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?

  • 1
    $\begingroup$ I don't understand what you mean with amplitude in this context? You should sum over probabilities, not amplitudes. Otherwise you will have to distribute your amplitudes properly. $\endgroup$
    – Bernhard
    Sep 17 '15 at 18:51
  • 1
    $\begingroup$ You might want to read up on quantum random walks. $\endgroup$ Sep 17 '15 at 18:57
  • $\begingroup$ No, you sum over amplitudes if the end results are indistinguishable. Sum over probabilities if the results are distinguishable. If you only ever summed over probabilities you wouldn't get interference patterns. This is quantum mechanics not classical mechanics. $\endgroup$
    – zooby
    Sep 17 '15 at 19:04
  • $\begingroup$ See Feynman checkerboard. $\endgroup$
    – Qmechanic
    Sep 17 '15 at 19:12
  • $\begingroup$ Yes, it looks like the Feynman checkerboard. But in that case should the amplitudes for moving left and right be matrices? And in which case (a) what are they? And (b) how do you take the modulus of a matrix to get a probability and does it add up to one? $\endgroup$
    – zooby
    Sep 17 '15 at 19:25

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.

  • $\begingroup$ Thanks for taking the time for this explanation. It really helps! So I think my toy model must be too simple to actually work! Just goes to show that constructing a unitary quantum theory is not trivial. Thank you so much! $\endgroup$
    – zooby
    Sep 18 '15 at 11:54
  • $\begingroup$ @zooby Welcome and good luck. $\endgroup$
    – udrv
    Sep 18 '15 at 19:22

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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.