# Operators

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We now examine the tensor product, $$U_n\otimes U^*_n$$ of the space $$U_n$$ with its dual $$U^*_n$$. Using Dirac notation we can write a vector in this product space as $$\ket{u}\bra{v}$$. Without any proof, just by using the rules of Dirac notation, we can see that this object can act on vectors, $$\ket{w}$$, from the space $$U_n$$ $$\ket{u}\bra{v}\ket{w}=\sprod{v}{w}\ket{u}. \label{simple_diad}$$ Equation \ref{simple_diad} defines a 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 $$\left( \begin{array}{c} v^1\\ v^2\\ \vdots \\ v^N \end{array} \right) \otimes (u_1^* u_2^* \ldots u_n^*) =\left( \begin{array}{cccc} u_1^*v^1 & u_2^*v^1 & \ldots & u_n^*v^1 \\ u_1^*v^2 & u_2^*v^2 & \ldots & u_n^*v^2 \\ \vdots & \vdots & \vdots & \vdots \\ u_1^*v^n & u_2^*v^n & \ldots & u_n^*v^n \end{array} \right).$$ 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 $$\hat{A}=\sum_{i,j=1}^na^i_{\;j}\ket{e_i}\bra{e_j}. \label{operator_in_basis}$$

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 $$\hat{A}^\dagger=\sum_{i,j=1}^n(a^{\;\;i}_{j})^*\ket{e_i}\bra{e_j}.$$ 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 $$\sprod{A^\dagger u}{v}=(\hat{A}^\dagger_\rightarrow\ket{u})^\dagger \ket{v}=(\bra{u}_\leftarrow (\hat{A}^\dagger)^\dagger)\ket{v}=\bra{u}_\leftarrow\hat{A}\ket{v}=\bra{u}\hat{A}_\rightarrow \ket{v}=\sprod{u}{Av}.$$ 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 $$[\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}.$$ 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 $$Tr\{\hat{A}\}=\sum_{i=1}^n\ev{e_i}{\hat{A}}{e_i}.$$ i.e. the sum of the diagonal elements of the matrix representation. The trace is cyclic $$Tr\{\hat{A}_1\hat{A}_2\ldots\hat{A}_p\}=Tr\{\hat{A}_p\hat{A}_1\ldots\hat{A}_{p-1}\},$$ and also has the property that the trace of the tensor product is the product of the factor traces $$Tr\{\hat{A}\otimes \hat{B}\}=Tr\{\hat{A}\}Tr\{\hat{B}\}.$$ 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 $$\hat{A}\ket{e_i}=\lambda_i \ket{e_i}.$$ 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 $$\hat{A}=\sum_{i=1}^n \lambda_i \ket{e_i}\bra{e_i}.$$

## 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 $$\ket{v}$$ $$\ket{e}\bra{e}\ket{v}=\sprod{e}{v}\ket{e}.$$ 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 $$\{\ket{e_i},\ldots,\ket{e_m}\}$$ $$\hat{P}_m=\sum_{i=1}^m \ket{e_i}\bra{e_i}.$$ 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 $$\hat{N}=\sum_{i=1}^m \lambda_i \hat{P}_i,\;\;\;\;\;\; (\hat{P}_i\hat{P}_j=\delta_{ij}\hat{P}_i). \label{eigen_decomposition}$$ 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 $$\hat{P}^2=\hat{P}\Rightarrow \hat{P}^n=\hat{P} \;\;\;\;\;\;\;\;\;\;\;\; \hat{P}^\dagger=\hat{P}. \label{projection_properties}$$

## Hermitian Operators

If an operator is identical to its adjoint, $$\hat{A}^\dagger=\hat{A}$$, we call it a Hermitian operator. For Hermitian operators there is no disctinciton in their action on the bra's and ket's, $$(\hat{A}\ket{u})^\dagger=\bra{u}_\leftarrow \hat{A}^\dagger=\bra{u}_\leftarrow \hat{A}$$. Their eignevalues are real and represent the possible outcomes of measurements in quantum mechanics $$\hat{H}=\sum_{i=1}^m h_i \hat{P}_i \;\;,\;\;\; h_i\in \rn.$$

