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Interpretations of Quantum Mechanics · Chapter 05

Everett and Relative States

What happens if the measurement interaction is never replaced by collapse?

In this chapter · 3 sections
  1. 5.1 Relative states
  2. 5.2 Branches and stable records
  3. 5.3 Probability in a unitary world
Reading guide · preparation and approach

Starting point: Chapters 1–4; familiarity with density operators and linear algebra.

Anchor example: A correlated observer record and unequal branch weights.

Method: Calculate a concrete experiment, identify the assumptions, then compare what different accounts explain.

Watch for: Counting branches is not a well-defined substitute for the Born rule.

This lesson includes explanations, worked examples, and problems with solutions. Advanced results whose proofs are outside the chapter are identified explicitly.

What happens if the measurement interaction is never replaced by collapse?

5.1 Relative states

5.1.1 Condition a description without deleting the global state

Return to the state \(|\Psi\rangle=\alpha|0,A_0\rangle+\beta|1,A_1\rangle\). Everett's proposal takes the universal quantum state and its unitary dynamics seriously, including the observer among the systems described. The theory does not introduce a new stochastic event that deletes one term. Instead, a system state can be specified relative to a record.

\[|\psi_{S|A_k}\rangle=\frac{\langle A_k|\Psi\rangle}{\|\langle A_k|\Psi\rangle\|}.\]

For nonzero α and β the relative states are |0⟩ and |1⟩, up to irrelevant phases. In each correlated record component, the apparatus agrees with the measured system. This is how relative-state reasoning addresses the appearance of definite correlated observations. The global state remains the superposition.

The formal conditional vector alone is not the whole interpretation: ordinary quantum mechanics also uses conditional states. The additional commitment is to the continued global state and an account of observers within it. Modern Everettian accounts commonly describe approximately autonomous decohered branches, rather than adding a physical splitting operation to the Hamiltonian.

There is therefore an explanatory task, not merely a change of notation: why should the patterns identified as branches support the experiences and expectations of observers?

Worked check 5.1. Find the relative system state for a record A₁ when β=i/2.

Reasoned solution

Projection gives i|1⟩/2; normalization gives i|1⟩, physically the same ray as |1⟩. Its branch weight is 1/4.

5.2 Branches and stable records

5.2.1 Use decoherence rather than an arbitrary basis expansion

A vector can be expanded in many bases. Why privilege one expansion as a description of records? The Bell state can be written in both the Z-correlated and X-correlated forms. Algebra alone cannot select a classical narrative. The environment and the dynamics matter because they stabilize some records and rapidly disperse interference between them.

Suppose |A₀⟩ and |A₁⟩ leave distinguishable records across many uncontrolled degrees of freedom. Within the observables accessible to an observer in one such record, interference with the other record becomes negligible. An approximately autonomous branch is a dynamical and coarse-grained notion. Its definition need not assign an exact integer number of worlds at every instant.

\[\alpha|0,A_0\rangle\longrightarrow \frac{\alpha}{\sqrt m}\sum_{j=1}^m|0,A_0,E_j\rangle\]

illustrates a harmless refinement of one outcome into m orthogonal environmental alternatives. Its total squared norm is still |α|². A rule that gives equal probability to every written term would change the probability of the macroscopic record merely because its environmental description was refined. Stable predictions require a measure insensitive to such arbitrary counting.

Decoherence thus helps explain robust record structure, but the role of branch weights still needs an argument. We should resist treating the word “branch” as a new primitive probability unit.

Worked check 5.2. Split a branch of weight 3/4 into three equal orthogonal components.

Reasoned solution

Each new component has weight 1/4, and their weights sum to 3/4. A macro-record probability based on total weight is unchanged.

5.3 Probability in a unitary world

5.3.1 Identify the assumptions behind a weight rule

For mutually orthogonal components, squared norm is additive: \(\|u+v\|^2=\|u\|^2+\|v\|^2\). It is also invariant under unitary refinement and normalized for a normalized state. These mathematical properties explain why squared amplitudes are natural weights. They do not alone establish what an observer should expect when every decohered outcome component remains in the universal state.

