1 Ket-bra operators

1.1 Kets and bras

Definition 1.1.1

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A ket operator $| \cdot ⟩$ on a Hilbert space $H$ is defined as the linear map $B (H, B (C, H))$ and is given by $x \mapsto (α \mapsto α x)$ .

Definition 1.1.2

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A bra operator $⟨ \cdot |$ on a Hilbert space $H$ is defined as the anti-linear map $H \to B (H, C)$ and is given by $x \mapsto (y \mapsto ⟨ x ∣ y ⟩)$ .

Lemma 1.1.3

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Given $x \in H$ , we get ${⟨ x |}^{*} = | x ⟩$ .

Proof ▶

Corollary 1.1.4

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Given $x, y \in H$ , we get $⟨ x | \circ | y ⟩ (1) = ⟨ x ∣ y ⟩$ .

Proof ▶

Corollary 1.1.5

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Let $A$ be an algebra and a Hilbert space. Then the unit map $η : C \to A$ (which is given by $α \mapsto α 1$ ) is exactly $| 1 ⟩$ .

Proof ▶

Corollary 1.1.6

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The adjoint of the unit map $η : C \to A$ in an algebra and Hilbert space $A$ is $⟨ 1 |$ .

Proof ▶

Lemma 1.1.7

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Let $f \in B (H_{1}, H_{2})$ and $x \in H_{1}$ . Then $f \circ | x ⟩ = | f (x) ⟩$ .

Proof ▶

Lemma 1.1.8

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Let $f \in B (H_{1}, H_{2})$ and $x \in H_{2}$ . Then $⟨ x | \circ f = ⟨ f^{*} (x) |$ .

Proof ▶

1.2 Ket-bras

Definition 1.2.1

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A ket-bra operator $| \cdot ⟩ ⟨ \cdot |$ is defined as the linear map from $E_{2}$ to the anti-linear map $E_{1} \to B (E_{1}, E_{2})$ and is given by

\begin{array}{r} x \mapsto (y \mapsto (u \mapsto ⟨ y ∣ u ⟩ x)) . \end{array}

This is exactly $| \cdot ⟩ ⟨ \cdot | = | \cdot ⟩ \circ ⟨ \cdot |$ .

Let $E_{1}, E_{2}, E_{3}$ be inner product spaces over $C$ . So given $x \in E_{2}$ and $y \in E_{1}$ , we write $| x ⟩ ⟨ y |$ to mean the map $u \mapsto ⟨ y ∣ u ⟩ x$ .

Lemma 1.2.2

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Given $x$ and $y$ in a finite-dimensional Hilbert space $A$ , we get $Tr (| x ⟩ ⟨ y |) = ⟨ y ∣ x ⟩$ .

Proof ▶

Corollary 1.2.3

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Given a linear map $T_{1} \in B (E_{2}, E_{3})$ and elements $x \in E_{2}$ , $y \in E_{1}$ , we get, $T_{1} \circ | x ⟩ ⟨ y | = | T_{1} (x) ⟩ ⟨ y |$

Proof ▶

Corollary 1.2.4

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Given a linear map $T_{2} \in B (E_{3}, E_{1})$ and elements $x \in E_{2}$ , $y \in E_{1}$ , we get $| x ⟩ ⟨ y | \circ T_{2} = | x ⟩ ⟨ T_{2}^{*} (y) |$ .

Proof ▶

Corollary 1.2.5

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Given $x \in E_{2}$ and $y \in E_{1}$ , we get $| x ⟩ {⟨ y |}^{*} = | y ⟩ ⟨ x |$ .

Proof ▶

Lemma 1.2.6

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Given an orthonormal basis $(u_{i})$ of a $C$ -inner product space $E$ , we get $\sum_{i} | u_{i} ⟩ ⟨ u_{i} | = id$ .

Proof ▶

Lemma 1.2.7

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Given a Hilbert space $H$ , we have ${B (H)}^{'} = {α id : α \in C}$ .
In other words, $x \in B (H)$ commutes with all operators $y \in B (H)$ if and only if $x = α id$ for some $α \in C$ .

