How to perform the unitary transformation $U|i,j_1\rangle = 1/\sqrt{k}(|i,j_1\rangle+|i,j_2\rangle+|i,j_3\rangle...+|i,j_k\rangle)$?

Question

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:

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

Answer 1

There is a question that has something in common with your question, see the answer by @DaftWullie.

Since all three input states after the operation, $U$, give the same result, then we can not decide which state is our input(it can not be recovered), this means that the operation is not reversible.

Maybe, after you appended some ancilla and enlarge the operation $U$ your requirement is possible.

Answer 2

To compute the values of $U$ matrix representation you should calculate its elements $<{i, j}|U|{i', j'}>$ for all values of $i, i', j, j'$. For instance:

$$ U_{i,j_1,i,j_1} = <{i, j_1}|U|{i, j_1}> = 1/\sqrt{k} $$ $$ U_{i,j_2,i,j_1} = <{i, j_2}|U|{i, j_1}> = 1/\sqrt{k} $$

and so on..

You should also define the state $U|i,j>$ for all $i$ and $j$ to completely determine the $U$ matrix. Otherwise, it will contain unknown elements.

However, your example shows that your transformation is not unitary:

If $U|x> = U |y>$ while $<x|y> = 0$, it implies that if $U$ is unitary, then $$ <x|U^\dagger U|y> = 0$$ which is contradiction since $U|x> = U |y>$.

Hence, your transformation is NOT unitary (which is obvious from the identical outputs which implies IR-REVERSIBILITY). You should include ENTANGLING operations with additional qubits to achieve a legal quantum gate.