How to prove that a naive quantum random walk is non-unitary

Question

A 2000 paper by Nayak and Vishwanath provides an analysis of the dynamics of quantum random walks. In this paper, they mention a "naive" approach to defining a walk. I include the quote as follows:

In direct analogy, one may naively try to define quantum walk on the line as follows: at every time step, the particle moves, in superposition, both left and right with equal amplitudes (perhaps with a relative phase difference). However, such a walk is physically impossible, since the global process is non-unitary.

Intuitively, I imagine this is the case because, without further qualification, it seems that the various probability amplitudes would interfere constructively and lead to a wavefunction that was not normalized. Although it still seems like it might be possible to induce a relative phase difference such that normalization is achieved.

How does one prove/verify the author's statement in a formal way that would be satisfying to theorists?

Answer 1

I'm going to define $|n\rangle$ to be "the walker is at site $n$". Now imagine the walk as specified: $$ |n\rangle\rightarrow (|n-1\rangle+|n+1\rangle)/\sqrt{2}. $$ You can put some phases in if you want to, it's not going to change my basic argument. Now, imagine this is implemented by a unitary operator. This means that we need $$ \langle n-1|U^\dagger U|n+1\rangle=0 $$ since $U$ is unitary and the two walker positions $n\pm 1$ are distinguishable.

However, calculate what is given by the desired walk: $$ \langle n-1|U^\dagger U|n+1\rangle=\frac{1}{\sqrt{2}}(\langle n-2|+\langle n|)\frac{1}{\sqrt{2}}(|n\rangle+|n+2\rangle)=\frac12\neq 0. $$ Hence, this desired operation is not unitary.