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I would like to calculate the probability of measuring some state $U\rho U^\dagger$ in the basis state $b \in (0,1)^{\otimes n}$, i.e. $<b|U\rho U^\dagger|b>$. Now, according to Gottesmann and Knill’s theorem, this can be efficiently calculated, if $\rho$ is a stabilizer state and U is some Clifford operation.

I made several attempts, figuring this out by myself. In order to achieve at least some result, I resorted to calculate the problem the old fashioned way, matrix multiplication style. But as one expects, the bigger the system size, the longer it takes to calculate. Also i am planning to calculate up to 5000 probabilities, each for different basis states and clifford operations.

In qiskit I tried the following:

from qiskit import QuantumCircuit
from qiskit.quantum_info import DensityMatrix, random_clifford, StabilizerState
from numpy import zeros

qc = ghz(3)                                      
rnd_clifford = random_clifford(3)                    #generates a random clifford operation
rho_ghz = DensityMatrix(qc)
rho_ghz.evolve(rnd_clifford)                         #evolves the GHZ state with random clifford operation
                                                     #this should be equivalent to U*rho*U†

prob_b = basis(3, b).transpose()@rho_ghz@basis(3, b)  #calculation of <b|U*rho*U†|b>


def ghz(num_qubits):
    qc = QuantumCircuit(num_qubits)
    qc.h(0)
    for k in range(1, num_qubits, 1):
        qc.cx(0,k)
    qc.barrier()
    return qc

def basis(num_qubits, bitstring):
    bas = zeros(2**num_qubits); bas[int(bitstring, 2)] = 1
    return bas

I want to iterate this procedure many times and save the results to a list. Especially for many iterations and high dimensionality (exponential in qubits) of the GHZ state, this can take a long while to calculate.

So my question is: Is there a more efficient way to calculate $<b|U\rho U^\dagger|b>$, given the prerequisites of $\rho$ being a stabilizer state and U being some Clifford Operator, i.e. making use of Gottesmann and Knills theorem?

Anoher idea i had, would be to maybe make use of the $\texttt{qiskit.quantum_info.StabilizerState.probabilities_dict}$ method, and to then get the dictionary entry for the desired basis b

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  • $\begingroup$ Why don't you just use the qasm simulator and take a lot of shots? And use those to estimate the frequencies? If you go down this route, it's better to use Stim, as it will yield even faster simulations. $\endgroup$ Feb 7, 2023 at 20:55
  • $\begingroup$ it's important that this step is in post processing, i.e. i dont want to retrieve the actual probability distribution of the evolved state, but rather calculate a number of single matrix elements <b|UrhoU†|b>, each for different matrices UrhoU† :) $\endgroup$
    – Coryn7
    Feb 7, 2023 at 22:51
  • $\begingroup$ Scott Aaronson's CHP performs / shows how to perform measurements on stabilizer states in $O(n^2)$ time. $\endgroup$
    – ChrisD
    Feb 8, 2023 at 3:42
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    $\begingroup$ You should edit your question to state that you are interested in "probability amplitudes", rather than "measurement probabilities". Some questions (1) $\rho$ drawn from a small set or can be an arbitrary stabilizer state? (2) Is $\rho$ a pure stabilizer state or mixture of them? (3) What is the largest $\rho$ you are going to be working with? $\endgroup$ Feb 8, 2023 at 21:41
  • $\begingroup$ you are right, i should have specified - rho is a pure, n-qubit ghz state, calculations are planned for n in [3, 10] qubits $\endgroup$
    – Coryn7
    Feb 8, 2023 at 22:59

1 Answer 1

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You have indicated that $\rho$ is a pure $n$-qubit GHZ state, $$ |G_n\rangle = \frac{|0\rangle^{\otimes n} + |1\rangle^{\otimes n}}{\sqrt{2}}. $$ Then, $$ \langle b| U\rho U^\dagger |b\rangle = \langle b| U |G_n\rangle\langle G_n| U^\dagger |b\rangle = |\langle b| U |G_n\rangle|^2. $$

You have further indicated that $|b\rangle$ is the standard computational basis, so there is a 1 in exactly one spot of the vector $|b\rangle$. Furthermore, $|G_n\rangle$ only has non-zero entries in the first and last spot. So, $U|G_n\rangle$ is just the sum of the first and last columns of $U$. And $\langle b| U |G_n\rangle$ is the $b$-th entry of $U|G_n\rangle$.

In other words, $\langle b| U |G_n\rangle$ is the sum of the matrix elements $U_{b,0} + U_{b,2^n-1}$.

So, once you generate your $U$, only a trivial computation is needed to compute $\langle b| U\rho U^\dagger |b\rangle$.

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  • $\begingroup$ thank you very very much, this has been extremely helpful (and also embarrassingly something i should have thought of myself haha) - nevertheless, this speeds the procedure up by a lot, thanks ! :) $\endgroup$
    – Coryn7
    Feb 9, 2023 at 16:55
  • $\begingroup$ No problem. Since you are an undergrad, my advice to you would be to always work out a small instance of the problem on paper before trying to write code. Much time will be saved this way. :-) $\endgroup$ Feb 9, 2023 at 20:20

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