PhyslibAlpha.ProbabilisticTheory.HilbertSpace.TraceClass.Basic

A bounded linear operator $T$ on a complex Hilbert space $H$ is said to be **trace class** if there exists a Hilbert basis $\{e_i\}_{i \in I}$ of $H$ such that the series $$\sum_{i \in I} \langle e_i, |T| e_i \rangle$$ converges, where $|T| = \sqrt{T^* T}$ denotes the absolute value of $T$ as defined by the continuous functional calculus.

Let $H$ be a complex Hilbert space and let $\{e_i\}_{i \in I}$ be a Hilbert basis for $H$. For any vector $y \in H$, the series of the squared magnitudes of the inner products of $y$ with the basis vectors converges to the squared norm of $y$, that is: $$\sum_{i \in I} |\langle e_i, y \rangle|^2 = \|y\|^2.$$

Let $H$ be a complex Hilbert space and $S: H \to H$ be a bounded self-adjoint operator. Let $\{e_i\}_{i \in I}$ and $\{f_j\}_{j \in J}$ be two Hilbert bases for $H$. If the series $\sum_{i \in I} \|S e_i\|^2$ is summable, then the series $\sum_{j \in J} \|S f_j\|^2$ is also summable and converges to the same value, i.e., $$\sum_{j \in J} \|S f_j\|^2 = \sum_{i \in I} \|S e_i\|^2.$$

Let $H$ be a complex Hilbert space and let $T: H \to H$ be a trace-class bounded linear operator. For any Hilbert basis $\{e_i\}_{i \in I}$ of $H$, the series of the real parts of the diagonal entries of the absolute value of $T$ is summable, i.e., the sum $$\sum_{i \in I} \text{Re} \langle e_i, |T| e_i \rangle$$ converges, where $|T| = \sqrt{T^* T}$ is the absolute value of $T$ defined via the continuous functional calculus.

Let $H$ be a complex Hilbert space and $T: H \to H$ be a bounded linear operator. $T$ is a trace-class operator if and only if for every Hilbert basis $\{e_i\}_{i \in I}$ of $H$, the series of diagonal inner products of its absolute value $$\sum_{i \in I} \langle e_i, |T| e_i \rangle$$ is summable, where $|T| = \sqrt{T^* T}$ is the absolute value of $T$ defined via the continuous functional calculus.

Let $H$ be a finite-dimensional complex Hilbert space. Every bounded linear operator $T: H \to H$ is trace-class.

Let $H$ be a finite-dimensional complex Hilbert space and $T: H \to H$ be a bounded linear operator. For any Hilbert basis $\{b_i\}_{i \in w}$ of $H$, the trace of $T$ (viewed as a linear map) is equal to the sum of the diagonal inner products: $$\operatorname{Tr}(T) = \sum_{i \in w} \langle b_i, T b_i \rangle$$ where $\langle \cdot, \cdot \rangle$ denotes the complex inner product on $H$.

Let $H$ be a complex Hilbert space and $T: H \to H$ be a bounded linear operator. For a trace-class operator $T$ (with a witness $h$ of `IsTraceClass T`), the **trace norm** $\|T\|_1$ is defined as the sum of the diagonal inner products of its absolute value $|T|$ with respect to a Hilbert basis $\{e_i\}_{i \in I}$: $$\|T\|_1 = \sum_{i \in I} \langle e_i, |T| e_i \rangle$$ where $|T| = \sqrt{T^* T}$ is the absolute value of $T$ defined via the continuous functional calculus. The value is a real number and is independent of the specific Hilbert basis chosen.

Let $H$ be a complex Hilbert space and $T: H \to H$ be a bounded linear operator. If $h_1$ and $h_2$ are two proofs (witnesses) that $T$ is a trace-class operator, then the trace norm of $T$ evaluated with respect to $h_1$ is equal to the trace norm of $T$ evaluated with respect to $h_2$: $$\|T\|_{1, h_1} = \|T\|_{1, h_2}$$ In other words, the value of the trace norm $\|T\|_1$ is independent of the specific proof used to establish that the operator is trace-class.

Let $H$ be a complex Hilbert space and $T: H \to H$ be a bounded linear operator. Given a proof $h$ that $T$ is a trace-class operator (which implies there exists a Hilbert basis $\{e_i\}_{i \in I}$ of $H$ such that $\sum_{i \in I} \langle e_i, |T| e_i \rangle$ converges), the **trace** of $T$, denoted $\operatorname{Tr}(T)$, is defined as the sum of its diagonal entries: $$\operatorname{Tr}(T) = \sum_{i \in I} \langle e_i, T e_i \rangle$$ where $\{e_i\}_{i \in I}$ is the specific Hilbert basis provided by the witness $h$.

