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From a 9×9 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 9×9 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 4×4 matrix, so how will it act on the unitary time evolution operator which is a 9×9 matrix.

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

Article

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  • $\begingroup$ Welcome to QCSE! I'd just like to point out that a $9\times 9$ Hamiltonian matrix lies in a $9\times 9$ space. Remember that a $n\times n$ matrix is $n^2$ dimensional. $\endgroup$ – Sanchayan Dutta Jun 30 '18 at 22:55
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    $\begingroup$ @Blue: I disagree, I think the question is okay the way it is. "Dimension" of a matrix is often used to mean the same as "rank" of a matrix, which in this case is 9, not 81 or 9x9. $\endgroup$ – user1271772 Jun 30 '18 at 23:08
  • $\begingroup$ @user1271772 Okay, I've never seen that convention used in any mathematics textbook. If it's standard in quantum computing or physics, I'm fine with it. $\endgroup$ – Sanchayan Dutta Jul 1 '18 at 7:44
  • $\begingroup$ It is a convention in mathematics that the dimension of V is the number of vectors in the basis of V. To avoid confusion, it might be better to say "rank of a matrix" and "dimension of the Hilbert space" but in the case I think the OP was clear enough from the context. $\endgroup$ – user1271772 Jul 2 '18 at 1:37
  • $\begingroup$ cross-posted from physics.stackexchange.com/q/414507/58382 $\endgroup$ – glS Jul 6 '18 at 12:21
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A $9\times 9$ matrix $H$ can act on a $9$ dimensional state vector, say something like:

$$|\Psi\rangle = a_0|0\rangle + a_1|1\rangle + .... + a_8|8\rangle$$

Now, say you want to find the matrix which only acts on the subspace spanned by the basis $\{|0\rangle,|1\rangle\,|2\rangle,|3\rangle\}$, but has the same effect as the original $H$ matrix.

Find, $H|0\rangle$ and $H|1\rangle$ (where $|0\rangle = [1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0\ 0 \ 0 ]^{T}$, $|1\rangle = [0 \ 1 \ 0 \ 0 \ 0 \ 0 \ 0\ 0 \ 0 ]^{T}$, $|2\rangle = [0 \ 0 \ 1 \ 0 \ 0 \ 0 \ 0\ 0 \ 0 ]^{T}$ and $|3\rangle = [0 \ 0 \ 0 \ 1 \ 0 \ 0 \ 0\ 0 \ 0 ]^{T}$)

Then find the unique scalars $\alpha_{j,i}$ such that

$H|0\rangle = \alpha_{0,0}|0\rangle + \alpha_{1,0}|1\rangle + \alpha_{2,0}|2\rangle+ \alpha_{3,0}|3\rangle$,

$H|1\rangle = \alpha_{0,1}|0\rangle + \alpha_{1,1}|1\rangle + \alpha_{2,1}|2\rangle + \alpha_{3,1}|3\rangle$,

$H|2\rangle = \alpha_{0,2}|0\rangle + \alpha_{1,2}|1\rangle + \alpha_{2,2}|2\rangle + \alpha_{3,2}|3\rangle$ and

$H|3\rangle = \alpha_{0,3}|0\rangle + \alpha_{1,3}|1\rangle + \alpha_{2,3}|2\rangle + \alpha_{3,3}|3\rangle$

Now, your required $4\times 4$ matrix is $\begin{bmatrix}\alpha_{0,0} & \alpha_{0,1} & \alpha_{0,2} & \alpha_{0,3}\\\alpha_{1,0} & \alpha_{1,1} & \alpha_{1,2} & \alpha_{1,3} \\ \alpha_{2,0} & \alpha_{2,1} & \alpha_{2,2} & \alpha_{2,3} \\ \alpha_{3,0} & \alpha_{3,1} & \alpha_{3,2} & \alpha_{3,3}\end{bmatrix}$

Let me know if you're confused about any particular step.

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  • $\begingroup$ Say $H$ has all 81 non-zero entries, then your first equation in the set seems to make contradiction although $\alpha_{i,j}$ can be determined. $H|0\rangle = [a_{00} \ a_{10} \ a_{20} \ a_{30} \ a_{40} \ a_{50} \ a_{60} \ a_{70} \ a_{80}]^T = [\alpha_{0,0} \ \alpha_{1,0} \ \alpha_{2,0} \ \alpha_{3,0} \ 0 \ 0 \ 0 \ 0 \ 0]^T = \alpha_{0,0}|0\rangle + \alpha_{1,0}|1\rangle + \alpha_{2,0}|2\rangle+ \alpha_{3,0}|3\rangle$. Where $a_{i0}$ are the non-zero values of first column of $H$. In this case, I think all the 9-component basis vectors will also have all non-zero entries. $\endgroup$ – Jitendra Jul 5 '18 at 22:33

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