int_{\mathbb{R}}f(u)\exp\left(-% \frac{\pi ix^{2}}{t}+\frac{2\pi iux}{t}-\frac{\pi iu^{2}}{t}\right)du\\ \displaystyle=&\displaystyle\sqrt{\frac{i}{t}}\int_{\mathbb{R}}f(u)e^{-\frac{% \pi i(x-u)^{2}}{t}}du\\ \displaystyle=&\displaystyle\int_{\mathbb{R}}f(u)\Gamma(t,x-u)du\\ \displaystyle=&\displaystyle\psi(f)(t,x).\end{split}$

In other words,

 $\displaystyle\psi(f)(t,x)$ $\displaystyle=\Gamma(t,x)\cdot\psi(\widehat{f})(-1/t,-x/t)$ $\displaystyle=\sqrt{\frac{i}{t}}e^{-\pi ix^{2}/t}\cdot\int_{\mathbb{R}}% \widehat{f}(\xi)\cdot\sqrt{-it}\exp\left(\pi it\left(-\frac{x}{t}-\xi\right)^{% 2}\right)d\xi$ $\displaystyle=\int_{\mathbb{R}}\widehat{f}(\xi)\exp\left(-\frac{\pi ix^{2}}{t}% +\frac{\pi ix^{2}}{t}+2\pi ix\xi+\pi it\xi^{2}\right)d\xi$ $\displaystyle=\int_{\mathbb{R}}\widehat{f}(\xi)e^{2\pi ix\xi+\pi it\xi^{2}}d\xi.$

## 6 The Schrödinger equation on 𝕋

Given $t$ and $x$, let $\gamma(y)=\Gamma(t,x-y)$. We calculate

 $\displaystyle\widehat{\gamma}(\xi)$ $\displaystyle=\int_{\mathbb{R}}\gamma(y)e^{-2\pi i\xi y}dy$ $\displaystyle=\int_{\mathbb{R}}\sqrt{\frac{i}{t}}e^{-\pi i(x-y)^{2}/t}e^{-2\pi i% \xi y}dy$ $\displaystyle=\int_{\mathbb{R}}\sqrt{\frac{i}{t}}\exp\left(-\frac{\pi ix^{2}}{% t}+\frac{2\pi ixy}{t}-\frac{\pi iy^{2}}{t}-2\pi i\xi y\right)dy.$

Using

 $\int_{\mathbb{R}}\exp\left(\frac{1}{2}iaw^{2}+iJw\right)dw=\sqrt{\frac{2\pi i}% {a}}\exp\left(-\frac{iJ^{2}}{2a}\right)$

with $a=-\frac{2\pi}{t}$ and $J=\frac{2\pi x}{t}-2\pi\xi$, for which $J^{2}=\frac{4\pi^{2}x^{2}}{t^{2}}-\frac{8\pi^{2}x\xi}{t}+4\pi^{2}\xi^{2}$,

 $\displaystyle\widehat{\gamma}(\xi)$ $\displaystyle=\sqrt{\frac{i}{t}}\exp\left(-\frac{\pi ix^{2}}{t}\right)\cdot% \sqrt{-it}\exp\left(\frac{it}{4\pi}J^{2}\right)$ $\displaystyle=\exp\left(-\frac{\pi ix^{2}}{t}\right)\exp\left(\frac{i\pi x^{2}% }{t}-2\pi ix\xi+\pi i\xi^{2}t\right)$ $\displaystyle=\exp\left(-2\pi ix\xi+\pi i\xi^{2}t\right).$

The Poisson summation formula tells us

 $\sum_{n\in\mathbb{Z}}\gamma(n)=\sum_{n\in\mathbb{Z}}\widehat{\gamma}(n),$

i.e.

 $\sum_{n\in\mathbb{Z}}\Gamma(t,x-n)=\sum_{n\in\mathbb{Z}}e^{-2\pi inx+\pi itn^{% 2}}=\sum_{n\in\mathbb{Z}}e^{2\pi inx+\pi itn^{2}}.$

Define

 $\Theta(t,x)=\sum_{n\in\mathbb{Z}}e^{\pi i(tn^{2}+2xn)}=\sum_{n\in\mathbb{Z}}e^% {\pi itn^{2}}e^{2\pi ixn}=\sum_{n\in\mathbb{Z}}\Gamma(t,x-n).$

For $\phi\in\mathscr{S}$, namely a Schwartz function, define

 $\Theta_{t}\phi(x)=\sum_{n\in\mathbb{Z}}\int_{\mathbb{R}}\phi(x)e^{\pi itn^{2}}% e^{2\pi ixn}dx,$

which satisfies

 $\Theta_{t}\phi(x)=\sum_{n\in\mathbb{Z}}\widehat{\phi}(-n)e^{\pi itn^{2}}=\sum_% {n\in\mathbb{Z}}\widehat{\phi}(n)e^{\pi itn^{2}}.$

If $f$ is $1$-periodic, for $n\in\mathbb{Z}$ let

 $\widehat{f}(n)=\int_{0}^{1}f(y)e^{-2\pi iny}dy.$

Define

 $\psi(f)(t,x)=\Theta_{t}*f(x)=\int_{0}^{1}\Theta(t,x-y)f(y)dy,$

which satisfies

 $\displaystyle\psi(f)(t,x)$ $\displaystyle=\int_{0}^{1}\sum_{n\in\mathbb{Z}}e^{\pi itn^{2}}e^{2\pi i(x-y)n}% f(y)dy$ $\displaystyle=\sum_{n\in\mathbb{Z}}e^{\pi itn^{2}}e^{2\pi ixn}\int_{0}^{1}f(y)% e^{-2\pi iny}dy$ $\displaystyle=\sum_{n\in\mathbb{Z}}e^{\pi itn^{2}}e^{2\pi ixn}\widehat{f}(n).$

We remind ourselves

 $\Theta(t,x)=\Theta_{t}(x)=\sum_{n\in\mathbb{Z}}e^{\pi itn^{2}}e^{2\pi ixn}$

and

 $\widehat{\Theta}_{t}(n)=e^{\pi itn^{2}}.$

Say $t=\frac{2M}{N}$. Then for $k\in\mathbb{Z}$,

 $\displaystyle\widehat{\Theta}_{t}(k+N)$ $\displaystyle=\exp\left(\pi i\cdot\frac{2M}{N}\cdot(k+N)^{2}\right)$ $\displaystyle=\exp\left(\pi i\cdot\frac{2M}{N}\cdot k^{2}\right).$