### Quantum computing - Hadamard gate

The Hadamard gate acts on a single qubit. It maps the basis state to \begin{eqnarray} && |0\rangle = \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle), \hspace{1cm}{\rm and}\\ &&|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle). \end{eqnarray} In the adiabatic STIRAP process, there are altogether 6 pulses that are applied to the system. The pulse order is shown in the enclosed figure.

$$A$$ is the pulse height, $$T$$ is the pulse width and $$\Delta T$$ is a time between the pulses. Three couplings $$\Omega_{10}$$, $$\Omega_{11}$$ and $$\Omega_{2}$$ connect the levels. The Hamiltonian which describes this system is

$$H(t) = \left[ \matrix{ 0 & 0 & \Omega_{10}(t) & 0 \cr 0 & 0 & \Omega_{11}(t) & 0 \cr \Omega_{10}(t) & \Omega_{11}(t) & 0 & \Omega_2(t) \cr 0 & 0 & \Omega_2(t) & 0 \cr } \right].$$ We consider first the counterintuitive process (first set of pulses). When choosing the couplings as

\begin{eqnarray} &&\Omega_{10}(t) = a_{10}(t)*A\sqrt{\frac{1+\tanh(t/T)}{2}} \hspace{1mm}{\rm sech }(t/T) , \phantom{\frac{A}{A}} \\ &&\Omega_{11}(t) = a_{11}(t)*A\sqrt{\frac{1+\tanh(t/T)}{2}} \hspace{1mm}{\rm sech }(t/T) , \phantom{\frac{A}{A}} \\ &&\Omega_2(t) = A\sqrt{\frac{1-\tanh(t/T)}{2}} \hspace{1mm}{\rm sech }(t/T). \phantom{\frac{A}{A}} \end{eqnarray} and using the relations \begin{eqnarray} && \psi_1(t) = a_{10}^*(t)\psi_{10} + a_{11}^*(t)\psi_{11}, \phantom{\frac{A}{A}}\\ && \Omega_{10}(t) = a_{10}(t)*\Omega_{1}(t), \phantom{\frac{A}{A}}\\ && \Omega_{11}(t) = a_{11}(t)*\Omega_{1}(t), \phantom{\frac{A}{A}}\\ &&|a_{10}(t)|^2 + |a_{11}(t)|^2 = 1,\phantom{\frac{A}{A}} \\ &&\partial_ta_{10}^* \psi_{10} + \partial_ta_{11}^* \psi_{11} = i(\varphi_1 \psi_{10}+ \varphi_2 \psi_{11}),\phantom{\frac{A}{A}} \end{eqnarray} the Halmiltonian reduces to the corresponding 3x3 matrix. This can be solved exactly

\begin{eqnarray} \psi_1(t) &=& \tanh\eta\sin\phi\sin(2I) +\cos\phi[\hspace{1mm}{\rm sech }^2\eta+ \tanh^2\eta\cos(2I)],\\ \psi_2(t) &=& -i2\tanh\eta\hspace{1mm}{\rm sech }\eta \sin^2(I),\\ \psi_3(t) &=& \tanh\eta\cos\phi\sin(2I) -\sin\phi[\hspace{1mm}{\rm sech }^2\eta+ \tanh^2\eta\cos(2I)], \end{eqnarray}
where $$I(t) = AT\cosh\eta \hspace{1mm} {\rm artan} [\exp(t/T)]$$ and $$\sinh\eta = 1/(2AT)$$. When the pulse areas are finite, Rabi cycle oscillations are present at all levels. Parameters $$AT$$ and $$\Delta T$$ can be used to tune the transition efficiency.

When applying the second set of pulses to the system (intuitive order), after some calculation one can show that the result of the final state is $$|\psi_f \rangle = a_{f0} | \psi_0 \rangle + a_{f1} | \psi_1 \rangle = U |\psi_i \rangle,$$ where $$U$$ is the Hadamard gate $$U = \frac{1}{\sqrt{2}} \left[ \matrix{ 1 & 1 \hspace{-1.5em} \phantom{\frac{\dot{1}}{\sqrt{1}}} \cr 1 & -1 \hspace{-1.5em} \phantom{\frac{\dot{1}}{\sqrt{1}}} \cr } \right]$$ It represents a rotation of $$\pi$$ about the axis at the Bloch sphere. More details and discussions can be found from the reference paper.

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