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.
Linearity is a fundamental property of quantum mechanics. If I’m trying to just describe it in wave mechanics terms, I would say that the linearity of quantum mechanics derives from the fact that the wave equation describes a linear system and thus solutions to it must obey the (general, mathematical) principle of superposition.
It’d be fine if it were linear in general, but it’s not for combinations that aren’t orthogonal. Suppose a is drawn from R^2. P(sqrt(2))=P(|(1,1)|)=P(1,1)=P(1,0)+P(0,1)=2*P(|(1,0)|)=2*P(1) which agrees with your analysis, but P(sqrt(5))=P(|(2,1)|)=P(2,1)/=P(1,0)+P(1,1)=3*P(1) doesn’t add up.
It’s been a while since I’ve done any wave mechanics, but I’ll try to take a crack at this. The Schrodinger equation describes a linear PDE such that the sum of any two solutions is also a solution and any constant multiple of a solution is also a solution. Furthermore, the Schrodinger equation just takes the form ^HΨ=EΨ, thus “solutions to the Schrodinger equation” is equivalent to “eigenfunctions of the Hamiltonian.” Thus, if ϕi are eigenfunctions of the Hamiltonian with eigenvalues Ei, then ∑iaiϕi must also be an eigenfunction of the Hamiltonian. This raises a problem for any theory with P nonlinear across a sum of eigenfunctions, however, because it lets me change bases into an equivalent form with a potentially different result.
If a1 is 2 and phi1 has eigenvalue 3, and a2 is 4 and phi2 has eigenvalue 5, then 2*phi1+3*phi2 is mapped to 6*phi1+20*phi2 and therefore not an eigenfunction.
Accepting that probability is some function of the magnitude of the amplitude, why should it be linear exactly under orthogonal combinations?
Everett argued in his thesis that the unitary dynamics motivated this:
He made the analogy with Liouville’s theorem in classical dynamics, where symplectic dynamics motivated the Lebesgue measure on phase space.
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.
Yeah, that’s a great argument—Everett’s thesis always has the answers.
Linearity is a fundamental property of quantum mechanics. If I’m trying to just describe it in wave mechanics terms, I would say that the linearity of quantum mechanics derives from the fact that the wave equation describes a linear system and thus solutions to it must obey the (general, mathematical) principle of superposition.
It’d be fine if it were linear in general, but it’s not for combinations that aren’t orthogonal. Suppose a is drawn from R^2. P(sqrt(2))=P(|(1,1)|)=P(1,1)=P(1,0)+P(0,1)=2*P(|(1,0)|)=2*P(1) which agrees with your analysis, but P(sqrt(5))=P(|(2,1)|)=P(2,1)/=P(1,0)+P(1,1)=3*P(1) doesn’t add up.
It’s been a while since I’ve done any wave mechanics, but I’ll try to take a crack at this. The Schrodinger equation describes a linear PDE such that the sum of any two solutions is also a solution and any constant multiple of a solution is also a solution. Furthermore, the Schrodinger equation just takes the form ^HΨ=EΨ, thus “solutions to the Schrodinger equation” is equivalent to “eigenfunctions of the Hamiltonian.” Thus, if ϕi are eigenfunctions of the Hamiltonian with eigenvalues Ei, then ∑iaiϕi must also be an eigenfunction of the Hamiltonian. This raises a problem for any theory with P nonlinear across a sum of eigenfunctions, however, because it lets me change bases into an equivalent form with a potentially different result.
If a1 is 2 and phi1 has eigenvalue 3, and a2 is 4 and phi2 has eigenvalue 5, then 2*phi1+3*phi2 is mapped to 6*phi1+20*phi2 and therefore not an eigenfunction.
Ah, I see the confusion. Since we’re in a wave mechanics setting, I should have written ^HΨ=i¯h∂∂tΨ rather than ^HΨ=EΨ.