Quantum computing -- deuteron

Article

Details

The article is based on the Unitary Coupled Clusters (UCC) ansatz (see link for details) which can be obtained using the following circuit:

This results in the following wave function after the first operations:

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

The UCC ansatz wave function looks like a parametrized Bell state.

Hamiltonian on two qubits

Once the ansatz is set, one can apply the Hamiltonian derived in the article:

\(\begin{equation} {H}_{2} = a I + b {Z}_{0} + c {Z}_{1} + d ({X}_{0} {X}_{1} + {Y}_{0} {Y}_{1}) \end{equation}\)
where \({ a }\), \({ b }\), \({ c }\), and \({ d }\) are constants given by the problem (see article). The identity \({ I }\) and the \({ Z }\) gates are trivial while the two-qubit gates \({ XX }\) and \({ YY }\) need to be computed before. The effect of each term of the Hamiltonian on the wave function is:

\begin{equation} \begin{aligned} b Z_0 \ket{ {\Psi}_{\text{UCC}} } &= b \left( \cos(\frac{\theta}{2}) \begin{pmatrix} 1 & 0 \\\
0 & -1 \end{pmatrix} \begin{pmatrix} 0 \\\
1 \end{pmatrix} \otimes \begin{pmatrix} 1 \\\
0 \end{pmatrix} + \sin(\frac{\theta}{2}) \begin{pmatrix} 1 & 0 \\\
0 & -1 \end{pmatrix} \begin{pmatrix} 1 \\\
0 \end{pmatrix} \otimes \begin{pmatrix} 0 \\\
1 \end{pmatrix} \right) \\\
&= -b \cos(\frac{\theta}{2}) \ket{10} + b \sin(\frac{\theta}{2}) \ket{01} \end{aligned} \end{equation}

\begin{equation} \begin{aligned} c Z_1 \ket{ {\Psi}_{\text{UCC}} } &= c \left( \cos(\frac{\theta}{2}) \begin{pmatrix} 0 \\\
1 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\\
0 & -1 \end{pmatrix} \begin{pmatrix} 1 \\\
0 \end{pmatrix} + \sin(\frac{\theta}{2}) \begin{pmatrix} 1 \\\
0 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\\
0 & -1 \end{pmatrix} \begin{pmatrix} 0 \\\
1 \end{pmatrix} \right) \\\
&= c \cos(\frac{\theta}{2}) \ket{10} -c \sin(\frac{\theta}{2}) \ket{01} \end{aligned} \end{equation}

\begin{equation} \begin{aligned} d X_0 X_1 \ket{ {\Psi}_{\text{UCC}} } &= d \begin{pmatrix} 0 & 0 & 0 & 1 \\\
0 & 0 & 1 & 0 \\\
0 & 1 & 0 & 0 \\\
1 & 0 & 0 & 0 \end{pmatrix} \left( \cos(\frac{\theta}{2}) \begin{pmatrix} 0 \\\
0 \\\
1 \\\
0 \end{pmatrix} \sin(\frac{\theta}{2}) \begin{pmatrix} 0 \\\
1 \\\
0 \\\
0 \end{pmatrix} \right) \end{aligned} \end{equation}

\begin{equation} \begin{aligned} d Y_0 Y_1 \ket{ {\Psi}_{\text{UCC}} } &= d \begin{pmatrix} 0 & 0 & 0 & -1 \\\
0 & 0 & 1 & 0 \\\
0 & 1 & 0 & 0 \\\
-1 & 0 & 0 & 0 \end{pmatrix} \left( \cos(\frac{\theta}{2}) \begin{pmatrix} 0 \\\
0 \\\
1 \\\
0 \end{pmatrix} \sin(\frac{\theta}{2}) \begin{pmatrix} 0 \\\
1 \\\
0 \\\
0 \end{pmatrix} \right) \end{aligned} \end{equation}

\begin{equation} \begin{aligned} d (X_0 X_1 + Y_0 Y_1) \ket{ {\Psi}_{\text{UCC}} } &= d \cos(\frac{\theta}{2}) \begin{pmatrix} 0 \\\
1 \\\
0 \\\
0 \end{pmatrix} +d\sin(\frac{\theta}{2}) \begin{pmatrix} 0 \\\
0 \\\
1 \\\
0 \end{pmatrix} +d\cos(\frac{\theta}{2}) \begin{pmatrix} 0 \\\
1 \\\
0 \\\
0 \end{pmatrix} +d\sin(\frac{\theta}{2}) \begin{pmatrix} 0 \\\
0 \\\
1 \\\
0 \end{pmatrix} \end{aligned} \end{equation}

Adding the contribution from the identity one obtains:

\[\begin{equation} H_2 \ket{ {\Psi}\_{\text{UCC}} } = \left[ (a+b-c) \sin(\frac{\theta}{2}) + 2d \cos(\frac{\theta}{2}) \right] \ket{01} + \left[ (a-b+c) \cos(\frac{\theta}{2}) + 2d \sin(\frac{\theta}{2}) \right] \ket{10} \end{equation}\]

One can then compute the average energy of the system as:

\begin{equation} \begin{aligned} & \bk{ {\Psi}_{\text{UCC}} }{ H_2 }{ {\Psi}_{\text{UCC}} } \\\
&= \left( \cos(\frac{\theta}{2}) \bra{10} + \sin(\frac{\theta}{2}) \bra{01} \right) \left( \left[ (a+b-c) \sin(\frac{\theta}{2}) + 2d \cos(\frac{\theta}{2}) \right] \ket{01} + \left[ (a-b+c) \cos(\frac{\theta}{2}) + 2d \sin(\frac{\theta}{2}) \right] \ket{10} \right) \\\
&= \cos(\frac{\theta}{2}) \left( (a-b+c) \cos(\frac{\theta}{2}) + 2d \sin(\frac{\theta}{2}) \right) + \sin(\frac{\theta}{2}) \left( (a+b-c) \sin(\frac{\theta}{2}) + 2d \cos(\frac{\theta}{2}) \right) \\\
&= a + (c-b) \left( { \cos(\frac{\theta}{2}) }^{2} - { \sin(\frac{\theta}{2}) }^{2} \right) + 4d \sin(\frac{\theta}{2}) \cos(\frac{\theta}{2}) \\\
&= a + (c-b) \cos(\theta) + 4d \sin(\frac{\theta}{2}) \cos(\frac{\theta}{2}) \end{aligned} \end{equation}