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 \(|f\rangle=f^i|i\rangle \Leftrightarrow \langle f|=f^{i*}\langle i|\).


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


  • \(a_i = a^{i*}\)
  • \(\langle a|b\rangle = a_ib^i = (|a\rangle^\dagger, |b\rangle)\)
  • \(|a\rangle = (a^1 a^2 \cdots)^T\) - 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)

Adjoint of an Operator

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


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


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).

Author: Christian Cunningham

Created: 2023-06-25 Sun 02:25