QuantumInfo.Finite.Unitary

Let $A$ be a Hermitian matrix and $U$ be a unitary matrix. The trace of the unitary conjugation of $A$ by $U$ is equal to the trace of $A$. That is, $\text{tr}(U A U^\dagger) = \text{tr}(A)$.

Let $A$ and $B$ be Hermitian matrices and let $U$ be a unitary matrix. Then the conjugation of $A$ by $U$, defined as $UAU^\dagger$, is less than or equal to the conjugation of $B$ by $U$, $UBU^\dagger$, if and only if $A \le B$ in the Loewner order.

Let $A$ and $B$ be Hermitian matrices and let $U$ be a unitary matrix. The inner product of the matrices transformed by the unitary conjugation $U A U^\dagger$ and $U B U^\dagger$ is equal to the inner product of the original matrices $A$ and $B$. That is, $\langle U A U^\dagger, U B U^\dagger \rangle = \langle A, B \rangle$.

Let $A$ be a Hermitian matrix and $U$ be a unitary matrix. The eigenvalues of the matrix formed by conjugating $A$ with $U$, denoted as $U A U^\dagger$, are equal to the eigenvalues of the original matrix $A$.

Given a mixed state $\rho$ represented as a density matrix in $\text{MState } d$ and a unitary matrix $U$ from the unitary group $\mathcal{U}(d)$, $U \rho U^\dagger$ is the mixed state obtained by conjugating the underlying matrix of $\rho$ by $U$. This transformation preserves the non-negativity and the trace of the density matrix.

For a quantum mixed state $\rho$ and a unitary operator $U$ (belonging to the unitary group $\mathbf{U}(d)$), the notation $U \triangleleft \rho$ represents the action of the unitary on the mixed state by conjugation. Mathematically, this corresponds to the transformation $\rho \mapsto U \rho U^\dagger$, where $U^\dagger$ is the adjoint of $U$.

Let $\rho$ be a mixed quantum state of dimension $d$ and let $U \in \mathbb{U}[d]$ be a unitary matrix. Let $\rho' = U \rho U^\dagger$ be the state obtained by the unitary conjugation of $\rho$ by $U$. Then the eigenvalue spectrum of $\rho'$ is equal to the eigenvalue spectrum of $\rho$. Here, the spectrum is represented as a probability distribution of eigenvalues, which are canonically sorted.

For any two quantum mixed states $\rho$ and $\sigma$ in a Hilbert space of dimension $d$, and for any unitary operator $U \in \mathcal{U}(d)$, the inner product of the states after undergoing a unitary transformation $U \rho U^\dagger$ and $U \sigma U^\dagger$ is equal to the inner product of the original states, i.e., $\langle U \rho U^\dagger, U \sigma U^\dagger \rangle_{\text{Prob}} = \langle \rho, \sigma \rangle_{\text{Prob}}$. Here, the inner product $\langle \cdot, \cdot \rangle_{\text{Prob}}$ is defined as the Hilbert-Schmidt inner product of the underlying density matrices.

Let $d$ be a finite dimension and $|\psi\rangle, |\phi\rangle, |f\rangle$ be quantum state vectors (kets) in a $d$-dimensional Hilbert space. Let $\rho_\psi = |\psi\rangle\langle\psi|$, $\rho_\phi = |\phi\rangle\langle\phi|$, and $\rho_f = |f\rangle\langle f|$ be the corresponding pure states. Suppose there exists a unitary operator $U$ acting on the tensor product space of dimension $d \times d$ such that it perfectly clones both $|\psi\rangle$ and $|\phi\rangle$ using the fiducial state $|f\rangle$, specifically: $$U (\rho_\psi \otimes \rho_f) U^\dagger = \rho_\psi \otimes \rho_\psi$$ $$U (\rho_\phi \otimes \rho_f) U^\dagger = \rho_\phi \otimes \rho_\phi$$ If the states $\rho_\psi$ and $\rho_\phi$ are not identical (their Hilbert-Schmidt inner product $\langle \rho_\psi, \rho_\phi \rangle_{\text{Prob}} < 1$), then they must be orthogonal, i.e., $\langle \rho_\psi, \rho_\phi \rangle_{\text{Prob}} = 0$.