Marginal Probabilities

A two-qubit state can be measured in many ways. Often you care about only one qubit — the other is not observed, or its result is ignored. The probability of getting a particular outcome on the qubit you do measure is called a marginal probability.

Where marginals come from

A general two-qubit state is written

where $c_{ij}$ are complex amplitudes satisfying $|c_{00}|^2 + |c_{01}|^2 + |c_{10}|^2 + |c_{11}|^2 = 1$.

If you measure only qubit 0 (the most significant bit) and ignore qubit 1, the Born rule says you must sum over every outcome of the unobserved qubit:

The two marginals sum to 1, as required for a valid probability distribution.

A worked example

The Bell state $|\Phi^+\rangle$ is

Its four amplitudes are $c_{00} = \tfrac{1}{\sqrt{2}}$, $c_{01} = 0$, $c_{10} = 0$, $c_{11} = \tfrac{1}{\sqrt{2}}$. Applying the marginal formula to qubit 0:

Each outcome is equally likely — exactly as you would expect from the symmetric form of $|\Phi^+\rangle$. Notice that you never needed to specify what happens to qubit 1; the marginal absorbs it by summing over all its values.

Marginals for product states

For a product state $|\psi\rangle = |\psi_0\rangle \otimes |\psi_1\rangle$, summing over one qubit simply returns the probability already carried by the other qubit's own state. If $|\psi_0\rangle = \alpha|0\rangle + \beta|1\rangle$, then $P(q_0 = 0) = |\alpha|^2$ regardless of $|\psi_1\rangle$. The two qubits are statistically independent. For entangled states this independence breaks down — the qubits' individual marginals give only a partial picture of the joint state.

Try it

The state is $|\Phi^+\rangle = \tfrac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$. Use the formula above to compute the marginal probability that qubit 0 measures 0, and return it.

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