The one-dimensional periodic Schrödinger equation

1 Translations and convolution

For $y\in ℝ$, let

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

To say that $f:ℝ\to ℂ$ is uniformly continuous means that ${\parallel {\tau }_{h}f-f\parallel }_b\to 0$ as $h\to 0$, where

$${\parallel g\parallel }_b=\underset{x\in ℝ}{sup}|g(x)|.$$

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

Lemma 1.

$y\mapsto {\tau }_y$ is continuous $\mathrm{R}\mathrm{\to }\mathrm{L}\mathit{}\mathrm{(}{L}^p\mathit{}\mathrm{(}\mathrm{R}\mathrm{)}\mathrm{)}$, using the strong operator topology.

Proof.

For $y\in ℝ$ and $f\in L^p(ℝ)$, ${\parallel {\tau }_{y+h}f-{\tau }_{y}f\parallel }_{L^p}={\parallel {\tau }_{h}f-f\parallel }_{L^p}$. Take $ϵ>0$ and let $\varphi \in C_c(ℝ)$ with . Say $\mathrm{supp}\varphi \subset [a,b]$. Let $K=[a-1,b+1]$. For $|h|\le 1$, if $x\notin K$ then $x-h,x\notin \mathrm{supp}\varphi$, and hence

${\parallel {\tau }_h\varphi -\varphi \parallel }_{L^p}^p$ $={\displaystyle {\int }_{ℝ}}{|\varphi (x-h)-\varphi (x)|}^p𝑑x$

$={\displaystyle {\int }_K}{|\varphi (x-h)-\varphi (x)|}^p𝑑x$

$\le (b-a+2){\parallel {\tau }_h\varphi -\varphi \parallel }_b^p$

$=(b-a+2){\parallel {\tau }_{\varphi }-{\tau }_y\varphi \parallel }_b^p.$

Because $\varphi \in C_c(ℝ)$, $\varphi$ is uniformly continuous on $ℝ$, whence ${\parallel {\tau }_h\varphi -\varphi \parallel }_{L^p}\to 0$ as $h\to 0$, say for $|h|\le h_{ϵ}$. Hence

${\parallel {\tau }_{y+h}f-{\tau }_{y}f\parallel }_{L^p}$ $={\parallel {\tau }_{h}f-f\parallel }_{L^p}$

$\le {\parallel {\tau }_{h}f-{\tau }_h\varphi \parallel }_{L^p}+{\parallel {\tau }_h\varphi -\varphi \parallel }_{L^p}+{\parallel \varphi -f\parallel }_{L^p}$

$=2{\parallel f-\varphi \parallel }_{L^p}+{\parallel {\tau }_h-\varphi \parallel }_{L^p}$

∎

Define $A:ℝ\times ℝ\to ℝ$ by

$$A(x_1,x_2)=x_1+x_2.$$

If ${\mu }_1,{\mu }_2$ are finite Borel measures on $ℝ$, let ${\mu }_1\otimes {\mu }_2$ be the product measure on ${ℝ}^2$, 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 $f:ℝ\to [0,\mathrm{\infty }]$ 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 $ℝ$ then applying the above with $f=1_B$,

$({\mu }_1*{\mu }_2)(B)$ $={\displaystyle \int 1_{B}d({\mu }_1*{\mu }_2)}$

$={\displaystyle \int \left(\int 1_B(x_1+x_2)𝑑{\mu }_1(x_1)\right)𝑑{\mu }_2(x_2)}$

$={\displaystyle \int {\mu }_1(B-x_2)𝑑{\mu }_2(x_2)}.$

2 Periodic functions

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

$$d_n(\varphi ,\psi )=\underset{x\in [0,1]}{sup}|{\varphi }^{(n)}(x)-{\psi }^{(n)}(x)|$$

and

$$d(\varphi ,\psi )=\sum _{n=0}^{\mathrm{\infty }}{2}^{-n}\frac{d_n(\varphi ,\psi )}{1+d_n(\varphi ,\psi )}.$$

With this metric, $𝒮(𝕋)$ is a Fréchet space.

