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Suppose you have a Hamiltonian of the form

$$ H = ZXXX + YXXX + XXXX $$

where $Z,X,Y$ are the usual Pauli matrices with $ZXXX = Z \otimes X \otimes X \otimes X$ and similar for the other two terms. Since the last three qubits are being acted upon by the same operator $XXX$ can we simplify the calculation of $\langle \psi | H | \psi \rangle = \langle \psi | (Z+Y+X) \otimes XXX| \psi \rangle$ more efficiently?

Here is a little more background to this question: Usually given a Hamiltonian $H = \sum_i \alpha_i P_i$, one can determine a Clifford unitary transformation through $\mathbb{Z}_2$ symmetries (and permutation) to transform the Hamiltonian so that the transformed Hamiltonian have the last $k$ qubits are either acted by $I$ or $X$ only. Once you get to such form then you can taper off these $k$ qubits by replacing them with their eigenvalues $\pm 1$ as stated here on page 14 of this paper Tapering off qubits to simulate fermionic Hamiltonian. I was able to find the $\mathbb{Z}_2$ symmetries to transformed the original Hamiltonian (Table B1 on pg. 13) to the transfomred Hamiltonian (Table B4 on page 14) in the linked paper, but I don't fully understand how to replace these operators with their $\pm 1$ eigenvalues to get rid of these qubits.

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2 Answers 2

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If my understanding of qubit tapering is correct,

then it cannot be used to reduce the number of qubits when measuring expectation values of any general state $|{\psi}\rangle$, but you can use it to reduce measurement cost and qubit requirements when finding eigenstates or approximations to those, because we already know the structure of the eigenstates for some of the qubits.

Let's take the example ${H}=(\alpha H_1\otimes I+\beta H_2\otimes X)$, where $H_1,H_2$ act on the same number of qubits. Because $I$ and $X$ commute and have the same eigenstates $|-\rangle$ and $|+\rangle$, we know that each eigenstate of $H$is of the form $|\psi\otimes-\rangle$ or $|\psi\otimes+\rangle$ for some $|\psi\rangle$. When acting on the minus state, we get $$\langle\psi\otimes-|(\alpha H_1\otimes I+\beta H_2\otimes X)|\psi\otimes-\rangle=\langle\psi|(\alpha H_1-\beta H_2)|\psi\rangle$$ and for the plus state $$\langle\psi\otimes +|(\alpha H_1\otimes I+\beta H_2\otimes X)|\psi\otimes+\rangle=\langle\psi|(\alpha H_1+\beta H_2)|\psi\rangle$$

we have thus replaced $I$ and $X$ by the values $\pm 1$ and tapered off one qubit. Observe how we now have to diagonalize two Hamiltonians in order to find which of them has the lowest eigenvalue, but usually, one can use physical insight to find which of the symmetry sectors of the Hamiltonian we are interested in. For a Hamiltonian of the form ${H}=(\alpha H_1\otimes II+\beta H_2\otimes XI+\gamma H_3\otimes IX+\delta H_4\otimes XX)$ we can then remove two qubits and end up with 4 new Hamiltonians etc.

In each of these cases, $H$ is block-diagonal in some representation, and qubit tapering simply corresponds to diagonalizing each of these blocks individually. From a VQE point of perspective, we do not need an ansatz for $N$ qubits, but an ansatz for $N-(\text{number of qubits tapered off})$ qubits.

In your example, it is particularly easy: In order to find the eigenstates of $H$, you would only have to diagonalize $Z+Y+X$, and you only need an ansatz $|{\psi}\rangle$ for one qubit.

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  • $\begingroup$ thanks for the answer. Ah.. they were aiming at finding the eigenstate of $H$ and NOT an arbitrary state. So given $H = \sum_i^M H_i \otimes P$ where $P$ is a Pauli string of length $k$ and $H_i$ are Pauli strings of length $w$. If $P $ has length $k$ then we can perform a parity check on each of the $2^k$ state and group the $\pm$ sign together. At the end you have $M$ different Hamiltonians of $w$ qubits instead of the original $w + k$ qubits. But in exchange, you are performing an additional $2^k$ parity check and $M$ more diagonaliations (over simpler Hamiltonian). Sounds right? $\endgroup$
    – KAJ226
    Apr 3 at 15:36
  • $\begingroup$ If P is a single Pauli string, then it is involutory, and we will only get two different eigenvalues (but still $2^k$ eigenstates for those k qubits). The tapering generalizes to Hamiltonians $H=\sum_i^M H_i \otimes P_i$ where $[P_i,P_j]=0$ for each i,j. You get, in general, $2^k$ different Hamiltonians of $w$ qubits , but each matrix has dimensionality $2^w\times2^w$ (instead of w+k), so it's still simpler, and in practice, you often know which of the $2^k$ Hamiltonians is the system you're interested in. $\endgroup$
    – cheetah
    Apr 3 at 15:57
  • $\begingroup$ Sure, you can group the commuting $P_i$ together. In these transformations you get $I$ or $X$ in $P_i$ so you can grouped them together. Oh yeah, sorry for stating the wrong thing earlier... right, you have $2^k$ different $w$-qubit Hamiltonians to diagonalize over due to the $2^k$ eigenstates and each of the Hamiltonian still have $M$ terms. So as $k$ gets large, we essentially performing full diagonalization... which makes sense. $\endgroup$
    – KAJ226
    Apr 3 at 16:19
  • $\begingroup$ Cheetah, thanks for the answer and discussion. +1 $\endgroup$
    – KAJ226
    Apr 5 at 14:05
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You can always write $|\psi\rangle$ in terms of the eigenvectors of the common Pauli matrices. Let me explain with an example. Suppose the Hamiltonian $H = (X+Z)Z = H_1 \otimes Z$. Then, for such two-qubit cases, any $|\psi\rangle$ can be expressed as $|\alpha\rangle|0\rangle + |\beta\rangle|1\rangle$ where $|0\rangle$ and $|1\rangle$ are the eigenvectors of $Z$ with eigenvalues $\pm 1$.

Now, $\langle \psi|H|\psi\rangle$ simplifies to $\langle \alpha|H_1|\alpha\rangle - \langle\beta|H_1|\beta\rangle$.

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  • $\begingroup$ @Gauruav, thanks for the answer. Sure, you can do that, but then there are $2^k$ different parity check you have to do, no? $\endgroup$
    – KAJ226
    Apr 3 at 15:19
  • $\begingroup$ But since those eigenstates are orthonormal, they'll all cancel out, leaving only $k$ elements to be computed, right? $\endgroup$ Apr 3 at 19:28
  • $\begingroup$ (with $ZZ$, the eigenvectors would be $|00\rangle$, $|01\rangle$, $|10\rangle$, and $|11\rangle$.) $\endgroup$ Apr 3 at 23:59
  • $\begingroup$ Hey Gaurav, thanks for the answer and discussion. +1 $\endgroup$
    – KAJ226
    Apr 5 at 14:05
  • $\begingroup$ @KAJ226 thanks for the interesting question and initiating the discussion. :) $\endgroup$ Apr 5 at 17:15

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