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:

1. the differential operators ${\partial }/{\partial }x$,${\partial }/{\partial }y$,${\partial }/{\partial }z$,${\partial }/{\partial }t$, such as the one which was considered above;

1. the operators of the form $f(\mathbf {r} ,t)$ whose action consists in multiplying the function $\Psi$ by the function $f(\mathbf {r} ,t)$

Starting from certain linear operators, one can form new linear operators by the following algebraic operations:

1. multiplication of an operator $A$ by a constant $c$:

$(cA)\Psi :=c(A\Psi )$

1. the sum $S=A+B$ of two operators $A$ and $B$:

$S\Psi :=A\Psi +B\Psi$

1. the product $P=AB$ of an operator $B$ by the operator $A$:

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.

References

[1] Messiah, Albert. "Quantum mechanics: volume I." Amsterdam, North-Holland Pub. Co.; New York, Interscience Publishers, 1961-62.

This entry is a derivative of the Public domain work [1].