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Consider \(O = \sigma _{z}\), the Pauli-Z operator. We show that
Given the identity \(\mathbb {I} \in \mathcal{U}(2)\),
Given the identity \(\mathbb {I} \in \mathcal{U}(2)\),
We prove this statement now using the abstract Schur-Weyl duality.
Consider the set of \(\mathcal{U}(2)^{\otimes k}\) defined over \(\mathbb {C}\). We show that
An operator \(X \in \mathcal{L}(\mathcal{X})\) is a diagonal operator if \(X_{a, b} = 0\) for all \(a, b \in [d]\) where \(a \neq b\). For a given vector \(u \in \mathcal{X}\), we can define a diagonal operator \(\text{Diag}(u) \in \mathcal{L}(\mathcal{X})\) to denote the diagonal operator whose diagonal entries are given by the entries of \(u\). That is, we have
Let \(G\) be a group of order \(N\) and \(R\) a representation of the group. Define the group average to be
The operator-vector correspondence is a bijective mapping between the set of linear operators \(\mathcal{L}(\mathcal{X})\) and the tensor product space \(\mathcal{X} \otimes \mathcal{X}\) defined as
One relevant identity we use is that for \(A_0, A_1, B \in \mathcal{L}(\mathcal{X})\),
Let \(\mu _{2}\) denote the Haar measure on a single qubit, represented with vectors in \(\mathbb {C}^{2}\). Define the single-qubit Haar measure in terms of the parameters given in Remark 1.4.7, where
is the unnormalized Haar measure. The normalized computation of moments using the Haar measure is given by
where the factor of \(\frac{1}{8\pi ^2}\) can be found by computing the integral and setting \(UOU^{\dagger } =1\).
The standard computational basis of linear operators \(\{ E_{i, j}\} _{i, j \in [d]}\) is defined as the set of \(d^2\) operators in \(\mathcal{L}(\mathcal{X})\) such that
for all \(i, j \in [d]\), where \(\{ |i\rangle \} _{i \in [d]}\) is the standard computational basis of \(\mathcal{X}\).
The symmetric subspace of linear operators \(\mathcal{L}(\mathcal{X})^{\mathbin {\bigcirc \mkern -15mu\vee }k}\) is the set of all operators \(X \in \mathcal{L}(\mathcal{X}^{\otimes k})\) that remain unchanged (invariant) under the permutation \(W_{\pi }\). That is,
Let \(\mathcal{X}\) be a complex Euclidean space, let \(\mathcal{V} \subseteq \mathcal{X}\) be a subspace, and let \(A \in \mathcal{L}(\mathcal{X})\) be an operator. The following statements are equivalent:
It holds that both \(A\mathcal{V} \subseteq \mathcal{V}\) and \(A^{*}\mathcal{V} \subseteq \mathcal{V}\).
It holds that \([A, \Pi _{\mathcal{V}}] = 0\).
Let \(d \in \mathbb {N}\) and \(k = 1\). Then
Let \(v, w \in \mathbb {C}^2\) be unit vectors (i.e., \(|v| = |w| = 1\)). Then there exists a unitary matrix \(U \in U(2)\) such that \(Uv = w\).
Let \(\mathcal{X}\) be a complex Euclidean space and \(k\) be a positive integer. Then the following statements are equivalent:
\(X \in \mathcal{L}(\mathcal{X})^{\mathbin {\bigcirc \mkern -15mu\vee }k}\).
For \(V \in U(\mathcal{X}^{\otimes k} \otimes \mathcal{X}^{\otimes k}, (\mathcal{X} \otimes \mathcal{X})^{\otimes k})\) being the isometry defined by
\begin{equation} V \text{vec}(Y_1 \otimes \dots \otimes Y_k) = \text{vec}(Y_1) \otimes \dots \text{vec}(Y_n) \end{equation}14holding for all \(Y_1, \dots , Y_k \in \mathcal{L}(\mathcal{X})\), one has that
\begin{equation} V \text{vec}(X) \in (\mathcal{X} \otimes \mathcal{X})^{\mathbin {\bigcirc \mkern -15mu\vee }k}. \end{equation}15\(X \in \text{Span}\{ Y^{\otimes k} : Y \in \mathcal{L}(\mathcal{X})\} \)
For any single-qubit quantum state \(\rho \), we can write it in terms of its Bloch vector,
where \(r = (r_x, r_y, r_z) \in \mathbb {R}^3\) is the coordinate of the quantum state on the Bloch sphere and \(\sigma = (\sigma _x, \sigma _y, \sigma _z)\) is a three-tuple of the Pauli operators.
Let \(\mathcal{A}\) be a self-adjoint, unital subalgebra of \(\mathcal{L}(\mathcal{X})\). Then the double commutant of \(\mathcal{A}\) is equal to \(\mathcal{A}\), i.e.,
Let \(\mathcal{X}\) be a complex Euclidean space and \(k\) be a positive integer. Then the symmetric subspace of linear operators is equivalent to the span of the unitary tensor products, i.e.,
Let \(\mathcal{X}\) be a complex Euclidean space, \(k\) be a positive integer, and \(X \in \mathcal{L}(\mathcal{X}^{\otimes k})\) be an operator. Then the set of all commutators of the unitary group is equivalent to the span of the permutation operators, i.e.,
Let \(k\) be a positive integer, and let \(X \in \mathbb {C}^[d]\). For any set \(\mathcal{A} \subseteq \mathbb {C}\) satisfying \(|\mathcal{A}| \geq k + 1\), it holds that
For an arbitrary single-qubit state \(\rho \),
the maximally mixed state (since it is proportional to the identity matrix).