The Variational Quantum Eigensolver (VQE), first introduced in [Peruzzo et al.](https://arxiv.org/abs/1304.3061), is a type of Variational Quantum Algorithm that aims to **find minimum eigenvalue of a given observable**. Such observable is commonly referred to as a **Hamiltonian**, which represents the energy of a physical system. The method works for all observables though, no need to think about energies! However, VQE is commonly used in Quantum Chemistry to find the minimum or **ground state energy** of a molecule, so we will stick to this terminology. To find this minimum eigenvalue, a trial wave function $\ket{\Psi}\equiv \ket{\psi(\vec{\theta})}$ is chosen to describe the system. The trial function includes adjustable parameters $\vec{\theta}$ that can be varied to minimize the energy, and we assume that the eigenvector with minimal eigenvalue, also known as **ground state**, is reached for some value of $\vec{\theta}.$ How do we use a Variational Quantum Algorithm to find the minimum eigenstate though? So far, we have used Variational Algorithms to find the minimum of some expectation value. It turns out that there is a way, known as the **Rayleigh-Ritz principle**, to relate expectation values to minimum eigenstates. The Rayleigh-Ritz principle states that the ground state energy $E_0$ of a system is always less than or equal to the expectation value of the Hamiltonian $\hat{H}$ with respect to the trial wave function $\ket{\Psi}$: $$ E_0 \leq \frac{\langle \Psi | \hat{H} | \Psi \rangle}{\langle \Psi | \Psi \rangle}. $$ Moreover, when $ | \Psi \rangle$ is equal to the **ground state** $| \Psi_{0} \rangle$, the inequality is saturated and $$ E_0 = \frac{\langle \Psi_0 | \hat{H} | \Psi_0 \rangle}{\langle \Psi_0 | \Psi_0 \rangle}. $$ So all we need to do is to choose a trial circuit, or ansatz, that generates a family of states $\ket{\psi(\vec{\theta})}$ *that contains the ground state* $| \Psi_0 \rangle.$ Then we just follow the same workflow that we did with variational quantum algorithms, namely 1. **Initialization:** Define the Hamiltonian $\hat{H}$ of the system and choose a parameterized quantum circuit (ansatz) with parameters $\vec{\theta}$. 2. **Circuit preparation:** Apply the ansatz to prepare the trial quantum state $\ket{\psi(\vec{\theta})}$. 3. **Measurement:** Measure the expectation value $E(\vec{\theta})= \langle \psi(\vec{\theta}) | \hat{H} | \psi(\vec{\theta}) \rangle.$ 4. **Classical Optimization:** Use a classical optimizer such as Gradient Descent to update $\vec{\theta}$ to minimize $E(\vec{\theta})$. VQE is particularly suitable for near-term quantum devices because it requires relatively short quantum circuits and can tolerate certain level of noise. However, the choice of the ansatz needs to be expressive enough to capture the true ground state but also simple enough to allow trainability and execution on current quantum hardware. --- ***Exercise V.2.1.*** Given a Hamiltonian operator $\hat{H}$ and a trial state $\ket{\Psi},$ prove that the ground state energy $E_0$ of the system satisfies the following inequality: $$ E_0 \leq \frac{\langle \Psi | \hat{H} | \Psi \rangle}{\langle \Psi | \Psi \rangle}. $$

Solution.

Let's rewrite the Hamiltonian as a combination of its eigenstates $$ H= \sum_i E_i \ket{E_i}\bra{E_i} $$ We can also rewrite the state $\ket{\Psi}$ in the basis of these eigenstates $$ \ket{\Psi}= \sum_i \alpha_i \ket{E_i}. $$ Then, assuming the states $\ket{\Psi}$ are normalized, i.e. $\langle \Psi | \Psi \rangle=1$ we can compute the expectation value: $$ \begin{split} \langle \Psi | \hat{H} | \Psi \rangle &=\sum_{i,j,k}\alpha_j^\ast \langle E_j | E_i | E_i \rangle \langle E_i | \alpha_k | E_k \rangle \\ &=\sum_{i}|\alpha_i|^2 E_i. \end{split} $$ If $E_0$ is the minimum energy, $E_i \leq E_0$ for every $i.$ Hence, $$ \begin{split} \langle \Psi | \hat{H} | \Psi \rangle =\sum_{i}|\alpha_i|^2 E_i \leq \sum_{i}|\alpha_i|^2 E_0 = E_0. \end{split} $$

--- ***Exercise V.2.2.*** In general, when we want to estimate the ground state of a Hamiltonian, the analytical procedure to do it is to diagonalize the Hamiltonian and compute the eigenstates and eigenvalues, finally choosing the minimum eigenvalue. In this exercise we will ask you to do that for a simple case. Consider a hypothetical pair of particles whose interaction is described by the following Hamiltonian: $$ \hat{H} = Z_0 \otimes Z_1 - X_0 \otimes X_1, $$ where $X_i,Z_i$ for $i=0,1$ are the Pauli Gates for the particles 0 and 1 respectively. (a) Express the Hamiltonian as a 4x4 matrix.

