1
$\begingroup$

From a $9\times9$ Hamiltonian lying 9D space, I choose a certain subspace of 4D for designing a two qubit gate. Now the original unitary time evolution operator also lies in 9D space and it's a $9\times9$ size matrix. For action of unitary time evolution operator on the two qubit gate made out of 4D subspace it is required to project the unitary time evolution operator in the 4D subspace. After reviewing literature, I came across an article doing same thing with the use of projection operator.

My question- How to find the projection operator on the subspace?

Also I guess projection operator will be $4\times4$ matrix, so how will it act on the unitary time evolution operator which is a $9\times9$ matrix.

P.S.- I took the definition of projection operator from "Quantum Computation and Quantum Information, Isaac Chuang and Michael Nielsen".

Screenshot of the article

$\endgroup$
2
$\begingroup$

If I understand your question properly this can be done as follows. Let $\newcommand{\ket}[1]{\left|#1\right>} \newcommand{\bra}[1]{\left<#1\right|} \{ \ket{e_i}\mid i=0,\ldots, 9\}$ be an orthonormal basis of the $9\times 9$ Hilbert space, $\mathcal{H}_9$. The Hamiltonian can then be written as: $$\hat H=\sum_{ij} H_{ij} \ket{e_i}\bra{e_j}$$ And unitary operator like: $$\hat U=\sum_{ij} U_{ij} \ket{e_i} \bra{e_j}$$ let $\{\ket{d_i}\mid i=1,\ldots 4\}$ be an orthonormal basis of the $4\times 4$ Hilbert space, $\mathcal{H}_4$. The projection operator is then: $$\hat P=\sum_i \ket{d_i}\bra{d_i}$$ an important fact is that since $\ket{d_i}\in \mathcal{H}_9$.

When we consider the projection of an operator $\hat{\mathcal{O}}$ (e.g. $\hat{{H}}$ or $\hat{{U}}$) onto $\mathcal{H}_4$ what we care actually want is the matrix: $$[\hat{\mathcal{O}}_4]_{ij}\equiv\bra{d_i} \hat{\mathcal{O}} \ket{d_j}$$ the operator itself is given by: $$\hat{\mathcal{O}}_4=\sum_{ij} \ket{d_i} \bra{d_i} \hat{\mathcal{O}} \ket{d_j}\bra{d_j}\tag{$\star$}$$ $$=\hat P \hat{\mathcal{O}} \hat P$$

It is the equation $\star$ you use to evaluate the projection. Simply write $\hat{\mathcal{O}}$ as your $9\times 9$ matrix and $\ket{d}_i$ as a $9$-component vector (in the same basis as the matrix). This can be done since as said above $\ket{d}_i \in \mathcal{H}_9(\supset\mathcal{H}_4)$.

$\endgroup$
  • $\begingroup$ Thanks a lot for answering! One doubt that if $\ket{d_i}$ lies in $\mathcal{H}_4$, then in computational basis $\ket{d_1} = [1 0 0 0]^T$, $\ket{d_2} = [0 1 0 0]^T$, $\ket{d_3} = [0 0 1 0]^T$ and $\ket{d_4} = [0 0 0 1]^T$. To write $\ket{d_i}$ as a 9-component vector should I append rest of five values as zer0. I mean should I write $\ket{d_1} = [1 0 0 0 0 0 0 0 0]^T$ and similarly rest of the vectors. $\endgroup$ – Jitendra Jul 1 '18 at 13:26
  • $\begingroup$ @Jitendra Yep exactly. $\endgroup$ – Quantum spaghettification Jul 1 '18 at 13:30

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.