Let $H$ be a complex Hilbert space and $T: H \to H$ be a bounded linear operator. If $h_1$ and $h_2$ are two proofs that $T$ is a trace-class operator, then the trace of $T$ evaluated with respect to $h_1$ is equal to the trace of $T$ evaluated with respect to $h_2$: $$\operatorname{Tr}(T)_{h_1} = \operatorname{Tr}(T)_{h_2}.$$ This result establishes that for a trace-class operator $T$, the value of the trace is independent of the specific witness (and thus the underlying Hilbert basis) used to define it.

Let $H$ be a complex Hilbert space and $T: H \to H$ be a bounded linear operator. If $T$ is a trace-class operator, then for any Hilbert basis $\{b_i\}_{i \in I}$ of $H$, the trace norm $\|T\|_1$ is equal to the sum of the real parts of the diagonal inner products of its absolute value $|T| = \sqrt{T^* T}$ with respect to that basis: $$\|T\|_1 = \sum_{i \in I} \operatorname{Re}(\langle b_i, |T| b_i \rangle)$$ where $|T|$ is defined via the continuous functional calculus.

Let $H$ be a complex Hilbert space and $T: H \to H$ be a bounded linear operator. If $T$ is a positive operator (denoted as $0 \leq T$), then for any vector $x \in H$, the real part of the inner product $\langle x, Tx \rangle$ is non-negative: $$\operatorname{Re}(\langle x, Tx \rangle) \ge 0.$$

Let $H$ be a complex Hilbert space and $T: H \to H$ be a bounded linear operator. If $T$ is a trace-class operator, then its trace norm $\|T\|_1$ is non-negative: $$\|T\|_1 \ge 0$$

Let $H$ be a complex Hilbert space and let $P, Q: H \to H$ be bounded linear operators. If $P \leq Q$ in the operator order, then for any vector $x \in H$, the real part of the inner product of $x$ and $Px$ is less than or equal to the real part of the inner product of $x$ and $Qx$: $$\text{Re}\langle x, Px \rangle \leq \text{Re}\langle x, Qx \rangle$$

Let $H$ be a complex Hilbert space and $T : H \to H$ be a bounded self-adjoint operator. If $T$ is trace-class, then its positive part $T^+$, defined via the continuous functional calculus, is also a trace-class operator.

Let $H$ be a complex Hilbert space and $T: H \to H$ be a bounded self-adjoint operator. If $T$ is trace-class, then its negative part $T^-$, defined via the continuous functional calculus, is also a trace-class operator.

Let $H$ be a complex Hilbert space and $T: H \to H$ be a bounded linear operator. If $T$ is a positive operator (i.e., $T \geq 0$) and $T$ is trace-class, then for any Hilbert basis $\{b_i\}_{i \in I}$ of $H$, the family of diagonal inner products $(\langle b_i, T b_i \rangle)_{i \in I}$ is summable.

Let $H$ be a complex Hilbert space and $T: H \to H$ be a bounded linear operator. If $T$ is a positive operator ($T \ge 0$) and is trace-class, then for any Hilbert basis $\{b_i\}_{i \in I}$ of $H$, the trace of $T$, denoted $\operatorname{Tr}(T)$, is equal to the sum of the diagonal inner products: $$\operatorname{Tr}(T) = \sum_{i \in I} \langle b_i, T b_i \rangle$$

Let $H$ be a complex Hilbert space and $T: H \to H$ be a bounded linear operator. If $T$ is self-adjoint and trace-class, then for any Hilbert basis $\{b_i\}_{i \in I}$ of $H$, the trace of $T$ is equal to the sum of the diagonal inner products with respect to that basis: $$\operatorname{Tr}(T) = \sum_{i \in I} \langle b_i, T b_i \rangle$$ where $\operatorname{Tr}(T)$ denotes the trace of $T$.

Let $H$ be a complex Hilbert space. A bounded linear operator $p: H \to H$ is said to have **finite multiplicity** if it is a trace-class operator. For a projection, this is equivalent to its range being finite-dimensional.

Let $H$ be a complex Hilbert space and $p: H \to H$ be a bounded linear operator. If $p$ has finite multiplicity, then its range is finite-dimensional over $\mathbb{C}$. (Note: In this context, an operator has finite multiplicity if it is a trace-class projection).

Let $H$ be a complex Hilbert space and $p: H \to H$ be a bounded linear operator. If $p$ has finite multiplicity, then its trace $\operatorname{Tr}(p)$ is equal to the dimension of its range as a complex vector space, denoted by $\dim_{\mathbb{C}}(\operatorname{ran} p)$.

Let $H$ be a finite-dimensional complex Hilbert space. For any bounded linear operator $p : H \to H$ that has finite multiplicity, its trace $\operatorname{Tr}(p)$ is equal to the complex dimension of its range: $$\operatorname{Tr}(p) = \dim_{\mathbb{C}}(\operatorname{range} p)$$