I don’t see how that relates to what I said. I was addressing why an amplitude-only measure that respects unitarity and is additive over branches has to use amplitudes for a mutually orthogonal set of states to make sense. Nothing in Everett’s proof of the Born rule relies on a tensor product structure.
Then I have misunderstood Everett’s proof of the Born rule. Because the tensor product structure seems absolutely crucial for this, as you just can’t get mixed states without a tensor product structure.
Everett’s proof that the Born rule measure (amplitude squared for orthogonal states) is the only measure that satisfies the desired properties has no dependence on tensor product structure.
Everett’s proof that a “typical” observer sees measurements that agree with the Born rule in the long term uses the tensor product structure and the result of the previous proof.
We don’t lose unitarity just by choosing a different basis to represent the mixed states in the tensor-product space.
I don’t see how that relates to what I said. I was addressing why an amplitude-only measure that respects unitarity and is additive over branches has to use amplitudes for a mutually orthogonal set of states to make sense. Nothing in Everett’s proof of the Born rule relies on a tensor product structure.
Then I have misunderstood Everett’s proof of the Born rule. Because the tensor product structure seems absolutely crucial for this, as you just can’t get mixed states without a tensor product structure.
I will amend my statement to be more precise:
Everett’s proof that the Born rule measure (amplitude squared for orthogonal states) is the only measure that satisfies the desired properties has no dependence on tensor product structure.
Everett’s proof that a “typical” observer sees measurements that agree with the Born rule in the long term uses the tensor product structure and the result of the previous proof.