Exact propagation

From Schrödinger equation, iℏ\frac{∂ψ}{∂t} = Hψ, we can get

ψ(t + dt) = ψ(t) + \frac{∂ψ}{∂t} dt = ψ(t) − \frac{i}{ℏ}Hψ(t)dt = e^{− \frac{i}{ℏ} Hdt}ψ(t)

This last form is unitary.

e^{− \frac{i}{ℏ} Hdt} ≈ e^{− \frac{i}{ℏ} \frac{V}{2} dt}e^{− \frac{i}{ℏ} Hdt}e^{− \frac{i}{ℏ} \frac{V}{2} dt}

|ψ(t)⟩ = e^{− \frac{i}{ℏ} HT} |ψ(0)⟩

⟨xy|ψ(t)⟩ = ⟨xy| e^{− \frac{i}{ℏ} HT} |ψ(0)⟩ = ⟨xy| e^{− \frac{i}{ℏ} HT} |ψ(0)⟩

My notes.

iℏ\frac{dΨ(𝔯, t)}{dt} = HΨ(𝔯, t)

Assuming Ψ(𝔯, t) = ψ(𝔯)χ(t) won’t help (Ψ is non-separable). However, the above can be understood as an equation in the Hilbert space of functions from coordinate space (ℝ^{2}) to ℂ; so taking Ψ(𝔯, t) ≡ [Ψ (t)] (𝔯) we get

iℏ\frac{dΨ (t)}{dt} = HΨ (t)

Ψ (t) = e^{−iHt/ℏ} Ψ (0)

(Simple insertion shows that this is the result we are looking for.) Taking ϵ = t/N, we get

Ψ (t) = (e^{−iHϵ/ℏ})^{N} Ψ (0)
≈ (e^{−iV ϵ/2ℏ} e^{−iKϵ/ℏ} e^{−iV ϵ/2ℏ})^{N} Ψ (0)

= (e^{\frac{−iV ϵ}{2ℏ}} e^{\frac{−iKϵ}{ℏ}} e^{\frac{−iV ϵ}{2ℏ}} ) (e^{\frac{−iV ϵ}{2ℏ}} e^{\frac{−iKϵ}{ℏ}} e^{\frac{−iV ϵ}{2ℏ}} ) … (e^{\frac{−iV ϵ}{2ℏ}} e^{\frac{−iKϵ}{ℏ}} e^{\frac{−iV ϵ}{2ℏ}} ) Ψ (0)

= e^{\frac{−iV ϵ}{2ℏ}} e^{\frac{−iKϵ}{ℏ}} e^{\frac{−iV ϵ}{ℏ}} e^{\frac{−iKϵ}{ℏ}} … e^{\frac{−iKϵ}{ℏ}} e^{\frac{−iV ϵ}{2ℏ}} Ψ (0)

where e^{−iV t/2ℏ}, e^{−iKt/ℏ} and e^{−iV t/ℏ} can be calculated by

e^{−iAt/ℏ} = ∑_{n=0}^{∞} \frac{(−iAt/ℏ)^{n}}{n!}

All the operators are functions from N-space to N-space, and could in a general discrete form be described by 2N-dimensional “matrices”. However, in coordinate form V̂ is just a multiplication of each element by V (𝔯) corresponding to its position,

Typeset by AMS-TEX

so e^{−iV̂ t/2ℏ} and e^{−iV̂ t/ℏ} have that form as well. Analogously, K̂ is in its momentum form is a multiplication of each element by its K(𝔭), and so is e^{−iK̂t/ℏ}.

Taylor sez:

Ψ(𝔯, t) = (2π)^{− \frac{3}{2}} ∫_{V} Φ(𝔨, t) exp i 𝔨·𝔯 d𝔨_{x}d𝔨_{y}d𝔨_{z}
Φ(𝔨, t) = (2π)^{− \frac{3}{2}} ∫_{V} Ψ(𝔯, t) exp −i 𝔨·𝔯 dxdydz

Numerical Recipes are, of course, geared towards evil electrical engineers, so they use non-angular frequencies, which corresopond to spatial frequenties (and not wave vector). We have 𝔨 ≡ 2π𝔣, so

\begin{aligned}Ψ(𝔯, t) & = (2π)^{+ \frac{3}{2}} ∫_{V} Φ(𝔣, t) e^{2πi 𝔣·𝔯}d𝔣_{x}d𝔣_{y}d𝔣_{z} \\ Φ(𝔣, t) & = (2π)^{− \frac{3}{2}} ∫_{V} Ψ(𝔯, t) e^{−2πi 𝔣·𝔯}dxdydz \end{aligned}