## Unitary Operators

For unitary operators their inverse is equal to their adjoint, $$\hat{U}^{-1}=\hat{U}^\dagger$$. Unitary operators preserve the norm of vectors: $$\norm{\hat{U}\ket{v}}=\sqrt{\ev{v}{\hat{U}^\dagger \hat{U}}{v}}=\sqrt{\ev{v}{\hat{U}^{-1} \hat{U}}{v}}=\sqrt{\sprod{v}{v}}=\norm{\ket{v}}.$$ The eigen-values of unitary operators are complex numbers of magnitude one, i.e. phase factors, $$\hat{U}=\sum_{k=1}^m e^{i \phi_k} \hat{P}_k \;\;,\;\;\; \phi_k \in \rn.$$ Unitary operators are nonsingular and define izomorphisms of the unitary space on which they act. They map the space into itself and preserve its geometry.

## Useful Analogies

The operation of finding the adjoint of operators is analogous to complex conjugation of complex numbers. Using this analogy hermitian and unitary operators are analogs of the real numbers and phase factors, respectively. If $$\hat{H}$$ and $$\hat{U}$$ are a hermitian and unitary operators, respectively, $$\hat{A}$$ an arbitrary operator, $$r, \phi \in \rn$$ and $$z \in \cn$$ then the follwoing analogies hold \begin{eqnarray} zz^* \in \rn &\longleftrightarrow& \hat{A}\hat{A}^\dagger\; \mbox{and} \; \hat{A}^\dagger\hat{A}\;\; \mbox{are always hermitian and equal only for normal operators}, \\ z=(z+z^*)+(z-z^*) &\longleftrightarrow& \hat{A}=(\hat{A}+\hat{A}^\dagger)+(\hat{A}-\hat{A}^\dagger), \nonumber \\ & \; &(z+z^*),i(z-z^*)\in \rn , (\hat{A}+\hat{A}^\dagger),i(\hat{A}-\hat{A}^\dagger)\; \mbox{are Hermitian}, \\ z=re^{i\phi}&\longleftrightarrow& \hat{A}=\hat{H}\hat{U}. \end{eqnarray} There are also geometric analogies. Unitary operators $$\hat{U}^{-1}=\hat{U}^\dagger$$, are analogous to rotations, $$\hat{R}^{-1}=\hat{R}^T$$, in real space that also preserve the norm of vectors. Hermitian operators, $$\hat{H}=\hat{H}^\dagger$$, are analogous to symmetric operator, $$\hat{S}=\hat{S}^T$$, in real space. Symmetric operators are the largest class of operators in real space that have an eigen-decomposition with real eigen-values.

## Operator Functions

For an ordinary function $$f(x)$$ that can be exapnded into a power series $$f(x)=\sum_{i=0}^\infty a_i x^i$$ we define the corresponding operator function $$\hat{f}(\hat{A})$$ as the power series $$\hat{f}(\hat{A})=\sum_{i=0}^\infty a_i \hat{A}^i.$$ In unitary spaces the resulting operator has the same eigenspaces as the argument of the function with eigen-values that are the values of the ordinary function evaluated at the eigen-values of the argument operator. We can show this using the eigen-decomposition \ref{eigen_decomposition} and the first property in \ref{projection_properties} \begin{eqnarray} \hat{f}(\hat{A})=\sum_{i=0}^\infty a_i \left( \sum_{j=1}^m \lambda_j \hat{P}_j \right)^i=\sum_{i=0}^\infty a_i \sum_{j=1}^m \lambda_j^i \hat{P}_j^i=\sum_{i=0}^\infty a_i \sum_{j=1}^m \lambda_j^i \hat{P}_j=\sum_{j=1}^m \left(\sum_{i=0}^\infty a_i\lambda_j^i \right) \hat{P}_j=\sum_{j=1}^m f(\lambda_j) \hat{P}_j \end{eqnarray}