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.

  • $\begingroup$ 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. $\endgroup$ Jul 13, 2023 at 21:02
  • $\begingroup$ That's true, yes, but the expectation values that are plotted in the paper are all Hermitian combinations of Pauli matrices, I believe. $\endgroup$
    – JoDraX
    Jul 13, 2023 at 22:34
  • $\begingroup$ 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. $\endgroup$ Jul 16, 2023 at 10:20
  • $\begingroup$ Where did you find that they are using this Ansatz for $|\psi(\theta) \rangle$ ? Going quickly through the paper, I did not find it. $\endgroup$ Jul 16, 2023 at 15:28
  • $\begingroup$ @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. $\endgroup$
    – JoDraX
    Jul 16, 2023 at 16:10

2 Answers 2


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.


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