Reitz Representation Theorem

Every bounded element of $V^{*}$ can be represented in this way. (This map is an isomorphism between the space and the dual space)

We say $⟨ f |$ is ``dual’’ to $| f ⟩$ and often write $⟨ f | = | f ⟩^{†}$ , the adjoint of $f$ .

What is the relation between them?

Constructive Proof. Choose an orthonormal basis ${| i ⟩}$ for $V$ and let ${⟨ i |}$ be corresponding dual basis $⟨ i | j ⟩ = δ_{j}^{i}$ . Note $(| i ⟩, | j ⟩) = δ_{j}^{i} = ⟨ i | j ⟩$ . (Then our map associates the basis dual to an orthonomal basis to the dual basis).

Every $⟨ f | \in V^{*}$ can be written as a linear combination of the dual basis and that $⟨ f | a ⟩ = f_{i} a^{i}$ . Write $(| f ⟩, | a ⟩) = f^{i *} a^{i}$ . Thus, $f_{i} = f^{i *}$ . Thus $| f ⟩ = f^{i} | i ⟩ \Leftrightarrow ⟨ f | = f^{i *} ⟨ i |$ .

Note

From now on, we will (almost) always use this equivalence to write the inner product in the form you’re accustomed to from QM.

Representations

$a_{i} = a^{i *}$
$⟨ a | b ⟩ = a_{i} b^{i} = (| a ⟩^{†}, | b ⟩)$
$| a ⟩ = (a^{1} a^{2} \dots)^{T}$ - Column vector
$⟨ a | = (a^{1 *} a^{2 *} \dots)$ - Row vector
$| a ⟩^{T *} = | a ⟩^{†} = ⟨ a |$ (Transpose notation is the tilde over the expression in PH562)

Adjoint of an Operator

Recall the notation $A | a ⟩ = | A a ⟩ == \sum_{i} a^{i} A | i ⟩ = \sum_{i} \sum_{j} A_{j}^{i} a^{j} | i ⟩$ .

Definition

The adjoint $A^{†}$ of an operator $A$ is the operator which satisfies $⟨ a | A^{†} | b ⟩ = ⟨ b | A | a ⟩^{*} = ⟨ b | A a ⟩^{*} ⟨ A a | b ⟩, \forall | a ⟩, | b ⟩ \in V$ .

Implications

If we define $⟨ a | B$ by $(⟨ a | B) | c ⟩ \equiv ⟨ a | B | c ⟩$ .

Then $⟨ A a | = (A | a ⟩)^{†}$ . Thus, $⟨ a | A = A^{†} | a ⟩$ .

So, $(A B)^{†} = B^{†} A^{†}$ . Proof. $⟨ b | B^{†} A^{†} | a ⟩ = (⟨ b | B^{†}) (A^{†} | a ⟩) = ⟨ a | A B | b ⟩^{*} = ⟨ b | (A B)^{†} | a ⟩$ .

Matrix Represenatation of Operators

Recall we defined $A_{j}^{i} = ⟨ i | A | j ⟩$ .

Then, $(A^{†})_{j}^{i} = ⟨ i | A^{†} | j ⟩ = ⟨ j | A | i ⟩^{*} = (A_{i}^{j})^{*}$

Riesz Lemma

$V$ is isomorphic to $V^{*}$ ( $V$ is an inner product space).