# What Hamiltonians generate Hadamard and CNOT? [closed]

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.

• I’m not sure I understand the question completely, how are the two H matrices related? Jan 4 at 8:03
• We have to make a Hadamard matrix from hamiltonian Jan 4 at 8:05
• It seems like homework. What have you tried already? Jan 4 at 8:32

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 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

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

• Thank you so much. Now I know what I was doing wrong. Jan 4 at 11:08