Local Unitary Equivalence

Two states are locally unitarily equivalent if one can be transformed into the other by unitary transformations applied independently to the named subsystems. For a bipartite split

$$\mathcal H_{AB} = \mathcal H_A\otimes\mathcal H_B,$$

a local unitary has the form

$$U_A\otimes U_B,$$

where $U_A$ acts only on subsystem $A$ and $U_B$ acts only on subsystem $B$.

Local unitaries can change bases, phases, and local measurement axes. They cannot create entanglement across the $A$ versus $B$ split, and they cannot destroy it. They preserve the entanglement type of a state across the same subsystem decomposition.

Helpful background. Schmidt decomposition supplies orthonormal pairs and nonnegative coefficients; unitary operators explains preservation of inner products.

Local unitary transformations

For normalized pure states on the same pair of finite-dimensional subsystem spaces, $\lvert\Psi\rangle$ and $\lvert\Phi\rangle$ are locally unitarily equivalent if there exist unitaries $U_A$ and $U_B$ such that

$$\lvert\Phi\rangle = (U_A\otimes U_B)\lvert\Psi\rangle.$$

For density operators, the corresponding relation is

$$\rho_{AB}' = (U_A\otimes U_B) \rho_{AB} (U_A^\dagger\otimes U_B^\dagger).$$

This relation is an equivalence relation: every state is equivalent to itself, the transformation can be inverted by $U_A^\dagger\otimes U_B^\dagger$, and successive local unitaries compose to another local unitary.

Passive and Active Readings

An active operation and a passive basis change both use unitary matrices, but their coordinate formulas must be distinguished.

A passive local basis change rewrites the same physical state using different local bases. For example, replacing the $A$-basis by another orthonormal basis is represented by a unitary change of coordinates on $\mathcal H_A$.

An active local unitary is a physical operation, such as applying a local spin rotation or a single-qubit gate to one subsystem. If the operation factors as $U_A\otimes U_B$, it is still local with respect to the $A$ versus $B$ split.

For example, if new basis vectors satisfy $|i'\rangle=\sum_k|k\rangle(V_A)_{ki}$ and similarly for $B$, a coefficient matrix changes passively as

$$C'=V_A^\dagger C V_B^*.$$

An active operation in a fixed basis instead gives $C'=U_A C U_B^{\mathsf T}$. Both preserve singular values and therefore bipartite pure-state entanglement.

Product States Stay Product

If

$$\lvert\Psi\rangle_{AB} = \lvert\psi\rangle_A\otimes\lvert\phi\rangle_B,$$

then

$$\begin{aligned} (U_A\otimes U_B)\lvert\Psi\rangle_{AB} &= (U_A\otimes U_B) (\lvert\psi\rangle_A\otimes\lvert\phi\rangle_B) \\ &= (U_A\lvert\psi\rangle_A) \otimes (U_B\lvert\phi\rangle_B). \end{aligned}$$

The result is again a product state. Therefore a local unitary cannot turn a product pure state into an entangled pure state.

The same logic works for separable mixed states. If

$$\rho_{AB} = \sum_k p_k\, \rho_A^{(k)}\otimes\rho_B^{(k)},$$

then after a local unitary

$$\rho_{AB}' = \sum_k p_k\, \bigl(U_A\rho_A^{(k)}U_A^\dagger\bigr) \otimes \bigl(U_B\rho_B^{(k)}U_B^\dagger\bigr),$$

which is still separable. Since the inverse transformation is also local, a local unitary cannot convert an entangled mixed state into a separable one either.

Joint unitaries need not preserve entanglement

CNOT maps $|+\rangle_A|0\rangle_B$ to $|\Phi^+\rangle$, so it cannot factor as $U_A\otimes U_B$. The Bell preparation circuit gives the step-by-step calculation.

The converse needs care: a joint unitary can fail to factor without entangling any product input. For equal-dimensional factors, SWAP sends $|a\rangle|b\rangle$ to $|b\rangle|a\rangle$. It exchanges subsystem information and is not a product of fixed local unitaries, but it preserves the Schmidt coefficients. Thus local unitaries preserve entanglement, while preservation alone does not characterize locality.

Schmidt Coefficients as Invariants

For finite-dimensional bipartite pure states, local unitary equivalence is completely classified by the Schmidt coefficients.

