Reitz Representation Theorem

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

We say $$\langle f|$$ is dual’’ to $$|f\rangle$$ and often write $$\langle f| = |f\rangle^\dagger$$, the adjoint of $$f$$.

What is the relation between them?

Constructive Proof. Choose an orthonormal basis $$\{|i\rangle\}$$ for $$\mathcal{V}$$ and let $$\{\langle i|\}$$ be corresponding dual basis $$\langle i|j\rangle = \delta^i_j$$. Note $$(|i\rangle, |j\rangle)=\delta^i_j=\langle i|j\rangle$$. (Then our map associates the basis dual to an orthonomal basis to the dual basis).

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

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

• $$a_i = a^{i*}$$
• $$\langle a|b\rangle = a_ib^i = (|a\rangle^\dagger, |b\rangle)$$
• $$|a\rangle = (a^1 a^2 \cdots)^T$$ - Column vector
• $$\langle a| = (a^{1*} a^{2*} \cdots)$$ - Row vector
• $$|a\rangle^{T*} = |a\rangle^\dagger = \langle a|$$ (Transpose notation is the tilde over the expression in PH562)

Recall the notation $$A|a\rangle=|Aa\rangle= = \sum_i a^i A|i\rangle = \sum_i\sum_j A^i_j a^j |i\rangle$$.

Definition

The adjoint $$A^\dagger$$ of an operator $$A$$ is the operator which satisfies $$\langle a|A^\dagger|b\rangle = \langle b|A|a\rangle^* = \langle b|Aa\rangle^*\langle Aa|b\rangle, \forall |a\rangle,|b\rangle\in\mathcal{V}$$.

Implications

If we define $$\langle a|B$$ by $$(\langle a|B)|c\rangle \equiv \langle a|B| c\rangle$$.

Then $$\langle Aa| = (A|a\rangle)^\dagger$$. Thus, $$\langle a|A = A^\dagger |a\rangle$$.

So, $$(AB)^\dagger = B^\dagger A^\dagger$$. Proof. $$\langle b|B^\dagger A^\dagger|a\rangle = (\langle b|B^\dagger)(A^\dagger|a\rangle) = \langle a|AB|b\rangle^* = \langle b|(AB)^\dagger|a\rangle$$.

• Matrix Represenatation of Operators

Recall we defined $$A^i_j = \langle i|A|j\rangle$$.

Then, $$(A^\dagger)^i_j = \langle i|A^\dagger|j\rangle = \langle j|A|i\rangle^* = (A^j_i)^*$$

Riesz Lemma

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

Created: 2023-06-25 Sun 02:25

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