How can I make this toy quantum random walk model unitary? Take a toy $(1+1)$-dimensional lattice model of the universe. A particle begins at $x=0$ at $t=0$. It has an amplitude ${1}/{\sqrt{2}}$ to move one step to the left and amplitude ${1}/{\sqrt{2}}$ to move one step to the right.
At time $t=1$ it will either be at $x=-1$ or $x=+1$ with probability $\frac{1}{2}$ each. At time $t=2$ it will either be at $x=-2$, $x=0$ or $x=2$.
Using sum over histories (it could have taken two paths) the probability it is at $x=0$ is:
$$
\begin{align}
P(x=0,t=2) &= \left|(\frac{1}{\sqrt{2}})^2+(\frac{1}{\sqrt{2}})^2\right|^2 = 1\\
P(x=-2,t=2) &= \left|(\frac{1}{\sqrt{2}})^2\right|^2 = \frac{1}{4}\\
P(x=+2,t=2) &= \left|(\frac{1}{\sqrt{2}})^2\right|^2 = \frac{1}{4}
\end{align}
$$
The total probability is $1.5$.  
Is there any way to make this theory Unitary? Or is it just a case of normalising it?
There is no way to set the amplitudes for moving left and right to get a probability adding up to 1. Or does it just mean this is not a viable model?
 A: Let $\psi(k,t)$ be the amplitude to locate the particle on site $k$ at time $t$. Also let $U_{j,k}$ be the matrix describing your process, such that 
$$
\psi(j, t+1) = \sum_k{U_{j,k}\psi(k,t)}
$$
Then the process you proposed is described by 
$$
U_{j,k} = \frac{1}{\sqrt{2}}\left( \delta_{j,k-1} + \delta_{j, k+1}\right)
$$
It can be verified immediately that applying process $U$ to an initial amplitude $\psi(k,t=0) = \delta_{k,0}$ produces exactly the states you described for $t=1$ and $t=2$. 
The problem you are having is that U is not unitary, since 
$$
\sum_l{U^*_{l,j}U_{l,k}} = \frac{1}{2}\sum_l{\left( \delta_{l,j-1} + \delta_{l, j+1}\right)\left( \delta_{l,k-1} + \delta_{l, k+1}\right)} = \frac{1}{2}\left( \delta_{j, k-2} + 2\delta_{j,k} + \delta_{j, k+2} \right) \neq \delta_{j,k}
$$
For a more formal approach, notice that U can be written as the sum 
$$
U = \frac{1}{\sqrt{2}}\left(W^\dagger + W \right)
$$
where
$$ 
W_{j,k} = \delta_{j,k+1} \;\;\;\text{and} \;\;\; \sum_l{W^*_{j,l}W_{l,k}} = \delta_{j+1,k+1}, \;\; \sum_l{W_{j,l}W^\dagger_{l,k}} = \delta_{j,k}
$$
This means that $W$ is unitary, since 
$$
W^\dagger W = WW^\dagger = I
$$ 
Unfortunately, the fact that $W^2 \neq W$, $\left(W^\dagger\right)^2  \neq W^\dagger$, leads to $U^\dagger U = \frac{1}{2}\left( W^2 + \left(W^\dagger\right)^2 + W W^\dagger + W^\dagger W \right) \neq I$, as we already know. 
How to fix: I don't think you need to take the amplitudes as matrices. What follows is not a definitive solution by any means, but it shows that things can be worked out in a variety of ways. 
We need a convenient unitary $U$ that involves only jumps between adjacent sites. We already know that a simple superposition of left and right jumps along the entire lattice does not work. One way to get around this is to use a simple dynamics on pairs of adjacent sites. For instance: split the lattice in non-overlapping pairs $(2k, 2k+1)$, $k = 0, \pm 1, \pm 2,...$ and for each pair define a SU(2) unitary matrix $U^{(k)}$, 
$$
U^{(k)}_{2k,2k} = cos(\theta_k),\;\;U^{(k)}_{2k,2k+1} = sin(\theta_k)e^{i\alpha_k} \\
U^{(k)}_{2k+1,2k} = -sin(\theta_k)e^{i\alpha_k},\;\;U^{(k)}_{2k+1,2k+1} = cos(\theta_k)
$$
The superposition 
$$
W^{(+)} = \sum_k{U^{(k)}}
$$
is unitary, but acts separately on each pair $(2k, 2k+1)$. Since we want the total process  to mix adjacent sites in both directions, let's construct also similar processes $V^{(k)}$ for pairs $(2k-1, 2k)$, $k = 0, \pm 1, \pm 2,...$, 
$$
V^{(k)}_{2k-1,2k-1} = cos(\theta_k),\;\;V^{(k)}_{2k-1,2k} = sin(\theta_k)e^{i\alpha_k} \\
V^{(k)}_{2k,2k-1} = -sin(\theta_k)e^{i\alpha_k},\;\;V^{(k)}_{2k,2k} = cos(\theta_k)
$$ 
and let us define the additional unitary superposition 
$$
W^{(-)} = \sum_k{V^{(k)}}
$$
Now we can define the unitary time step process U as
$$
U = W^{(-)} W^{(+)}
$$
To see how it works on a simple example, take $\theta_k = \frac{\pi}{4}$, $\alpha_k = 0$ and $\psi(j,t=0) = \delta_{j,0}$. Then if
$$
{\bar \psi}(j, 0) = \left[W^{(+)}\psi(t=0)\right]_j = \left[U^{(0)}\psi(t=0)\right]_j = \frac{1}{\sqrt{2}}\left( \delta_{j,0} - \delta_{j,1}\right) 
$$
we have for $t=1$
$$
\psi(j, t=1) = \left[U\psi(t=0)\right]_j = \left[ W^{(-)}{\bar \psi}(t=0) \right]_j = \left[ \left(V^{(0)} + V^{(1)}\right) {\bar \psi}(t=0)\right]_j = \frac{1}{2}\left(\delta_{j,-1} + \delta_{j,0} - \delta_{j,1} + \delta_{j,2} \right)
$$
It is a slightly asymmetrical propagation, but it goes in both directions on neighboring sites and it does conserve probability. Hope it helps.
A: What you are describing is a Hadamard walk on a line.
The Hadamard walk is a discrete quantum walk. To make it unitary, the walker state has to be defined not in one but two spaces: coin (possible directions of a walker) and position. Then, the walk itself is performed by applying two unitary transformations.
You can find more on that here in the section 3.1.
A: I have found a set of four amplitudes that seem to work. You have to assume the particle has two states.
L  Move left (amplitude = +1/2)
L' Move left and flip polarisation (amplitude = +1/2)
R  Move right (amplitude = +1/2)
R' Move right and flip polarisation (amplitude = -1/2)
I've tested it up to t=3 and it seems to work. Although I haven't proved it works always. It appears to model a particle travelling on the light cone. As the probability for $x^2\neq t^2$ seems always to be zero.