Everettian probability programs add assumptions about rational preferences, self-locating uncertainty, or typicality. Decision-theoretic arguments, for example, constrain how an agent values quantum acts and identify conditions under which preferences can be represented using Born weights. A theorem derived from those conditions is not a derivation from the Schrödinger equation alone. Whether its assumptions adequately explain observed probability is part of the interpretive debate.

If Born weights are used for N repeated independent preparations with outcome weight p, total weight of the k-success record subspace is

\[w(k)=\binom Nk p^k(1-p)^{N-k}.\]

The mean frequency is p and its variance is p(1−p)/N. This concentration is mathematically clear. The remaining issue is why branch weight should govern empirical expectation or typicality, rather than merely being a number attached to components. Keep that distinction visible when assessing any proposed solution.

Worked check 5.3. For N=100 and p=1/4, find the standard deviation of the relative frequency.

Reasoned solution

It is √[p(1−p)/N]=√0.001875≈0.0433. This calculation assumes the Born-weight assignment.

Sources and further reading

Summary

Questions and worked problems

These exercises draw on the whole chapter. Attempt each question before opening a hint, explain your reasoning, then compare it with the worked solution.

1. Quick check §

Is a relative-state projection necessarily a new physical collapse?

Hint 1

Distinguish extracting a conditional description from global evolution.

Worked solution

No. In the relative-state account the projection defines a conditional state while the universal vector is retained.

Revisit the relevant explanation

2. Worked problem §

What happens when the conditioning record has zero amplitude?

Hint 1

Inspect the denominator.

Worked solution

The normalized relative state is undefined because the conditioning event has zero norm; it is not an additional realized branch.

Revisit the relevant explanation

3. Misconception §

What extra interpretive claim distinguishes Everett from ordinary conditional-state calculations?

Hint 1

Ask what happens to the global vector.

Worked solution

Everett retains universal unitary evolution and treats the full state as describing the physical situation, including observer correlations.

Revisit the relevant explanation

4. Extension §

Write |Φ+⟩ in the X basis.

Hint 1

Expand |++⟩ and |−−⟩.

Worked solution

It equals (|++⟩+|−−⟩)/√2. This demonstrates that a basis expansion alone does not identify a classical record basis.

Revisit the relevant explanation

5. Quick check §

Why is environmental refinement a problem for branch counting?

Hint 1

Let only one macro-outcome acquire more subrecords.

Worked solution

Counting changes its relative frequency of terms although its total norm and macro-observable predictions are unchanged.

Revisit the relevant explanation

6. Worked problem §

Is decoherence an exact count of worlds?

Hint 1

Recall the use of coarse-graining and approximate autonomy.

Worked solution

No. It supplies dynamical criteria for robust approximately independent records, generally without a unique exact branch count.

Revisit the relevant explanation

7. Misconception §

Prove additivity of squared norm for orthogonal vectors.

Hint 1

Expand the inner product.

Worked solution

The cross terms ⟨u|v⟩ and its conjugate vanish, leaving ||u||²+||v||².

Revisit the relevant explanation

8. Extension §

For N=2 and p=1/4 find weights of k=0,1,2.

Hint 1

Use the binomial distribution.

Worked solution

They are 9/16, 6/16, and 1/16. The two one-success sequences contribute separately to the middle weight.

Revisit the relevant explanation

9. Quick check §

Does binomial concentration alone solve the probability problem?

Hint 1

Identify what measure defined the distribution.

Worked solution

No. It establishes concentration with respect to Born weight; the interpretive justification for using that weight remains an assumption or further argument.

Revisit the relevant explanation

10. Worked problem §

How should a decision-theoretic Born-rule claim be evaluated?

Hint 1

Inspect its premises, not just its conclusion.

Worked solution

State the allowed acts, preference axioms, continuity or equivalence assumptions, and conclusion. Then examine whether those assumptions are justified for observers in a branching state.

Revisit the relevant explanation

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