# Quantum walk on the line

1. Sep 25, 2009

### mvillagra

I'm reproducing the paper "Quantum walk on the line" by Nayak and Vishwanath (http://arxiv.org/abs/quant-ph/0010117), and cannot understand one part of the procedure. What they do is to Fourier transform the walk, then analyze the walk in Fourier space and then transform it back in order to get the original wave. These are the equations in Fourier space

$$\widetilde{\psi}_L(k,t)=\frac{1}{2}(1+\frac{\cos k}{\sqrt{1+\cos^2k}})e^{-iw_kt}+\frac{(-1)^t}{2}(1-\frac{\cos k}{\sqrt{1+\cos^2k}})e^{iw_kt}$$

$$\widetilde{\psi}_R(k,t)=\frac{ie^{ik}}{2\sqrt{1+\cos^2k}}(e^{-iw_kt}-(-1)^te^{iw_kt})$$

and after the inverse Fourier transform

$$\psi_L(n,t)=\frac{1+(-1)^{n+t}}{2}\int_{-\pi}^\pi \frac{dk}{2\pi}(1+\frac{\cos k}{\sqrt{1+\cos^2k}})e^{-(iw_kt+kn)}$$

$$\psi_R(n,t)=\frac{1+(-1)^{n+t}}{2}\int_{-\pi}^\pi \frac{dk}{2\pi}\frac{e^{ik}}{\sqrt{1+\cos^2k}}e^{-(iw_kt+kn)}$$

Remeber that the inverse quantum Fourier transform is
$$\psi(n,t)=\frac{1}{2\pi}\int_{-\pi}^\pi \widetilde{\psi}(k,t)e^{-ikn}dk$$

How did they put the n as an exponent on the -1? Also, the imaginary constant in the first term of $$\widetilde{\psi}_R$$ disappeared, and the minus sign in the second term in $$\widetilde{\psi}_L$$ is now a +. I've been fighting with this several days already and cannot solve it.