# Periodicity of Observables of Variational Ansatz

I am following along with the contents of this paper on a rudimentary quantum computation of the ground state energy of the deuterium nucleus. A variational quantum eigensolver is used with a wide array of angles in the variational ansatz shown here: $$|\psi(\theta)\rangle = \exp\left( \theta \left( a^\dagger_0a_1 - a^\dagger_1a_0 \right) \right)|10\rangle = \cos{\theta}|10\rangle - \sin{\theta}|01\rangle,$$ where $$a_j=\frac{1}{2}\left(X_j+iY_j\right),$$ $$a^\dagger_j=\frac{1}{2}\left(X_j-iY_j\right),$$ and $$X_j,Y_j,Z_j$$ are the respective Pauli matrices acting on qubit $$j$$. Obviously this ansatz is periodic with period $$2\pi$$, however, it also has the property $$|\psi(\theta+\pi)\rangle = -|\psi(\theta)\rangle$$, since $$\sin(\theta+\pi)=-\sin{\theta}$$ and $$\cos(\theta+\pi)=-\cos{\theta}$$. This means that for any observable $$\hat{\mathcal{O}}$$, $$\langle \hat{\mathcal{O}} \rangle_{\theta+\pi}=\langle \psi(\theta+\pi)|\hat{\mathcal{O}}|\psi(\theta+\pi)\rangle=\left(-\langle\psi(\theta)|\right)\hat{\mathcal{O}}\left(-|\psi(\theta)\rangle\right)=\langle \psi(\theta)|\hat{\mathcal{O}}|\psi(\theta)\rangle=\langle\hat{\mathcal{O}}\rangle_\theta,$$ hence the observables should not only be $$2\pi$$-periodic, but $$\pi$$-periodic. However, the expectation values of the Pauli matrices (which are Hermitian, and hence can be considered "observable") that are listed in the paper appear very clearly to be only $$2\pi$$-periodic. I feel that maybe I'm misunderstanding something fundamental here, but I'm unsure what. I would appreciate any suggestions or direction as to what I may be misunderstanding.

• The combinations of Pauli matrices to define the $a$'s and $a^{\dagger}$ are not Hermitian. It may be misleading in this sense to think in terms of the properties of the Pauli matrices. Jul 13, 2023 at 21:02
• That's true, yes, but the expectation values that are plotted in the paper are all Hermitian combinations of Pauli matrices, I believe. Jul 13, 2023 at 22:34
• The exponential is a unitary operator, $e^{i\theta \sigma_2}= I \cos\theta+i\sin\theta ~ \sigma_2$. It provides a conjugation for any hermitian operator it conjugates. As a bilinear, it has the well known half angle symmetry. Jul 16, 2023 at 10:20
• Where did you find that they are using this Ansatz for $|\psi(\theta) \rangle$ ? Going quickly through the paper, I did not find it. Jul 16, 2023 at 15:28
• @AdrienMartina When they define the unitary operator entangling the two qubits in eq 7. After some examination, it appears that this unitary operator and the variational circuit shown in figure 1 do give different results for the ansatz (notably, the circuit has a period of $4\pi$, and $2\pi$ for the observables). However, I also reached out to one of the authors of the paper, who sent along a portion of his code used in writing the paper, and it appears that the unitary operator defined in eq 7 was used to prepare the ansatz. I may also share the code if I can get permission from the author. Jul 16, 2023 at 16:10

I'll be very thorough in this answer to make sure there's nowhere else the factor of 2 could be hiding. First let's retrace the steps in the paper to get Eq. (7) for completeness.

$$a_0^\dagger a_1 = \frac{1}{2} (X_0 - iY_0) \cdot \frac{1}{2} (X_1 + iY_1)\\ = \frac{1}{4}(X_0 X_1 +iX_0Y_1 - iY_0X_1 + Y_0Y_1)$$ and similarly $$a_1^\dagger a_0 = \frac{1}{4}(X_0X_1 + iY_0X_1-iX_0Y_1 + Y_0Y_1)$$ so $$\theta(a_0^\dagger a_1 - a_1^\dagger a_0) = i\frac{\theta}{2}(X_0 Y_1 - Y_0 X_1)$$ This matches the right side of Eq. (7), so far everything is consistent. I could not find it explicitly in the text, but I assume $$U(\theta)$$ is supposed to be applied to an $$\lvert\uparrow\downarrow\rangle$$ or $$\lvert\downarrow\uparrow\rangle$$ state ($$\downarrow$$ means $$1$$ in this context). Because the exponent in $$U$$ always flips both qubits, never just one of them, it is block diagonal with one block being $$\{\lvert\uparrow\downarrow\rangle, \lvert\downarrow\uparrow\rangle\}$$. Within that block, $$(X_0 Y_1 - Y_0 X_1)/2$$ is involutory$$^1$$ (i.e. its square is the identity), therefore Euler's identity applies and we get$$^2$$ (I'll start indexing from the right, i.e. the right qubit is number 0)

