Preliminary remarks and notation

Let us introduce some basic notation here. The quantity $\vert\psi\rangle$ refers to the ket of the wavefunction, which is an object part of an abstract space called Hilbert space. Kets can be transformed by the action of operators, $\hat{A}$, which are also objects of the Hilbert space. One can define a corresponding space of bra's $\langle \phi \vert$, and correspondingly adjoint operators ${\hat{A}}^\dag$ such that the bra corresponding to $\hat{A}\vert\psi\rangle$ is $\langle \psi \vert \hat{A}^\dag$. A bra and a ket can form a scalar product, also said to be projected onto each other. The projection $\langle \phi \vert \psi \rangle$ is a complex number, and we have $\langle \psi \vert \phi \rangle = \langle \phi \vert \psi \rangle^*$. Let $\vert\psi^a_j\rangle$ be an eigenstate of $\hat{A}$ so that $\hat{A} \vert \psi^a_j\rangle = a_j \vert\psi^a_j\rangle$. We see that the eigenvalue $a_j$ can be obtained by projecting the ket $\hat{A} \vert \psi^a_j\rangle$ onto the bra $\langle \psi^a_j \vert$, $a_j = \langle \psi_j^a \vert \hat{A} \vert \psi_j^a \rangle$, because the eigenvectors are normalised $\langle \psi_j^a \vert \psi_j^a \rangle = 1$. The bra corresponding to $\hat{A} \vert \psi^a_j\rangle$ is $\langle \psi^a_j \vert \hat{A}^\dag$ and if we project this onto the ket $\vert\psi^a_j\rangle$ we obtain $\langle \psi_j^a \vert \hat{A}^\dag \vert \psi_j^a \rangle= \langle \psi_j^a \vert \hat{A} \vert \psi_j^a \rangle^*$, and we can write $\langle \psi_j^a \vert \hat{A}^\dag = a_j^* \langle \psi_j^a \vert$. We see, therefore, that if $\hat{A}$ has real eigenvalues then $\hat{A}^\dag = \hat{A}$, and the operator is called self-adjoint. If the eigenvalues of an operator $\hat{T}$ are of the the type $e^{i \theta_j}$, with $\theta_j$ a real number, then we have $\hat{T} \vert\psi^t_j \rangle = e^{i \theta_j} \vert\psi^t_j \rangle$, and since $\langle \psi^t_j \vert \hat{T}^\dag = e^{-i \theta_j} \langle \psi^t_j \vert$, by forming the product $\langle \psi^t_j \vert \hat{T}^\dag\hat{T} \vert\psi^t_j \rangle$ we see that we must have $\hat{T}^\dag\hat{T} = \mathds{1}$, where $\mathds{1}$ is the identity operator. This also implies that the inverse operator is equal to its adjoint: $\hat{T}^{-1} = \hat{T}^\dag$. An operator that satisfies these properties is called unitary.

A wavefunction can be represented by projecting it onto a complete set of eigenvectors of any operator. If this set is formed by the position eigenstates we write $\langle {\bf r} \vert \psi \rangle = \psi({\bf r})$. Similarly, any operator can be represented by projections on any complete set, only that in this case one obtains matrices of the type $A_{ij} = \langle \psi_i \vert A \vert \psi_j \rangle$ because every transformed wavefunction $A\vert\psi_j \rangle$ needs to be represented. If the set is formed by the eigenstates of the operator then obviously the matrix is diagonal. A real space representation would have the form $\langle {\bf r^\prime} \vert A \vert {\bf r} \rangle = A({\bf r},{\bf r^\prime})$.