"Twisted" traces of Pauli strings

Question

Consider a system of $L$ qubits. A generic string of Pauli operators acting on such a system can be written as (neglecting phases for simplicity) $$\mathcal{O}(\vec{v},\vec{w}) = \bigotimes_{j=1}^{L}Z^{v_j}X^{w_j}$$

where $\vec{v}, \vec{w}$ are length $L$ binary vectors. The Pauli strings are trace-orthogonal with inner product

$$\text{Tr}\left[\mathcal{O}(\vec{\mu},\vec{\nu})^{\dagger}\mathcal{O}(\vec{v},\vec{w})\right] = 2^{L}\delta(\vec{v},\vec{\mu})\delta(\vec{\nu},\vec{w})$$

I'm interested in computing a closely related property of Pauli strings that I'm calling the "twisted trace". To define this, we need to define the shift operator, $T$, which increments spin labels by one. More precisely, let $|s_1,\cdots,s_L\rangle$ be a state of the system in the computational basis. Then $T|s_1,\cdots,s_L\rangle = |s_2,s_3,\cdots s_L,s_1\rangle$. I am interested in computing the traces

$$\text{Tr}\left[\mathcal{O}(\vec{v},\vec{w})T^n\right]$$

for some integer $0<n<L$. This calculation seems quite non-trivial in general and I expect a closed form solution for arbitrary $n$ and $L$ would be... annoying to get. I would be satisfied with some sort of efficient strategy for doing this calculation (programmatically is fine), perhaps by mapping Paulis onto binary strings in some manner. But this would involve finding a decomposition of $T$ into Pauli strings which also seems difficult.

Anyway, this is the kind of calculation that I imagine others have done before so I'm curious if there are any recommended strategies for this. Thanks in advance!

Answer 1

Below I give a solution for the $n=1$ case. Let $P = P_1 \otimes \cdots \otimes P_N$ denote a Pauli string acting on $N$ qubits and let $T$ be the right-shift-by-one operator $T |a_1, \dots, a_N\rangle = |a_N, a_1, \dots, a_{N_1}\rangle$. Then, we claim that $$Tr(PT) = Tr(P_NP_{N-1}\cdots P_1)$$ Observe that \begin{align} Tr(PT) &= Tr((P_1 \otimes \cdots \otimes P_N)T)\\ &= \sum_{i_1, \dots, i_N=0}^1 \langle i_1, \dots, i_N| P_1 \otimes \cdots \otimes P_NT|i_1, \dots, i_N\rangle\\ &= \sum_{i_1, \dots, i_N=0}^1 \langle i_1, \dots, i_N| P_1 \otimes \cdots \otimes P_N|i_N, i_1, \dots, i_{N-1}\rangle\\ &= \sum_{i_1, \dots, i_N=0}^1 \langle i_1| P_1 |i_N\rangle \langle i_2| P_2 |i_1 \rangle \cdots \langle i_{N-1}| P_{N-1}| i_{N-2} \rangle \langle i_N | P_N | i_{N-1}\rangle\\ &= \sum_{i_1, \dots, i_N=0}^1 \langle i_N | P_N | i_{N-1}\rangle \langle i_{N-1}| P_{N-1}| i_{N-2} \rangle \cdots \langle i_2| P_2 |i_1 \rangle \langle i_1| P_1 |i_N\rangle\\ &= \sum_{i_N=0}^1 \langle i_N | P_NP_{N-1} \cdots P_1 | i_N\rangle\\ &= Tr(P_NP_{N-1}\cdots P_1). \end{align}

For the general case, observe that $T^n$ maps index $i \to i+n$. Consider the partition of $N$ into subsets of the form $I_i = \{i+kn\}_{k=1}^m$, where $m = N/\mathrm{gcd}(n,N)$. We can reduce $T^n$ to acting as a single right-shift on each of the subsets of qubits with indices in $I_i$. Applying the $n=1$ result to each subcycle gives the desired result. The final result is (if I've got my notation right) $$Tr(PT^n) = \prod_{i=1}^m Tr(P_{N-m+i}\cdots P_i)$$

I should also add that a friend of mine showed me a way to compute this using tensor networks and its super neat and elegant. Really trivializes a lot of the algebra and makes it easy to see why $T^n$ both breaks into cycles and why the product in each cycles reverses order.