To simplify life a bit, we can change to Φ′ ≡ (2π)^{\frac{3}{2}} Φ:

\begin{aligned}Ψ(𝔯, t) & = ∫_{V} Φ′(𝔣, t) e^{2πi 𝔣·𝔯}d𝔣_{x}d𝔣_{y}d𝔣_{z} \\ Φ′(𝔣, t) & = ∫_{V} Ψ(𝔯, t) e^{−2πi 𝔣·𝔯}dxdydz \end{aligned}

For discrete r and 𝔣, this becomes

\begin{aligned}Ψ(𝔯, t) & = ∑_{k_{n}=0}^{N_{n}−1} ⋯ ∑_{k_{1}=0}^{N_{1}−1} Φ′(𝔣, t) e^{2πi 𝔣·𝔯}d𝔣_{x}d𝔣_{y}d𝔣_{z} \\ Φ′(𝔣, t) & = \frac{1}{N^{n}} ∑_{k_{n}=0}^{N_{n}−1} ⋯ ∑_{k_{1}=0}^{N_{1}−1} Ψ(𝔯, t) e^{−2πi 𝔣·𝔯}dxdydz \end{aligned}

Let us assume we’re sampling the positions at intervals Δ, i.e.

𝔯_{mnp} = mΔi⃗ + nΔj⃗ + pΔk⃗ m, n, p = …, −1, 0, 1, … Then the spatial frequencies at points of Fourier transform will be

𝔣_{mnp} = \frac{m}{NΔ}i⃗ + \frac{n}{NΔ}j⃗ + \frac{p}{NΔ}k⃗ m, n, p = …, −1, 0, 1, …

𝔨_{mnp} = 2π𝔣_{mnp}
(𝔭_{mnp} = ℏ 𝔨_{mnp})

K(𝔭) = \frac{|𝔭|^{2}}{2m}

= \frac{𝔭_{x}^{2} + 𝔭_{y}^{2} + 𝔭_{z}^{2}}{2m}

= \frac{(2πℏ)^{2}}{2m} \frac{m^{2} + n^{2} + p^{2}}{Δ^{2}N^{2}}

e^{−iKt/ℏ} = exp (−it\frac{ℏ}{Δ^{2}} \frac{2π^{2}}{mN^{2}} (m^{2} + n^{2} + p^{2}))

For a harmonic oscillator potential:

V (𝔯) = \frac{1}{2}k|𝔯|^{2}

= \frac{k}{2} (𝔯_{x}^{2} + 𝔯_{y}^{2} + 𝔯_{z}^{2})

= \frac{k}{2} Δ^{2}(m^{2} + n^{2} + p^{2})

e^{−iV t/ℏ} = exp (−it\frac{k}{2} \frac{Δ^{2}}{ℏ} (m^{2} + n^{2} + p^{2}))

The maximum values of V t/ℏ and Kt/ℏ for a d-dimensional matrix, N ×…×N, are

tΔ^{2} \frac{k}{2ℏ} \frac{dN^{2}}{4} and \frac{t}{Δ^{2}} \frac{2ℏπ^{2}}{mN^{2}} \frac{dN^{2}}{4}

For a “good” calculation, the exponentials should stay close to linear, thus the exponents should be < μ ≪ 1. Still working with the harmonic oscillator potential, we get

t^{2} \frac{kπ^{2}}{m} (\frac{dN}{4} )^{2} < μ^{2}

t < \frac{4μ}{dNπ} \sqrt{\frac{m}{k}} = \frac{2μ}{dNπ^{2}} T_{c}

where T_{c} = 2π\sqrt{m/k} is the classical period of a harmonic oscillator.

For the Nelson potential, we have

V = μ\frac{x^{2}}{2} + (y − \frac{x^{2}}{2} )^{2}

“Good” values of parameters seem to be [Baranger]

\begin{aligned}μ & = 0.1 \\ m & = 1 \\ ℏ & = 0.05 \\ Eclas & = 0.5 \end{aligned}

for HOP:

Ĥ = T̂ + V̂ = \frac{1}{2m} (p_{x}^{2} + p_{y}^{2}) + k \frac{x^{2}}{2} + k \frac{y^{2}}{2}

for Nelson:

Ĥ = T̂ + V̂ = \frac{1}{2m} (p_{x}^{2} + p_{y}^{2}) + (μ\frac{x^{2}}{2} + (y − x^{2}/2)^{2})