The one-dimensional periodic Schrödinger equation

Jordan Bell
April 23, 2016

1 Translations and convolution

For y, let

τyf(x)=f(x-y).

To say that f: is uniformly continuous means that τhf-fb0 as h0, where

gb=supx|g(x)|.

Let 1p< and let (Lp()) be the Banach algebra of bounded linear operators Lp()Lp(), with the strong operator topology: a net Ti converges to T in the strong operator topology if and only if for each fLp(), Tif-TfLp0.

Lemma 1.

yτy is continuous RL(Lp(R)), using the strong operator topology.

Proof.

For y and fLp(), τy+hf-τyfLp=τhf-fLp. Take ϵ>0 and let ϕCc() with f-ϕLp<. Say suppϕ[a,b]. Let K=[a-1,b+1]. For |h|1, if xK then x-h,xsuppϕ, and hence

τhϕ-ϕLpp =|ϕ(x-h)-ϕ(x)|p𝑑x
=K|ϕ(x-h)-ϕ(x)|p𝑑x
(b-a+2)τhϕ-ϕbp
=(b-a+2)τϕ-τyϕbp.

Because ϕCc(), ϕ is uniformly continuous on , whence τhϕ-ϕLp0 as h0, say τhϕ-ϕLp<ϵ for |h|hϵ. Hence

τy+hf-τyfLp =τhf-fLp
τhf-τhϕLp+τhϕ-ϕLp+ϕ-fLp
=2f-ϕLp+τh-ϕLp
<3ϵ.

Define A:× by

A(x1,x2)=x1+x2.

If μ1,μ2 are finite Borel measures on , let μ1μ2 be the product measure on 2, and let

μ1*μ2=A*(μ1μ2)

be the pushforward of μ1μ2 by A, called the convolution of μ1 and μ2. If f:[0,] is measurable then applying the change of variables formula and then Tonelli’s theorem we obtain

fd(μ1*μ2) =fAd(μ1μ2)
=(fA(x1,x2)𝑑μ1(x1))𝑑μ2(x2)
=(f(x1+x2)𝑑μ1(x1))𝑑μ2(x2).

If B is a Borel set in then applying the above with f=1B,

(μ1*μ2)(B) =1Bd(μ1*μ2)
=(1B(x1+x2)𝑑μ1(x1))𝑑μ2(x2)
=μ1(B-x2)𝑑μ2(x2).

2 Periodic functions

Let 𝕋=/, and let 𝒮(𝕋) be the collection of C functions ϕ: satisfying ϕ(x+1)=ϕ(x) for all x𝕋. For ϕ,ψ𝒮(𝕋), for n1 let

dn(ϕ,ψ)=supx[0,1]|ϕ(n)(x)-ψ(n)(x)|

and

d(ϕ,ψ)=n=02-ndn(ϕ,ψ)1+dn(ϕ,ψ).

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

For n, define

en(x)=e2πinx,x.

For fL1(𝕋), define f^:, for n, by

ϕ^(n)=01ϕ(x)e-n(x)𝑑x=01ϕ(x)e-2πinx𝑑x.

Denote by 𝒮(𝕋) the dual space of 𝒮(𝕋), the collection of continuous linear maps 𝒮(𝕋). For L𝒮(𝕋), define L^: by

L^(n)=Le-n.

For x, define δx:𝒮(𝕋) by

δxϕ=ϕ(x).

δx belongs to 𝒮(𝕋), and

δ^x(n)=δxe-n=e-n(x)=e-2πinx.

For fL1(𝕋), define Lf𝒮(𝕋) by

Lfϕ=01f(x)ϕ(x)𝑑x,ϕ𝒮(𝕋).

For n,

Lf^(n)=Lfe-n=01f(x)e-n(x)𝑑x=f^(n).

3 The Poisson summation formula

If f1(),

01n|f(x+n)|dx =n01|f(x+n)|𝑑x
=nnn+1|f(x)|𝑑x
=|f(x)|𝑑x.

This implies that there is a Borel set Nf in with λ(Nf)=0 such that for xNfc,

n|f(x+n)|<.

We define Pf(x)=nf(x+n) for xNfc and Pf(x)=0 for xNf. Thus it makes sense to define P:L1()L1() by

Pf(x)=nf(x+n),

in other words,

Pf=nτ-nf.

Then

01Pf(x)e-2πimx𝑑x =01(nf(x+n))e-2πimx𝑑x
=n01f(x+n)e-2πimx𝑑x
=nnn+1f(x)e-2πimx𝑑x
=f(x)e-2πimx𝑑x
=f^(m).

That is,

Pf^(m)=f^(m).

Supposing that Pf(x)=nPf^(n)e2πinx,

Pf(x)=nf^(n)e2πinx

and supposing Pf(x)=nf(x+n),

nf(x+n)=nf^(n)e2πinx,

the Poisson summation formula.

For N1, let

LN=1Nj=0N-1δj/N.

For n,

L^N(n)=1Nj=0N-1δj/Ne-n=1Nj=0N-1e-n(j/N)=1Nj=0N-1e-2πinj/N.

If nN then L^N(n)=1 and otherwise L^N(n)=0. That is,

LN=1Nj=0N-1δj/NkL^N(k)ek=keNk.

