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Given a set of stabilizer generators, is it always possible to write down the state corresponding to it? Is there a way to write down the quantum state corresponding to a stabilizer generator?

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2 Answers 2

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Given an $n$-qubit system and $n$ generators $g_i$ (which commute and square to identity), then the state that you are after satisfies $$ g_i|\psi\rangle=|\psi\rangle. $$ (This defines it uniquely, up to a global phase and normalisation.) One straightforward way to construct this directly is just $$ |\psi\rangle\langle\psi|=\frac{1}{2^n}\prod_i(I+g_i). $$ Thus, you can compute the matrix using the $\{g_i\}$ and find the vector $|\psi\rangle$ by determining the one eigenvector that does not have 0 eigenvalue. In practice, you can do this simply by reading off any non-trivial column of the matrix, and normalising.

For example, let $n=2$ and take $$ g_1=X\otimes X,\qquad g_2=Z\otimes Z. $$ We have $$ |\psi\rangle\langle\psi|=\frac14(I+XX)(I+ZZ)=\frac14(I+XX+ZZ-YY). $$ If I read off the first column, this is equivalent to computing $$ |\psi\rangle\langle\psi|00\rangle=\frac14(I+XX+ZZ-YY)|00\rangle=\frac12(|00\rangle+|11\rangle). $$ Thus, $$ |\psi\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle). $$

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You can do this easily using Qiskit as follows:

from qiskit.quantum_info import StabilizerState, Statevector
 
stabilizer_list = ["XX", "ZZ"]
stabilizer_state = StabilizerState.from_stabilizer_list(stabilizer_list)
sv = Statevector.from_label('0'*stabilizer_state.num_qubits).evolve(stabilizer_state)
sv.draw('latex')

The result: $$\frac{\sqrt{2}}{2} |00\rangle+\frac{\sqrt{2}}{2} |11\rangle$$

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