Recall that the commutator of two linear operators, \(S\) and \(T\), is \([S,T]=ST-TS\) and that they are said to commute if \([S,T]=0\).

Theorem Two normal operators, \(S\) and \(T\), commute if and only if there is an orthonormal basis consisting of common eigenvectors of \(S\) and \(T\). In this case \(S\) and \(T\) are said to be simultaneously diagonalisable since their respective matrix representations with respect to such a basis are diagonal.

Proof The if is trivial so we focus on the only if. \(S\) and \(T\) have spectral decompositions of the form,
\begin{equation*}
S=\sum_{i=1}^r\mu_iQ_{\mu_i}\quad\text{and}\quad T=\sum_{i=1}^s\lambda_iP_{\lambda_i},
\end{equation*}
where the \(\mu_i\) and \(\lambda_i\) are the distinct eigenvalues of \(S\) and \(T\) respectively and \(Q_{\mu_i}\) and \(P_{\lambda_i}\) the projectors onto the corresponding eigenspaces. Now, if any operator, \(A\), commutes with a normal operator, \(T\), then the eigenspaces, \(V_{\lambda_i}\), of \(T\) are \(A\)-invariant since if, \(\ket{\lambda}\in V_\lambda\), then, \(TA\ket{\lambda}=AT\ket{\lambda}=\lambda A\ket{\lambda}\), so, \(A\ket{\lambda}\in V_\lambda\). This means that both, \(V_{\lambda_i}\), and its orthogonal complement, \(V_{\lambda_i}^\perp\), are \(A\)-invariant, so we have a direct sum decomposition, \(V=V_{\lambda_i}\oplus V_{\lambda_i}^\perp\), with corresponding orthogonal projectors, \(P_{\lambda_i}\) and \(P_{\lambda_i}^\perp\), such that for any \(\ket{v}\in V\) we can write, \(P_{\lambda_i}A\ket{v}=P_{\lambda_i}A(P_{\lambda_i}\ket{v}+P_{\lambda_i}^\perp\ket{v})=AP_{\lambda_i}\ket{v}\). That is, \(A\) commutes with each of the projection operators of the spectral decomposition of \(T\). It follows that if the two normal operators, \(S\) and \(T\), commute then so too must their respective projectors, \(Q_{\mu_i}P_{\lambda_j}=P_{\lambda_j}Q_{\mu_i}\). Now define \(R_{ij}=Q_{\mu_i}P_{\lambda_j}\), clearly \(R_{ij}^2=R_{ij}\), \(R_{ij}R_{kl}=0\) unless \(i=k\) and \(j=l\), and \(R_{ij}^\dagg=R_{ij}\) so the \(R_{ij}\) are orthogonal projectors which moreover satisfy,
\begin{equation}
\sum_{i,j}R_{ij}=\sum_{i=1}^rQ_{\mu_i}\sum_{j=1}^sP_{\lambda_j}=\id_V,
\end{equation}
because, \(\sum_{i=1}^rQ_{\mu_i}=\id_V\) and \(\sum_{j=1}^sP_{\lambda_j}=\id_V\). Thus, since, \(\img R_{ij}=V_{\mu_i}\cap V_{\lambda_j}\), we have
\begin{equation}
V=\bigoplus_{i,j}V_{\mu_i}\cap V_{\lambda_j},
\end{equation}
and choosing an orthonormal basis for each \(V_{\mu_i}\cap V_{\lambda_j}\) we obtain, as desired, an orthonormal basis for \(V\) consisting of common eigenvectors of \(S\) and \(T\). Equivalently, we note that \(S\) and \(T\) have spectral decompositions with respect to the projectors \(R_{ij}\), \(S=\sum_{i=1}^r\mu_iQ_{\mu_i}=\sum_{i,j}\mu_iR_{ij}\) and \(T=\sum_{j=1}^s\lambda_jP_{\lambda_j}=\sum_{ij}\lambda_jR_{ij}\).\(\blacksquare\)

Throughout this section we assume our vector space \(V\) is complex Hermitian with positive definite inner product.

Dirac invented a notation for linear algebra particularly well suited to quantum mechanics. In this notation, vectors \(\psi,\phi\in V\) are denoted by the kets, \(\ket{\psi}\) and \(\ket{\phi}\), and their inner product, \((\psi,\phi)\), by the bra-ket \(\braket{\psi|\phi}\). The bra, \(\bra{\psi}\), and corresponding ket, \(\ket{\psi}\), are viewed as being distinct objects, with \(\bra{\psi}\) being precisely the image in \(V^*\) under the Riesz antiisomorphism, of \(\ket{\psi}\). A linear combination of vectors, \(a\psi+b\phi\), where \(a,b\in\CC\), would be denoted either as \(\ket{a\psi+b\phi}\) or, more commonly, as, \(a\ket{\psi}+b\ket{\phi}\). Likewise, the corresponding bra is usually denoted, \(a^*\bra{\psi}+b^*\bra{\phi}\), though sometimes it may be convenient to write it equivalently as \(\bra{a\psi+b\phi}\).

