by claude on

Short answer: Hamming weight itself can't be basis independent, because counting 1s needs a preferred "0" state on every qubit. What is basis independent is the tensor-product structure, i.e. which qubits something touches. "Basis independent" here really means "invariant under local unitaries U_1 x U_2 x ... x U_n", and there are a few good definitions that satisfy it:

1. Operator weight. The weight of an operator O is the smallest number of qubits k such that O acts as the identity on all the others (O = O_S x I_rest, with |S| = k). It uses no basis, only the split into qubits, and local unitaries don't change it. This is the "weight" used in quantum error correction; the Hamming weight of a Pauli string is the special case.

2. Weight of a state relative to a reference. Fix a product reference state |r> = |r_1>|r_2>...|r_n>. The weight of |psi> is the size of the smallest set S such that |psi> = |phi>_S x |r>_rest, i.e. how many qubits are not in their reference state. With |r> = |0...0> and |psi> = |x>, this is exactly the Hamming weight of x. Choosing |r> is the one choice you can't avoid; once it's made, nothing else depends on a basis.

3. Number operator. With the same reference, N = sum_i (I - |r_i><r_i|). Its eigenvalues are 0..n, and on basis states it returns the Hamming weight. A superposition gets a distribution over weights rather than one number, and <psi|N|psi> is the expected weight.

4. Distance version. For a Hamming-style distance between arbitrary states, see "The Quantum Wasserstein Distance of Order 1" (De Palma, Marvian, Trevisan, Lloyd, 2021). It reduces to Hamming distance on computational basis states and doesn't single out any basis.

Rule of thumb: anything defined only by which qubits are involved is basis independent; anything that counts 1s needs a reference state on each qubit.
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