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

$$e_n(x)=e^{2\pi inx},x\in ℝ.$$

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

$$\widehat{\varphi }(n)={\int }_0^1\varphi (x)e_{-n}(x)𝑑x={\int }_0^1\varphi (x)e^{-2\pi inx}𝑑x.$$

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

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

For $x\in ℝ$, define ${\delta }_x:𝒮(𝕋)\to ℂ$ by

$${\delta }_x\varphi =\varphi (x).$$

${\delta }_x$ belongs to ${𝒮}^{\prime }(𝕋)$, and

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

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

$$L_f\varphi ={\int }_0^{1}f(x)\varphi (x)𝑑x,\varphi \in 𝒮(𝕋).$$

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 $f\in {ℒ}^1(ℝ)$,

${\displaystyle {\int }_0^1}{\displaystyle \sum _{n\in \mathbb{Z}}}|f(x+n)|dx$ $={\displaystyle \sum _{n\in \mathbb{Z}}}{\displaystyle {\int }_0^1}|f(x+n)|𝑑x$

$={\displaystyle \sum _{n\in \mathbb{Z}}}{\displaystyle {\int }_n^{n+1}}|f(x)|𝑑x$

$={\displaystyle {\int }_{ℝ}}|f(x)|𝑑x.$

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

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 $P:L^1(ℝ)\to L^1(ℝ)$ by

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

in other words,

$$Pf=\sum _{n\in \mathbb{Z}}{\tau }_{-n}f.$$

Then

${\displaystyle {\int }_0^1}Pf(x)e^{-2\pi imx}𝑑x$ $={\displaystyle {\int }_0^1}\left({\displaystyle \sum _{n\in \mathbb{Z}}}f(x+n)\right)e^{-2\pi imx}𝑑x$

$={\displaystyle \sum _{n\in \mathbb{Z}}}{\displaystyle {\int }_0^1}f(x+n)e^{-2\pi imx}𝑑x$

$={\displaystyle \sum _{n\in \mathbb{Z}}}{\displaystyle {\int }_n^{n+1}}f(x)e^{-2\pi imx}𝑑x$

$={\displaystyle {\int }_{ℝ}}f(x)e^{-2\pi imx}𝑑x$

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

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 $N\ge 1$, 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 $x\in ℝ$ and $t>0$ define

$$H_t(x)={\int }_{ℝ}{e}^{-4{\pi }^{2}t{\xi }^2}{e}^{2\pi i\xi x}𝑑\xi .$$

Using

$${\int }_{ℝ}\exp \left(\frac{1}{2}ia{w}^2+iJw\right)𝑑w=\sqrt{\frac{2\pi i}{a}}\exp \left(-\frac{i{J}^2}{2a}\right),$$

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

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

$=\sqrt{{\displaystyle \frac{1}{4\pi t}}}\exp \left(-{\displaystyle \frac{x^2}{4t}}\right).$

By the Fourier inversion theorem,

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

For $f\in L^1(ℝ)$,

$$\widehat{{\tau }_{y}f}(\xi )={\int }_{ℝ}f(x-y)e^{-2\pi i\xi x}𝑑x=e^{-2\pi i\xi y}\widehat{f}(\xi )=e_{-n}(y)\widehat{f}(\xi ).$$

5 The Schrödinger equation on ℝ

Let

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

which satisfies

$${\partial }_x\mathrm{\Gamma }(t,x)=-\frac{2\pi ix}{t}\mathrm{\Gamma }(t,x),{\partial }_x^2\mathrm{\Gamma }(t,x)=-\frac{4{\pi }^2{x}^2}{t^2}\mathrm{\Gamma }(t,x)-\frac{2\pi i}{t}\mathrm{\Gamma }(t,x)$$

and

$${\partial }_t\mathrm{\Gamma }(t,x)=-\frac{1}{2}{t}^{-1}\mathrm{\Gamma }(t,x)+\pi i{x}^2{t}^{-2}\mathrm{\Gamma }(t,x).$$