Suppose

$$\lvert\Psi\rangle = \sum_{r=1}^R s_r\, \lvert u_r\rangle_A\lvert v_r\rangle_B$$

is a Schmidt decomposition. Applying a local unitary gives

$$(U_A\otimes U_B)\lvert\Psi\rangle = \sum_{r=1}^R s_r\, (U_A\lvert u_r\rangle_A) (U_B\lvert v_r\rangle_B).$$

Because unitaries preserve inner products, the transformed vectors remain orthonormal sets. The coefficients $s_r$ are unchanged.

Conversely, if two pure states on the same finite-dimensional factor spaces have the same Schmidt coefficients, including multiplicities, map each Schmidt vector to its partner. Complete both Schmidt sets to orthonormal bases to extend these maps to local unitaries on the full spaces. This works also when some coefficients vanish or are degenerate. Therefore:

$$\text{same Schmidt coefficients} \quad\Longleftrightarrow\quad \text{same local-unitary class}$$

for finite-dimensional bipartite pure states.

The Schmidt decomposition derives the decomposition itself. Its full coefficient list classifies pure-state local unitary equivalence; one scalar entropy generally does not.

Reduced States and Entanglement Measures

For any bipartite density operator, the reduced density operator transforms as

$$\rho_A' = \operatorname{Tr}_B \bigl[ (U_A\otimes U_B) \rho_{AB} (U_A^\dagger\otimes U_B^\dagger) \bigr] = U_A\rho_AU_A^\dagger.$$

Thus $\rho_A$ and $\rho_A'$ have the same eigenvalues. For a pure joint state, those eigenvalues are the squared Schmidt coefficients, so any pure-state entanglement measure built only from them is invariant under local unitaries.

For example, the entanglement entropy

$$S(\rho_A) = -\operatorname{Tr}(\rho_A\log_2\rho_A)$$

is unchanged by local unitaries, because unitary conjugation preserves the spectrum of $\rho_A$.

The same invariance principle is required of entanglement measures more generally: changing local bases or applying reversible local operations should not change the amount of entanglement assigned to the state.

Two-Qubit Examples

All four Bell states are locally unitarily equivalent: $I\otimes Z$, $I\otimes X$, and $I\otimes XZ$ map $|\Phi^+\rangle$ to $|\Phi^-\rangle$, $|\Psi^+\rangle$, and $|\Psi^-\rangle$, respectively. The Bell-state identities fix the operator order and signs.

For a general two-qubit pure state, local unitaries can put the state into Schmidt form

$$\lvert\Psi_\theta\rangle = \cos\theta\,\lvert00\rangle +\sin\theta\,\lvert11\rangle, \qquad 0\le \theta\le \frac{\pi}{4}.$$

The angle $\theta$ encodes the Schmidt coefficients. The endpoints have distinct meanings:

$$\theta=0 \quad\Longrightarrow\quad \text{product state},$$

while

$$\theta=\frac{\pi}{4} \quad\Longrightarrow\quad \text{maximally entangled two-qubit state}.$$

Local unitaries can change which local basis states appear as $\lvert0\rangle$ and $\lvert1\rangle$, but they cannot change $\theta$.

As a worked example, take

$$|\psi\rangle=\frac{\sqrt3}{2}|0\rangle|+\rangle +\frac{i}{2}|1\rangle|-\rangle.$$

Apply $H$ to $B$, sending $|+\rangle\mapsto|0\rangle$ and $|-\rangle\mapsto|1\rangle$. Then apply $\operatorname{diag}(1,-i)$ to $A$ to remove the relative phase. The result is $\sqrt3|00\rangle/2+|11\rangle/2$, with $\theta=\pi/6$. Its squared Schmidt coefficients, $3/4$ and $1/4$, were unchanged throughout.

What Local Unitary Equivalence Is Not

Local unitary equivalence is not the same as equality. The states

$$\lvert00\rangle \qquad \text{and} \qquad \lvert+\rangle\lvert1\rangle$$

are different vectors, but they are locally unitarily equivalent because each is a product state and one can rotate the local factors independently.

