*linear operator*that maps the space \(U_n\) (\(\ket{w} \in U_n\)) to itself (\(\ket{u} \in U_n\)). The linearity follows from the linearity of the scalar product. Linear operators on columns are ordinary matrices and we can see this using the same rules that we used to get the representation of the tensor products of two columns. Remembering that the dual vector is represented by a column that is transposed and conjugated (a row of complex conjugate values) we get that \begin{equation} \left( \begin{array}{c} v^1\\ v^2\\ \vdots \\ v^N \end{array} \right) \otimes (u_1^* u_2^* \ldots u_n^*) =\left( \begin{array}{cccc} u_1^*v^1 & u_2^*v^1 & \ldots & u_n^*v^1 \\ u_1^*v^2 & u_2^*v^2 & \ldots & u_n^*v^2 \\ \vdots & \vdots & \vdots & \vdots \\ u_1^*v^n & u_2^*v^n & \ldots & u_n^*v^n \end{array} \right). \end{equation} The tensor product space \(U_n\otimes U^*_n\) is the space of linear operators on \(U_n\). Since the uncorellated vectors \(\ket{e_i}\bra{e_j}\) form a basis, every operator, \(\hat{A}\), that acts on \(U_n\) can be expressed as their linear combination \begin{equation} \hat{A}=\sum_{i,j=1}^na^i_{\;j}\ket{e_i}\bra{e_j}. \label{operator_in_basis} \end{equation}

## The Adjoint Operator

As we can see from \ref{operator_in_basis} every operator can act on both the parent space, \(U_n\), and its dual, \(U_n^*\) \begin{eqnarray} \hat{A}_\rightarrow\ket{u}&=&\sum_{i,j=1}^na^i_{\;j}\ket{e_i}\bra{e_j}\ket{u}=\sum_{i,j=1}^na^i_{\;j}\sprod{e_j}{u}\ket{e_i}\in U_n, \\ \bra{u}_\leftarrow \hat{A}&=&\sum_{i,j=1}^na^i_{\;j}\bra{u}\ket{e_i}\bra{e_j}=\sum_{i,j=1}^na^i_{\;j}\sprod{u}{e_i}\bra{e_j}\in U_n^\dagger. \end{eqnarray} Note that this action of \(\hat{A}\) on the dual vectors is not its proper image in the dual space \begin{eqnarray} (\hat{A}_\rightarrow\ket{u})^\dagger&=&\left( \sum_{i,j=1}^na^i_{\;j}\sprod{e_j}{u}\ket{e_i} \right)^\dagger=\sum_{i,j=1}^n(a^i_{\;j})^*\sprod{e_j}{u}^*\bra{e_i}=\\ &&\sum_{i,j=1}^n(a^i_{\;j})^*\sprod{u}{e_j}\bra{e_i}\neq \sum_{i,j=1}^na^i_{\;j}\sprod{u}{e_i}\bra{e_j}=\bra{u}_\leftarrow \hat{A}. \label{adjoint_inequality} \end{eqnarray} From \ref{adjoint_inequality} we can identify the operator the produces the correct image of \(\hat{A}\) in \(U_n^*\) as \begin{equation} \hat{A}^\dagger=\sum_{i,j=1}^n(a^{\;\;i}_{j})^*\ket{e_i}\bra{e_j}. \end{equation} This operator is called the adjoint operator of \(\hat{A}\). It's matrix representation is the transposed and complex conjugated matrix of \(\hat{A}\), \((a^i_{\;j})^\dagger=(a^{\;\;i}_{j})^*\). The adjoint of the adjoint is the starting operator, \((\hat{A}^\dagger)^\dagger=\hat{A}\), i.e. \(\dagger\) is an involution. The standard definition of the adjoint operator is that \(\sprod{A^\dagger u}{v}=\sprod{u}{Av}\) which in Dirac notation is \begin{equation} \sprod{A^\dagger u}{v}=(\hat{A}^\dagger_\rightarrow\ket{u})^\dagger \ket{v}=(\bra{u}_\leftarrow (\hat{A}^\dagger)^\dagger)\ket{v}=\bra{u}_\leftarrow\hat{A}\ket{v}=\bra{u}\hat{A}_\rightarrow \ket{v}=\sprod{u}{Av}. \end{equation} PICTURE We also note that (\(\diad{u}{v})^\dagger=\diad{v}{u}\).## Comutator and Trace

