Vanishing expectation value $|\langle Z_1Z_2...Z_N \rangle|$

Question

I'm doing a research involving expectation values of different observables.

I've observed that, given a random Quantum Circuit $U$ with $N$ qubits acting on an inital state $|0\rangle$ in such a way that $|\psi\rangle = U|0\rangle$, the average $|\langle \psi|Z_1\otimes Z_2 \otimes...\otimes Z_N|\psi\rangle|$ across the random quantum circuits goes to 0 as the number of qubits of the Quantum Circuit increases.

EDIT: The main problem actually is not for the average between all the expectation values, but all the expectation values that I get are close to 0 for large N (and therefore also the average).

Is there any explanation behind this thing? Because this kind of expectation value (without the absolute value) is important for many well known quantum algorithms such as the VQE. I've added the absolute value because the expectation values can be positive or negative, but they are all close to 0 for large N.

Below the code showing this fact:

import numpy as np
from scipy.stats import unitary_group

n_samples = 1000

for N in range(1,8):
    valcounts = []
    for i in range(n_samples):
    
        state = np.zeros(2**N)
        state[0] = 1
        
        
        ru = unitary_group.rvs(2**N)
        
        state = np.matmul(ru, state)
        
        key = []
        for i in range(2**N):
            bit = [*"{0:b}".format(i)]
            bitemp = ['0' for i in range((N-len(bit)))]
            bit = bitemp + bit
            key.append(bit)
        key = np.array(key)
        
        valcount = 0
        for i in range(len(key)):
            numbones = len(np.where(key[i]=='1')[0])
            if numbones%2 == 0:
                valcount += np.abs(state[i])**2
            else:
                valcount -= np.abs(state[i])**2
                
        print(valcount)
        valcounts.append(valcount)

    print(np.average(np.abs(valcounts)))

Answer 1

Taking another stab. If $|\psi\rangle = U|0\rangle$ for a fixed $U$, then $|\langle \psi |Z_1\otimes \dots \otimes Z_N |\psi\rangle| = |\langle 0 | U^\dagger(Z_1\otimes \dots \otimes Z_N)U |0\rangle|$. Then $U^\dagger(Z_1\otimes \dots \otimes Z_N)U$ is itself a random unitary, and the value you care about is simply the $0,0$ entry. Since the column-space of a unitary is an orthonormal basis, the absolute of value of the $0,0$ element should be around $2^{-N}$ for a fixed, randomly generated $N$-qubit unitary.

Answer 2

If we factor $|\psi\rangle = U|0\rangle$ into the $N$ individual qubit states as $|\psi\rangle = |\psi_1\rangle \otimes \dots \otimes |\psi_N\rangle$, then $|\langle \psi| Z_1 \otimes \dots \otimes Z_N|\psi\rangle| = |\langle \psi_1 |Z_1|\psi_1\rangle \dots \langle \psi_N|Z_N|\psi_N\rangle|$.

The expected value of $|\langle \psi_j|Z_j|\psi_j\rangle|$ over all possible values of $U$ is $0$ with some variance, which is why for a single $U$ you see some small nonzero $|\langle \psi_j|Z_j|\psi_j\rangle|$. But as you multiply many such $|\langle \psi_j|Z_j|\psi_j\rangle|$ together, you converge to $0$ because these small values compound (and you can think that as $N$ increases, you're more likely to get a $|\langle \psi_j|Z_j|\psi_j\rangle|$ that is very close to $0$ which would then make the whole product $0$).