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In physics, an operator is a function acting on the space of physical states. As a result of its application on a physical state, another physical state is obtained, very often along with some extra relevant information.

The simplest example of the utility of operators is the study of symmetry. Because of this, they are a very useful tool in classical mechanics. In quantum mechanics, on the other hand, they are an intrinsic part of the formulation of the theory.

Operators in classical mechanics
Let us consider a classical mechanics system led by a certain Hamiltonian $$H(q,p)$$, function of the generalized coordinates $$q$$ and its conjugate momenta. Let us consider this function to be invariant under the action of a certain group of transformations $$G$$, i.e., if $$S\in G, H(S(q,p))=H(q,p)$$. The elements of $$G$$ are physical operators, which map physical states among themselves.

An easy example is given by space translations. The hamiltonian of a translationally invariant problem does not change under the transformation $$q\to T_a q=q+a$$. Other straightforward symmetry operators are the ones implementing rotations.

If the physical system is described by a function, as in classical field theories, the translation operator is generalized in a straightforward way:


 * $$f(x) \to T_a f(x)=f(x-a).$$

Notice that the transformation inside the parenthesis should be the inverse of the transformation done on the coordinates.

Concept of generator
If the transformation is infinitesimal, the operator action should be of the form


 * $$ I + \epsilon A $$

where $$I$$ is the identity operator, $$\epsilon$$ is a small parameter, and $$A$$ will depend on the transformation at hand, and is called a generator of the group. Again, as a simple example, we will derive the generator of the space translations on 1D functions.

As it was stated, $$T_a f(x)=f(x-a)$$. If $$a=\epsilon$$ is infinitesimal, then we may write


 * $$T_\epsilon f(x)=f(x-\epsilon)\approx f(x) - \epsilon f'(x).$$

This formula may be rewritten as


 * $$T_\epsilon f(x) = (I-\epsilon D) f(x)$$

where $$D$$ is the generator of the translation group, which in this case happens to be the derivative operator. Thus, it is said that the generator of translations is the derivative.

The exponential map
The whole group may be recovered, under normal circumstances, from the generators, via the exponential map. In the case of the translations the idea works like this.

The translation for a finite value of $$a$$ may be obtained by repeated application of the infinitesimal translation:


 * $$T_a f(x) = \lim_{N\to\infty} T_{a/N} \cdots T_{a/N} f(x)$$

with the $$\cdots$$ standing for the application $$N$$ times. If $$N$$ is large, each of the factors may be considered to be infinitesimal:


 * $$T_a f(x) = \lim_{N\to\infty} (I -(a/N) D)^N f(x).$$

But this limit may be rewritten as an exponential:


 * $$T_a f(x)= \exp(-aD) f(x).$$

To be convinced of the validity of this formal expression, we may expand the exponential in a power series:


 * $$T_a f(x) = \left( I - aD + {a^2D^2\over 2!} - {a^3D^3\over 3!} + \cdots \right) f(x).$$

The right-hand side may be rewritten as


 * $$f(x) - a f'(x) + {a^2\over 2!} f(x) - {a^3\over 3!} f'(x) + \cdots$$

which is just the Taylor expansion of $$f(x-a)$$, which was our original value for $$T_a f(x)$$.

The mathematical properties of physical operators are a topic of great importance in itself. For further information, see C*-algebra and Gelfand-Naimark theorem.

Operators in quantum mechanics
The mathematical description of quantum mechanics is built upon the concept of an operator.

Physical pure states in quantum mechanics are unit-norm vectors in a certain vector space (a Hilbert space). Time evolution in this vector space is given by the application of a certain operator, called the evolution operator. Since the norm of the physical state should stay fixed, the evolution operator should be unitary. Any other symmetry, mapping a physical state into another, should keep this restriction.

Any observable, i.e., any quantity which can be measured in a physical experiment, should be associated with a self-adjoint linear operator. The operators must yield real eigenvalues, since they are values which may come up as the result of the experiment. Mathematically this means the operators must be Hermitian. The probability of each eigenvalue is related to the projection of the physical state on the subspace related to that eigenvalue. See below for mathematical details.

Linear operators on a wave function
Let ψ be the wave function for a quantum system, and $$\hat{A}$$ be any linear operator for some observable A (such as position, momentum, energy, angular momentum etc), then


 * $$\hat{A} \psi = a \psi ,$$

where a is the eigenvalue of the operator. The eigenvalue corresponds to the measured value of the observable, i.e. observable A has a measured value a. If this relation holds the wave function is said to be an eigenfunction. If ψ is an eigenfunction, then the eigenvalue can be found and so the observable can be measured, conversely if ψ is not an eigenfunction then the eigenfunction can't be found and the observable can't be measured for that case. For a discrete basis of the eigenstates $$ | \psi_i \rangle $$, the corresponding eigenvalues ai will also be discrete. Likewise, for a continuous basis there is a continuum of eigenstates $$ | \psi \rangle $$ and accordingly a continuum of eigenvalues a.