Local unitary equivalence is also not the same as having the same reduced density matrices in a chosen basis. Local unitaries can rotate the eigenvectors of reduced states while preserving their spectra. For pure bipartite states, the invariant data are the Schmidt coefficients, not the matrix entries of $\rho_A$ in one arbitrary basis.

For mixed states, reduced spectra are necessary invariants but not a complete test. A Bell projector and $I_4/4$ both have marginals $I/2$, yet a unitary cannot map the globally pure state to the globally mixed one.

Finally, local unitary equivalence is not the same as convertibility by irreversible local operations and classical communication. Local unitaries are reversible and preserve entanglement exactly; more general local protocols can lose information or probabilistically transform states.

Common pitfalls

Equal entanglement entropy need not imply local equivalence. Beyond two nonzero Schmidt coefficients, different coefficient lists can have the same entropy. Compare the full spectrum.

Active and passive matrices use different adjoints. State which basis is held fixed before applying a transformation to the coefficient matrix.

General local operations can destroy entanglement. Reversibility is essential to local unitary invariance. Measurements and noise are different operations.

References

• Ingemar Bengtsson and Karol Życzkowski, Geometry of Quantum States: An Introduction to Quantum Entanglement, 2nd ed., Cambridge University Press, 2017.
• Michael A. Nielsen and Isaac L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, section 2.5.
• John Watrous, The Theory of Quantum Information, Cambridge University Press, 2018, chapters 1 and 6. Author-hosted text.

Exercises

1. Let $\lvert\Psi\rangle=\lvert\psi\rangle_A\lvert\phi\rangle_B$. Show that $(U_A\otimes U_B)\lvert\Psi\rangle$ is product.

Solution

Use the defining action of a tensor-product operator:

$$(U_A\otimes U_B) (\lvert\psi\rangle_A\otimes\lvert\phi\rangle_B) = (U_A\lvert\psi\rangle_A) \otimes (U_B\lvert\phi\rangle_B).$$

The result is a tensor product of a state in $\mathcal H_A$ and a state in $\mathcal H_B$, so it is product.

2. Show that a local unitary does not change the Schmidt coefficients of a pure bipartite state.

Solution

Start from the Schmidt decomposition

$$\lvert\Psi\rangle = \sum_r s_r \lvert u_r\rangle_A\lvert v_r\rangle_B.$$

Then

$$(U_A\otimes U_B)\lvert\Psi\rangle = \sum_r s_r (U_A\lvert u_r\rangle_A) (U_B\lvert v_r\rangle_B).$$

Since $U_A$ and $U_B$ preserve inner products, the transformed local vectors remain orthonormal. This is again a Schmidt decomposition with the same coefficients $s_r$.

3. Find a local unitary that maps $\lvert\Phi^+\rangle$ to $\lvert\Psi^+\rangle$.

Solution

Act with the Pauli $X$ operator on the second qubit:

$$(I\otimes X) \frac{1}{\sqrt2} \bigl( \lvert00\rangle+\lvert11\rangle \bigr) = \frac{1}{\sqrt2} \bigl( \lvert01\rangle+\lvert10\rangle \bigr) = \lvert\Psi^+\rangle.$$

Thus $I\otimes X$ is one such local unitary.

4. Explain why SWAP on two qubits cannot equal a fixed product $U_A\otimes U_B$, even though it preserves every product input.

Solution

Hold the first input fixed at $|0\rangle$. SWAP sends $|00\rangle$ to $|00\rangle$ but $|01\rangle$ to $|10\rangle$, so the first output changes when only the second input changes. Under fixed $U_A\otimes U_B$, the first output would always be $U_A|0\rangle$. This contradiction proves SWAP is not a product of local unitaries.

5. Decide whether the states

$$\lvert\psi\rangle = \sqrt{\frac34}\,\lvert00\rangle +\frac12\lvert11\rangle$$

and

$$\lvert\phi\rangle = \sqrt{\frac34}\,\lvert01\rangle +\frac12\lvert10\rangle$$

are locally unitarily equivalent.

Solution

They have the same Schmidt coefficients, $\sqrt{3/4}$ and $1/2$, so they must be locally unitarily equivalent. Explicitly,

$$(I\otimes X)\lvert\psi\rangle = \sqrt{\frac34}\,\lvert01\rangle +\frac12\lvert10\rangle = \lvert\phi\rangle.$$