\begin{align} \lvert\psi(\theta)\rangle_U = &\exp\left(i\frac{\theta}{2}(X_0 Y_1 - Y_0 X_1)\right) \lvert\downarrow\uparrow\rangle \\ &= \cos(\theta) \lvert\downarrow\uparrow\rangle + \frac{i}{2}\sin(\theta)(X_0Y_1-Y_0X_1)\lvert\downarrow\uparrow\rangle \\ &= \cos(\theta) \lvert\downarrow\uparrow\rangle + \frac{i}{2}\sin(\theta)(-i\lvert\uparrow\downarrow\rangle - i\lvert\uparrow\downarrow\rangle))\\ &= \cos(\theta) \lvert\downarrow\uparrow\rangle + \sin(\theta)\lvert\uparrow\downarrow\rangle \end{align}

Then the authors claim the following.

We note that $$U(\eta)$$ [...] can be simplified further because a single-qubit rotation about the $$Y$$ axis implements the same rotation as Eq. (7) within the twodimensional subspace $$\{\lvert\downarrow\uparrow\rangle, \lvert\uparrow\downarrow\rangle\}$$. [...] The resulting gate decomposition for the UCC operations are illustrated in Fig. 1.

I think that $$U(\eta)$$ is a typo, because there is no $$\eta$$ in Fig. 1 or Eq. (7), but there is a $$\theta$$ in both.

Figure 1 starts with the state $$\lvert\uparrow\uparrow\rangle$$ and applies an $$X_1$$ and a $$R_0^y(\theta)$$, followed by a CNOT where the control is on qubit 0, which I will write as $$C_0[X_1]$$. Let's check its action.

\begin{align} \lvert\psi(\theta)\rangle_\mathrm{C} &= C_0[X_1] R_0^y(\theta) X_1\lvert\uparrow\uparrow\rangle\\ &=C_0[X_1] R_0^y(\theta) \lvert\downarrow\uparrow\rangle\\ &= C_0[X_1] \big(\cos(\theta/2) \lvert\downarrow\uparrow\rangle + \sin(\theta/2) \lvert\downarrow\downarrow\rangle\big)\\ &= \cos(\theta/2) \lvert\downarrow\uparrow\rangle + \sin(\theta/2) \lvert\uparrow\downarrow\rangle \end{align}

This is almost the same as above, but now we see where the mismatch comes from. The $$\theta$$ in Eq. (7) is not the same $$\theta$$ as in Fig. 1; they differ by a factor of 2. Circuit and equation are equivalent, but not identical. This seems to be sloppy notation by the authors, and they most likely used the $$\theta$$ from the circuit to plot their graphs. Using two different variables for operator and circuit would have definitely been better.

1. It's important that the factor of 1/2 is part of the involutory part, otherwise it would go inside the arguments of $$\sin$$ and $$\cos$$ together with $$\theta$$, adding to the confusion.

2. Note that this is slightly different from what you have, our labelling systems are probably different, i.e. I used a different state from you. I'll stick with mine as defined to keep it consistent, it seems to match up with the circuit in the paper.

Even though Noah provided a great detailed answer, the $$\pi$$-periodicity of the mean value of the observables can be shown with a simpler reasoning. Let's recall beforehand, just for the sake of intuition, that $$\langle\hat{\mathcal{O}}\rangle_\psi = \langle\psi(\theta)| \hat{\mathcal{O}} |\psi(\theta)\rangle$$ is somewhat quadratic in $$|\psi(\theta)\rangle$$, so that $$\langle\hat{\mathcal{O}}\rangle_\psi \sim \cos^2\theta$$ at fisrt glance, which is indeed $$\pi$$-periodic.

Now, with a view to be more rigorous, let's fix the basis to $$\mathcal{B} = \{|00\rangle, |01\rangle, |10\rangle, |11\rangle\}$$, with respect to which $$\hat{\mathcal{O}}$$ will be represented by a $$4 \times 4$$ matrix. The scalar product $$\langle\psi(\theta)| \hat{\mathcal{O}} |\psi(\theta)\rangle$$ can be then computed component-wise. However, since all physicists are lazy ;), this computation can be circumvented by the use of the density matrix, namely $$\hat{\rho} = |\psi(\theta)\rangle\langle\psi(\theta)| = \cos^2\theta\,|00\rangle\langle00| + \sin^2\theta\,|11\rangle\langle11| - \sin\theta\cos\theta\,(|01\rangle\langle10|+|10\rangle\langle01|)$$ in the present case, so that $$\langle\hat{\mathcal{O}}\rangle_\psi = \mathrm{Tr}(\hat{\rho}\hat{\mathcal{O}})$$. This last expression will generate a linear combination of the density matrix components, whose scalar coefficients are precisely the components of the operator $$\hat{\mathcal{O}}$$. Yet, since $$\sin\theta\cos\theta = \frac{1}{2}\,\sin2\theta$$, all the components of $$\hat{\rho}$$ are $$\pi$$-periodic, hence the observed behaviour.