I reply with the same point about orthogonality: Why should (2,1) split into one branch of (2,0) and one branch of (0,1), not into one branch of (1,0) and one branch of (1,1)? Only the former leads to probability equaling squared amplitude magnitude.
(I’m guessing that classical statistical mechanics is invariant under how we choose such branches?)
Why should (2,1) split into one branch of (2,0) and one branch of (0,1), not into one branch of (1,0) and one branch of (1,1)?
Again, it’s because of unitarity.
As Everett argues, we need to work with normalized states to unambiguously define the coefficients, so let’s define normalized vectors v1=(1,0) and v2=(1,1)/sqrt(2). (1,0) has an amplitude of 1, (1,1) has an amplitude of sqrt(2), and (2,1) has an amplitude of sqrt(5).
(2,1) = v1 + sqrt(2) v2, so we need M[sqrt(5)] = M[1] + M[sqrt(2)] for the additivity of measures. Now let’s do a unitary transformation on (2,1) to get (1,2) = −1 v1 + 2 sqrt(2) v2 which still has an amplitude of sqrt(5). So now we need M[sqrt(5)] = M[2 sqrt(2)] + M[-1] = M[2 sqrt(2)] + M[1]. This can only work if M[2 sqrt(2)] = M[sqrt(2)]. If one wanted a strictly monotonic dependence on amplitude, that’d be the end. We can keep going instead and look at the vector (a+1, a) = v1 + a sqrt(2) v2, rotate it to (a, a+1) = -v1 + (a+1) sqrt(2) v2, and prove that M[(a+1) sqrt(2)] = M[a sqrt(2)] for all a. Continuing similarly, we’re led inevitably to M[x] = 0 for any x. If we want a non-trivial measure with these properties, we have to look at orthogonal states.
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.
I reply with the same point about orthogonality: Why should (2,1) split into one branch of (2,0) and one branch of (0,1), not into one branch of (1,0) and one branch of (1,1)? Only the former leads to probability equaling squared amplitude magnitude.
(I’m guessing that classical statistical mechanics is invariant under how we choose such branches?)
Again, it’s because of unitarity.
As Everett argues, we need to work with normalized states to unambiguously define the coefficients, so let’s define normalized vectors v1=(1,0) and v2=(1,1)/sqrt(2). (1,0) has an amplitude of 1, (1,1) has an amplitude of sqrt(2), and (2,1) has an amplitude of sqrt(5).
(2,1) = v1 + sqrt(2) v2, so we need M[sqrt(5)] = M[1] + M[sqrt(2)] for the additivity of measures. Now let’s do a unitary transformation on (2,1) to get (1,2) = −1 v1 + 2 sqrt(2) v2 which still has an amplitude of sqrt(5). So now we need M[sqrt(5)] = M[2 sqrt(2)] + M[-1] = M[2 sqrt(2)] + M[1]. This can only work if M[2 sqrt(2)] = M[sqrt(2)]. If one wanted a strictly monotonic dependence on amplitude, that’d be the end. We can keep going instead and look at the vector (a+1, a) = v1 + a sqrt(2) v2, rotate it to (a, a+1) = -v1 + (a+1) sqrt(2) v2, and prove that M[(a+1) sqrt(2)] = M[a sqrt(2)] for all a. Continuing similarly, we’re led inevitably to M[x] = 0 for any x. If we want a non-trivial measure with these properties, we have to look at orthogonal states.
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.