Correlated Measurements

The Bell state $|\Phi^+\rangle = \tfrac{1}{\sqrt{2}}\big(|00\rangle + |11\rangle\big)$ has an extraordinary property: each qubit individually appears to be random, yet the two qubits are perfectly correlated whenever measured together.

What the probabilities look like

Applying the Born rule to the two-qubit state $|\Phi^+\rangle$ , the four computational-basis outcomes have probabilities

P(00) = \left|\tfrac{1}{\sqrt{2}}\right|^2 = \tfrac{1}{2}, \quad P(01) = 0, \quad P(10) = 0, \quad P(11) = \left|\tfrac{1}{\sqrt{2}}\right|^2 = \tfrac{1}{2}.

Only the "same-outcome" events $|00\rangle$ and $|11\rangle$ have nonzero amplitude. The cross terms $|01\rangle$ and $|10\rangle$ are completely absent — they cannot occur. This is what perfect correlation means: once you know the result of one qubit, you know the other with certainty.

Why each qubit looks random on its own

If you ignore qubit 1 and look only at qubit 0, you observe 0 half the time and 1 half the time. Qubit 1 behaves the same way. There is no local information in either qubit that reveals the other's value — the correlation is a joint, global property of the entangled state. This is the feature that makes Bell-state correlations stronger than any pre-shared classical strategy could produce, as Bell's inequality makes precise.

Try it

Prepare $|\Phi^+\rangle$ and measure both qubits. The grader checks the probability distribution: it should be exactly $\tfrac{1}{2}$ on $|00\rangle$ and $\tfrac{1}{2}$ on $|11\rangle$ , with zero weight on the mixed-outcome strings $|01\rangle$ and $|10\rangle$ .

Run your code to see the quantum state.

Open the Probabilities tab and verify that only the two bars at "00" and "11" are visible, each at height 0.5. The correlation is not a coincidence — it is enforced by the entangled state's structure, and no local operation on a single qubit can produce this pattern.