In bra-ket notation the above can be written;


 * $$\begin{align} & \hat{A} \psi = \hat{A} \psi ( \mathbf{r} ) = \hat{A} \langle \mathbf{r} | \psi \rangle = \langle \mathbf{r} | \hat {A} | \psi \rangle \\

& a \psi = a \psi ( \mathbf{r} ) = a \langle \mathbf{r} | \psi \rangle = \langle \mathbf{r} | a | \psi \rangle \\ \end{align} $$

Linear operators work in any number of dimensions. That is why an operator can take the form of a vector, as each component of the vector acts on the function separately due to linearity. One mathematical example is the del operator, which is itself a vector. This is useful in other quantum operators, as illustrated below.

An operator in n-dimensional space can be written:


 * $$ \mathbf{\hat{A}} = \sum_{j=1}^n \mathbf{e}_\mathrm{j} \hat{A}_j $$

where ej are basis vectors, corresponding to each component operator Aj. Each component will yield a corresponding eigenvalue. Acting this on the wave function ψ:


 * $$ \mathbf{\hat{A}} \psi = \left ( \sum_{j=1}^n \mathbf{e}_\mathrm{j} \hat{A}_j \right ) \psi = \sum_{j=1}^n \left ( \mathbf{e}_\mathrm{j} \hat{A}_j \psi \right ) = \sum_{j=1}^n \left ( \mathbf{e}_\mathrm{j} a_j \psi \right ) $$

in which


 * $$ \hat{A}_j \psi = a_j \psi .$$

In bra-ket notation:


 * $$\begin{align} & \mathbf{\hat{A}} \psi = \mathbf{\hat{A}} \psi ( \mathbf{r} ) = \mathbf{\hat{A}} \langle \mathbf{r} | \psi \rangle = \langle \mathbf{r} | \mathbf{\hat{A}} | \psi \rangle \\

& \left ( \sum_{j=1}^n \mathbf{e}_\mathrm{j} \hat{A}_j \right ) \psi = \left ( \sum_{j=1}^n \mathbf{e}_\mathrm{j} \hat{A}_j \right ) \psi ( \mathbf{r} ) = \left ( \sum_{j=1}^n \mathbf{e}_\mathrm{j} \hat{A}_j \right ) \langle \mathbf{r} | \psi \rangle = \left \langle \mathbf{r} \Bigg | \sum_{j=1}^n \mathbf{e}_\mathrm{j} \hat{A}_j \Bigg | \psi \right \rangle \\

\end{align} \,\!$$

Commutation of operators on Ψ
If two observables A and B have linear operators $$ \hat{A} $$ and $$ \hat{B} $$, the commutator is defined by,


 * $$ \left [ \hat{A}, \hat{B} \right ] = \hat{A} \hat{B} - \hat{B} \hat{A} $$

The commutator is itself a (composite) operator. Acting the commutator on ψ gives:


 * $$ \left [ \hat{A}, \hat{B} \right ] \psi = \hat{A} \hat{B} \psi - \hat{B} \hat{A} \psi . $$

If ψ is an eigenfunction with eigenvalues a and b for observables A and B respectively, and if the operators commute:


 * $$ \left [ \hat{A}, \hat{B} \right ] \psi = 0, $$

then the observables A and B can be measured at the same time with measurable eigenvalues a and b respectively. To illustrate this:


 * $$ \begin{align}\left [ \hat{A}, \hat{B} \right ] \psi & = \hat{A} \hat{B} \psi - \hat{B} \hat{A} \psi \\

& = \hat{A} \left ( b \psi \right ) - \hat{B} \left ( a \psi \right ) \\ & = b \left ( \hat{A} \psi \right ) - a \left ( \hat{B} \psi \right ) \\ & = b \left ( a \psi \right ) - a \left ( b \psi \right ) \\ & = ab \psi - ab \psi \\ & = 0 .\\ \end{align} $$

If the operators do not commute:


 * $$ \left [ \hat{A}, \hat{B} \right ] \psi \neq 0, $$

they can't be measured simultaneously to arbitrary precision, and there is an uncertainty relation between the observables, even if ψ is an eigenfunction. Notable pairs are position and momentum, and energy and time - Hiesenberg's uncertainty relations, and the angular momenta (spin, orbital and total) about any two orthogonal axes (such as Lx and Ly, or sy and sz etc).

