If you have a pure composite system whose two subsystems are in a product state, then the outcomes of measuring the subsystems (in any basis) are statistically independent. If the subsystems are entangled, then the measurement outcomes will generically be correlated.

Is there an example of an entangled state of a composite system of two isomorphic qudits, such that if you measure both subsystems in one basis, then the subsystems' measurement outcomes in that basis are independent, but if you measure both subsystems in another basis, then the outcomes are correlated?

For example, is there an entangled state $|\psi\rangle$ of two qubits such that for the joint probability mass function for a measurement in the $Z$-basis $$\left\{ P(\uparrow, \uparrow) = |\langle \uparrow \uparrow | \psi\rangle|^2, P(\uparrow, \downarrow) = |\langle \uparrow \downarrow | \psi\rangle|^2, P(\downarrow, \uparrow) = |\langle \downarrow \uparrow | \psi\rangle|^2, P(\downarrow, \downarrow) = |\langle \downarrow \downarrow | \psi\rangle|^2 \right\},$$ the measurement outcomes of the two qubits are independent, but for the joint probability mass function for a measurement of the same state in the $X$-basis $$\left\{ P(+, +) = |\langle + + | \psi\rangle|^2, P(+, -) = |\langle + - | \psi\rangle|^2,\\ P(-, +) = |\langle - + | \psi\rangle|^2, P(-, -) = |\langle - - | \psi\rangle|^2 \right\},$$ the measurement outcomes of the two qubits are dependent?

I don't see why this shouldn't be possible, but I can't think of an example.

  • $\begingroup$ "I don't see why this shouldn't be possible, but I can't think of an example." -- This likely has to do with complex numbers. Real examples are much trickier, if existent (for one thing, |00>+|11> is invariant under real rotations $U\otimes U$, but not under complex ones, since it is in fact invariant under $U\otimes \bar U$, so it behaves the same way under X and Z, but not under X and Y.). So using either X and Y instead of X and Z, or states with complex amplitudes as in Craig's example, is the way to go. $\endgroup$ Commented Aug 28, 2020 at 10:28

2 Answers 2


The state $|00\rangle + i|11\rangle$ has deterministic Z parity and random X parity.

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Basically the trick is to rotate $|00\rangle + |11\rangle$ around the Z axis to mess up the agreement of the X observables. Instead of ZZ, YY, and XX being deterministic; ZZ, XY, and YX are.


Consider two copies of a $d$-dimensional system, $\mathcal X_1,\mathcal X_2$, and take two mutually unbiased bases (MUBs) in each space. Denote these bases with $$\newcommand{\ket}[1]{\lvert #1\rangle} U\equiv \{\ket{u_k}\}_k,\,V\equiv \{\ket{v_k}\}_k.$$

Consider the state $\mathcal X_1\otimes\mathcal X_2\ni\ket\psi\equiv\sum_k \ket{u_k}\otimes\ket{u_k}$ (apply normalisations at will).

Then, measuring in the basis $U$ on both sides, the measurement outcomes are fully correlated: if $\ket{u_k}$ is found in $\mathcal X_1$, then with certainty $\ket{u_k}$ will be found in $\mathcal X_2$.

On the other hand, using different bases on different sides, we get fully uncorrelated outcomes. For example, measuring with $U$ on $\mathcal X_1$, the post-measurement state on $\mathcal X_2$ is some $\ket{u_k}$. Measuring this state in the basis $V$ we can get any $\ket{v_j}$ with equal probability.

In the particular case of qubits, you can use the eigenstates of the Pauli matrices as mutually unbiased bases. If for some reason you want to completely change bases, i.e. change the measurement basis in both spaces at the same time, you just need three MUBs, so that the state give correlated outcomes in $U\otimes U$ but uncorrelated ones in $V\otimes W$, with $W$ the third MUB.

Of course, which bases you want to use will also depend on the initial state. E.g., the example in this other answer works by using the same measurement basis on both sides (the Pauli $X$ basis) because the initial state includes some phases.


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