4 The heat kernel

For x and t>0 define

Ht(x)=e-4π2tξ2e2πiξx𝑑ξ.

Using

exp(12iaw2+iJw)𝑑w=2πiaexp(-iJ22a),

for 12ia=-4π2t we get a=8iπ2t and J=2πx, and we calculate

Ht(x) =2πi8π2itexp(-i16π2it4π2x2)
=14πtexp(-x24t).

By the Fourier inversion theorem,

Ht^(ξ)=e-4π2tξ2.

For fL1(),

τyf^(ξ)=f(x-y)e-2πiξx𝑑x=e-2πiξyf^(ξ)=e-n(y)f^(ξ).

5 The Schrödinger equation on ℝ

Let

Γ(t,x)=ite-πix2/t,

which satisfies

xΓ(t,x)=-2πixtΓ(t,x),x2Γ(t,x)=-4π2x2t2Γ(t,x)-2πitΓ(t,x)

and

tΓ(t,x)=-12t-1Γ(t,x)+πix2t-2Γ(t,x).

This satisfies

tΓ(t,x) =12(-1t+2πix2t2)Γ(t,x)
=14πi(-2πit-4π2x2t2)Γ(t,x)
=14πix2Γ(t,x).

For f:, let

ψ(f)(t,x)=f*Γ(t,)(x)=f(y)Γ(t,x-y)𝑑y.

This satisfies

tψ(f)(t,x) =f(y)tΓ(t,x-y)𝑑y
=f(y)14πix2Γ(t,x-y)𝑑y
=14πix2ψ(f)(t,x).

We also calculate

ψ(f)(t,x) =f(y)Γ(t,x-y)𝑑y
=f(y)ite-πi(x-y)2/t𝑑y
=f(y)itexp(-πix2t+2πixyt-πiy2t)𝑑y
=Γ(t,x)f(y)exp(-πit(y2-2xy))𝑑y.

Let

f^(y)=f(x)e-2πixy𝑑x.

Using

exp(12iaw2+iJw)𝑑w=2πiaexp(-iJ22a),

we get, with a=2πt and J=2πu,

Γ(t,x)ψ(f^)(-1/t,-x/t)=Γ(t,x)f^(y)Γ(-1t,-xt-y)𝑑y=Γ(t,x)f^(-xt-y)Γ(-1t,y)𝑑y=ite-πix2/t(f(u)e-2πiu(-xt-y)𝑑u)-iteπity2𝑑y=e-πix2/tf(u)e2πiux/t(e2πiuy+πity2𝑑y)𝑑u=e-πix2/tf(u)e2πiux/t2πi2πtexp(-i4πt(2πu)2)𝑑u=e-πix2/titf(u)e2πiux/texp(-πiu2t)𝑑u=itf(u)exp(-πix2t+2πiuxt-πiu2t)𝑑u=itf(u)e-πi(x-u)2t𝑑u=f(u)Γ(t,x-u)𝑑u=ψ(f)(t,x).

In other words,

ψ(f)(t,x) =Γ(t,x)ψ(f^)(-1/t,-x/t)
=ite-πix2/tf^(ξ)-itexp(πit(-xt-ξ)2)𝑑ξ
=f^(ξ)exp(-πix2t+πix2t+2πixξ+πitξ2)𝑑ξ
=f^(ξ)e2πixξ+πitξ2𝑑ξ.

6 The Schrödinger equation on 𝕋

Given t and x, let γ(y)=Γ(t,x-y). We calculate

γ^(ξ) =γ(y)e-2πiξy𝑑y
=ite-πi(x-y)2/te-2πiξy𝑑y
=itexp(-πix2t+2πixyt-πiy2t-2πiξy)𝑑y.

Using

exp(12iaw2+iJw)𝑑w=2πiaexp(-iJ22a)

with a=-2πt and J=2πxt-2πξ, for which J2=4π2x2t2-8π2xξt+4π2ξ2,

γ^(ξ) =itexp(-πix2t)-itexp(it4πJ2)
=exp(-πix2t)exp(iπx2t-2πixξ+πiξ2t)
=exp(-2πixξ+πiξ2t).

The Poisson summation formula tells us

nγ(n)=nγ^(n),

i.e.

nΓ(t,x-n)=ne-2πinx+πitn2=ne2πinx+πitn2.

Define

Θ(t,x)=neπi(tn2+2xn)=neπitn2e2πixn=nΓ(t,x-n).

For ϕ𝒮, namely a Schwartz function, define

Θtϕ(x)=nϕ(x)eπitn2e2πixn𝑑x,

which satisfies

Θtϕ(x)=nϕ^(-n)eπitn2=nϕ^(n)eπitn2.

If f is 1-periodic, for n let

f^(n)=01f(y)e-2πiny𝑑y.

Define

ψ(f)(t,x)=Θt*f(x)=01Θ(t,x-y)f(y)𝑑y,

which satisfies

ψ(f)(t,x) =01neπitn2e2πi(x-y)nf(y)dy
=neπitn2e2πixn01f(y)e-2πiny𝑑y
=neπitn2e2πixnf^(n).

We remind ourselves

Θ(t,x)=Θt(x)=neπitn2e2πixn

and

Θ^t(n)=eπitn2.

Say t=2MN. Then for k,

Θ^t(k+N) =exp(πi2MN(k+N)2)
=exp(πi2MNk2).