Quantum computing

The coupling of two spins is the easiest quantum problem to map on a quantum computer.

Hamiltonian

\[\begin{equation} \hat{H} = \vec{S}^2 = {( \vec{s}_1 + \vec{s}_2 )}^2 = \frac{ {\hbar}^2 }{4} {( \vec{\sigma}_1 + \vec{\sigma}_2 )}^2 = \frac{ {\hbar}^2 }{4} ( \vec{\sigma}_1^2 + \vec{\sigma}_2^2 + 2 \vec{\sigma}_1.\vec{\sigma}_2 ) \end{equation}\]

Each Pauli vector writes \({ \vec{\sigma} = {\sigma}_{x} \vec{u}_{x} + {\sigma}_{y} \vec{u}_{y} + {\sigma}_{z} \vec{u}_{z} }\) where the vectors \({ \vec{u}_{i} }\) are normalized to one and the operators \({ {\sigma}_{i} }\) are represented by the Pauli matrices. It follows that the representation of the Hamiltonian using quantum gates is given by:

\[\begin{equation} H = \frac{ {\hbar}^{2} }{4} ( 3 {I}_{1} + 3 {I}_{2} + 2 ( {X}_{1} {X}_{2} + {Y}_{1} {Y}_{2} + {Z}_{1} {Z}_{2} ) ) \end{equation}\]

In matrix representation one has:

\begin{equation} \begin{aligned} H = \frac{ 3 {\hbar}^2 }{4} \begin{pmatrix} 1 & 0 & 0 & 0 \\\
0 & 1 & 0 & 0 \\\
0 & 0 & 1 & 0 \\\
0 & 0 & 0 & 1 \end{pmatrix} +\frac{ {\hbar}^2 }{2} \left[ \begin{pmatrix} 0 & 0 & 0 & 1 \\\
0 & 0 & 1 & 0 \\\
0 & 1 & 0 & 0 \\\
1 & 0 & 0 & 0 \end{pmatrix} +\begin{pmatrix} 0 & 0 & 0 & -1 \\\
0 & 0 & 1 & 0 \\\
0 & 1 & 0 & 0 \\\
-1 & 0 & 0 & 0 \end{pmatrix} +\begin{pmatrix} 1 & 0 & 0 & 0 \\\
0 & -1 & 0 & 0 \\\
0 & 0 & -1 & 0 \\\
0 & 0 & 0 & 1 \end{pmatrix} \right] \end{aligned} \end{equation}

Unitary Coupled Clusters (UCC) ansatz

See link for details. The UCC wave function writes:

\[\begin{equation} \ket{ {\Psi}_{\text{UCC}} } = \cos\left(\frac{\theta}{2}\right) \ket{10} + \sin\left(\frac{\theta}{2}\right) \ket{01} \end{equation}\]

Applying the UCC wave function on the Hamiltonian gives:

\begin{equation} \begin{aligned} H \ket{ {\Psi}_{\text{UCC}} } &= \frac{ 3 {\hbar}^2 }{4} \ket{ {\Psi}_{\text{UCC}} } \\\
&+ \frac{ {\hbar}^2 }{2} \begin{pmatrix} 0 & 0 & 0 & 1 \\\
0 & 0 & 1 & 0 \\\
0 & 1 & 0 & 0 \\\
1 & 0 & 0 & 0 \end{pmatrix} \left[\cos\left(\frac{\theta}{2}\right) \begin{pmatrix} 0 \\\
0 \\\
1 \\\
0 \end{pmatrix} +\sin\left(\frac{\theta}{2}\right) \begin{pmatrix} 0 \\\
1 \\\
0 \\\
0 \end{pmatrix} \right] \\\
&+\frac{ {\hbar}^2 }{2} \begin{pmatrix} 0 & 0 & 0 & -1 \\\
0 & 0 & 1 & 0 \\\
0 & 1 & 0 & 0 \\\
-1 & 0 & 0 & 0 \end{pmatrix} \left[\cos\left(\frac{\theta}{2}\right) \begin{pmatrix} 0 \\\
0 \\\
1 \\\
0 \end{pmatrix} +\sin\left(\frac{\theta}{2}\right) \begin{pmatrix} 0 \\\
1 \\\
0 \\\
0 \end{pmatrix} \right] \\\
&+\frac{ {\hbar}^2 }{2} \begin{pmatrix} 1 & 0 & 0 & 0 \\\
0 & -1 & 0 & 0 \\\
0 & 0 & -1 & 0 \\\
0 & 0 & 0 & 1 \end{pmatrix} \left[\cos\left(\frac{\theta}{2}\right) \begin{pmatrix} 0 \\\
0 \\\
1 \\\
0 \end{pmatrix} +\sin\left(\frac{\theta}{2}\right) \begin{pmatrix} 0 \\\
1 \\\
0 \\\
0 \end{pmatrix} \right] \\\
&= \frac{ 3 {\hbar}^2 }{4} \left[ \cos\left(\frac{\theta}{2}\right) \ket{10} + \sin\left(\frac{\theta}{2}\right) \ket{01} \right] \\\
&+ \frac{ {\hbar}^2 }{2} \left[ 2 \cos\left(\frac{\theta}{2}\right) \ket{01} + 2\sin\left(\frac{\theta}{2}\right) \ket{10} - \cos\left(\frac{\theta}{2}\right) \ket{10} - \sin\left(\frac{\theta}{2}\right) \ket{01} \right] \\\
&= \frac{ {\hbar}^2 }{4} \left[ \left( \cos\left(\frac{\theta}{2}\right) + 4 \sin\left(\frac{\theta}{2}\right) \right) \ket{10} + \left( \sin\left(\frac{\theta}{2}\right) + 4 \cos\left(\frac{\theta}{2}\right) \right) \ket{01} \right] \end{aligned} \end{equation}

Energy of the system

The ground state energy of the two-spins system can be found by minimizing its energy given by:

\begin{equation} \begin{aligned} \bk{ {\Psi}_{\text{UCC}} }{ H }{ {\Psi}_{\text{UCC}} } &= \frac{ {\hbar}^2 }{4} \left[ \cos\left(\frac{\theta}{2}\right) \bra{10} + \sin\left(\frac{\theta}{2}\right) \bra{01} \right] \left[ \left( \cos\left(\frac{\theta}{2}\right) + 4 \sin\left(\frac{\theta}{2}\right) \right) \ket{10} + \left( \sin\left(\frac{\theta}{2}\right) + 4 \cos\left(\frac{\theta}{2}\right) \right) \ket{01} \right] \\\
&= \frac{ {\hbar}^2 }{4} \left( {\cos}^2\left(\frac{\theta}{2}\right) + 4 \cos\left(\frac{\theta}{2}\right) \sin\left(\frac{\theta}{2}\right) + {\sin}^{2}\left(\frac{\theta}{2}\right) + 4 \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right) \right) \\\
&= \frac{ {\hbar}^2 }{4} + {\hbar}^2 \cos\left(\frac{\theta}{2}\right) \sin\left(\frac{\theta}{2}\right) \end{aligned} \end{equation}

The minimum is obtained for \({ \theta = -\pi/2 }\) and gives \({ E = - {\hbar}^{2}/4 }\) as expected. One notes that the excited state energy can be obtained for \({ \theta = +\pi/2 }\) and one has \({ E = + 3 {\hbar}^{2}/4 }\).