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

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

4

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.

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.

If $$U|x> = U |y>$$ while $$ = 0$$, it implies that if $$U$$ is unitary, then $$ = 0$$ 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.

2Trace preservation is not enough, $U$ needs to take orthogonal states to orthogonal states, which it doesn't because it has repeated columns. – chrysaor4 – 2020-11-20T14:40:08.560

Yes I didn't see the examples ! Thank you – Appo – 2020-11-20T18:36:43.320

But what do you mean by trace preservation? @chrysaor4 – Appo – 2020-11-20T18:48:36.240

Please, can you tell me what do you mean by including entangling operations? – Binshumesh sachan – 2020-11-21T06:42:59.207

@AbdellahTounsi For pure states, $\langle \psi | \psi \rangle = Tr(\langle \psi | \psi \rangle) = Tr(\rho)$, so preserving the trace is the same as preserving the norm. – chrysaor4 – 2020-11-21T11:10:05.103

I am mean by entangling operation produces a state which cannot be separated by tensor product such as Bell state. In the question, additional qubits should be entangled with the main qubits. Otherwise, they remain separated by tensor product and nothing would be yielded. @Binshumeshsachan – Appo – 2020-11-21T18:06:34.607

For instance, Bell sates are entangled states such: $1/\sqrt{2}(|00> + |11>)$ which cannot be written as $(\alpha_0 |0> + \beta _0|1>) \otimes (\alpha_1 |0> + \beta _1|1>)$ where $\otimes$ is the tensor product. @Binshumeshsachan – Appo – 2020-11-21T18:12:05.230