2
$\begingroup$

I came across Lecture 12 here https://viterbi-web.usc.edu/~tbrun/Course/ that does this but I was not able to understand. An example would be very helpful

$\endgroup$
4
  • $\begingroup$ Does this answer help? quantumcomputing.stackexchange.com/a/8863/15820 $\endgroup$ Jun 8 at 18:03
  • $\begingroup$ Use householder transformation $\endgroup$
    – KAJ226
    Jun 8 at 18:16
  • 2
    $\begingroup$ you can simply write the operator $|\psi\rangle\!\langle\phi|$ and complete it to be a unitary. This can be done by completing in an arbitrary way $|\phi\rangle$ and $|\psi\rangle$ into orthonormal bases, call them $\{|\phi_i\rangle\}$ and $\{|\psi_i\rangle\}_i$ with $\psi_1=\psi$ and $\phi_1=\phi$. Any corresponding unitary $U=\sum_i |\psi_i\rangle\!\langle\phi_i|$ sends the input to the output. I'm pretty sure this was already asked and answered on the site, but I can't find the post right now $\endgroup$
    – glS
    Jun 8 at 23:39
  • $\begingroup$ related: quantumcomputing.stackexchange.com/q/5167/55 $\endgroup$
    – glS
    Jun 8 at 23:49
3
$\begingroup$

I didn't go through the attached pdf. But if you want to find a unitary matrix $U$ that maps a quantum state $|\psi \rangle$ to $|\phi\rangle$ then you can use the Householder transformation as I commented. Here the two vectors have the same length (they are unit vectors) because we are thinking of them as a quantum state, so there will always exist a Householder transformation that can do this.

For example: If you want to find a unitary, $U$, that maps $|00\rangle = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}$ to $|11\rangle = \begin{pmatrix} 0\\0\\0\\1\end{pmatrix}$ then you can construct it as: $$ U = I - 2vv^T $$ where $v$ is the normalized vector of $|00\rangle - |11\rangle = \begin{pmatrix} 1 \\ 0 \\ 0 \\ -1 \end{pmatrix}$. That is, $ v = \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ 0 \\ -1/\sqrt{2} \end{pmatrix} $.

From here, we can write $U$ out explicitly as:

\begin{align} U &= \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix} - 2 \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ 0 \\ -1/\sqrt{2} \end{pmatrix} \begin{pmatrix} 1/\sqrt{2} & 0 & 0 &-1/\sqrt{2} \end{pmatrix} \\ &= \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix} - 2 \begin{pmatrix} 1/2 & 0 & 0 & -1/2\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -1/2 & 0 & 0 & 1/2 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & 1\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0\end{pmatrix} \end{align}

You can check that this is infact unitary since $U\cdot U^\dagger = I$ and that

$$ U|00\rangle = \begin{pmatrix} 0 & 0 & 0 & 1\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0\end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} = |11\rangle $$


The question now is about how to decompose this unitary matrix into a quantum circuit with certain set of gates... This can be done in different ways... look up KAK decomposition if you are interested.

$\endgroup$

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.