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.

Definition

For pure bipartite states, $\lvert\Psi\rangle$ 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

The same formula appears in two related ways.

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.

Both readings preserve entanglement because neither introduces a coupling between $A$ and $B$.

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.

Global Unitaries Can Change Entanglement

A general unitary on $\mathcal H_A\otimes\mathcal H_B$ need not factor as $U_A\otimes U_B$. Such a global unitary can change entanglement.

For two qubits, let

$$\lvert+\rangle = \frac{1}{\sqrt2} \bigl( \lvert0\rangle+\lvert1\rangle \bigr).$$

The product state

$$\lvert+\rangle_A\lvert0\rangle_B = \frac{1}{\sqrt2} \bigl( \lvert00\rangle+\lvert10\rangle \bigr)$$

is sent by the controlled-NOT gate, with $A$ as control and $B$ as target, to

$$\operatorname{CNOT}_{A\to B} \lvert+\rangle_A\lvert0\rangle_B = \frac{1}{\sqrt2} \bigl( \lvert00\rangle+\lvert11\rangle \bigr) = \lvert\Phi^+\rangle.$$

The input is product, while the output is a Bell state. Thus CNOT is not a local unitary across the $A$ versus $B$ split.

This distinction is central in circuit language: single-subsystem gates are local for that split; two-subsystem entangling gates are global.

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 bipartite pure states have the same Schmidt coefficients, local unitaries can map the Schmidt basis of one state to the Schmidt basis of the other. Therefore:

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

for finite-dimensional bipartite pure states.

This is the classification statement. The Schmidt Decomposition page is the canonical home for proving the decomposition itself.

Reduced States and Entanglement Measures

For a pure bipartite state, 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. Since those eigenvalues are the squared Schmidt coefficients, any pure-state entanglement measure built only from them is invariant under local unitaries.

For example, the entanglement entropy

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

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. Acting on the second qubit with Pauli operators gives

$$\begin{aligned} (I\otimes I)\lvert\Phi^+\rangle &= \lvert\Phi^+\rangle,\\ (I\otimes Z)\lvert\Phi^+\rangle &= \lvert\Phi^-\rangle,\\ (I\otimes X)\lvert\Phi^+\rangle &= \lvert\Psi^+\rangle,\\ (I\otimes XZ)\lvert\Phi^+\rangle &= -\lvert\Psi^-\rangle. \end{aligned}$$

The final minus sign is a global phase, so it does not change the physical state. This is why all Bell states are equally entangled.

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 $\theta$.

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.

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 Mistakes

• Treating every change of coordinates on the joint Hilbert space as a local basis change.
• Assuming a global unitary preserves entanglement across a fixed subsystem split.
• Thinking a local unitary can change Schmidt coefficients.
• Using the matrix entries of a reduced state instead of its spectrum as the invariant.
• Confusing “same entanglement entropy” with full local-unitary equivalence in higher Schmidt rank. The full list of Schmidt coefficients matters.
• Forgetting that local means local with respect to a specified subsystem decomposition.

Cross-Links

• Product States
• Separable Mixed States
• Entangled States
• Classical Correlation versus Entanglement
• Bell States
• Singlet and Triplet States
• Entanglement Depends on a Decomposition
• Schmidt Decomposition
• Schmidt Rank
• Entanglement Entropy
• Concurrence for Two Qubits
• LOCC Preview
• Reduced Density Operators
• Local Measurement Statistics
• Operators on Composite Systems
• Unitary Operators

References

• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.
• R. Horodecki, P. Horodecki, M. Horodecki, and K. Horodecki, “Quantum entanglement”, Reviews of Modern Physics 81, 865-942, 2009.
• I. Bengtsson and K. Zyczkowski, Geometry of Quantum States: An Introduction to Quantum Entanglement, 2nd ed., Cambridge University Press, 2017.
• J. Preskill, Lecture Notes for Physics 219: Quantum Computation, California Institute of Technology.

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 CNOT can create entanglement even though single-qubit gates cannot.

Solution

Single-qubit gates have the local form $U_A\otimes U_B$ with one factor possibly equal to the identity. They preserve product states and Schmidt coefficients.

CNOT does not factor as $U_A\otimes U_B$. It maps

$$\lvert+\rangle_A\lvert0\rangle_B$$

to

$$\frac{1}{\sqrt2} \bigl( \lvert00\rangle+\lvert11\rangle \bigr),$$

which has Schmidt coefficients $1/\sqrt2,1/\sqrt2$ and is entangled. Therefore CNOT is a global two-qubit unitary across the $A$ versus $B$ split.

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,

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