Proof ▶

Let $x \in B (H)$ . Obviously, if $x = α id$ for some $α \in C$ , then it commutes with every other operator. Now suppose $x$ commutes with every operator in $B (H)$ . So this means $| a ⟩ ⟨ x^{*} (b) | = | a ⟩ ⟨ b | x = x | a ⟩ ⟨ b | = | x (a) ⟩ ⟨ b |$ for all $a, b \in H$ . Suppose there exists some non-zero $a \in H$ , otherwise this is trivial. Then, for any $b \in H$ , we have

x (b) = \frac{{‖ a ‖}^{2}}{{‖ a ‖}^{2}} x (b) = \frac{1}{{‖ a ‖}^{2}} | x (b) ⟩ ⟨ a | (a) = \frac{1}{{‖ a ‖}^{2}} | b ⟩ ⟨ x^{*} (a) | (a) = \frac{⟨ x^{*} (a) ∣ a ⟩}{{‖ a ‖}^{2}} b .

Thus $x = α id$ where $α = ⟨ x^{*} (a) ∣ a ⟩ / {‖ a ‖}^{2}$ .

Proposition 1.2.8

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Let $H$ be a Hilbert space, and let $x, y \in H$ . Then if $| x ⟩ ⟨ x | = | y ⟩ ⟨ y |$ , then there exists some $0 \neq α \in C$ such that $x = α y$ (i.e., they are co-linear).

Proof ▶

Suppose $| x ⟩ ⟨ x | = | y ⟩ ⟨ y |$ . Then it is clear that we get $x = 0$ if and only if $y = 0$ . So we assume $x \neq 0$ (and so $y \neq 0$ ), otherwise this is trivial. Then we have

{‖ x ‖}^{2} x = | x ⟩ ⟨ x | (x) = | y ⟩ ⟨ y | (x) = ⟨ y ∣ x ⟩ y .

And as $x \neq 0$ , we get $x = \frac{⟨ y ∣ x ⟩}{{‖ x ‖}^{2}} y$ . We have $⟨ y ∣ x ⟩ \neq 0$ (otherwise, this would mean $x = 0$ which is a contradiction). Thus we can let $α = ⟨ y ∣ x ⟩ / {‖ x ‖}^{2} \neq 0$ such that $x = α y$ .

Proposition 1.2.9

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Let $H_{1}, H_{2}$ be Hilbert spaces, and let $a, c \in H_{1} ∖ {0}$ and $b, d \in H_{2} ∖ {0}$ . Then if $| a ⟩ ⟨ b | = | c ⟩ ⟨ d |$ , then there exists some $0 \neq α, β \in C$ such that $a = α c$ and $b = α β d$ .

Proof ▶

As $| a ⟩ ⟨ b | = | c ⟩ ⟨ d |$ , we get $⟨ b ∣ b ⟩ a = ⟨ d ∣ b ⟩ c$ (so $a = \frac{⟨ d ∣ b ⟩}{{‖ b ‖}^{2}} c$ ). Taking adjoints of the hypothesis, we get $| b ⟩ ⟨ a | = | d ⟩ ⟨ c |$ , and so

\begin{aligned} b & = \frac{⟨ c ∣ a ⟩}{{‖ a ‖}^{2}} d \\ = \frac{⟨ d ∣ b ⟩ {‖ c ‖}^{2}}{{‖ a ‖}^{2} {‖ b ‖}^{2}} d . \end{aligned}

Clearly, $⟨ d ∣ b ⟩ \neq 0$ (otherwise, we get $a = 0$ ). So then we can let $α = ⟨ d ∣ b ⟩ / {‖ b ‖}^{2}$ and $β = {‖ c ‖}^{2} / {‖ a ‖}^{2}$ .

Lemma 1.2.10

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Given a finite-dimensional inner product space $E$ over $C$ and $T \in B (E)$ , we get
$T$ is positive semi-definite $\Leftrightarrow$ $T = \sum_{i} | v_{i} ⟩ ⟨ v_{i} |$ for some tuple $(v_{i})$ in $E$ .