Expectation values of operators on Ψ
The expectation value (equivalently the average or mean value) is the average measurement of an observable, for particle in region R. The expectation value $$\langle \hat{A} \rangle $$ of the operator $$ \hat{A} $$ is calculated from :


 * $$\langle \hat{A} \rangle = \int_R \psi^{*}\left( \mathbf{r} \right ) \hat{A} \psi \left( \mathbf{r} \right ) \mathrm{d}^3\mathbf{r} = \langle \psi | \hat{A} | \psi \rangle .$$

This can be generalized to any function F of an operator:


 * $$ \langle F ( \hat{A} ) \rangle = \int_R \psi(\mathbf{r})^{*} \left [ F ( \hat{A} ) \psi(\mathbf{r}) \right ] \mathrm{d}^3 \mathbf{r} = \langle \psi | F ( \hat{A} ) | \psi \rangle, $$

An example of F is the 2-fold action of A on ψ, i.e. squaring an operator or doing it twice:


 * $$\begin{align}

& F(\hat{A}) = \hat{A}^2 \\ & \Rightarrow \langle \hat{A}^2 \rangle = \int_R \psi^{*} \left( \mathbf{r} \right ) \hat{A}^2 \psi \left( \mathbf{r} \right ) \mathrm{d}^3\mathbf{r} = \langle \psi \vert \hat{A}^2 \vert \psi \rangle \\ \end{align}\,\!$$

Hermiticity of QM operators
The definition of a Hermitian operator is :


 * $$\hat{A} = \hat{A}^\dagger$$

Following from this, in bra-ket notation:


 * $$\langle \phi_i | \hat{A} | \phi_j \rangle = \langle \phi_j | \hat{A} | \phi_i \rangle^*.$$

Matrix representation of quantum operators
An operator can be written in matrix form to map one basis vector to another. Since the operators and basis vectors are linear, the matrix is a linear transformation (aka transition matrix) between bases. Each basis element $$\phi_j $$ can be connected to another $$A_{ij} = \langle \phi_i | \hat{A} | \phi_j \rangle $$ using vector and matrix indices ,


 * $$A_{ij} = \langle \phi_i | \hat{A} | \phi_j \rangle,$$

in which,


 * $$\hat{A} = \begin{pmatrix}

A_{11} & A_{12} & \cdots & A_{1n} \\ A_{21} & A_{22} & \cdots & A_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ A_{n1} & A_{n2} & \cdots & A_{nn} \\ \end{pmatrix} $$

A further property of a hermitain operator is that eigenfunctions corresponding to different eigenvalues are orthogonal. In matrix form, operators allow real eigenvalues to be found, corresponding to measurements. Orthogonality allows a suitable basis set of vectors to represent the state of the quantum system. The eigenvalues of the operator are also evaluated in the same way as for the square matrix, by solving the characteristic polynomial:


 * $$ \det\left ( \hat{A} - a \hat{I} \right ) = 0 ,$$

where I is the n × n identity matrix, as an operator it corresponds to the identity operator.

Table of QM operators
The operators used in quantum mechanics are collected in the table below (see for example, ). The bold-face vectors with circumflexes are not unit vectors, they are 3-vector operators; all three spatial components taken together.

Examples of applying quantum operators
The procedure for extracting information from a wave function is as follows. Consider the momentum p of a particle as an example. The momentum operator in one dimension is:


 * $$\hat{p} = -i\hbar\frac{\partial }{\partial x}$$

Letting this act on ψ we obtain:


 * $$\hat{p} \psi = -i\hbar\frac{\partial }{\partial x} \psi ,$$

if ψ is an eigenfunction of $$\hat{p}$$, then the momentum eigenvalue p is the value of the particle's momentum, found by:


 * $$ -i\hbar\frac{\partial }{\partial x} \psi = p \psi.$$

For three dimensions the momentum operator uses the nabla operator to become:


 * $$\mathbf{\hat{p}} = -i\hbar\nabla .$$

In Cartesian coordinates (using the standard Cartesian basis vectors ex, ey, ez) this can be written;


 * $$\mathbf{e}_\mathrm{x}\hat{p}_x + \mathbf{e}_\mathrm{y}\hat{p}_y + \mathbf{e}_\mathrm{z}\hat{p}_z = -i\hbar\left ( \mathbf{e}_\mathrm{x} \frac{\partial }{\partial x} + \mathbf{e}_\mathrm{y} \frac{\partial }{\partial y} + \mathbf{e}_\mathrm{z} \frac{\partial }{\partial z} \right ),$$

that is:


 * $$ \hat{p}_x = -i\hbar \frac{\partial}{\partial x}, \quad \hat{p}_y = -i\hbar \frac{\partial}{\partial y}, \quad \hat{p}_z = -i\hbar \frac{\partial}{\partial z} \,\!$$

The process of finding eigenvalues is the same. Since this is a vector and operator equation, if ψ is an eigenfunction, then each component of the momentum operator will have an eigenvalue corresponding to that component of momentum. Acting $$ \mathbf{\hat{p}} $$ on ψ obtains:


 * $$ \begin{align}

\hat{p}_x \psi & = -i\hbar \frac{\partial}{\partial x} \psi = p_x \psi \\ \hat{p}_y \psi & = -i\hbar \frac{\partial}{\partial y} \psi = p_y \psi \\ \hat{p}_z \psi & = -i\hbar \frac{\partial}{\partial z} \psi = p_z \psi \\ \end{align} \,\!$$