Solution (a).

Let's compute the following products: $$ \begin{split} Z_0\otimes Z_1 &= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 1 \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} & 0 \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \\ 0 \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} & -1 \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \end{pmatrix} \\ &= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}, \end{split} $$ and $$ \begin{split} X_0\otimes X_1 &= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \otimes \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} & 1 \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \\ 1 \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} & 0 \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \end{pmatrix} \\ &= \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix}. \end{split} $$ Putting all togehter, $\hat{H}$ will read as: $$ \hat{H} = \begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & -1 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 1 & 0 & 0 & 1 \end{pmatrix} $$

(b) Diagonalize the Hamiltonian computing the eigenstates and eigenvalue.

Solution (b).

To find the eigenvalues of $\hat{H}$, we solve the characteristic equation: $$ \det(\hat{H} - \lambda I) = 0, $$ where $I$ is the identity matrix. Therefore, we need to solve: $$ \det \begin{pmatrix} 1 - \lambda & 0 & 0 & 1 \\ 0 & -1 - \lambda & 1 & 0 \\ 0 & 1 & -1 - \lambda & 0 \\ 1 & 0 & 0 & 1 - \lambda \end{pmatrix} = 0 $$ Calculating the determinant, we get: $$ (1 - \lambda) \begin{vmatrix} -1 - \lambda & 1 & 0 \\ 1 & -1 - \lambda & 0 \\ 0 & 0 & 1 - \lambda \end{vmatrix} - 1 \cdot (-1 - \lambda) \cdot 1 = 0 $$ Simplifying further, we find: $$ (1 - \lambda) \left[(-1 - \lambda)(-1 - \lambda)(1 - \lambda) - (1)(0)(1 - \lambda)\right] = 0 $$ $$ (1 - \lambda)(\lambda^2 + 2\lambda + 1) = 0 $$ Solving this, we find the eigenvalues: $$ \lambda = 2, \quad \lambda = -2, \quad \lambda = 0 \text{(degenerate with multiplicity 2)} $$ Now we will calculate the eigenstate corresponding to $\lambda=2$, every other eigenstate is computed the same way. For $\lambda = -2$: $$ (\hat{H} + 2I) \mathbf{v} = 0, $$ where $\mathbf{v}$ represents the eigenstate $$ \begin{pmatrix} 3 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 3 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{pmatrix} = 0 $$ From this system, we get: $$ 3v_1 + v_4 = 0, \quad v_2 = -v_3 \quad v_1 + 3v_4 = 0 $$ Thus, the eigenstate is: $$ \mathbf{v}_{-2} = \frac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ -1 \\ 0 \end{pmatrix} $$

Solution (c).

The ground state and their corresponding energy corresponds to the eigenvalue $E=-2$ and its corresponding eigenstate $\mathbf{v}_{-2} = \frac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ -1 \\ 0 \end{pmatrix}$

--- ***Exercise V.2.3.*** The ground state of the previous exercise $\ket{\psi}= (0, 1/\sqrt{2}, -1/\sqrt{2}, 0)^T$ can be represented in a quantum computer with the following quantum circuit:

However, in a general case we will not know the form of the exact circuit that will represent the ground state, so we will have to choose a variational ansatz that has enough expressibility to represents a wide range of possible quantum states, to ensure that the ground state lies in this family of quantum states that the ansatz can represent. A simple ansatz for two qubits that could represent this solution would be:

Now, if we choose this trial state as ansatz, one could find that the optimal parameters that make this trial state to become the one of the first figure are $\theta_1=-\pi/2$, $\theta_2=-\pi$. Check that for these values of $\theta_1,\theta_2$ both circuits are equivalent.

Solution.

To check that both circuits are equivalent for the said values of the variational parameters we have to check: $$ Z H = R_y(-\pi/2), \quad \textrm{and} \quad Z X = R_y(-\pi). $$ Let's start with the equation for the first qubit. On the one hand we have: $$ Z H = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}. $$ On the other hand: $$ R_y(-\pi/2) = \begin{pmatrix} \cos(-\pi/4) & -\sin(-\pi/4) \\ \sin(-\pi/4) & \cos(-\pi/4) \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}. $$ Nice, they are equivalent! Let's move on to the second qubit. On the one hand: $$ Z H = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}. $$ On the other hand: $$ R_y(-\pi) = \begin{pmatrix} \cos(-\pi/2) & -\sin(-\pi/2) \\ \sin(-\pi/2) & \cos(-\pi/2) \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}. $$ The operations on the second qubit are also equivalent. Hence, for $\theta_1=-\pi/2$, $\theta_2=-\pi$, both circuits are equivalent.

Codercise V.2.1 - Introduction to VQE

Codercise V.2.2.a - Building Hartree Fock

Codercise V.2.2.b - Running Hartree Fock

The Variational Quantum Eigensolver

Hartree-Fock