A binary operation very important for quantum mechanics is the comutator \begin{equation} [\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}. \end{equation} Since operators do not necessarily commute, the commutator measures how much and in what way two operators do not commute. In quantum mechanics expectation values are calculated using the*trace*. The trace is a linear map from the space of linear operators to the complex numbers and is one way to define the norm of an operator. It is defined as \begin{equation} Tr\{\hat{A}\}=\sum_{i=1}^n\ev{e_i}{\hat{A}}{e_i}. \end{equation} i.e. the sum of the diagonal elements of the matrix representation. The trace is cyclic \begin{equation} Tr\{\hat{A}_1\hat{A}_2\ldots\hat{A}_p\}=Tr\{\hat{A}_p\hat{A}_1\ldots\hat{A}_{p-1}\}, \end{equation} and also has the property that the trace of the tensor product is the product of the factor traces \begin{equation} Tr\{\hat{A}\otimes \hat{B}\}=Tr\{\hat{A}\}Tr\{\hat{B}\}. \end{equation} The trace is independent of the basis in which it is calculated.

## The Eigen-problem

The eigen-problem of an operator, \(\hat{A}\), consists of finding a set of \(n\) pairs \(\{\ket{e_i},\lambda_i\}\), where \(\ket{e_i}\in U_n\) and \(\lambda_i \in \cn\), for which the following equations holds \begin{equation} \hat{A}\ket{e_i}=\lambda_i \ket{e_i}. \end{equation} The vectors \(\ket{e_i}\) (eigen-vectors) form the eigen-basis of the operator and the numbers \(\lambda_i\) are the corresponding eigen-values. If more then one vector hah the same eigen-value \(\lambda_i\) that eigen-value is called degenerate. The vectors corresponding to that eigen-value form an eigen-space. Every eigen-space is orthogonal to every other eignespace (every vector in one eignespace has a zero scalar product with every vector in any other eigen-space). The operator acts as a scalar in it's eigen-spaces and can be written as \begin{equation} \hat{A}=\sum_{i=1}^n \lambda_i \ket{e_i}\bra{e_i}. \end{equation}## Normal Operators

Operators that commute with their adjoint operator, \(\hat{N}\hat{N}^\dagger =\hat{N}^\dagger \hat{N}\), form the largest class of operators that have an eigen-decomposition. All operators that will be discussed in this section are normal operators.## Projection Operators

If \(\ket{e}\) is a normalized (unit norm) vector the operator \(\ket{e}\bra{e}\) has some interesting properties. Its trace is one since \(\ket{e}\) and if it acts on an arbitrary vector \(\ket{v}\) \begin{equation} \ket{e}\bra{e}\ket{v}=\sprod{e}{v}\ket{e}. \end{equation} it produces the vector component of the linear decomposition of \(\ket{v}\) parallel to \(\ket{e}\). We say that this operator*projects*to the one dimensional space defined by \(\ket{e}\). One can also construct a projection operator that projects onto an arbitrary \(m\)-dimensional subspace spanned by the basis vectors \(\{\ket{e_i},\ldots,\ket{e_m}\}\) \begin{equation} \hat{P}_m=\sum_{i=1}^m \ket{e_i}\bra{e_i}. \end{equation} In this case the trace of the projector equals the dimension of the subspace to which it projects to. The eigen-decomposition of a normal operator can be written using mutually orthogonal projection operators to its eigen-spaces as \begin{equation} \hat{N}=\sum_{i=1}^m \lambda_i \hat{P}_i,\;\;\;\;\;\; (\hat{P}_i\hat{P}_j=\delta_{ij}\hat{P}_i). \label{eigen_decomposition} \end{equation} where \(\hat{P}_i\) are projection operators to the eigen-spaces. Projection operators have two eigen-values \(1\), for the space that they project onto, and \(0\), for the space orthogonal to the space that they poject onto. Projections are idempotent and hermitian \begin{equation} \hat{P}^2=\hat{P}\Rightarrow \hat{P}^n=\hat{P} \;\;\;\;\;\;\;\;\;\;\;\; \hat{P}^\dagger=\hat{P}. \label{projection_properties} \end{equation}