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Find a $2 \times 2$ Hamiltonian $H_H$ such that $e^{iH_H}$ equals the Hadamard matrix and a $4 \times 4$ Hamiltonian $H_{CNOT}$ such that $e^{-iH_{CNOT}}$ equals the matrix of the CNOT gate.

I have been trying to solve this but couldn't come to any conclusion. Any help would be really appreciated.

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  • $\begingroup$ I’m not sure I understand the question completely, how are the two H matrices related? $\endgroup$ – rjh324 Jan 4 at 8:03
  • $\begingroup$ We have to make a Hadamard matrix from hamiltonian $\endgroup$ – Anurag Singla Jan 4 at 8:05
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    $\begingroup$ It seems like homework. What have you tried already? $\endgroup$ – nippon Jan 4 at 8:32
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Let us denote Hadamard with $H$ and the two Hamiltonians as $H_H$ and $H_{CNOT}$, i.e.

$$ H = \exp (-iH_H) \\ CNOT = \exp (-iH_{CNOT}). $$

We will make use of the fact that for any normal matrix $A$ with eigendecomposition

$$ A = \sum_i \lambda_i |i\rangle\langle i| $$

and for any analytic function $f$, we can compute $f(A)$ by applying it to eigenvalues

$$ f(A) = \sum_i f(\lambda_i) |i\rangle\langle i|. $$

Hadamard

Hadamard has zero trace and determinant $-1$, so its eigenvalues are $-1$ and $+1$. Therefore, we can write it as

$$ \begin{align} H &= |a\rangle\langle a| - |b\rangle\langle b| \\ &= e^0 |a\rangle\langle a| + e^{-i\pi} |b\rangle\langle b| \\ &= \exp \left (-i\pi |b\rangle\langle b|\right) \end{align} $$

so

$$ H_H = \pi |b\rangle\langle b| $$

where $|b\rangle$ is the normalized eigenvector of Hadamard associated with eigenvalue $-1$.

Controlled-NOT

CNOT leaves $|00\rangle$ and $|01\rangle$ states unchanged, so these are two eigenvectors associated with eigenvalue $1$. Also, $X$ is

$$ X = |+\rangle\langle +| - |-\rangle\langle -| $$

so

$$ \begin{align} CNOT &= |00\rangle\langle 00| + |01\rangle\langle 01| + |{1+}\rangle\langle{1+}| - |{1-}\rangle\langle{1-}| \\ & = e^0|00\rangle\langle 00| + e^0|01\rangle\langle 01| + e^0|{1+}\rangle\langle{1+}| + e^{-i\pi}|{1-}\rangle\langle{1-}| \\ & = \exp(-i\pi |{1-}\rangle\langle{1-}|) \end{align} $$

and

$$ H_{CNOT} = \pi |{1-}\rangle\langle{1-}|. $$

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    $\begingroup$ Thank you so much. Now I know what I was doing wrong. $\endgroup$ – Anurag Singla Jan 4 at 11:08

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