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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)))
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2 Answers 2

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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.

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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 $0$).

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    $\begingroup$ Hi. Why do you assume that $|\psi\rangle$ can be factorized? I think that OP had in mind an arbitrary $n$-qubit unitary that creates possibly a large amount of entanglement. $\endgroup$ Nov 3, 2022 at 15:54
  • $\begingroup$ @MarcoFellous-Asiani Yes, exactly. $\endgroup$
    – stopper
    Nov 3, 2022 at 16:17
  • $\begingroup$ Actually your answer can be used also for my problem, since the expected value is 0. But I've stated it in a wrong way, I am not only interested on the average over many random quantum circuits. In general, if I have 1000 random circuits, all of them have expectation value close to 0, not only the average between them. I edit the orginal post $\endgroup$
    – stopper
    Nov 3, 2022 at 16:20
  • $\begingroup$ @stopper my suspicion is that if you take a random $U$, then your density matrix will converge to identity. Hence the expectation value for a given observable $O$ satisfies: $Tr(\rho O)=Tr(O)$, which in your case will be equal to $0$. Now this results comes when averaged over several $U$. In practice I guess that each of the random element you pick is "arbitrary enough" such that on each instance you will have something close to $0$. To not have something close to $0$ "at the single trajectory level", you need some "very unusual" $U$ I suppose (for instance $U=Z_1 \otimes ... \otimes Z_N$). $\endgroup$ Nov 3, 2022 at 16:27
  • $\begingroup$ @MarcoFellous-Asiani Thank you. But why "if you take a random U, then your density matrix will converge to identity"? $\endgroup$
    – stopper
    Nov 3, 2022 at 16:30

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