This satisfies

${\partial }_t\mathrm{\Gamma }(t,x)$ $={\displaystyle \frac{1}{2}}\left(-{\displaystyle \frac{1}{t}}+{\displaystyle \frac{2\pi i{x}^2}{t^2}}\right)\mathrm{\Gamma }(t,x)$

$={\displaystyle \frac{1}{4\pi i}}\left(-{\displaystyle \frac{2\pi i}{t}}-{\displaystyle \frac{4{\pi }^2{x}^2}{t^2}}\right)\mathrm{\Gamma }(t,x)$

$={\displaystyle \frac{1}{4\pi i}}{\partial }_x^2\mathrm{\Gamma }(t,x).$

For $f:ℝ\to ℂ$, let

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

This satisfies

${\partial }_t\psi (f)(t,x)$ $={\displaystyle {\int }_{ℝ}}f(y)\cdot {\partial }_t\mathrm{\Gamma }(t,x-y)𝑑y$

$={\displaystyle {\int }_{ℝ}}f(y)\cdot {\displaystyle \frac{1}{4\pi i}}{\partial }_x^2\mathrm{\Gamma }(t,x-y)𝑑y$

$={\displaystyle \frac{1}{4\pi i}}{\partial }_x^2\psi (f)(t,x).$

We also calculate

$\psi (f)(t,x)$ $={\displaystyle {\int }_{ℝ}}f(y)\cdot \mathrm{\Gamma }(t,x-y)𝑑y$

$={\displaystyle {\int }_{ℝ}}f(y)\cdot \sqrt{{\displaystyle \frac{i}{t}}}{e}^{-\pi i{(x-y)}^2/t}𝑑y$

$={\displaystyle {\int }_{ℝ}}f(y)\cdot \sqrt{{\displaystyle \frac{i}{t}}}\exp \left(-{\displaystyle \frac{\pi i{x}^2}{t}}+{\displaystyle \frac{2\pi ixy}{t}}-{\displaystyle \frac{\pi i{y}^2}{t}}\right)𝑑y$

$=\mathrm{\Gamma }(t,x)\cdot {\displaystyle {\int }_{ℝ}}f(y)\exp \left(-{\displaystyle \frac{\pi i}{t}}(y^2-2xy)\right)𝑑y.$

Let

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

Using

$${\int }_{ℝ}\exp \left(\frac{1}{2}ia{w}^2+iJw\right)𝑑w=\sqrt{\frac{2\pi i}{a}}\exp \left(-\frac{i{J}^2}{2a}\right),$$

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

In other words,

$\psi (f)(t,x)$ $=\mathrm{\Gamma }(t,x)\cdot \psi (\widehat{f})(-1/t,-x/t)$

$=\sqrt{{\displaystyle \frac{i}{t}}}{e}^{-\pi i{x}^2/t}\cdot {\displaystyle {\int }_{ℝ}}\widehat{f}(\xi )\cdot \sqrt{-it}\exp \left(\pi it{\left(-{\displaystyle \frac{x}{t}}-\xi \right)}^2\right)𝑑\xi$

$={\displaystyle {\int }_{ℝ}}\widehat{f}(\xi )\exp \left(-{\displaystyle \frac{\pi i{x}^2}{t}}+{\displaystyle \frac{\pi i{x}^2}{t}}+2\pi ix\xi +\pi it{\xi }^2\right)𝑑\xi$

$={\displaystyle {\int }_{ℝ}}\widehat{f}(\xi )e^{2\pi ix\xi +\pi it{\xi }^2}𝑑\xi .$