Proof ▶

$(\Rightarrow)$: Suppose $0 \leq T$ . We use the spectral theorem and let $(v_{i})$ be the eigenbasis of $T$ in $E$ with corresponding eigenvalues $(λ_{i})$ . Note that, as $0 \leq T$ , we also get each $0 \leq λ_{i}$ . So then let each $x_{i} = \sqrt{λ_{i}} u_{i}$ . Then we have $\sum_{i} | x_{i} ⟩ ⟨ x_{i} | = \sum_{i} \sqrt{λ_{i}} \overset{―}{\sqrt{λ_{i}}} | u_{i} ⟩ ⟨ u_{i} | = \sum_{i} λ_{i} | u_{i} ⟩ ⟨ u_{i} | = T$ , where the last equality comes from Corollary 1.2.6.
$(\Leftarrow)$: Suppose we have some tuple $(v_{i})$ in $E$ such that $T = \sum_{i} | v_{i} ⟩ ⟨ v_{i} |$ . Then, for any $x \in E$ , we get $⟨ x ∣ T (x) ⟩ = \sum_{i} ⟨ x ∣ v_{i} ⟩ ⟨ v_{i} ∣ x ⟩ = \sum_{i} {| ⟨ x ∣ v_{i} ⟩ |}^{2} \geq 0$ . Thus $T$ is positive semi-definite.

Given an orthonormal basis $b = (b_{i})$ of a finite-dimensional Hilbert space $H$ , we define $R_{b}$ to be the linear isomorphism $H ≅ C^{\dim H}$ given by $R_{b} (x)_{i} = ⟨ b_{i} ∣ x ⟩$ with its inverse given by $x \mapsto \sum_{i} x_{i} b_{i}$ .

Lemma 1.2.11

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Let $e = (e_{i})$ be an orthonormal basis of a finite-dimensional Hilbert space $H$ . Then $R_{e}^{*} = R_{e}^{- 1}$ .

Proof ▶

Let $x \in H$ and $y \in C^{\dim H}$ . Then we compute,

\begin{aligned} {⟨ x ∣ R_{e}^{*} (y) ⟩}_{H} & = {⟨ R_{e} (x) ∣ y ⟩}_{C^{\dim H}} = \sum_{i} {⟨ R_{e} (x)_{i} ∣ y_{i} ⟩}_{C} = \sum_{i} {⟨ {⟨ e_{i} ∣ x ⟩}_{H} ∣ y_{i} ⟩}_{C} \\ = \sum_{i} {⟨ x ∣ e_{i} ⟩}_{H} y_{i} = \sum_{i} {⟨ x ∣ y_{i} e_{i} ⟩}_{H} = {⟨ x ∣ R_{e}^{- 1} (y) ⟩}_{H} . \end{aligned}

Thus $R_{e}^{*} = R_{e}^{- 1}$ .

Given orthonormal bases $b = (b_{i})$ and $c = (c_{j})$ of Hilbert spaces $H_{1}, H_{2}$ , then we let $M$ denote the identification from $B (H_{1}, H_{2})$ to $M_{\dim H_{2}, \dim H_{1}}$ , which is given by $M_{b, c} (T)_{k p} = ⟨ c_{k} ∣ T (b_{p}) ⟩$ .

Lemma 1.2.12

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Given orthonormal bases $b = (b_{i})$ and $c = (c_{j})$ of finite-dimensional Hilbert spaces $H_{1}, H_{2}$ , and elements $x \in H_{1}$ and $y \in H_{2}$ , we have $M_{c, b} (| x ⟩ ⟨ y |) = R_{b} (x) {R_{c} (y)}^{*}$ .

Proof ▶

For any $i \in [\dim H_{1}], j \in [\dim H_{2}]$ , we compute,

M_{c, b} (| x ⟩ ⟨ y |)_{i j} = ⟨ b_{i} ∣ | x ⟩ ⟨ y | (c_{j}) ⟩ = ⟨ b_{i} ∣ x ⟩ ⟨ y ∣ c_{j} ⟩ = R_{b} (x)_{i} \overset{―}{R_{c} (y)_{j}} = {(R_{b} (x) {R_{c} (y)}^{*})}_{i j} .

Thus $M_{c, b} (| x ⟩ ⟨ y |) = R_{b} (x) {R_{c} (y)}^{*}$ . □