For any linear operator, \(T\in\mathcal{L}(V)\), the vector \(T\psi\) is denoted either as \(\ket{T\psi}\) or \(T\ket{\psi}\). The corresponding bra, \(\bra{T\psi}\), is such that \(\braket{T\psi|\phi}=\braket{\psi|T^\dagger\phi}\) for any \(\ket{\phi}\in V\). This is just the inner product of, \(\ket{\psi}\), and, \(\ket{T^\dagger\phi}=T^\dagger\ket{\phi}\). As such, it it is typically denoted, \(\braket{\psi|T^\dagger|\phi}\), which is itself then equal to, \(\braket{\phi|T|\psi}^*\), and we can think of the bra corresponding to the action of \(T\) (from the left) on a ket \(\psi\), \(T\ket{\psi}\), as the action of \(T^\dagg\) from the right on the bra \(\bra{\psi}\), \(\bra{\psi}T^\dagg\). In particular, Hermitian operators, \(T\), are defined to satisfy
\begin{equation}
\braket{\psi|T|\phi}=\braket{\phi|T|\psi}^*.\label{equ:Dirac Hermitian}
\end{equation}

It’s worth noting that even though one rarely sees scalars or operators within the bras or kets of Dirac notation, there is no reason why we shouldn’t and indeed sometimes it may be convenient. Dirac notation’s particular elegance is in its handling of a construction known as the outer product.

For any vectors \(\psi,\phi\in V\) we can define an operator, \(A_{\psi,\phi}\in\mathcal{L}(V)\), such that for any \(\chi\in V\)
\begin{equation}
A_{\psi,\phi}\chi=(\phi,\chi)\psi.
\end{equation}
The operator \(A_{\psi,\phi}\) is called the outer product of \(\psi\) and \(\phi\). In Dirac notation, this operator would simply be the ‘butterfly product’, \(\ket{\psi}\bra{\phi}\), with its action on \(\ket{\chi}\), \(\ket{\psi}\braket{\phi|\chi}\).

In particular, if \(\ket{\psi}\in V\) is a normalised vector then the projection onto the subspace spanned by \(\ket{\psi}\), \(P_\psi\), is just,
\begin{equation}
P_\psi=\ket{\psi}\bra{\psi}.
\end{equation}

When using Dirac notation, it is typical to denote orthonormal basis vectors, \(e_i\), by indexed kets, \(\ket{i}\). Then the expansion of an arbitrary vector in this basis is given by
\begin{equation}
\ket{\psi}=\sum_{i=1}^{N}\ket{i}\braket{i|\psi}=\sum_{i=1}^{N}P_i\ket{\psi},
\end{equation}
where \(P_i\) is the projector on to the one-dimensional subspace spanned by \(\ket{i}\). This must hold for every \(\ket{\psi}\) so we see that the identity operator for the space may be written as
\begin{equation}
I=\sum_{i=1}^N\ket{i}\bra{i}=\sum_{i=1}^NP_i.
\end{equation}
This is known as the resolution of the identity associated with the given basis set.

The adjoint of an operator of the form \(\ket{u}\bra{v}\), \((\ket{u}\bra{v})^\dagg\) is simply \(\ket{v}\bra{u}\) since for any vectors \(\ket{\psi}\) and \(\ket{\phi}\),
\begin{equation*}
\bra{\psi}(\ket{u}\bra{v})^\dagg\ket{\phi}=\left(\braket{\phi|u}\braket{v|\psi}\right)^*=\braket{\psi|v}\braket{u|\phi}=\bra{\psi}(\ket{v}\bra{u})\ket{\phi}.
\end{equation*}

We know that if \(T\in\mathcal{L}(V)\) is a normal operator then there exists an orthonormal basis for \(V\) of eigenvectors of \(T\). If \(T\) has \(r\) distinct eigenvalues, \(\lambda_1,\dots,\lambda_r\), each with geometric multiplicity, \(d_i\), then an eigenvalue \(\lambda_i\) for which \(d_i>1\) is said to be degnerate with its degree of degeneracy, \(d_i\). Let us denote this orthonormal basis, \(\ket{i,j}\), with \(i=1,\dots,r\) and \(j=1,\dots,d_i\). That is,
\begin{equation}
T\ket{i,j}=\lambda_i\ket{i,j},\quad i=1,\dots,r,\; j=1,\dots,d_i.
\end{equation}
If we denote by \(P_\lambda\) the projector onto the eigenspace, \(V_\lambda\), so that
\begin{equation}
P_{\lambda_i}=\sum_{j=1}^{d_i}\ket{i,j}\bra{i,j},
\end{equation}
then \(T\) may be written as the spectral decomposition,
\begin{equation}
T=\sum_{i=1}^r\lambda_iP_{\lambda_i}.
\end{equation}