I am trying to find the ground state of a Hamiltonian using VQE. I have decomposed the Hamiltonian into a set of Pauli strings. To decrease the number of actual measurements that has to be done, can I just measure the generators of this set of Pauli strings? For example, say if the set is $\{I, Z_1Z_2, Z_2Z_3, Z_1Z_3\}$, can I just measure $Z_1Z_2$, and $Z_2Z_3$, and then find the expectation value of $Z_1Z_3$ from the expectation values of $Z_1Z_2$ and $Z_2Z_3$?


1 Answer 1


Here I am going to show why $\langle Z_1 Z_3 \rangle$ generally cannot be estimated from $\langle Z_1 Z_2 \rangle$ and $\langle Z_2 Z_3 \rangle$. Let's start with an arbitrary three-qubit state:

\begin{align*} |\psi \rangle = c_{000} &|000\rangle + c_{001} |001\rangle + c_{010} |010\rangle + c_{011} |011\rangle + \\ +c_{100} &|100\rangle + c_{101} |101\rangle + c_{110} |110\rangle + c_{111} |111\rangle \end{align*}

And because of this answer the expectation value for $Z_1 Z_2$, $Z_2 Z_3$ and $Z_1 Z_3$ (I am using this convention for qubit indexes $|q_1 q_2 q_3 \rangle$):

$$ \langle Z_1 Z_2 \rangle = |c_{000}|^2 + |c_{001}|^2 - |c_{010}|^2 - |c_{011}|^2 - |c_{100}|^2 - |c_{101}|^2 + |c_{110}|^2 + |c_{111}|^2 \\ \langle Z_2 Z_3 \rangle = |c_{000}|^2 - |c_{001}|^2 - |c_{010}|^2 + |c_{011}|^2 + |c_{100}|^2 - |c_{101}|^2 - |c_{110}|^2 + |c_{111}|^2 \\ \langle Z_1 Z_3 \rangle = |c_{000}|^2 - |c_{001}|^2 + |c_{010}|^2 - |c_{011}|^2 - |c_{100}|^2 + |c_{101}|^2 - |c_{110}|^2 + |c_{111}|^2 $$

From this expressions one can see that $\langle Z_1 Z_3 \rangle$ generally cannot be estimated from $\langle Z_1 Z_2 \rangle$ and $\langle Z_1 Z_3 \rangle$. For prove let's consider this conterexample when $|c_{000}| = |c_{010}|$, $|c_{001}| = |c_{011}|$:

\begin{equation*} \langle Z_1 Z_2 \rangle = - |c_{100}|^2 - |c_{101}|^2 + |c_{110}|^2 + |c_{111}|^2 \\ \langle Z_2 Z_3 \rangle = |c_{100}|^2 - |c_{101}|^2 - |c_{110}|^2 + |c_{111}|^2 \\ \langle Z_1 Z_3 \rangle = 2|c_{000}|^2 - 2|c_{001}|^2 - |c_{100}|^2 + |c_{101}|^2 - |c_{110}|^2 + |c_{111}|^2 \end{equation*}

So for the same $\langle Z_1 Z_2 \rangle$ and $\langle Z_2 Z_3 \rangle$ there can be many different possible values for $\langle Z_1 Z_3\rangle$ because by varying the values $|c_{000}| = |c_{010}|$ and $|c_{001}| = |c_{011}|$ the expectations $\langle Z_1 Z_2 \rangle$ and $\langle Z_2 Z_3 \rangle$ will not be changed but $\langle Z_1 Z_3 \rangle$ will be changed. So $\langle Z_1 Z_3 \rangle$ cannot be estimated from $\langle Z_1 Z_2 \rangle$ and $\langle Z_2 Z_3 \rangle$ in this conterexample.

  • $\begingroup$ This makes sense. But if I were to just make a measurement, instead of finding the expectation value, then could I have done so? For example, in this paper arxiv.org/abs/1907.13623, they mention that they can measure $Y_1Y_2$ by measuring $X_1X_2$ and $Z_1Z_2$. $\endgroup$
    – e-eight
    Oct 29, 2020 at 9:07
  • $\begingroup$ I think I understand it now. If we can do a qubit-wise multiplication then we can obtain the expectation value, i.e. qubit-wise multiplication of $XX$ and $ZZ$ is proportional to $YY$, but qubit-wise multiplication of $Z_1Z_2$ and $Z_2Z_3$ is not $Z_1Z_3$. $\endgroup$
    – e-eight
    Oct 29, 2020 at 9:30
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    $\begingroup$ @e-eight, in paragraph 7.2 they show how one can group $XX$, $YY$, and $ZZ$ in order to measure their expectation values together if I understood right. That can be done because they have simultaneous (common) eigenvectors...this can happen if operators commute with each other. This answer might be interesting in this regard where a simpler example is discussed. This answer also is about grouping Pauli strings. $\endgroup$ Oct 29, 2020 at 9:30
  • $\begingroup$ @e-eight, I might be wrong about $\langle YY \rangle =? -\langle XX \rangle \langle ZZ \rangle$...maybe this is not what the paper implies. I need to think about this. $\endgroup$ Oct 29, 2020 at 9:38
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    $\begingroup$ @e-eight, if by measurement they mean finding if the state is in $+1$ or $-1$ eigenspace of the Pauli string operator, then note that for Bell state $|\Phi^-\rangle$ that is in $+1$ eigenspace of $YY$ we have $YY |\Phi^- \rangle = -(XX) (ZZ) |\Phi^- \rangle = -(-1) (+1) |\Phi^- \rangle = (+1)|\Phi^- \rangle$, because $|\Phi^- \rangle$ is in the $+1$ eigenspace for $ZZ$ operator and $-1$ eigenspace for $XX$ operator. So, after Bell basis measurement if we know in what eigenspace is our measured state for $XX$ and $ZZ$ we will be able to find what is the corresponding eigenspace for $YY$. $\endgroup$ Oct 29, 2020 at 12:21

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