# PlanetPhysics/Quantum Operator Concept

Consider the function ${\frac {\partial \Psi }{\partial t}}$, the derivative of $\Psi$ with respect to time; one can say that the operator ${\frac {\partial }{\partial t}}$ acting on the function $\Psi$ yields the function ${\frac {\partial \Psi }{\partial t}}$. More generally, if a certain operation allows us to bring into correspondence with each function $\Psi$ of a certain function space, one and only one well-defined function $\Psi ^{\prime }$ of that same space, one says the $\Psi ^{\prime }$ is obtained through the action of a given operator $A$ on the function $\Psi$, and one writes
$\Psi ^{\prime }=A\Psi .$
By definition $A$ is a linear operator if its action on the function $\lambda _{1}\Psi _{1}+\lambda _{2}\Psi _{2}$, a linear combination with constant (complex) coefficients, of two functions of this function space, is given by
$A\left(\lambda _{1}\Psi _{1}+\lambda _{2}\Psi _{2}\right)=\lambda _{1}\left(A\Psi _{1}\right)+\lambda _{2}\left(A\Psi \right).$
Among the linear operators acting on the wave functions
$\Psi :=\Psi (\mathbf {r} ,t):=\Psi (x,y,z,t)$
associated with a particle, let us mention:
Starting from certain linear operators, one can form new linear operators by the following algebraic operations:
$(cA)\Psi :=c(A\Psi )$
$S\Psi :=A\Psi +B\Psi$
Note that in contrast to the sum, the product of two operators is not commutative . Therein lies a very important difference between the algebra of linear operators and ordinary algebra.
The product $AB$ is not necessarily identical to the product $BA$; in the first case, $B$ first acts on the function $\Psi$, then $A$ acts upon the function $(B\Psi )$ to give the final result; in the second case, the roles of $A$ and $B$ are inverted. The difference $AB-BA$ of these two quantities is called the commutator of $A$ and $B$; it is represented by the symbol $[A,B]$:
   $[A,B]:=AB-BA$
If this difference vanishes, one says that the two operators commute:
$AB=BA$
As an example of operators which do not commute, we mention the operator $f(x)$, multiplication by function $f(x)$, and the differential operator ${\partial }/{\partial x}$. Indeed we have, for any $\Psi$,
${\frac {\partial }{\partial x}}f(x)\Psi ={\frac {\partial }{\partial x}}(f\Psi )={\frac {\partial f}{\partial x}}\Psi +f{\frac {\partial \Psi }{\partial x}}=\left({\frac {\partial f}{\partial x}}+f{\frac {\partial }{\partial x}}\right)\Psi$
In other words
   $\left[{\frac {\partial }{\partial x}},f(x)\right]={\frac {\partial f}{\partial x}}$
and, in particular
   $\left[{\frac {\partial }{\partial x}},x\right]=1$
However, any pair of derivative operators such as ${\partial }/{\partial }x$,${\partial }/{\partial }y$,${\partial }/{\partial }z$,${\partial }/{\partial }t$, commute.
A typical example of a linear operator formed by sum and product of linear operators is the Laplacian operator
$\nabla ^{2}:={\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {\partial ^{2}}{\partial y^{2}}}+{\frac {\partial ^{2}}{\partial z^{2}}}$
which one may consider as the scalar product of the vector operator gradient $\nabla :=\left({\frac {\partial }{\partial x}},{\frac {\partial }{\partial y}},{\frac {\partial }{\partial z}}\right)$, by itself.