# Measuring Pauli strings using generators

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 $$Z_2Z_3$$?

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.

• 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 $Z_1Z_2$. Oct 29 '20 at 9:07
• 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 $Z_1Z_3$. Oct 29 '20 at 9:30
• @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. Oct 29 '20 at 9:30
• @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. Oct 29 '20 at 9:38
• @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 $YY$. Oct 29 '20 at 12:21