6 The Schrödinger equation on 𝕋

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

$\widehat{\gamma }(\xi )$ $={\displaystyle {\int }_{ℝ}}\gamma (y)e^{-2\pi i\xi y}𝑑y$

$={\displaystyle {\int }_{ℝ}}\sqrt{{\displaystyle \frac{i}{t}}}{e}^{-\pi i{(x-y)}^2/t}{e}^{-2\pi i\xi y}𝑑y$

$={\displaystyle {\int }_{ℝ}}\sqrt{{\displaystyle \frac{i}{t}}}\exp \left(-{\displaystyle \frac{\pi i{x}^2}{t}}+{\displaystyle \frac{2\pi ixy}{t}}-{\displaystyle \frac{\pi i{y}^2}{t}}-2\pi i\xi y\right)𝑑y.$

Using

$${\int }_{ℝ}\exp \left(\frac{1}{2}ia{w}^2+iJw\right)𝑑w=\sqrt{\frac{2\pi i}{a}}\exp \left(-\frac{i{J}^2}{2a}\right)$$

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

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

$=\exp \left(-{\displaystyle \frac{\pi i{x}^2}{t}}\right)\exp \left({\displaystyle \frac{i\pi x^2}{t}}-2\pi ix\xi +\pi i{\xi }^{2}t\right)$

$=\exp \left(-2\pi ix\xi +\pi i{\xi }^{2}t\right).$

The Poisson summation formula tells us

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

i.e.

$$\sum _{n\in \mathbb{Z}}\mathrm{\Gamma }(t,x-n)=\sum _{n\in \mathbb{Z}}{e}^{-2\pi inx+\pi it{n}^2}=\sum _{n\in \mathbb{Z}}{e}^{2\pi inx+\pi it{n}^2}.$$

Define

$$\mathrm{\Theta }(t,x)=\sum _{n\in \mathbb{Z}}{e}^{\pi i(t{n}^2+2xn)}=\sum _{n\in \mathbb{Z}}{e}^{\pi it{n}^2}{e}^{2\pi ixn}=\sum _{n\in \mathbb{Z}}\mathrm{\Gamma }(t,x-n).$$

For $\varphi \in 𝒮$, namely a Schwartz function, define

$${\mathrm{\Theta }}_t\varphi (x)=\sum _{n\in \mathbb{Z}}{\int }_{ℝ}\varphi (x)e^{\pi it{n}^2}{e}^{2\pi ixn}𝑑x,$$

which satisfies

$${\mathrm{\Theta }}_t\varphi (x)=\sum _{n\in \mathbb{Z}}\widehat{\varphi }(-n)e^{\pi it{n}^2}=\sum _{n\in \mathbb{Z}}\widehat{\varphi }(n)e^{\pi it{n}^2}.$$

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

$$\psi (f)(t,x)={\mathrm{\Theta }}_t*f(x)={\int }_0^1\mathrm{\Theta }(t,x-y)f(y)𝑑y,$$

which satisfies

$\psi (f)(t,x)$ $={\displaystyle {\int }_0^1}{\displaystyle \sum _{n\in \mathbb{Z}}}{e}^{\pi it{n}^2}{e}^{2\pi i(x-y)n}f(y)dy$

$={\displaystyle \sum _{n\in \mathbb{Z}}}{e}^{\pi it{n}^2}{e}^{2\pi ixn}{\displaystyle {\int }_0^1}f(y)e^{-2\pi iny}𝑑y$

$={\displaystyle \sum _{n\in \mathbb{Z}}}{e}^{\pi it{n}^2}{e}^{2\pi ixn}\widehat{f}(n).$

We remind ourselves

$$\mathrm{\Theta }(t,x)={\mathrm{\Theta }}_t(x)=\sum _{n\in \mathbb{Z}}{e}^{\pi it{n}^2}{e}^{2\pi ixn}$$

and

$${\widehat{\mathrm{\Theta }}}_t(n)=e^{\pi it{n}^2}.$$

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

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

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