1

Is the following unitary transformation possible? If so, what will be the value of $U$?

$$U|i,j_1\rangle = 1/\sqrt{k}(|i,j_1\rangle+|i,j_2\rangle+|i,j_3\rangle...+|i,j_k\rangle)$$

Here, $i$ is a node in a graph and $j_1,j_2....j_k$ are the nodes to which node $i$ is attached. $k$ is the degree of node.

For example: Consider we have 8 nodes in the graph and Node 0($|000\rangle$) is attached to node 1($|001\rangle$) ,node 7($|111\rangle$) and node 5($|101\rangle$). So, what I want is a single $U$ operator which does this:

$$U|000,001\rangle=1/\sqrt{3}(|000,001\rangle+|000,111\rangle+|000,101\rangle)$$ $$U|000,111\rangle=1/\sqrt{3}(|000,001\rangle+|000,111\rangle+|000,101\rangle)$$ $$U|000,101\rangle=1/\sqrt{3}(|000,001\rangle+|000,111\rangle+|000,101\rangle)$$

Shouldn't it be $U|00\rangle = \dfrac{1}{\sqrt{2}} \bigg(|00\rangle + |01\rangle$ ? And if I am not mistaken, then for 3 qubit, it would be $U|000\rangle = \dfrac{1}{\sqrt{3}} \bigg( |000 \rangle + |001\rangle + |010\rangle \bigg)$? – KAJ226 – 2020-11-20T05:40:52.473

No. As I mentioned in question I want equal probability to all the links attached to the node. – Binshumesh sachan – 2020-11-20T05:44:58.087

Have you tried computing a small example of what you want by hand? Maybe a graph with $3$ or $4$ nodes? Write the transformation as some arbitrary matrix $A = \sum_{ij} a_{ij} |i \rangle \langle j |$ and solve for the $a_{ij}$ then check if the resulting matrix is unitary. – Rammus – 2020-11-20T10:02:22.910